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May 18

Strange-looking cherries

Hello all, I've been meaning to ask this for some time. Lately, when shopping at one particular small grocery in my area (Central California), I've been seeing some cherries that look like two or three berries fused together, sometimes with more than one stone. Last time I shopped there, they also had heart-shaped cherries mixed in with these. Are these new varieties, or some kind of rare genetic mutation, or maybe something else? Has anyone seen anything like this in other places? 24.23.196.85 (talk) 19:15, 18 May 2013 (UTC)[reply]

Did they actually have more than one stone? WP:OR, but it wouldn't be surprising if sellers would market larger cherries when they get them, larger usually appeals to customers. μηδείς (talk) 19:23, 18 May 2013 (UTC)[reply]

There are many, many varieties of cherries. Many of them are heart-shaped. Many of them have fused fruit. Browsing the list of early California cherries here, the ones that I think of from your description are the Early Burlat (big, dark cherries), but again, there are lots of cherries. --Mr.98 (talk) 20:10, 18 May 2013 (UTC)[reply]

I've many times seen a few "Siamese twin" cherries mixed in with a batch of ordinary ones. Presumably the ones you found have mutated in some way that makes that the norm rather than the exception. Almost all the fruit we find in stores is genetically freakish, so things like that aren't incredibly surprising. Looie496 (talk) 02:17, 19 May 2013 (UTC)[reply]

Bonobo endocrinology

Does Bonobo Chimpanzee have more oxytocin \ other "friendship hormones" compared to Common Chimpanzee? 109.64.101.74 (talk) 21:20, 18 May 2013 (UTC)[reply]

First, bonobos are no longer generally referred to as chimpanzees -- they are a different (but very closely related) species. Concerning the question, I can't spot any evidence for that, though I don't think the topic has been studied very deeply. There is apparently a difference between bonobo and chimpanzee in the form of receptors for vasopressin, another peptide that is involved in social behavior. Apparently bonobos and humans share the same form, which is different from the form in chimps. See PMID 17118932 for discussion. Looie496 (talk) 02:08, 19 May 2013 (UTC)[reply]

We've heard about beneficial bacteria, perhaps there are beneficial cancers?

Perhaps some slow growing cancers that have in some way been "bred" to be somewhat differentiated, could be beneficial? For one thing they might keep immune system things like interferon on their toes? Thanks, Rich76.218.104.120 (talk) 22:57, 18 May 2013 (UTC)[reply]

What do you mean be "bred"? Plasmic Physics (talk) 00:20, 19 May 2013 (UTC)[reply]

..And, in what relavent way are bacteria like cancer? Plasmic Physics (talk) 00:53, 19 May 2013 (UTC)[reply]

What relevant way does "in what relavent way are bacteria like cancer?" have to do with my question? I'm sure you're reasonably intelligent, but your question seems off the point, perhaps you should explain your thinking first, bto motivate me to try to give a dissertation on how bacteria are like cancer. Did I ever say they were alike? Let's see both have some carbon atoms as do pencils and diamonds. What does that have to do with anything?76.218.104.120 (talk) 01:30, 19 May 2013 (UTC)[reply]

From the question header, it appears as though you are basing your logic on the assumption that they are. I'm just trying to follow the logic you used to derive your proposition. Plasmic Physics (talk) 01:43, 19 May 2013 (UTC)[reply]

The question makes sense, but I've never heard of such a thing. Cancer is defined as tissue growth that has escaped from normal genetic control, and it's hard to see how anything that has escaped from control can be good. Looie496 (talk) 02:09, 19 May 2013 (UTC)[reply]

I could imagine it occurring paradoxically on a TV show like House where, for example, someone's hypoglycemia is masked by the diabetes he develops as a secondary effect of his pancreatic cancer. But I am making up that scenario and I have never heard of it happening. μηδείς (talk) 02:34, 19 May 2013 (UTC)[reply]

(edit conflict) Have you heard of a cursed blessing? Certain types of cancer may have positive side effects, especially brain tumours, they are known to occasionally cause savant like changes. Plasmic Physics (talk) 02:37, 19 May 2013 (UTC)[reply]

ec "No violence, gentlemen -- no violence, I beg of you! Consider the furniture!". Let's not start the flame war now, shall we. The question before us, if I understand it correctly, is the following. We and our gut flora have evolved to lead at least commensal and, in some aspects, outright mutually-beneficial existence. So, the question goes, is it possible that some slow-growing cancers have evolved to be beneficial, as well? Now, to the best of my knowledge, let me try to give a (partial) answer. First of all, we did not evolve specific cancers; rather, we evolved predisposition to specific cancers. There are directly transmittable cancers, yes, such as the one decimating the tasmanian devil populations for several years now; but none AFAIK are common among humans. (Melanoma can be transmitted from mother to fetus, though; but this is off-topic right now). Human cancers occur by two primary mechanisms: random damage to the DNA due to environmental events (exposure to carcinogens, smoking, ionizing radiation, etc.), and damage to the DNA due to viral infections. These are not hereditary. Hereditary effects determine predisposition to particular types of cancer: breast cancer in BRCA1 gene carriers, colon cancer in Ashkenazi Jews, etc.; but we do not, to the best of my knowledge, carry extra genetic instructions that would make some of our stem cells differentiate into a specifically "cancerous" (but still beneficial) cell or tissue type. That of course doesn't mean that this does not happen. To understand why, let's first look into what makes a cancerous cell different from a normal cell. Cancerous cell, you see, can divide indefinitely: each one divides to make two new ones, with largely the same DNA (except usually for some mistakes), and unless there are too many of these mistakes, the two new cells can divide again, and again, and again. In fact, HeLa cancer cell line is still dividing, in vitro, since 1950s. The "normal" tissue cells, OTOH, can't do that. Instead, they have a so-called Hayflick limit on how many times a cell can divide; typically 35-60. The mechanism is simple: each time a cell divides, a small repetitive piece called telomere at the end of the DNA strand gets lost. Lose too many, and the cell can no longer divide. Now here is the tricky bit: all of our cells have genes that code for an enzyme called telomerase. This enzyme, when active, can regrow the lost telomeres. This enzyme is indeed active in most human malignant tumors. However, this also means that it is impossible to strictly define a "slow-growing" cancer: a cell with some telomerase activity may, theoretically, divide more times (100? 1000?) before reaching its Hayflick limit. Would you call such a cell a slow-growing cancerous cell or a normal cell with delayed senescence? I think at this point the question becomes purely semantic. I hope this helps. --Dr Dima (talk) 02:37, 19 May 2013 (UTC)[reply]

Thanks, that does help. Nice to have my question taken seriously rather than being browbeaten bwith questions from knowitalls.76.218.104.120 (talk)

I fail to find the provocative post neccesetated by the definition of 'flame war'. Plasmic Physics (talk) 02:58, 19 May 2013 (UTC) [reply]

Who are you to complain about provocative posts?76.218.104.120 (talk) 05:09, 22 May 2013 (UTC)[reply]

I stopped reading once I got to that phrase, lol. μηδείς (talk) 03:08, 19 May 2013 (UTC) [reply]

Incidentally, I also stopped reading at that same point. OsmanRF34 (talk) 12:37, 19 May 2013 (UTC)[reply]

Too bad :) --Dr Dima (talk) 03:31, 19 May 2013 (UTC)[reply]

Never mind me, I am sure your comments were of benefit to your intended audience. μηδείς (talk) 03:37, 19 May 2013 (UTC)[reply]

Is this what they call small talk? Plasmic Physics (talk) 03:49, 19 May 2013 (UTC)[reply]

Can small talk be so big?OsmanRF34 (talk) 12:37, 19 May 2013 (UTC) [reply]

Funny how nobody says they're engaging in "big talk" when it's just normal conversation. -- Jack of Oz ^[Talk] 23:37, 19 May 2013 (UTC) [reply]

Aye, but there is such a thing as big talk, as in "He talks big!" Plasmic Physics (talk) 00:07, 20 May 2013 (UTC)[reply]

Animals are slowly growing cancers. Count Iblis (talk)

Except, of course, that that conceit is entirely false, that animals follow a developmental program, with a set adult form, some like nematodes (see Caenorhabditis elegans} even having a determinate number of cells in the adult organism, while a cancer is defined as the opposite, an undefined, unregulated mass. μηδείς (talk) 23:44, 19 May 2013 (UTC)[reply]

I see, but surely there is at least some regulation, tumors need to tell the immune system to not attack them, they need to make bloodvessels to supply them etc. etc. Count Iblis (talk) 00:01, 20 May 2013 (UTC)[reply]

They don't 'need' to do those things, they just do therefore they are. Plasmic Physics (talk) 00:07, 20 May 2013 (UTC)[reply]

May 19

Quantum foam, zero-point energy, the fabric of the cosmos

Various articles in Wiki point mention the paradox of a cosmological constant that should be small, but when calculated, is many orders of magnitude too large. Wiki questions: Why doesn't the zero-point energy density of the vacuum (sometimes called 'quantum foam') change with changes in the volume of the universe? And related to that, why doesn't the large constant zero-point energy density of the vacuum cause a large cosmological constant? My question is: what if the 'quantum foam' at the Planck scale doesn't change because it is the real universe (supposedly infinite and eternal), and the Big Bang was only a one-off local outburst? Could that put to rest the paradox of the large cosmological constant? A related question is: If this is a possibility, what property or condition in the 'quantum foam' could have triggered the Big Bang?Robert van der Hoff (talk) 01:49, 19 May 2013 (UTC)[reply]

It's hard to see the value of inventing some random bizarre idea and then asking why it isn't valid. Looie496 (talk) 02:12, 19 May 2013 (UTC)[reply]

How would you test this conjecture of yours? I think you're delving past science and wandering into philosophy. Praemonitus (talk) 02:17, 19 May 2013 (UTC)[reply]

Thanks for your comments, points taken, but I'm still wandering and wonderingRobert van der Hoff (talk) 10:18, 19 May 2013 (UTC)[reply]

By definition a cosmological constant doesn't change its energy density with the universe's scale factor, that's just one of it's characteristics. By observation, we can say that dark energy appears consistent with a cosmological constant, but we don't know for sure that it's energy density is independent of scale factor. The observational parameter w measuring the equation of state of dark energy has a value of -0.98 +/- 0.05, where -1 means independent of scale factor and something like matter whose energy density scales in direct response to changes in the volume of the universe has a value of 0. As you note, attempts to explain a cosmological constant based on quantum mechanical properties of the vacuum give results that are much too large. At present, no one really knows why that is. Dragons flight (talk) 10:21, 19 May 2013 (UTC)[reply]

Thank you. The mystery remains Robert van der Hoff (talk) 23:02, 19 May 2013 (UTC)[reply]

What time of day and date of the year would most people on earth be in darkness?

My guess would be on the Winter Solstice (Dec 21), which has the shortest day in the higher populated Northern hemisphere. I would also think it would be when the sun was over the Pacific ocean, but before dawn in Asia, so perhaps around sunset on the US west coast?

Could we also calculate (roughly) what percentage of people would be in darkness? Jaseywasey (talk) 06:50, 19 May 2013 (UTC)[reply]

Your logic seems to be correct. It would probably be easier to calculate the number of people with daylight in those circumstances. You would also need to define darkness more precisely. Are you ignoring twilight?--Shantavira|^{feed me} 08:16, 19 May 2013 (UTC)[reply]

Could take into account this map... AnonMoos (talk) 17:14, 19 May 2013 (UTC)[reply]

Ha, I made a map of this once, with the winter solstice day/night curve, the "earth at night" light data, and, well, I made a guess about what time would be the "most dark". If the time isn't exactly right, it must be close. I guessed 0:30 GMT, Dec 22: http://www.flickr.com/photos/pfly/4133160304/sizes/o/ ...although I made this map with the idea of "most dark", in terms of nighttime lights being on, rather than population. Nighttime lights and population are related, but not the same. I suspect a slightly earlier time than my map shows would put more people in darkness—trading more daylight in the North America for more night in Indonesia and Japan. Pfly (talk) 08:45, 20 May 2013 (UTC)[reply]

Dark band between two lasers?

At the top right of this image, two laser beams are spreading away from the building and there is a noticeable dark band between them. Is this a form of Alexander's band? – Kerαunoςcopia^◁_galaxies 07:20, 19 May 2013 (UTC)[reply]

It is a retouched photo, the original is at File:The Shard, Inauguration Lightshow, 2012.jpg, there seems to be a little of the same effect but it is much less apparent. Personally I prefer the original. Dmcq (talk) 08:09, 19 May 2013 (UTC)[reply]

I kind of do too, it's what attracted me to it in the first place. Thanks for the mention, I've replaced the photo in the article with the original. Ok back on topic :) The dark band is still there. The image on the right is the original photo now, and it's still visible. Very curious what's causing it. – Kerαunoςcopia^◁_galaxies 09:37, 19 May 2013 (UTC)[reply]

I think this is an example of the optical illusion known as the contrast effect -- see http://www.michaelbach.de/ot/lum_dynsimcontrast/index.html for an especially vivid demonstration. Looie496 (talk) 14:42, 19 May 2013 (UTC)[reply]

It is clearly not any optical illusion as displying the picture full screen or blowing it up shows that the pixels between the two laser lines are quite a bit darker. The mostly likely explanation is that the image was "improved" by digital retouching and the person doing it messed it up. They may have wanted to lighten the sky by chroma keying and forgot the bit isolated between the lines. Wickwack 60.230.253.45 (talk) 15:57, 19 May 2013 (UTC)[reply]

Interesting. Ok thank you, I thought maybe this was a real-life occurrence. I ran the original image through "Jeffrey's Exif viewer" online and I see that Microsoft Pro Photo Tools was used, but seeing as how that's a metadata editor (maybe more, i don't know), I guess I really don't know what I'm looking at. I'd simply have to ask the original author if he did any touch-ups himself; that'd more than likely be the explanation I'm looking for. Thanks for all the replies. – Kerαunoςcopia^◁_galaxies 19:50, 19 May 2013 (UTC)[reply]

I disagree entirely with wickwack. I blew this up (on a 17" HD monitor) and looked at it before I read his comment, and the pixels between and outside the rays are the same color when the laser lines are obscured. μηδείς (talk) 22:02, 19 May 2013 (UTC)[reply]

Your eyes are mistaken, or they're just not sensitive enough, or your monitor isn't appropriately adjusted for this comparison. If you measure the sky's brightness using the tool of your choice, you can see a clear, quantifiable reduction in the area between the two beams in the upper right corner of the image. (For a free tool, download ImageJ, select a line of pixels, and choose Analyze...Plot Profile. The mean gray value is about 20% lower between the beams.) Whether this is an artifact of image processing in-camera or afterwards I can't say. TenOfAllTrades(talk) 02:38, 20 May 2013 (UTC)[reply]

Here is a screenshot (imgur.com) of an 11x11 pixel Photoshop analysis (on the original image as seen above) of three points, two on either side of the band. Using Lab, one can easily see that the band between the lasers is four points darker in the luminosity channel than on the sky outside of the lasers. There is also a slight shift in the a (red-green) and b (yellow-blue) color channels. It's definitely not an optical illusion. The photographer is busy and I'll have to bother him about it at a later date, but I'm not really seeing anything that might give away the fact that the image was edited outside of the lasers only. I can't read FotoForensics.com results either, if that site would even turn anything up. – Kerαunoςcopia^◁_galaxies 03:23, 20 May 2013 (UTC)[reply]

You don't need to go to all that trouble. You can just display the image on your monitor at full size. Then, take a piece of cardboard (eg from a business card or a cereal box) and cut two holes in it, about 8 mm diamter and about 25 mm apart. Then place the card against the screen and position it so that pixels from between the laser lines are viewed thru one hole, and pixels elsewhere in the sky are viewed thru the other hole. You will clealy see that the pixels between the laser lines are darker. The card prevents any possibilty of an optical illusion. Medeis obviously was just being silly. Wickwack 60.228.233.13 (talk) 04:18, 20 May 2013 (UTC)[reply]

Should I ask whether you are just being an idiot, wickwack? Personal comments are not necessary. I did the same thing, blowing up the image to max on a large screen laptop, blocking out the lasers as I said above, and the sky between the beams then appeared no different from outside to me. μηδείς (talk) 08:35, 20 May 2013 (UTC)[reply]

So who is being an idiot then? First you posted "I dissagree entirely with Wickwack." Now, after more than one person said you are wrong, you have again asserted what is clearly ridiculous. You have made several silly posts recently, resulting in several people attacking you. Are you on drugs or something? Wickwack 60.228.233.13 (talk) 09:02, 20 May 2013 (UTC)[reply]

(unindent)

Please guys...this is a silly argument. Just measure it with any halfway reasonable image tool - and the debate is over. Measuring a patch of sky from just outside of the two lasers got me an average brightness of 91/256, measuring an area between the lines got me 75/256 - so the image is DEFINITELY darker between the lasers. It's not an optical illusion. Furthermore, where the blue laser crosses the green ones, the blue laser is also attenuated...since the atmosphere between the two lasers would have to block that light somehow for this to be a "real" effect, we can rule out things like the two green lasers scattering light in all directions except right between them. For the blue laser to be attenuated, the air would have to become more opaque between the two lasers...which simply isn't reasonable. Hence this isn't any kind of peculiar real-world effect either. Furthermore, doing a Google Images search on "Shard inauguration laser show" produced a gazillion other photos of the event - many showing two lasers right next to each other like that - none that I could see showed this darkening effect, despite many shots from many different angles showing pairs of closely-spaced lasers.

Odds are very good then that this is an image processing artifact. If I had to bet on a cause, I'd say that someone used a "despeckle" filter on the image in an effort to eliminate the obvious noise in the background of the sky...but it's really hard to tell for sure. SteveBaker (talk) 16:10, 20 May 2013 (UTC)[reply]

Are dogs ever allergic to wheat gluten ?

I just saw an ad for dog food without wheat gluten, and they are obviously hoping people will buy it because they think it's healthier (although I notice they never actually made this claim). So, is it really healthier or is this just a scam ?

Or perhaps they are appealing to people who remember the 2007 pet food recalls, where Chinese poisoned wheat gluten killed many pets ? StuRat (talk) 18:45, 19 May 2013 (UTC)[reply]

A trivial Google search show many pages devoted to discussing food allergies in dogs in general, and wheat allergies in particular. Such as [1]. Based on that, it seems to be a real thing. Dragons flight (talk) 19:49, 19 May 2013 (UTC)[reply]

Exocrine_pancreatic_insufficiency is very common in some breeds, and can pop up in any dog. Many sources report gluten sensitivity, and recommend avoiding feeding gluten to dogs with EPI. SemanticMantis (talk) 19:53, 19 May 2013 (UTC)[reply]

Machine made of Human parts

I have a question regarding the name of a concept, and thus far have found no answers, save for some delightful articles on related subjects. Seeing as how Wikipedia knows everything, i thought this was the logical course of action.

if it is understood that a Cyborg is a hybrid between a CYBernetic being and an ORGanism, and an Android is a synthetic organism, a 'human' made of machine parts, then what would one call a 'machine' made of human parts? — Preceding unsigned comment added by 71.59.51.225 (talk) 19:02, 19 May 2013 (UTC)[reply]

"Frankenstein". ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:03, 19 May 2013 (UTC)[reply]

"Frankenstein's monster", not Victor. Clarityfiend (talk) 22:59, 19 May 2013 (UTC)[reply]

Yes. Hence the quote marks. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:06, 20 May 2013 (UTC)[reply]

Interpreting 'machine' loosely, dentures were at one time sometimes made using human teeth. AndyTheGrump (talk) 20:09, 19 May 2013 (UTC)[reply]

Even more loosely: soy sauce, apparently.. AndrewWTaylor (talk) 11:52, 20 May 2013 (UTC)[reply]

The film Existenz has an interesting take on this concept. Astronaut (talk) 17:56, 20 May 2013 (UTC)[reply]

I don't think this concept exists outside of sci-fi. Making a machine out of human parts would have the disadvantages of both and the advantages of neither. For example, I believe in The Matrix movie series, human "brain energy" powers the Matrix. This is the most absurd source of energy imaginable. It would take many times as much energy to keep the people alive as you would get from them. StuRat (talk) 06:53, 21 May 2013 (UTC)[reply]

May 20

Throwing one's hands in the air

When humans get excited, they tend to put their arms up. Three prominent examples are in worship services (e.g., File:Zhromko.jpg), at music festivals, and at sporting events. Why do they do this? Magog the Ogre (t • c) 00:21, 20 May 2013 (UTC)[reply]

When humans get excited, they do lots of things, they dance, they cheer, they clap, it's partially cultural, but I can't actually see what needs an explanation? What else can humans do when they get excited? All we have are arms, legs and voices. Unless you also have a lighter, a phone or a vuvuzella... Vespine (talk) 01:13, 20 May 2013 (UTC)[reply]

I've inserted a picture showing another type of excitement. Looie496 (talk) 01:26, 20 May 2013 (UTC)[reply]

@Vespine, I do not think it is cultural. I have observed it across cultures, and it comes quite naturally in some situations. Magog the Ogre (t • c) 04:27, 20 May 2013 (UTC)[reply]

I said partially cultural. There is a cultural component which would help determine exactly how people will react to exciting situations. Look at the history of Applause for example: Within each culture, however, it is usually subject to conventions.. Watching the world cup soccer staged in different countries can also reveal how different cultures react in exciting situations, they're not all precisely the same. Of course I'm NOT saying there aren't common components which aren't cross-cultural, of which lifting one's hands certainly might be one of the most "basic". Vespine (talk) 04:53, 20 May 2013 (UTC)[reply]

I once read the following, describing a certain unnamed culture: When speaking, they gesture frantically with their hands in an attempt to distract your gaze from their ugly faces, upon which are clearly etched the marks of their moral and intellectual degeneracy. -- Jack of Oz ^[Talk] 05:03, 20 May 2013 (UTC)[reply]

I once read in a scientific text that the reason rattlesnakes evolved rattles was because all snakes naturally shake their tails when excited, given they have nothing else to shake. μηδείς (talk) 08:30, 20 May 2013 (UTC)[reply]

I suppose it's possible that this is a gesture that's intended to show the opposite of "defensiveness". You're much more vulnerable to attack with your arms waving up in the air - and you can more clearly see that a person isn't armed. Perhaps this is a way for body-language to say "I'm excited, but not in an angry or threatening way." ?? SteveBaker (talk) 15:35, 20 May 2013 (UTC)[reply]

Or an urge to be associated with the win or celebration. Like "I am also part of this victory"; "I was on/cheering for the winning side" ; "I voted for you so don't put me in jail" etc.165.212.189.187 (talk) 18:28, 20 May 2013 (UTC)[reply]

Just don't get carried away with it.[2] ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:18, 20 May 2013 (UTC)[reply]

I remember reading a paper claiming that having an "open" or "closed" posture had an effect on behaviour and self-esteem. The study claimed that people who were more confident had an open posture, and that having an open posture led to increased confidence. It had people stay in closed or open postures for 10 minutes prior to being given a job interview, and the people who had a closed posture performed much worse than those having an open posture. Personally I found the methodology dubious for this particular study, but it is indicative of an attempt to link posture and confidence. It could be that when one is celebrating, they are feeling more confident and thus adopting an open posture is natural, according to this way of thought. This is merely idle speculation but it might open some avenues of research. Look up scientific papers on posture and self esteem / confidence. Superficially, Google will yield a lot of pop-psychology self help on the subject as well. — Preceding unsigned comment added by 64.201.173.145 (talk) 21:00, 21 May 2013 (UTC)[reply]

Decay product

Why does decay product is referred as 'daughter', not 'son' ? Concepts of Physics (talk) 13:00, 20 May 2013 (UTC)[reply]

The most likely reason is that in many cases it can go on to be the mother atom of a subsequent round of decay. -- 71.35.111.68 (talk) 16:17, 20 May 2013 (UTC)[reply]

No reason, just somebody started doing it that way and it caught on. Gender language, when assigned to anything regardless of it's actual sex (if any) is often arbitrary. StuRat (talk) 06:43, 21 May 2013 (UTC)[reply]

Here is a reason: only the female gender can give birth, thus it would volate that principle to call the decay product a son, as opposed to a daughter. Plasmic Physics (talk) 11:54, 21 May 2013 (UTC)[reply]

Mysterious lack of airspeed

This 1958 advertisement for Delta announced their nonstop DC-7 service, Atlanta to New York, took 2 hours, 39 minutes. My calculator tells me that's an average speed (assuming 760 statute miles from Atlanta Hartsfield to New York LaGuardia, per Google Earth) of about 286 statute miles per hour. Well and good, but a google search for today's nonstop flights reveals that 2 hours 10 minutes is the fastest scheduled time on a Boeing 737, or a speed of about 350 statute miles per hour. WTF? The article DC-7 gives a cruising speed of 359 mph for that 4-engine prop job, but the latest Boeing 737s have a cruising speed of 511 mph. Another couple of clicks shows that the DC-7 made the journey at an average of 80% of its cruising speed; yet the Boeing 737 travels at an average of only 68% of its cruising speed for the same journey. Thus the trip is only half an hour faster in 2013 than in 1958. Why so slow? Textorus (talk) 15:02, 20 May 2013 (UTC)[reply]

Some possibilities:

Airliners aren't at cruising speed for the entirety of the flight. The approach phase, particularly, is slow and often circuitous. However, that "less than cruising speed" speed is probably comparable between the DC-7 and a modern jetliner (at least more comparable than their cruising speeds), which means the modern jetliner will proportionally experience more delay relative to its theoretical minimum travel time.
Airliner routing is different over a 55-year span. The DC-7 didn't have to contend with either the volume of other aircraft in the airspace or the post-9/11 security measures that a modern jetliner encounters, both of which restrict the path an aircraft can take. More restrictions mean a longer flight, generally.
The definitions of "departure time" and "arrival time" may have shifted (this one is just speculation on my part). What is "departure time"? When no new passengers are admitted aboard? When the cabin door is shut? When the plane begins to move? When it actually leaves the ground? That's 15 to 30 minutes of possible range right there, and if the 1958 definition doesn't match the 2013 definition, it's potentially a significant input. — Lomn 13:56, 20 May 2013 (UTC)[reply]

(ec) Travel times are usually derived from gate-to-gate times. So undocking, pushing back, taxing, runway wait times, takeoff and acceleration times, plus the equivalent on landing all add to the total time without adding commensurately to average speed. Add that to the much higher traffic densities and larger taxiing distances for modern airports, and you should see why it is at least plausible that on shorter trips the times may not have improved much, despite the cruising speed increasing substantially. The large difference between your calculated speeds and the aircraft's cruising speeds tips one off to this. — Quondum 14:04, 20 May 2013 (UTC)[reply]

Departure from Atlanta was relatively straightforward in 1958: roll away (no pushback, since there were no jetways), taxi a mile or so to the end of the active runway, wait for maybe one or two take-offs and landings, and go. Nowadays it's possible to taxi five miles at Atlanta and wait for a dozen planes ahead of you. It's not as bad in terms of distance at LGA, but it's more congested - waiting for a gate (no pulling up on an empty ramp space anymore). Acroterion (talk) 14:19, 20 May 2013 (UTC)[reply]

As Quondum and Acroterion both note, the gate-to-gate time includes an awful lot of time spent not cruising, particularly on a shorter flight. Try working the math the other direction, and see what you get. A 760-mile journey at the DC-7's 359 mph gives 2:07 cruising time; if the total time is 2:39, then that's 32 minutes of taxiing, approaching, and so forth. A 760-mile journey at the 737's 511 mph gives 1:29 cruising time; given a total travel time of 2:10, there's 41 minutes of 'non-cruising' time.

Granted, that sort of back-of-the-envelope isn't quite realistic – aircraft obviously don't jump instantaneously from the runway threshold to full cruising speed and altitude – but it's not a ridiculous result. Nine minutes difference in the 'non-cruising' time can be down to differences in the way departure and arrival times are defined, increased traffic and delays at the two airports, changes in permitted routes, and the fact that many airlines have slightly reduced their cruising speeds to save fuel. TenOfAllTrades(talk) 15:14, 20 May 2013 (UTC)[reply]

I fly on a regular basis between London and Amsterdam. The flight is timetabled at a little over an hour. However, at Schiphol the plane will often take 20 minutes to taxi between terminal and runway (so much that it sometimes seems like we are going all the way there on the ground), and then spend less than 40 minutes in the air. Heathrow is a busier airport with fewer runways and sometimes long waits for other aircraft ahead. Astronaut (talk) 18:14, 20 May 2013 (UTC)[reply]

Another possibility that nobody so far has mentioned is that the Boeing's 511-knot cruising speed is only achievable at high altitude (above 25,000 feet) -- below that level, it drops off sharply so that near sea level, the V(ne) (that's the MAXIMUM safe speed) is only 330 knots! (This is true of ALL jetliners, as a matter of fact -- the engines produce less power at low altitude, and the denser, more turbulent air puts unacceptable stress on the airframe.) And since for such a relatively short flight, the plane spends a greater part of the journey at low altitudes, even its average FLYING speed will be lower because of this. (Not to mention the time spent in the holding pattern at the outer marker while waiting to land.) 24.23.196.85 (talk) 04:59, 21 May 2013 (UTC)[reply]

Thanks, guys, for all the responses. I suppose the relatively slow speed (to my mind) of the jet flight is due to some combination of all the points mentioned above. I see that for comparison, today's best flight time between Atlanta and Dallas (730 statute miles) is 2:25, or 61% of the 737's cruising speed, on average - even worse than to New York! Though from EWR to LAX (~2455 miles), the Boeing gets up to 88% of cruising speed, speaking on average once again - so it seems that the longer the flight, the more efficien the jet plane is.

NEW QUESTION then: Aren't we wasting a lot of fuel and money flying jets on short-to-medium routes? Wouldn't it be better to go back to propeller craft, which - correct me if I'm wrong - don't use nearly as much fuel and get you from A to B in nearly the same time? Wouldn't that help save the airlines, the economy, the trees, the whales, etc.?? Textorus (talk) 19:35, 21 May 2013 (UTC)[reply]

Yes, that's all true, but the key issue is whether consumers will accept turboprops; industry experience has shown that consumers prefer jets, all other things being equal. A casual google search suggests that the industry may be moving back towards turboprops for regional routes. — Lomn 19:53, 21 May 2013 (UTC)[reply]

Note that, while earlier jet engines were not fuel efficient, modern high-bypass jet engines compare pretty well with turboprops. Wickwack 121.215.140.49 (talk) 10:13, 22 May 2013 (UTC)[reply]

Human body

What is it called the upper part of the human foot? Thank you.175.157.180.68 (talk) 15:03, 20 May 2013 (UTC)[reply]

The top is the dorsal, unlike the underside 'plantar' surface (where one gets plantar warts – better known as verrucas). As it the foot, it is therefore the dorsum pedis. Does that help?Aspro (talk) 15:19, 20 May 2013 (UTC)[reply]

Wikipedia has an article on foot.--Shantavira|^{feed me} 15:29, 20 May 2013 (UTC)[reply]

It's called the instep. μηδείς (talk) 18:41, 20 May 2013 (UTC)[reply]

Original Earth-Moon distance

Hi: I looked up the entry for the Moon, trying to read what was the original distance between the Earth and Moon. I did not see it in the story. Did I miss it or is it not there? Thanks for you help. Red.leaf.flyers (talk) 16:53, 20 May 2013 (UTC)[reply]

It depends on which model for the Origin of the Moon you are working with. One of the dominant models is the Giant impact hypothesis, in which case the original distance between the Earth and Moon was 0 km, insofar as they were once the same body. --Jayron32 17:06, 20 May 2013 (UTC)[reply]

In a giant impact, material would be ejected into Earth orbit, where it would eventually coalesce to a ball. It is a bit hard to say where the orbit would be, but simulations suggest 3-5 Earth radii (20.000 - 30.000 km). An impact would have a hard time lifting enough material higher than five radii. Closer than about three Earth radii would put the impact ejecta inside the Roche limit, giving the Earth rings rather than a moon. 88.112.41.6 (talk) 17:39, 20 May 2013 (UTC)[reply]

Hang on, are you saying that since its formation, the moon has migrated from 3-5 earth radii away, to its current position 60 earth radii away? I'm not saying you are wrong, I just always imagined that the moon would have had to form very roughly around it's present position. However I can see now how that doesn't quite make as much sense as I thought it did... Vespine (talk) 06:54, 21 May 2013 (UTC)[reply]

According to Italo Calvino, it used to be close enough to climb up to it from a small boat, by way of a ladder. See La distanza della luna, the first vignette in Le cosmicomiche. It's available in English translation if necessary. Seriously, of course it's off-topic for this desk, but I very strongly recommend it. --Trovatore (talk) 06:57, 21 May 2013 (UTC) [reply]

On the topic of the Moon in fiction, Domingo Gonsalves harnessed large geese and tagged along for their annual migratory flight to the Moon circa 1638. Nimur (talk) 13:12, 21 May 2013 (UTC) [reply]

At it's current rate of motion, 3.8 cm / yr, the moon would migrate 25 Earth radii in the age of the Earth. However, that is almost certainly a lower bound as the rate of migration should have slowed over time. Dragons flight (talk) 07:57, 21 May 2013 (UTC)[reply]

It is actually possible to very roughly estimate this initial distance of roughly 30,000 km using a back of the envelope calculation. It boils to the fact that you have an impactor that had a similar orbit as the Earth, so it comes in from infinity at zero relative speed, and therefore the impact happens at escape velocity. If you where to give the ejecta that velocity, it would escape at infinity but, of course, a significant fraction of the impact energy is not available as kinetic energy. Then because the Earth radius is the quantity with the diemnsions of length, what happens is that if you make some rough approximations then whatever fraction is available, you find that the distance ends up being the Earth radius times some dimensionless factor, and that factor is not going to be very large like 100 or very small like 0.01. Count Iblis (talk) 11:48, 21 May 2013 (UTC)[reply]

As I am an amateur purveyor of all lunar literature, I have a fascinating text, Strange World of the Moon (authored by V. A. Firsoff in 1959) that makes an excellent chapter of dimensional analysis and inference about the origin of the moon. Needless to say, the book and its science predates manned spaceflight to the moon - or in fact, any spaceflight to the moon - so many of its conclusions have since been refined or refuted by better selenological evidence. In addition to the Giant Impact hypothesis, the author considers several other possibilities: tandem formation from a primordial nebula; capture of a separate celestial body; massive ejection by volcanic or other paleo-Earth processes; or simply large-scale fluid flow during Earth's formatory molten-rock era. From first principles of physics, and based on knowledge of orbital mechanics and basic facts of gravity, none of these alternate formation theories seem to sit well with the author; it quickly becomes clear why Giant Impact hypothesis gained traction in the following decades. However, even in 1959, it was easy to see that this was no ordinary impactor; the momentum necessary to eject a moon-sized object would have to be planetary in size. Evidence of the geochemistry of moon rocks - only possible after our first sample return missions in the late 1960s - strengthens the case; and many scientists now believe that the impactor may have been Mars. This is difficult to prove; but is widely accepted as "more plausible" than a mysterious impactor that has long since disappeared. For example, NASA's current science webpage at Solar System Exploration: Earth and Moon origin, asserts that the impactor would be "Mars-sized," without naming any names. Nimur (talk) 13:30, 21 May 2013 (UTC)[reply]

As an ex-aspiring astronomer who still reads popular accounts of the current art, I think you're slightly misinterpreting the proposed scenario. It isn't that another planetary body performed a hit-and-run on Earth, knocked off the Moon material, and went on its way largely intact (perhaps as Mars). Rather, the idea is that a (Mars-sized) proto-planet (sometimes dubbed Theia), stuck the proto-Earth, rendering both largely molten, and merged with the Earth: a good deal of the material splashed off from the impact then coalesced to form the Moon with much of the rest falling back to the now somewhat enlarged Earth. Only a relatively little debris would have escaped the Earth-Moon system, and the resulting materials of both the Earth and Moon would be similar but not identical blends of the original two colliders. {The poster formerly known as 87.81.230.195} 212.95.237.92 (talk) 14:05, 21 May 2013 (UTC)[reply]

Yes, and the impactor could plausibly have come from L4 or L5 as pointed out here. Count Iblis (talk) 15:16, 21 May 2013 (UTC)[reply]

I'm familiar with several variations on the theme. I think, based on factual evidence alone, there's still a lot of wiggle-room for scientific disagreement. It is my opinion that there are not very many conclusive facts about the early formation of the Earth-Moon system; rather, there are several competing theories and hypotheses supported by our sparsely-available evidence; each scenario varies in degree of plausibility. Nimur (talk) 22:09, 21 May 2013 (UTC)[reply]

Did I hear someone mention Immanuel Velikovsky? -- Jack of Oz ^[Talk] 23:35, 21 May 2013 (UTC)[reply]

A billion years from now interesting things could happen. Count Iblis (talk) 00:23, 22 May 2013 (UTC)[reply]

As explained here, General Relativity makes the solar system relatively stable: "These results also answer to the question raised more than 300 years ago by Newton, by showing that collisions among planets or ejections are actually possible within the life expectancy of the Sun, that is, in less than 5 Gyr. The main surprise that comes from the numerical simulations of the recent years is that the probability for this catastrophic events to occur is relatively high, of the order of 1%, and thus not just a mathematical curiosity with extremely low probability values. At the same time, 99% of the trajectories will behave in a similar way as in the recent past millions of years, which is coherent with our common understanding that the Solar System has not much evolved in the past 4 Gyr. What is more surprising is that if we consider a pure Newtonian world, the probability of collisions within 5 Gyr grows to 60 %, which can thus be considered as an additional indirect confirmation of general relativity." Count Iblis (talk) 00:41, 22 May 2013 (UTC)[reply]

May 21

what is the most amazing tornado footage ever filmed?

I would like to see some clear, compelling footage of a tornado destroying human habitats. Not a Hollywood movie, but real footage. I could just type tornado footage into Youtube and go fishing, but can you recommend a specific video?--Jerk of Thrones (talk) 04:34, 21 May 2013 (UTC)[reply]

Counter-intuitively, a smaller tornado may provide more graphic destruction. The larger ones tend to have a large dust cloud around them, obscuring the action, and the scale also makes it hard to pick out details. (What appear to be dots on the screen might be cars thrown about, for example.) Also, huge tornadoes don't twist much, and that makes them less interesting.

Double or triple tornadoes, where they twist around one another, are also visually interesting. StuRat (talk) 06:41, 21 May 2013 (UTC)[reply]

Start at National Severe Storm Laboratory's tornado education website, maintained by NOAA. They link to several videos. Nimur (talk) 11:54, 21 May 2013 (UTC)[reply]

Or watch CNN late afternoon US central time this week and keep your DVD recorder on standby. Count Iblis (talk) 13:02, 21 May 2013 (UTC)[reply]

Or watch this. Count Iblis (talk) 13:25, 21 May 2013 (UTC)[reply]

You might consider also waterspouts and fire devils. μηδείς (talk) 00:19, 22 May 2013 (UTC)[reply]

Spoiled Tabasco

Will my tabasco sauce spoil if I don't use it fast enough. Will mold and bacteria infest my precious tabasco sauce? — Preceding unsigned comment added by 128.214.48.186 (talk) 09:56, 21 May 2013 (UTC)[reply]

Unless you let some foreign matter into the bottle, vinegar and salt is a very unwelcoming place for anything to grow. An FAQ at tabasco.com gives a shelf life of five years for the regular variety, after which harmless discoloration may occur, but it shouldn't really spoil. You may want to shake an old bottle in case the ingredients have separated. Spices tend to lose their potency over time, so a really old bottle may taste different. 88.112.41.6 (talk) 11:17, 21 May 2013 (UTC)[reply]

To make it last longer:

1) Refrigerate it (even if the label doesn't say you need to, this will still extend it's life).

2) Keep bacteria out by keeping the lid on. When you use it, the lid should be off for seconds, not minutes or hours. And definitely don't pour it into a bowl, use the bowl for dipping, then pour it back into the bottle. StuRat (talk) 14:40, 21 May 2013 (UTC)[reply]

This is purely anecdotal, but I once found an old bottle of tabasco (in my grandparents' larder) on which the lid had split and no longer sealed, most probably some considerable time before. The sauce had lost its red colour (it was a sort of pale green - and no, it wasn't the green tabasco), most of its flavour and all of its heat. The advice about keeping the lid on is good, as is the advice about not pouring it back into the bottle, although the size of the opening on most sizes of tabasco bottle make that nigh on impossible anyway. Equisetum (talk | contributions) 15:02, 21 May 2013 (UTC)[reply]

Yes, it would involve use of a funnel. StuRat (talk) 21:59, 21 May 2013 (UTC)[reply]

One of the main benefits of spices is that they are preservatives, which rubs off on spicy condiments. μηδείς (talk) 00:16, 22 May 2013 (UTC)[reply]

Just to verify specifically what Medeis has alluded to, Capsicum varieties have been shown to have mildly anti-microbial properties, see this paper for example, mostly due to the capsaicin itself. So, even beyond the salt and vinegar, the Tabasco peppers themselves should be somewhat preservative. Which doesn't mean that peppers never rot, but that they do have compounds within them that inhibit bacterial growth. --Jayron32 03:19, 22 May 2013 (UTC)[reply]

The vinegar and salt would be the strongest of the three preservatives here. But I have never seen jarred hot peppers go bad. Oil also works as a preservative, since most bacteria require a more aqueous environment to thrive. Indeed, I have never had even half and half go bad on me due to the fat content. μηδείς (talk) 04:44, 22 May 2013 (UTC)[reply]

OK, we seem to be adequately answered, time for a distraction:

Q: How can you tell that an Iowa couple has been married for a REALLY long time?

A: They're on their second bottle of Tabasco!

--184.100.92.44 (talk) 01:19, 22 May 2013 (UTC)[reply]

why atomic silver clusters catalyze ionic silver reduction?

It's about photographic film development. General background like, silver halide is photosensitive and when it's hit by photon, few but electroconductive silver atoms formed on the silver halide crystals, then silver atom bearing sites become very sensitive to reducing agent and get reduced faster. Now the question, what is behind this atomic silver catalyzator? Why it catalyzes the redox reaction? I don't understand why atomic silver turns to be a catalyzator for silver ions. — Preceding unsigned comment added by 134.130.94.148 (talk) 13:53, 21 May 2013 (UTC)[reply]

This article seems to be a good introduction to the process. --Jayron32 14:52, 21 May 2013 (UTC)[reply]

Deriving the Saha equation Using statistical Mechanics

For a system consisting of hydrogen atoms and hydrogen ions, (i.e. where each particle is in one of two possible states: unoccupied (no electron present) and occupied (one electron present, in the ground state)), the Saha equation says that ${\frac {P_{p}}{P_{H}}}={\frac {kTn_{Q}}{P_{e}}}e^{-I/kT}$

where $P_{p}$ is the partial pressure of the ionized hydrogens, $P_{H}$ is the partial pressure of un-ionized hydrogens, and $P_{e}$ is the electrons pressure. $I$ is the ionization energy.

Is it possible to derive this equation, using statistical mechanics? AnalysisAlgebra (talk) 15:35, 21 May 2013 (UTC)[reply]

Yes, you can start with deriving the fact that in an equilibrium reaction where particles A1, A2, A3react and you get particles B1, B2, B3, according to the reaction formula

a1 A1 + a2 A2 + a3 A3.... <---> b1 B1 + b2 B2 + b3 B3 + ....

the chemical potentials of the particles A1, A2, A3,...B1, B2, B3 satisfy the equation:

a1 muA1 + a2 muA2 + a3 muA3 +.... = b1 muB1 + b2 muB2 + b3 muB3 + ....

Then, assuming that the particles are weakly interacting and that you therefore have an ideal gas, the chemical potential can be expressed in terms of the one particle partition function. The above formula for the chemical potential then implies:

(Na1/Za1)^a1 (Na2/Za2)^a2 (Na3/Za3)^a3...= (Nb1/Zb1)^b1 (Nb2/Zb2)^b2 (Nb3/Zb3)^b3...

where Zar and Zbr are the single particle partition function for particles of type ar and br. Count Iblis (talk) 16:01, 21 May 2013 (UTC)[reply]

Are you trying to say

Z_{H}=Z_{p}Z_{e}

?? But what are these quantities? What is the energy of an isolated electron? AnalysisAlgebra (talk) 16:53, 21 May 2013 (UTC)[reply]

I forgot to inser the particle numbers in the formula, the correct expression is

${\frac {N_{H}}{Z_{H}}}={\frac {N_{p}}{Z_{p}}}{\frac {N_{e}}{Z_{e}}}$ . The partition functions are quite easy to calculate, you only have to take into account the binding energy in Z_H. If we forget about the binding energy, we have for a particle of mass m that $Z=V\left({\frac {2\pi mkT}{h^{2}}}\right)^{\frac {3}{2}}$ . Here V is the volume. Now you have to multiply this by the spin degenracy, whch for hydrogen is 4 and for the proton and electron is 2,. but then these factors cancel out in the above equation. Then, Z_H gets an extra factor of exp[-Eb/(k T)] where Eb = -13.6 eV is the binding energy.

So, to summarize, what you have to do is to use the fact that in an isolated system the entropy is maximla in thermal equilibrium to prove the relation between the chemical potentials. Then you can do the same for a system kept at constant temperature and volume (there the Helmholtz free eenrgy is minimal, which you have to be able to prove also), or for a system kept at constant pressure and volume where the Gibbs free energy is minimal. In all these cases you get the same relation between the chemical potentials.

Next, you derrive the equation for the chemical potential e.g. by usung the fact that the Helholtz free energy is F = -k T Log(Z) where now Z is the full partition function of the system and that dF = -S dT - P dV + mu dN, so the partial derivative of F w.r.t. N at constant T and V is equal to mu. Then the partition function for an ideal dilute gas consisting of Nj molecules of type j is given by Z = Z1^N1/N1! Z2^N2/N2! ...., where the Zj are the single particle partition functions. You can to derive this by using the definition fo the partition function and the fact that permuting the dientical particles doesn't yield a new state. Then if the gas is dilute you only have a negligible contribution to the partition function where more than one particle of the same type are in the same state. This allows you to compute the partition function by taking the products of the powers of the one particle partition function and then divide by the factorials of the particle numbers to compensate for the overcounting. Count Iblis (talk) 18:08, 21 May 2013 (UTC)[reply]

So

{\frac {N_{H}}{m_{H}^{3/2}e^{I/kT}}}={\frac {N_{p}}{m_{p}^{3/2}}}{\frac {N_{e}}{V\left({\frac {2\pi m_{e}kT}{h^{2}}}\right)^{3/2}}}

. But that doesn't imply the statement I want to prove, does it?? For example, there's no extra factor of

kT

. AnalysisAlgebra (talk) 03:03, 22 May 2013 (UTC)[reply]

What you get using the ideal gas law and approximating the ratio of the proton and hydrogen mass as 1 is

{\frac {P_{e}P_{p}}{P_{H}}}=kT\exp \left(-{\frac {I}{kT}}\right){\frac {Z_{e}}{V}}

Ze/V is also sometimes denoted as nQ. You can interpret the single particle partition function per volume as the effective number of quantum states available per particle per unit volume, so, it's a density of quantum states for the electron hence the symbol nQ. To see this, consider that the partition function for a single particle is just the sum over exp[-E/(kT)] over all the available quantum states. While this is dimensionless, it is proportional to the volume because the number of quantum states increases as the volume increases (the spacing between energe levels gets less as the volume increases). So, you can express this dimensionless number as the volume divided by an effective volume VQ. Obviously the physical interpreation of VQ is the volume at which one particle would have effectively just one state available when taking into account the penalty in the form the Boltzmann factor that disfavours states with energy much higher than k T, so 1/Vq is the effective density of available quantum states per particle. Count Iblis (talk) 12:36, 22 May 2013 (UTC)[reply]

Teen Invents Super fast battery charger

Is this real? Can anybody shed some more light on this than the poorly written article? It seems highly unlikely that this girl has done what multi-bollion dollar energy companies cant (or just dont want to?)165.212.189.187 (talk) 17:27, 21 May 2013 (UTC)[reply]

Seems legit. Here is the original press release announcing her invention from the Society for Science and the Public and the International Science and Engineering Fair. She won $50,000 for her efforts, and that doesn't seem like the kind of scratch someone just gives away for fun. If you wanted the full details of her presentation or paper, I'm pretty sure you can contact that organization. --Jayron32 17:37, 21 May 2013 (UTC)[reply]

As a side note, it does happen sometimes that rank amateur will invent something on their own, rather than a team of employees in an R&D department for some multinational corp. For as many Wallace Carothers there are in the world, there's likely as many Philo Farnsworth's; Farnsworth had essentially invented modern television in his barn at 15 years old. Also, Steve Wozniak, Lee De Forest, Erasto Mpemba, etc. for examples of people who made significant scientific and technological advances completely independent of any large organization or corporation. --Jayron32 17:44, 21 May 2013 (UTC)[reply]

Isn't Mpemba regarded as more of a quasi-mythological culture hero these days? Evanh2008 ^{(talk|contribs)} 00:52, 22 May 2013 (UTC)[reply]

Erasto Mpemba is known for his student paper of the Mpemba effect, which counter intuitively says that warm water placed in a cold chamber freezes faster than cold water. Trouble is, his results appear to be based on faulty lab work. Also, Steve Wozniak was a 26-year old engineering employee of Hewlett-Packard when he designed his first computer, which was not particularly innovative and certainly no engineering breakthrough. Stories have been repeated about him single-handedly designed it at home, which given that there was no PC-based CAD at the time was certainly an achievement, but not a remarkable one. Wickwack 121.215.10.17 (talk) 01:03, 22 May 2013 (UTC) [reply]

Fine, forget Wozniak then. He did nothing of import. I suppose that destroys my thesis that sometimes people can invent something at home. I guess no one has ever done that. --Jayron32 01:12, 22 May 2013 (UTC)[reply]

Not at all. You just picked a couple of poor examples. A number of famous internet and cellphone applications were develoiped by teenagers at home. As a competent Engineer I stand in awe of them. Nor is Wozniack unimportant. In linking up with Steve Jobs, what Wozniack did later led to a profound change in how we think about PC's and what we expect of them. Wickwack 121.215.10.17 (talk) 01:24, 22 May 2013 (UTC)[reply]

Not that it's on topic, but I verified the Mpemba effect under controlled conditions in 2006. It's very easily reproducible, and the fact that it's languished so long on the fringes of thermodynamics, almost like a piece of pseudoscience, depresses me. Not that it's on topic... Evanh2008 ^{(talk|contribs)} 01:18, 22 May 2013 (UTC)[reply]

You might check out the Slashdot discussion on the subject. Briefly, capacitors are not a new invention. They suffer from a couple of fundamental problems: their power density is tiny compared to batteries; and unlike batteries, their output voltage drops as you draw power from them. No fundamental discovery is in evidence here, more like what has been done before was demonstrated by an attractive young lady, which is catnip to popular media. 88.112.41.6 (talk) 17:53, 21 May 2013 (UTC)[reply]

She developed an improved supercapacitor using rutile titanium dioxide crystals which provides 238.5 Farads per gram, with energy density of 20.1 Wh/kg, and power density of 20540 W/kg, and a loss of about 1/3 the storage after 10,000 cycles. Pretty damned impressive stats, since Battery (electricity) says the highest energy density for present rechargeable batteries is 0.46 MJ/kg or 128wh/kg See http://www.technewsworld.com/story/Teenagers-Power-Storage-Project-Lights-Up-Science-World-78085.html . Efficient circuitry to provide a constant output voltage to drive electronic circuits would be needed before we can throw away our celphone batteries and chargers. Maybe the idea is to dump energy into the supercapacitor in a few seconds, then charge the battery from the capacitor through some sort of voltage dropping circuit with an invertor, transformer, and rectifier. The putdowns of her just being a an "attractive young lady" who has done nothing special seem uncalled for. I also see websites where various anonymous persons are claiming first, that the energy storage device is noting special, and second that the real work must have been done by her parents or the professor. That just makes them look envious. She showed quite a bit of initiative in contacting dozens of college professors with a request for the use of lab facilities before she was allowed to use space at UC Santa Cruz. Edison (talk) 18:52, 21 May 2013 (UTC)[reply]

I tried to read our Supercapacitor article, but it seems to have been written by someone who is not fluent in English, as demonstrated by "All this first electrochemical capacitors used a cell design of two aluminum foils covered with activated carbon coins the electrodes which are soaked with an electrolyte and separated by a thin porous insulator implemented in a common housing." Anyone out there with an understanding of material science who could take the time to copyedit that article, which is both important and an embarrassment to the project? Edison (talk) 19:42, 21 May 2013 (UTC)[reply]

The article Electric double-layer capacitor largely duplicates what is in Supercapacitor, but gives a higher energy density for rechargeable batteries than does Battery (electricity). It says supercapacitors store up to 85 wh/kg, in a lab prototype. Ms. Khare's device might or might not scale up successfully. Edison (talk) 20:06, 21 May 2013 (UTC)[reply]

Melting point of a diamond

What temperature does diamond itself melt at? is it even possible for humans to melt? what device would be used to melt a diamond? Does it crack or burn first before it melts? Are there any scientific processes that involve the melting of a diamond? Thank you Horatio Snickers (talk) 19:28, 21 May 2013 (UTC)[reply]

You can't melt a diamond (or carbon in general) at any ordinary pressure. It sublimes rather than melting. The lowest pressure where you can get liquid carbon, if the diagram is correct, is around 10 MPa, which is about 100 atmospheres. --Trovatore (talk) 19:33, 21 May 2013 (UTC)[reply]

(edit conflict) Read up on diamond and Carbon#Characteristics. At atmospheric pressures carbon - which is what diamond is - sublimes at about 3,900 K, ie.: it goes straight from a solid to a gas. At much higher pressures and temperatures it is theoretically possible to turn carbon into a liquid, see phase change diagram. WegianWarrior (talk) 19:51, 21 May 2013 (UTC)[reply]

Why do you say "theoretically"? 10 MPa is a lot of pressure and 4000 K is hot, but both are obtainable in the laboratory. At least one of the refs in the carbon article has a title suggesting that the properties of liquid carbon have been experimentally measured. --Trovatore (talk) 20:01, 21 May 2013 (UTC)[reply]

Perhaps, because no one has tried? Plasmic Physics (talk) 21:06, 21 May 2013 (UTC)[reply]

If you read my last sentence, it appears that someone has in fact tried. --Trovatore (talk) 21:10, 21 May 2013 (UTC)[reply]

My appologies. Plasmic Physics (talk) 03:09, 22 May 2013 (UTC)[reply]

See http://www-als.lbl.gov/index.php/holding/276-time-resolved-study-of-bonding-in-liquid-carbon.html. Looie496 (talk) 22:54, 21 May 2013 (UTC)[reply]

Meowing of male and female cat

Is there a difference in meowing between male and female domestic cat, such as pitch, etc (assuming both are of the same age)?--93.174.25.12 (talk) 21:00, 21 May 2013 (UTC)[reply]

See Cat_communication for starters. Note that "A "caterwaul" is the cry of a cat in estrus (or "in heat")." --that is a very distinctive, loud, yowling call that only females make. As the article mentions, adult cats don't really meow to eachother, it is a kitten-to-mother call that got co-opted for cat-to-human communication somewhere in the domestication process. Anyway, aside from a cat in heat, my WP:OR is that there is a great deal of variation in meowing sounds among among individuals, with no clear sex-based differences. I have known some surly tom cats with tiny, weak "mew", as some with loud commanding "MEOOOW" -- and also the same for females. SemanticMantis (talk) 21:21, 21 May 2013 (UTC)[reply]

On average I would expect male cats to have deeper meows than females, since they are slightly larger and presumably have longer vocal chords. Of course, there will be exceptions. StuRat (talk) 00:26, 22 May 2013 (UTC)[reply]

Speaking purely impressionistically, but with a great deal of experience with cats, the males do tend to exhibit vocalizations of lower auditory frequency on average, but of course there is a great deal of variation; bear in mind that although all domestic cats belong to Felis catus by virtue of being able to produce viable and sexually productive offspring, there is a great deal of phenotypical differences between them, and the "voice" is no exception. That being said, the animal's size, health, and of course state of mind will all influence the tone of their meowing. I will say that while I understand the point that Mantis is getting at, it is not entirely true that adults do not meow at each other - they indeed do it a great deal, even in feral colonies and other circumstances outside domestication -- they simple do it under generally different circumstances than the greetings or attention-grabbing (for example, feeding) purposes they use it for with humans. Snow (talk) 01:19, 22 May 2013 (UTC)[reply]

I'm doing some digging for references, but so far this is what I came up with: This paper titled "The effects of articulation on the acoustical structure of feline vocalizations" seems to say, from the abstract, that the qualities of "feline vocalizations" is tied to the physical structures that produce those vocalizations, and that cats could be used as analogues for differences in human voices. While that does not directly answer the question, it does point in the following directions: differences in human voices are tied to differences in the structure of the voice-making apparatus in humans; that is the sexual dimorphism in human voices between males and females is directly related to the differences between the physical structures that make the voices in males and females and b) this paper indicates that similar processes may be at work in cat voices. To make the final connection, we'd need to show that cats directly display the same sort of sexual dimorphism, which this paper does not seem to. Still looking tho. --Jayron32 01:45, 22 May 2013 (UTC)[reply]

Not cats, but rats: This paper titled "Rat 22 kHz ultrasonic vocalizations as alarm cries" states, in the abstract, that alarm cries in rats "show gender differences", that is you can identify a rat as male or female from its alarm cry. Again, not cats, but still, it does show that gender differences in voice is exhibited outside of humans. If this has been done for rats, perhaps a study has been done for cats. Still digging. --Jayron32 01:50, 22 May 2013 (UTC)[reply]

No idea if it's a good match, but I did find This book, titled Your Ideal Cat: Insights into Breed and Gender Differences in Cat Behavior. Since it notes gender differences right in the title, it's a possibility it may cover meowing and other vocalization difference between male and female cats. --Jayron32 01:57, 22 May 2013 (UTC)[reply]

Well, after an fairly exhaustive search through every subset of Google, I can't find anything one way or another. It seems that, with the notable exception of "caterwauling" (the calls of a female cat in estrous), there are not significant differences between male and female cats in terms of their vocalizations. Or, at least, there is no evidence of anything in the literature to indicate such a difference (with the very important caveat that the absence of evidence is NOT the evidence of absence). --Jayron32 02:11, 22 May 2013 (UTC)[reply]

Spelling note: estrous is an adjective; the noun is estrus. I'll have to add that to my mental list of this particular spelling bugbear that gets a lot of people. More common pairs to confuse are mucus–mucous and callus–callous. Another slightly exotic one is phosphorus—phosphorous. In each case, the word that ends in -ous is the adjective. --Trovatore (talk) 02:24, 22 May 2013 (UTC)[reply]

Mi scuso, mio amico... --Jayron32 02:29, 22 May 2013 (UTC)[reply]

That's normally mi scusi as one is asking the other to excuse, not forgiving oneself. μηδείς (talk) 04:49, 22 May 2013 (UTC)[reply]

Pardonnez moi --Jayron32 04:55, 22 May 2013 (UTC)[reply]

Actually I thought that part was fine. Scusarsi can mean to apologize. Mi scusi (formal), or (more likely in this sort of exchange) the informal scusami, is a direct command, and these are more usual when talking to the interested party, but if you were talking to a third party in an high register, mi scuso col signor Trovatore would be entirely acceptable.

The part that didn't quite work was mio amico. Should be amico mio. --Trovatore (talk) 06:43, 22 May 2013 (UTC)[reply]

I don't know if the references unearthed by Jayron (well done!) mention this, but altered males (i.e. those who have had their testicles removed) will have higher-pitched meows than entire males, especially if the castration was done at an early age (4 months). OR - my male cat emits a higher-pitched squeak than either of my two female cats, and higher than my late male cat: I believe this is because he was neutered at an earlier age than my late male cat. In other words, castration has the same effect on cats as it does on humans. (No shit Sherlock, I hear you say!) --TammyMoet (talk) 10:10, 22 May 2013 (UTC)[reply]

I also wonder whether neutering alters the cat's voice, couldn't find an RS at first glance. 93.174.25.12 (talk) 11:23, 22 May 2013 (UTC)[reply]

May 22

Which types of dogs lives longer, larger dogs or smaller dogs?

The Chinese article says smaller dogs lives longer. Are there any differences?--朝鲜的轮子 (talk) 03:18, 22 May 2013 (UTC)[reply]

Large dogs are notoriously short-lived, with Irish Wolfhounds and Great Danes living 6 to 8, and sometimes 10 years, as noted in those articles. Generally, the smaller the dog, the longer the lifespan. Acroterion (talk) 03:21, 22 May 2013 (UTC)[reply]

Here's a good source backing up small dogs living longer: Which dogs live longest? Red Act (talk) 03:41, 22 May 2013 (UTC)[reply]

(EC)I just found the same article:) this article is interesting. It says that it's the weight that matters not necessarily the height and that generally dogs under 30 lbs live the longest. Vespine (talk) 03:45, 22 May 2013 (UTC)[reply]

Measuring entropy

How can one actually measure the total entropy of a system? 130.56.235.221 (talk) 05:04, 22 May 2013 (UTC)[reply]

Entropy is often defined thermodynamically; you don't measure it directly, you derive it from other measurements of the system. A fairly easy way to define it would be from the Gibbs free energy equation: the entropy of a system is simply S = (H-G)/T; that is you subtract the free energy from the enthalpy and divide that result by the temperature. In practical terms, you don't measure any of these values directly, you measure them as incremental changes as, for example, a chemical reaction progresses. So, we generally say then that ΔS =(ΔH-ΔG)/T, where the Δ values are the changes in those values you get as a result of changes to the system; so for a chemical reaction you can get ΔH by calorimetry, knowing that for a constant pressure system, ΔH is equal to the heat evolved or absorbed during the chemical reaction. ΔG can be calculated from the equilibrium constant, ΔG = -RT ln (K) which itself is calculated from the relative concentrations of the substances involved in the reaction. T is measured directly. So, you can calculate ΔS by going through several sets of calculations based on measurements you can make from first principles (i.e. concentrations, temperatures, etc.) --Jayron32 05:35, 22 May 2013 (UTC)[reply]

Special Relativity: follow up

In order for Special Relativity to develop its equations (e.g. Lorentz transformations), it's essential to assume that the speed of light does not depent on the inertial system measuring that speed; Is it essential to assume that the speed of light does not change over time/space either? HOOTmag (talk) 19:21, 12 May 2013 (UTC)[reply]

It's not essential to start from the constancy of the speed of light. Einstein did it that way in 1905 because at that time everyone knew that the speed of light was c but couldn't figure out what that speed was relative to, and he wanted to point out that you can just take the speed to be c, full stop, not relative to anything in particular, without any logical contradiction. There are other ways of motivating special relativity, though. For example, you can derive it from the reciprocity of redshifts: if two rocket ships move inertially away from a common starting point, each one sees the other redshifted by the same factor. That gets you a theory with the same mathematical structure as Einstein's theory that doesn't say anything about the speed of light as such. You can then, in the course of defining your system of units, take the speed of light to be constant. You don't have to, but you can without any contradiction—that's what distinguishes special relativity from Newtonian physics.

So it doesn't really make sense to ask whether the speed of light varies with position or time since that depends on how you define your units of measurement. A real time variation of physical constants would show up as a change in some other measurable quantity, such as the electron-proton mass ratio. -- BenRG 21:02, 12 May 2013 (UTC)

Admittedly, I'm rather surprised by your response. As far as I understand, Lorentz transformations are results of Special relativity, aren't they? Whereas assuming them is mathematically equivalent to assuming that the speed of light does not depend on the inertial system measuring that speed, isn't it? Hence, assuming that the speed of light does depend on the inertial system, contradicts Special Relativity, doesn't it? I'm just asking whether assuming that the speed of light changes over time - contradicts Special Relativity. HOOTmag (talk) 21:20, 12 May 2013 (UTC)[reply]

On the first point, it would be more correct to say that special relativity is a result (or an application) of the Lorentz transformation, rather than the other way round - see History of Lorentz transformations. Tevildo (talk) 23:24, 12 May 2013 (UTC)[reply]

From a historical retrospective point of view - you're right, but I'm talking from a relativistic point of view. Einstein concluded Lorentz transformations from the constancy of speed of light, not vice versa. HOOTmag (talk) 07:20, 13 May 2013 (UTC)[reply]

Minkowski spacetime is homogeneous in space and time, and you need some assumption to distinguish that from some other spacetime geometry that isn't, but I'm not sure that gets at the core of your question. The meaning of "speed" in Einstein's postulate is not obvious a priori, since the paper argued that the seemingly obvious notions of distance and time that physicists had had until then were actually wrong. Einstein uses the postulate to justify his method of synchronizing clocks, and it's not until the synchronized clocks are introduced that the inertial reference frames are defined and the "speed" in the original postulate has a clear meaning. That's okay because all physical theories are circular in that way (see this thread, the final reply beginning "There is an unavoidable circularity...", where I think I explained this better than I'm doing it here). The postulates in the original paper are a jumping-off point for the argument, but because of the circularity they aren't really postulates in a formal mathematical sense, and there's no useful distinction to be made between assumptions and conclusions. -- BenRG 07:33, 13 May 2013 (UTC)

Logically speaking - every assumption is a conclusion, while for the practical purpose of the current thread - I don't distinguish between assumptions and conclusions. By saying that A is an assumption/conclusion of B, I just mean that the negation of A contradicts B. As for Minkowski spacetime: My original question refers to Minkowski's relativistic equations as well. HOOTmag (talk) 08:23, 13 May 2013 (UTC)[reply]

I think the word for which you're looking is consequence. Conclusions are whatever propositions that are designated as conclusions (i.e, meant to be proved). If an assumption is not designated as a conclusion (and they usually aren't), then it isn't one. --Atethnekos (Discussion, Contributions) 17:23, 13 May 2013 (UTC)[reply]

As explained here, the speed of light is not a real physical constant. Count Iblis (talk) 23:51, 12 May 2013 (UTC)[reply]

Other users are giving IMO unnecessarily complicated answers, so to keep it simple the answer to OP's question is yes. In his 1905 paper, Einstein mentions explicitly (albeit briefly) that "it is clear that the [Lorentz transformations] must be linear on account of the properties of homogeneity which we attribute to space and time", which basically implies that the speed of light doesn't change in space or time. 65.92.6.9 (talk) 03:01, 13 May 2013 (UTC)[reply]

And, by "homogeneity," Einstein means to say that the value of the permittivity of free space and the permeability of free space - commonly, ε₀ and μ₀ - are well-defined and always constant at all positions. From this postulate, the equation of retarded time is trivially found by solving Maxwell's equations. The Lorentz transform is a mere algebraic simplification of the more general form.

By coincidence, I had been reading The Sign of the Four this evening - published 1890 - and it referenced (very indirectly) the Elements of the Philosophy of Newton by Voltaire (Holmes is quoting a pithy bit of French, and alluding to shedding some light on the case). Naturally, my inclination was to track down the text, and read as much of it as I could... it is available online (but not at Project Gutenberg, unfortunately, nor in English translation - here it is at Elementi della filosofia di Newton). It is absolutely amazing! To read Voltaire succinctly express Isaac Newton's optics - to talk about the constancy of the speed of light, and to talk of light as both a ray and as a particle... published in the year 1738 as a regurgitation of Newton's earlier and far more technical writings on optics - a thought crossed my mind, which I will summarize here... "those who think Einstein's work was really amazing have not spent enough time reading the works of his predecessors." Nimur (talk) 04:04, 13 May 2013 (UTC)[reply]

See Standing on the shoulders of giants, a quote not created by, but often attributed to, Newton on his own work. It applies to every scientist in history since the first caveman tried to make fire. --Jayron32 04:28, 13 May 2013 (UTC)[reply]

The giants on whose shoulders the cavemen stood were, of course, the nephilim. I'm not exactly sure how standing on a nephil's shoulder helps you make fire, but there you are. --Trovatore (talk) 08:11, 17 May 2013 (UTC) [reply]

OP's comment: I still wonder which relativistic equation must result in the constancy of speed of light in space/time. The equation of retarded time - just results in the equation:

c={\frac {|\mathbf {r} -\mathbf {r} '|}{t-t_{r}}}

, which does not tell us whether the very value:

{\frac {|\mathbf {r} -\mathbf {r} '|}{t-t_{r}}}

is constant. Note that Lorentz transformations do tell us that the speed of light does not depend on the inertial system measuring that speed. HOOTmag (talk) 07:20, 13 May 2013 (UTC)[reply]

You are confusing a result with a derivation. The derivation proceeds along a line of mathematical reasoning that starts by solving Maxwell's equations for a moving source. This gives a description of the electric and magnetic fields at all points in space and time. From that, a lot of algebraic manipulation gives you a wave equation with a propagation speed, independent of the motion of the source. Our articles cover these topics, but this is fairly advanced mathematics and physics. My recommendation is to begin studying the wave equation in its classical form, until it is so intimately familiar to you that you recognize it, even when obfuscated by multiple independent variables. Then you will be able to identify propagation velocity by inspection. More bluntly: even if you are an autodidact, you require a many years of mathematical preparation before you can reasonably interpret and use the equations that govern the relativistic behavior of light. Commonly, this means three to five years of rigorous study of introductory calculus and physics at a university level. It's a bit unreasonable to think that an encyclopedia can expedite that process to just a few days or hours. Nimur (talk) 14:56, 13 May 2013 (UTC)[reply]

No, I was not confusing a result with a derivation: I just wondered "which relativistic equation must result [by a mathematical derivation] in the constancy of speed of light in space/time". Anyways, as opposed to what you've claimed, I don't think one needs "three to five years of rigorous study of introductory calculus and physics at a university level" in order to answer my original question. HOOTmag (talk) 08:22, 16 May 2013 (UTC)[reply]

I said that mathematical preparation is usually necessary. You might be a genius beyond everyone's wildest expectations. But you're still asking the same question, which has already been answered incredibly thoroughly, leading me to believe that you don't have the prerequisite context so that you can understand the answer. Which part are you still stumbling on? A constant speed of light is a direct consequence of the assumption that the permittivity of free space and the permeability of free space - commonly, ε₀ and μ₀ - are well-defined and always constant at all positions. These parameters define the speed of light in our best theories of electromagnetics, and this premise matches physical experiment. Are you unable to see the link between physical observation and the equation that models it? If so, I suggest you start reading about electrostatics, and then electrodynamics. Nimur (talk) 13:58, 16 May 2013 (UTC)[reply]

The well-known constants ε₀ and μ₀ are really assumed to be constant at all positions, and we undoubtedly shouldn't forget that Einstein was highly inspired by their constancy when he developed his relativistic mechanics, even without us mentioning the important role played in electromagnetism by Special Relativity (e.g. by its motivating the "manifestly covariant" tensor form, and by giving formulas for how the electric and magnetic fields are altered under a Lorentz transformation from one inertial frame of reference to another, and by showing that the frame of reference determines whether the observation follows electrostatic or magnetic laws). However, I was not asking about relativistic electromagnetism, nor about the psychological inspiration of the constants ε₀ and μ₀ in Einstein's mind when he developed his relativistic mechanics. I was just interested in relativistic mechanics per se, i.e. in its explicit assumptions (e.g. the constancy of speed of light) and in its results (e.g. Lorentz transforms). Relativistic mechanics involves concepts like: time, space, mass, force, momentum, energy and the like, yet not concepts like electric charge or magnetic field. Relativistic mechanics assumes/concludes some claims, e.g. the equation of time dilation (and likewise), yet not any claim about ε₀ and μ₀, so that one can study relativistic mechanics, without having studied Maxwell's theory, and still wonder whether - one must assume/conclude the constancy of speed of light in time/space - in order to fully grasp the fundamental principles of relativistic mechanics.

As for mathematics: as a mathematician I can assure you that one needs no advanced mathematics for understanding whether the constancy of speed of light in time/space is needed for relativistic mechanics. HOOTmag (talk) 17:51, 16 May 2013 (UTC)[reply]

ds^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2. It's really not different from the factors of 1 multiplying dx^2, dy^2 and dz^2, which you can relate to Pythagoras' theorem (and which you can change to arbitrary values by using different units for measuring distances in the x, y and z directions). Count Iblis (talk) 13:12, 13 May 2013 (UTC)[reply]

The invariant ds^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2, only shows that c can't depend on the inertial frame. However, this invariant may vary in time, hence - one can suppose c itself varies in time - without contradicting the very invariant mentioned above. HOOTmag (talk) 08:22, 16 May 2013 (UTC)[reply]

Yes, but the c in this equation doesn't have any physical meaning whatsoever, you can just set it equal to 1. It's similar to writing the first law of thermodynamics as dE = dQ - p dW instead of dE = dQ - dW, because we could insist on using different units for measuring work and heat; we could have assigned work a different dimension from heat making the units for work incompatible with the unit for heat. You could then ask if you could prove using the laws of thermodynamics that p is a constant, if it can depend on time etc. etc.

Just like E = m c^2 says that mass is equivalent to energy in the sense that a box containing an amount of mass m in rest will have a total energy content of m c^2, we would say that an amount of work dW done by a system results in removing an amount of energy of p dW from the system. Advanced books on thermodynamics would work in p = 1 units. But at high school level, teachers would insist that work is not energy, as they have different units. Count Iblis (talk) 14:01, 17 May 2013 (UTC)[reply]

No, you can't "just set it equal to 1", because it may equal to 1 meter per second - today, but may equal 2 meters per second - tomorrow. Let's take your simple example of E = m c^2, whereby m is measured in units of mass (e.g. kg), c in units of velocity (e.g. meter / second) and E in units of mass X velocity X velocity (e.g. as above). Please notice that this equation is still consistent with the assumption that c varies in time, so that c may equal 1 meter per second - today, and will equal 2 meters per second - tomorrow. HOOTmag (talk) 16:03, 17 May 2013 (UTC)[reply]

But then you should first remove all the extraneous non-physcal degrees of freedom to make sure that any change is indeed a change in the physics. Units are by definition such non-physical degrees of freedom which should be removed in these sort of discussions. What you should do instead is consider the equations of special relativity in natural units and then make these time dependent. Otherwise, what you are doing now is equivalent to actually taking these equations in natural units, insert c considered as a scaling constant in various places see here for details and then saying that this c could be time dependent which then by construction won't affect the physics. Count Iblis (talk) 16:37, 17 May 2013 (UTC)[reply]

Let's take Newton's equation: E_k=mv²/2. The units of E are defined to be: mass X velocity X velocity, and nobody claims that we can set v equal to 1. Newton didn't have to use any constant , say p, for making the link between the E_k and the mv²/2, because if he had used such a constant, say p, then we could set p equal to 1!

250 years later, Einstein replaced Newton's equation by the equation: E_k=(m-m₀)c²; Please notice that m₀ is an invariant (i.e. does not depend on the frame of reference), but may still vary in time (e.g. when an electron and a positron collide)! So why can't c (whose unit is velocity) be an invariant which varies in time - just like m₀, while the constant p - which could be supposed to make the link between the E_k and the (m-m₀)c² - is set to 1!

To sum up, my question is very simple: if we assumed that the units of E are still (as in Newton's theory): mass X velocity X velocity, and that the invariant c varies in time/space - just as m₀ varies in time - and just as v (in Newton's equation mentioned above) varies in time, would that contradict any of the (well known) relativistic equations, e.g. Lorentz transforms, or the Minkowski invariants (and the like)? HOOTmag (talk) 19:18, 18 May 2013 (UTC)[reply]

If c(x,t) is such that the Riemann curvature tensor is zero, then you get the usual relativistic equations after a suitable coordinate transform. Count Iblis (talk) 22:59, 18 May 2013 (UTC)[reply]

You could get the usual relativistic equations - even if c(x,t) could depend on (x,t). HOOTmag (talk) 23:49, 18 May 2013 (UTC)[reply]

Introducing a c(x,t) doesn't make c(x,t) the speed of light that you would measure. I think you fail to understand that the units can't be fixed in the way you are assuming, because they are always unphysical gauge degrees of freedom. A simple example would be to consider the Schwarzschild metric. It has an effective c(r) in there, but the speed of light is simply constant and the physics is not the same as what is in a flat space-time. Count Iblis (talk) 13:12, 19 May 2013 (UTC)[reply]

I do understand that the units are always unphysical gauge degrees of freedom. However, I have already explained - in my response preceding my previous response - why all of this stuff has nothing to do with my original question. So let me remind you what my original question has been about:

1. As I have already shown in this thread (on 17 May at 16:03), it's mathematically provable - from some (well known) equations of Relativistic mechanics - that c does not depend on the frame of reference, i.e. that every pair of velocities v,v' satisfies c(v)=c(0)=c(v'). So I've only been asking, whether it's also mathematically provable - from any set of (well known) equations of Relativistic mechanics - that c does not depend on time either, i.e. that every pair of moments t,t' satisfies c(t)=c(0)=c(t').

2. All of experimental information we have accumulated - really approves of the assumption that c does not depend on time, i.e. that every pair of moments t,t' satisfies c(t)=c(t'). So I've only been asking whether this experimental fact can also be mathematically inferred from any set of (well known) equations of Relativistic mechanics, just as we can mathematically infer a parallel conclusion about the invariance of c with respect to frames of reference. HOOTmag (talk) 18:33, 19 May 2013 (UTC)[reply]

To address the OP's original question, if you require that a Lorentz boost has the same mathematical structure that it does now, and that boosts be invertible such that a boost by velocity

{\vec {v}}

followed by a boost by velocity

-{\vec {v}}

should return the original coordinate system then it follows that the speed of light field must be a Lorentz invariant, i.e.

c(x,y,z,t)=c(x',y',z',t')

measures the same value for all possible Lorentz transforms. This generally implies that the speed of light is a constant independent of space and time. Now, one could replace a Lorentz boost by an integral composition of differential boosts in such a way that one could self-consistently describe a world where the measured value of the speed of light varied in space and time; however, the math then describing a change of inertial reference frame would be more complicated then the current Lorentz transformation. Dragons flight (talk) 19:44, 13 May 2013 (UTC)[reply]

On reflection, one can also have a solution where only allows the speed of light field to transform with the boost. That version is also fairly natural, but would require forgoing the idea that all observers can agree on the way the speed of light changes in space and time. Dragons flight (talk) 23:18, 13 May 2013 (UTC)[reply]

I still wonder about how you derive that

c(x,y,z,t)=c(x',y',z',t')

in the same inertial frame. HOOTmag (talk) 09:31, 16 May 2013 (UTC)[reply]

That's a trivial consequence of spacetime homogeneity. Dauto (talk) 14:18, 17 May 2013 (UTC)[reply]

You haven't showed directly (rigorously) how the constancy of speed of light in spacetime is a consequence of spacetime homogeneity, although the constancy of my velocity is not a consequence of spacetime homogeneity.

Just to let you figure out what I mean by direct rigorous proof, I will prove you now - directly-rigorously - why the relativistic equation of time dilation alone, must assume - and must result in - the constancy of speed of light in all inertial frames, i.e. I will prove directly and rigorously that - assuming the speed of light depends on the inertial frame - contradicts the relativistic equation of time dilation, and the proof for that is quite simple:

Special Relativity results in the equation of time dilation:

\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c^{2}}}}

. If

c

had depended on the inertial frame, then that equation would have meant:

\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}

, so we would have received:

(1)

{\frac {\Delta t_{v}{\sqrt {c_{v}^{2}-v^{2}}}}{\Delta t_{0}}}=c_{v}

.

However, if we hadn't used the equation of time dilation nor Lorentz transforms nor Special Relativity, but rather had used pure mathematics only, then we would have concluded that:

c_{0}\Delta t_{0}=\Delta L={\sqrt {c_{v}^{2}-v^{2}}}\Delta t_{v}

, so we would have received:

(2)

{\frac {\Delta t_{v}{\sqrt {c_{v}^{2}-v^{2}}}}{\Delta t_{0}}}=c_{0}

.

Combining (1) with (2) gives:

c_{v}=c_{0}

, i.e.

c

does not depend on the inertial frame.

HOOTmag (talk) 16:03, 17 May 2013 (UTC)[reply]

Your math formulation is wrong. Consider these two diagrams. Let the case where the two panels appear at rest be denoted case 0, and the case where they are moving as case v. In the first case, light travels:

2L=c_{0}\Delta t_{0}

In the second case, it travels:

2D=c_{v}\Delta t_{v}=2{\sqrt {\left({\frac {1}{2}}v\Delta t_{v}\right)^{2}+L^{2}}}

Giving:

\Delta t_{v}={\frac {\frac {2L}{c_{v}}}{\sqrt {1-v^{2}/c_{v}^{2}}}}

Which ultimately implies:

\Delta t_{v}={\frac {c_{0}}{c_{v}}}{\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}

That's the correct relativistic description of time dilation if you want to assume that c depends on the frame of reference. By not including the prefactor, you implicitly assumed that

c_{0}=c_{v}

, which makes your proof circular. Dragons flight (talk) 09:51, 19 May 2013 (UTC)[reply]

No, I didn't assume that

c_{0}=c_{v}

, but your diagram is not mine, so both your calculation and my calculation are correct, since each one refers to a different diagram! Your diagram involves two (right) triangles, whereas my diagram involves one (right) triangle only. Additionally, your calculation uses the Pythagorean theorem for calculating the length of path, whereas my calculation uses the Pythagorean theorem for calculating the length of velocity. Anyways, both my diagram and my calculation are simpler than yours. If you follow them according to my current explanation, you'll realize that my calculation does not assume that

c_{0}=c_{v}

and is definitely correct. HOOTmag (talk) 21:36, 19 May 2013 (UTC)[reply]

As I said, the proof is trivial. It requires no math. By definition, spacetime homogeneity means that all properties of spacetime are homogeneous. In other words, all parameters necessary to describe the vacuum are constant - they cannot depend on the coordinates of specific place where you're chose to do your experiments. The constant "c" (known by the misnomer "speed of light") is one such parameter and therefore cannot depend on the coordinates. Dauto (talk) 17:31, 17 May 2013 (UTC)[reply]

Why is it a misnomer? -- Jack of Oz ^[Talk] 20:31, 17 May 2013 (UTC)[reply]

c is a constant related to the geometry of space-time. Photons have a mass of zero and travel at the speed of c. But to say that c is the speed of light is similar to saying that the number 0 is the mass of light :) Count Iblis (talk) 20:40, 17 May 2013 (UTC)[reply]

So, what in fact is the speed of light (commonly denoted c), if not what we say in our article speed of light? -- Jack of Oz ^[Talk] 20:58, 17 May 2013 (UTC)[reply]

Did you read the article? The first paragraph makes it clear that c is a physical constant and light - as all massless objects - travels at that speed. We might just as well call it "speed of gravity", or "square-root-of-mass-energy-conversion-factor", but the best choice would be "spacetime conversion factor" because that's what c is: a factor that allows us to convert between two units. To say that c = 299,792,458 m/s is equivalent to say that the conversion factor between inches and millimeters is 25.4 mm/inches. Dauto (talk) 23:32, 17 May 2013 (UTC)[reply]

Not the entire article, but enough. It seems to be saying that the name of the physical constant is "the speed of light" (from which the article derives its title), and its symbol is "c" (this would certainly square with how people call c "the speed of light"). Are you saying that it should be saying something different? If so, what? -- Jack of Oz ^[Talk] 01:17, 18 May 2013 (UTC)[reply]

No it should not be saying something different. Speed of light is the universally accepted name for the constant. the point is that if you want to really understand the meaning of that constant than you should try to avoid seeing it simply as a description of how fast light moves. It has a much deeper meaning than that. As Count Iblis pointed out, c plays a very important role in the understanding of the nature of spacetime itself. Dauto (talk) 02:59, 18 May 2013 (UTC)[reply]

Thanks. I certainly understand that it's not just light, but all EM radiation etc that travels at c. However, if you can say that "Speed of light is the universally accepted name for the constant", then how can you argue that that very expression is a misnomer? What would you prefer we all call it? -- Jack of Oz ^[Talk] 03:15, 18 May 2013 (UTC)[reply]

It is a misnomer because being the speed of light is not the only fact (or even the most important fact) about that constant. As Count Iblis said, nobody thinks of the number zero as "the mass of light", why should we think of c as the speed of light. Nevertheless, "speed of light" IS the universally accepted term and that's what we all should call it. Misnomer or not, that's its name. Dauto (talk) 15:40, 18 May 2013 (UTC)[reply]

No argument here. I was just looking for any substance to your original point, and we now agree there wasn't any. Thanks for clearing that up. -- Jack of Oz ^[Talk] 22:47, 18 May 2013 (UTC)[reply]

There is a point. You just missed it. Sorry I couldn't help you. Dauto (talk) 17:53, 22 May 2013 (UTC)[reply]

@Dauto, you assume what was to be demonstrated. I'm asking: does any (well known) relativistic equation (e.g. Lorentz transforms, or the Minkowski invariants) need to assume that the invariant c is (as you put it): "one of such parameters necessary to describe the vacuum"? In other words, if we assumed that the invariant c is not "one of such parameters necessary to describe the vacuum", but rather varies in time/space, would that contradict any of the (well known) relativistic equations, e.g. Lorentz transforms, or the Minkowski invariants (and the like)? This is my question! HOOTmag (talk) 18:20, 18 May 2013 (UTC)[reply]

Yes, a varying c would mess up the invariants. For instance, the invariant mass

m^{2}c^{4}=(E^{2}+c^{2}p^{2})

would change overtime unless either the energy or the momentum of the particle isn't conserved. Dauto (talk) 20:07, 18 May 2013 (UTC)[reply]

But the invariant mass (i.e. the "rest mass") does change over time (e.g. when an electron and a positron collide)! Einstein has never claimed that the invariant mass (i.e. the "rest mass") does not change over time, but rather that the invariant mass (i.e. the "rest mass") does not depend on the frame of reference. HOOTmag (talk) 20:19, 18 May 2013 (UTC)[reply]

Irrespective of any particle's average speed, for closed systems, energy conservation requires that c = sqrt(E/m) be constant (AKA mass-energy equivalence). --Modocc (talk) 21:16, 18 May 2013 (UTC)[reply]

As for the principle of "energy conservation": it means that for closed systems, E=mc² is constant: This does not mean that for closed systems c is constant - but rather that for closed systems mc² is constant, while m and c may vary. As for the principle of "mass-energy equivalence": If c changes over time, then this principle just means that - the mass existent at any given moment - is equivalent to the energy existent at that very moment, but still this principle does not let you conclude that - the mass existent at any given moment - is equal to the mass at the following moment. HOOTmag (talk) 22:34, 18 May 2013 (UTC)[reply]

Mass-energy equivalence and, more generally, its mass-energy conservation are well-established. A changing mass results in an exact corresponding change in energy. With conservation, should c and m change but not E then you don't have an equivalence relation.--Modocc (talk) 22:46, 18 May 2013 (UTC)[reply]

First of all, in order to prove the equation E_k=(m-m₀)c² - one does not need to assume that c does not change over time - but rather to only assume that c does not depend on the frame of reference; Whereas the simpler equation E=mc² is only an (unprovable) generalization of the (provable) equation E_k=(m-m₀)c². As for the very principle of Mass-energy equivalence: It is one possible interpretation - among some others - for the (unprovable) generalized equation: E=mc². Of course, the (unprovable) generalized equation itself does not claim that E is equivalent to m, unless we assume that the invariant c - which does not depend on the frame of reference - does not change over time either. Another possible interpretation for that (unprovable) generalized equation could claim, that c - which does not depend on the frame of reference - does vary in time, thus no "equivalence principle" would arise.

Second, even if we accept the "equivalence interpretation", it may help us in closed systems only. How about open ones - in which c may (apparently) change over time? HOOTmag (talk) 23:49, 18 May 2013 (UTC)[reply]

The constancy of c = sqrt(E/m) has been validated by numerous experiments that have included open systems. See the articles I linked to for more on this. --Modocc (talk) 00:14, 19 May 2013 (UTC)[reply]

I haven't claimed that c changes over time, but rather that c may change over time. I'm talking from a hypothetical point of view, and my original question is hypothetical: Which relativistic equation contradicts the hypothetical assumption that c may change over time? HOOTmag (talk) 00:23, 19 May 2013 (UTC)[reply]

Its been speculated that c has changed, could change, or is perhaps different in other parallel universes. There is nothing inherently contradictory about that kind of speculation other than the fact that its fictitious because there is no evidence for it. --Modocc (talk) 00:58, 19 May 2013 (UTC)[reply]

In other words, there is a difference between the constancy of c among the frames of reference, and the constancy of c during time: one can mathematically infer the first constancy - from some relativistic equations (as I've showed above by the equation of time dilation), whereas one can't mathematically infer the second constancy - from any relativistic equation. This means that the second constancy is an experimental fact only, yet not a substantial/essential property of the fundamentals of Relativistic Mechanics. Is this what you claim? HOOTmag (talk) 01:10, 19 May 2013 (UTC)[reply]

Its incorrect to infer invariance of c from time dilation. -Modocc (talk) 02:24, 19 May 2013 (UTC)[reply]

What do you mean by "incorrect"? I have correctly inferred - the invariance of c - from the equation of time dilation, in my response on 17 May at 16:03, as you can see above in this thread. On the other hand, if you just mean that I was not methodologically allowed to make this inference - although it's a mathematically correct inference (as I have already shown in this thread on 17 May at 16:03), then you have probably missed my original question. Please notice that what I have been asking about (along this thread), is whether Special Relativity must assume the constancy of c in spacetime - in order to develop the relativistic equations, just as Special Relativity must assume the invariance of c - in order to develop the relativistic equations, i.e. I've been asking whether there is any (well known) relativistic equation, e.g. any Lorentz transform, or any Minkowski invariant (and the like), which contradicts the assumption that c varies in spacetime. I haven't been asking that about the invariance of c, because I had already known that Special Relativity must assume the invariance of c - in order to develop the relativistic equations, i.e. I had already known that there were some relativistic equations, e.g the equation of time dilation, which contradicted the assumption that c depended on the frame of reference, as I had already proved in this thread on 17 May at 16:03. HOOTmag (talk) 06:12, 19 May 2013 (UTC)[reply]

Sorry, but substituting c(x,t) in f(c) only shows that c(x,t)=c is a solution of f(c). It doesn't prove invariance. Modocc (talk) 09:41, 19 May 2013 (UTC)[reply]

Sorry, but I have never proved anything about c(x,t), so I don't know what you're talking about. As you can see in this thread (on 17 May at 16:03), I have only proved - from a (well known) relativistic equation (of time dilation) - that every velocity v satisfies c(v)=c(0). Hence, every pair of velocities v,v' satisfies c(v)=c(0)=c(v'). Hence, c is invariant (i.e. does not depend on the frame of reference). QED. HOOTmag (talk) 10:10, 19 May 2013 (UTC)[reply]

You show that c(v)=c(dx/dt)=c are solutions of the time dilation equation. Such an invariance of c (whether it be with respect to space or time) is certainly consistent with the fact that this time dilation is true, but it does not prove what was actually assumed to begin with, that c(v) = c. Except for the substitution you made, your second equation is merely the first equation. Modocc (talk) 10:45, 19 May 2013 (UTC)[reply]

If you follow my proof, you can realize that - not only have I proved for every velocity v that the identity c(v)=c(0) is a possible solution of the equation of time dilation - but also that I have proved for every velocity v that the identity c(v)=c(0) is the only possible solution of the equation of time dilation (because the proof is based on identities, whereas every identity in the world must point at a unique solution - since "two" different things can't be equal in any identity). Hence (by substituting v' for v) , it's also provable for every velocity v' that the identity c(v')=c(0) is the only possible solution of the equation of time dilation. By combining both identities mentioned above, we infer that every pair of velocities v,v' satisfies c(v)=c(v'). Hence c does not depend on the frame of reference. QED. HOOTmag (talk) 11:21, 19 May 2013 (UTC)[reply]

The constancy of c can be obtained in a manner that doesn't depend on light speed invariance, but since my critique (which I believe to be valid) of your proof is only valid if its understood, I will desist on any further comment about it than I've written above and further down, below. Modocc (talk) 12:01, 19 May 2013 (UTC)[reply]

Why do you think your critique may not be understood? Since I've understood you up to now (although I haven't agreed with you - which is another matter), so I don't see why I won't understand you from now on. HOOTmag (talk) 12:21, 19 May 2013 (UTC)[reply]

See my reply below. --Modocc (talk) 14:33, 19 May 2013 (UTC)[reply]

Suppose $c(x,t)$ is not constant, and a boost in the $x$ direction with velocity $v$ can still be represented by a Lorentz-like form:

{\begin{aligned}t'&=\gamma \left(t-{\frac {vx}{c(x,t)^{2}}}\right)\\x'&=\gamma \left(x-vt\right)\\\gamma &={\frac {1}{\sqrt {1-{\left({v \over c(x,t)}\right)^{2}}}}}\end{aligned}}

Consider the inverse boost applied to the new frame, which implies:

{\begin{aligned}t&=\gamma '\left(t'+{\frac {vx'}{c'(x',t')^{2}}}\right)\\x&=\gamma '\left(x'+vt'\right)\\\gamma '&={\frac {1}{\sqrt {1-{\left({v \over c'(x',t')}\right)^{2}}}}}\end{aligned}}

Simpliying we get:

x=\gamma '\left(\gamma \left(x-vt\right)+v\gamma \left(t-{\frac {vx}{c(x,t)^{2}}}\right)\right)

\Rightarrow {\gamma ' \over \gamma }=1

\Rightarrow c'(x',t')=c(x,t)

The easiest way to deal with this, is simply to give up and say that $c$ was in fact a constant. However, one can adopt the alternative point of view, in which $c'(x',t')=c(x,t)$ defines how the speed of light field transforms under a change of reference frame. If you do that, then different observers will necessarily have qualitatively different perspectives on those changes. For example, there will be a privileged frame where ${\partial c \over \partial x}=0$ at the origin, implying that the change in $c$ appears locally to be purely timelike, while all other reference frame believe the change occurs over both space and time. These sorts of things play havoc with the notion that all observers are equal, which you pretty much have to throw away if you want to propose that $c$ varies in space and time. Dragons flight (talk) 01:08, 19 May 2013 (UTC)[reply]

Unfortunately, you didn't prove what was to be proven: I've asked whether it's provable that - every x,t,x',t' satisfy c(x,t)=c(x',t'), whereas you have only proved that - every x,t,x',t' satisfying the equations you have presented at the beginning of your last response - satisfy c(x,t)=c'(x',t'); In other words: you just proved, that c was not dependent on the frame of reference, as I did in this thread - more simply - on 17 May at 16:03. So, just to make things clearer, let's put aside the complex case - involving some frames of reference - you have dealt with, and let's discuss the simplest case - involving one frame of reference only. My question is (and has always been) whether you can infer, from any set of (well known) relativistic equations, that every x,t,x',t' satisfy c(x,t)=c(x',t') in that frame of reference. HOOTmag (talk) 08:14, 19 May 2013 (UTC)[reply]

Of course that can't be proved! "For all x, prove f(x) = g(x) ... assuming f ≠ g" - and if you've had any inkling of mathematics, you know how preposterous that is! It's trivial to construct a counterexample! This is called an unphysical equation, and it's a hallmark of Aristotelian physics - rather, the development of a theory from axioms without ever stopping to check whether the theory corresponds to experimental observation. In physics, we do not prove equations. We use them, and only when they are useful.

This discussion has gone on for a long time and is getting you no closer to the answer you seek. What reference material are you looking for - a book, a website, a college course guide - that will help you find what you need? Nimur (talk) 09:34, 19 May 2013 (UTC)[reply]

You ignore two facts:

1. As I have already shown in this thread (on 17 May at 16:03), it's mathematically provable - from some (well known) equations of Relativistic mechanics - that c does not depend on the frame of reference, i.e. that every pair of velocities v,v' satisfies c(v)=c(0)=c(v'). So I just wanted to know whether it's also mathematically provable - from any set of (well known) equations of Relativistic mechanics - that c does not depend on time either, i.e. that every pair of moments t,t' satisfies c(t)=c(0)=c(t').

2. All of experimental information we have accumulated - really approves of the assumption that c does not depend on time, i.e. that every pair of moments t,t' satisfies c(t)=c(t'). So I just wanted to know whether this experimental fact can also be mathematically inferred from any set of (well known) equations of Relativistic mechanics, just as we can mathematically infer a parallel conclusion about the invariance of c with respect to frames of reference. HOOTmag (talk) 09:55, 19 May 2013 (UTC)[reply]

With relativity, invariance of c (or the Lorentz transformations) is usually assumed with respect to both space and time. As I've said above, you show that this invariance/constancy is consistent with the formula which already denote c to be a constant and from what I can discern, this consistency is your only result. Its generally known that relativity begins with its postulated invariance(s) to arrive at models contrary to any simpler model that would assume prerelativistic velocity addition, because measurements of light waves show that the non-relativistic Doppler has an apparent extraneous second-order term and therefore published non-relativistic models have not ever accurately modeled the matter-waves' wavelengths, frequencies and energy (and that sorry track record doesn't mean we cannot ever do so). Gravitational theories involving light speed force carriers that at one-time competed with relativity also didn't get the energies involved correct either (for basically the same reason) and therefore were lacking. The thing about paradigms though, such as relativity, is that one can show sets of statements are consistent with other statements, but that doesn't actually prove that the paradigm's statements (some of which are assumed and/or are held to be "fact" based) are correct, which is why Occam's razor and various evidence is important. -Modocc (talk) 11:43, 19 May 2013 (UTC)[reply]

Yes, I know that "invariance of c (or the Lorentz transformations) is usually assumed with respect to both space and time". However I have been asking whether it's necessary to assume this for developing all of the (well known) equations of Relativistic mechanics.

No, I have shown - not only consistence - but necessity as well, i.e. I have proved that the identity c(v)=c(0) is the only solution of the equation of time dilation. You can realize that by just following my strict proof, but if you don't want to review it, then let me test you: Can you give me another solution of the equation of time dilation? If c(v) is not equal to c(0), then it's equal to what? Give another value, other than c(0) (e.g. c(0)+1 and likewise), and I will show you that such a value contradicts the equation of time dilation! HOOTmag (talk) 12:14, 19 May 2013 (UTC)[reply]

As I mentioned up the page, the time dilation formula you used was derived under the assumption that c is invariant with respect to space, time, and frame. It's not really a proof if you start your analysis by assuming that your conclusion is true. I already gave you a modified version of the time dilation formula that you would need to use if you actually want to consider the case that

c_{v}\neq c_{0}

. Dragons flight (talk) 19:44, 19 May 2013 (UTC)[reply]

As I have responded up the page, I hadn't assumed that

c_{0}=c_{v}

. You thought I had, because you thought I had used the same diagram you used, and also because you thought I had used the Pythagorean theorem for calculating the length of path - as you used, whereas I really used the Pythagorean theorem for calculating the length of velocity. To sum up, my calculation is correct and assumes nothing in advance. For more details, see my direct response to your detailed response up the page. Anyways, my original question was not about whether one can prove the invariance of c (i.e. with respect to frames of reference), but rather about whether one can prove the constancy of c (i.e. with respect to time and space). HOOTmag (talk) 21:36, 19 May 2013 (UTC)[reply]

This is not really a proper venue, but its sufficient to say that an apparent (non-relativistic) invariance c = c'(v) can be a result of conflating distinct, reference frame dependent, derivations for c: {c, c', c' ',...} with an appropriate model which involves velocity addition. In other words, each observer computes c even though the underling physics is not relativistic. -Modocc (talk) 14:33, 19 May 2013 (UTC)[reply]

You claim that there are also some kinds of non-relativistics mechanics which result in the invariance c = c'(v); As I understand, you claim this new claim - not in order to reject my claim that the invariance c = c'(v) is the only possibility for c under Relativistic mechanics (because your new claim does not contradict my claim about the invariance of c) - but rather in order to claim that Relativistic mechanics must assume the constancy of c in spacetime because invariance alone is consistent with some kinds of non-relativistic mechanics as well. So, let me remind you, that Einstein did infer his Relativistic mechanics (e.g. Lorentz transforms and the like) - from the invariance c = c'(v) alone, so I suspect your new claim is against Einstein's opinion. To make things clearer, would you like to give any new equation of the "non-relativistic mechanics" you propose, so that the new equation may contradict any of the (well known) relativistic equations? HOOTmag (talk) 18:33, 19 May 2013 (UTC)[reply]

Yikes, to be clear, there are two claims to address: 1)that a different value for c contradicts time dilation (I agree with this assertion) and 2)a different value for c contradicts light speed invariance and time dilation. Both of these claims do not necessarily get contradicted because the mere correlation of these two different claims does not imply that they must logically follow from each other. With Einstein's work, light speed invariance implies time dilation, however, time dilation does not imply light speed invariance if we expand this discourse to non-relativistic assumptions since with a non-relativistic model a value different from c should not get inserted into the time dilation equation (and to do so would contradict it), yet in this context, its an apparent invariant c that gets derived and not an actual invariant c. The reason is simple: it matters whether different observers actually arrive at their measurement for c using identical yardsticks or not. If observers' data vary very slightly with humidity (acceleration in our case), we generally infer that the yardsticks are not identical and have changed in some subtle way, such as by expanding or contracting. However, according to relativity, proper yardsticks (proper time and proper distance) are identical, and with the alternative model, the proper yardsticks used for different inertial frames are not identical in general. Showing, within the proper venue of course, why yardsticks are not always identical and how my derivations entail time dilation and other relations involving the Lorentz factor turns out to be only somewhat difficult (obviously its not a cakewalk) and enlightening (for me at least). Modocc (talk) 20:08, 19 May 2013 (UTC)[reply]

You claim: "With Einstein's work, light speed invariance implies time dilation". I think your current claim contradicts your response (of 19 May at 14:33) preceding your last response, unless I misinterpret you in the expression "time dilation". When I say "time dilation" I just mean the relativistic equation of time dilation: $\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c^{2}}}}$ , and if $c$ had depended on the inertial frame - then that equation would have meant: $\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}$ .
You claim: "time dilation does not imply light speed invariance if we expand this discourse to non-relativistic assumptions". When I say "time dilation" I just mean the relativistic equation of time dilation: $\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c^{2}}}}$ , and if $c$ had depended on the inertial frame - then that equation would have meant: $\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}$ . Please notice that I have already proved (on 17 May at 16:03) that this relativistic equation of time dilation necessarily implies the invariance of c (i.e. with respect to frames of reference), as opposed to your current claim.
You claim: "with a non-relativistic model a value different from c should not get inserted into the time dilation equation". of course! The time dilation equation must be of the form $\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c^{2}}}}$ , yet if $c$ had depended on the inertial frame - then that equation would have meant: $\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}$ . However, you had claimed that the identity c(v)=c(0) is not a necessary solution for the equation of time dilation, so I requested to give me another solution, that's all. If you think it's impossible to give another solution, then this proves that the identity c(v)=c(0) is the only solution for the equation of time dilation - just as I have always claimed.
As for the difference between the relativistic model and the non-relativistic models satisfying the invariance of c, you claim that "according to relativity, proper yardsticks (proper time and proper distance) are identical, and with the alternative [non-relativistic] model, the proper yardsticks used for different inertial frames are not identical in general". I'm asking you again: Would you like to give here any new equation of the "non-relativistic models", so that the new equation may contradict any of the (well known) relativistic equations? HOOTmag (talk) 22:37, 19 May 2013 (UTC)[reply]

As pointed out above

\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}

is simply NOT the correct form of the time-dilation equation for an assumption of frame dependence. You need to reexamine the derivation of the time dilation equation if you want to accurately consider the consequences of allowing a frame-dependent speed of light. In that case the correct form of the time dilation equation becomes

\Delta t_{v}={\frac {c_{0}}{c_{v}}}{\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}

. Dragons flight (talk) 22:51, 19 May 2013 (UTC)[reply]

I think you confuse the relativistic "time-dilation equation" (after adapting it to an assumption of frame dependence) with a purely mathematical equation which does not assume any relativistic assumption (nor Lorentz transformation). As for the so-called "time dilation equation" (i.e. the relativistic one), it must be of the form

\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c^{2}}}}

, whereas adapting it to an assumption of frame dependence - by simply substituting c_v for c - makes it:

\Delta t_{v}={\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}

. On the other hand, as for the purely mathematical equation which does not assume any relativistic assumption (nor Lorentz transformation): It's simply

\Delta t_{v}={\frac {c_{0}\Delta t_{0}}{\sqrt {c_{v}^{2}-v^{2}}}}

, which is mathematically equivalent to a more complex equation:

\Delta t_{v}={\frac {c_{0}}{c_{v}}}{\frac {\Delta t_{0}}{\sqrt {1-v^{2}/c_{v}^{2}}}}

. For more details, see my direct response to your detailed response in which you presented this complex form for the first time. Btw, I used (on 17 May at 16:03) the combination - of the purely mathematical equation - with the relativistic "time-dilation equation" (after adapting it to an assumption of frame dependence), in order to infer the invariance of c (i.e with respect to reference frames), whereas my original question has been whether one can similarly infer also the constancy of c (i.e with respect to spacetime). HOOTmag (talk) 00:05, 20 May 2013 (UTC)[reply]

Steroids and bodybuilding

Is it possible to get bodies like these [3][4] without steroids? --Yoglti (talk) 06:41, 22 May 2013 (UTC)[reply]

Oh yes, it takes a little skill, but you can use this. 86.4.181.3 (talk) 07:00, 22 May 2013 (UTC)[reply]

Biology

Give the biological significance of Van der Waal's forces? — Preceding unsigned comment added by Titunsam (talk • contribs) 11:03, 22 May 2013 (UTC)[reply]

I'll say it is a kind of weak force, similar to that that tie students to their homeworks, it's a weak link. OsmanRF34 (talk) 12:31, 22 May 2013 (UTC)[reply]

Have you tried reading Van der Waals force (and gecko)?--Shantavira|^{feed me} 13:12, 22 May 2013 (UTC)[reply]

Name the following psychiatric symptoms

What are the psychological terms for the following symptoms:

1. Destroying objects that reminds an unpleasant past incident 2. Performing mental ritual such as touching the door repeatedly before leaving home believing it will be lucky --Yoglti (talk) 11:14, 22 May 2013 (UTC)[reply]

1. Displacement (psychology)?

2. OCD? OsmanRF34 (talk) 12:24, 22 May 2013 (UTC)[reply]

3. Ignoring the 'no medical advice' rule? AlexTiefling (talk) 11:14, 22 May 2013 (UTC)[reply]

3 . Selective perception? OsmanRF34 (talk) 12:25, 22 May 2013 (UTC)[reply]

This question is either a request for medical advice, to whit a diagnosis of the possible symptoms he described, or a homework question. Either way it's not something we should answer, unless the OP shows us what effort he has made to find the answer himself, and where he got stuck. Wickwack 120.145.48.50 (talk) 13:00, 22 May 2013 (UTC)[reply]

Butterfly effect and scientific method

How can someone test scientifically the existence of a butterfly effect in weather prediction or other complex dynamic systems? I understand that it's not about an unknown butterfly somewhere causing havoc by simply flapping its wings, but about how minimal alteration of the starting system can alter completely a whole system. It's clear that if it were a die, you just could change something really small and see what happens, reaching the conclusion that a butterfly effect exits in tossing dice. But, where is the hypothesis-experiment-conclusion when it comes down to bigger stuff like the weather? OsmanRF34 (talk) 13:03, 22 May 2013 (UTC)[reply]

Half the problem is that the butterfly effect is a plain-english description of a phenomenon in chaos theory; and therefore, it is a mathematical development, and not a scientific fact. So, it's not subject to "hypothesis-experiment-conclusion" cycle any more than "2+2" is. It is a mathematical result that certain functions have immensely variable results; applied mathematicians can be very precise and specific in describing these characteristics. For example, the Lyapunov exponent is one quantitative measure of stability. In Control theory, we use phase margin and frequency response as a measure of system instability.

So this is more a matter of whether a mathematical model is applicable to a physical phenomenon. When physicists or engineers study air flow in experimental conditions, they deduce mathematical equations that appropriately model the observations. We can therefore apply the stability measurements to those models, allowing us to understand the limitations of the model's predictive power. This is sometimes formalized as sensitivity analysis. More theoretical physicists use the phrase "the calculus of variations" to refer to the same style of analysis when performed without an electronic calculator.

No reasonable scientist attempts an actual controlled experiment gauging behavior differences with- and without- the controlled release of a butterfly. Nimur (talk) 13:36, 22 May 2013 (UTC)[reply]

All Nimur's points are spot-on, but I'll add that there are some "experiments" that people perform to detect if a real system is chaotic (or perhaps more pedantically, is well-described by a model that has chaotic properties). So, if we measure a system, and show evidence that its dynamics are chaotic (with a positive Lyapunov exponent), then in a sense you could say that someone has demonstrated the butterfly effect in a real system. For a feel for how this works, google /detecting chaos in time series/, like so [5]. For another real-world "test of the butterfly effect" (please, understand this is loose phrasing), you might be interested in industrial application of chaotic mixing. The mixing properties of some chaotic systems are basically the same as "the butterfly effect", in that they both are results of the divergence of initially nearby states, due to a positive Lyapunov exponent. See e.g. this professor's web-bio [6], which has a nice overview of applications of chaotic mixing. The mixing is only effective in the real world because the math that describes it also displays the butterfly effect. Finally, the Baker's map is a chaotic process that you can test at home -- start with two similar arrangements of dough, iterate the map on each slab n times, see how differently they turn out! SemanticMantis (talk) 15:22, 22 May 2013 (UTC)[reply]

The above answers are quite good. I would also add, that one definitely does run experiments on global climate models in order to understand the effect of chaos, specifically, the role of sensitive dependence on initial conditions in influencing simulation results. It is now often routine to repeat computer model experiments subject to small perturbations in initial conditions in order to see how the results evolve differently. This allows one to differentiate which parts of the behavior may be predictable from the parts that are heavily impacted by chaos and may only be described in a broadly statistical sense using probabilities. Of course, one needs to make a small leap of faith that the computer models are an accurate representation of our planet's weather / climate, but it is certainly true that the computer models demonstrate the butterfly effect. Dragons flight (talk) 15:44, 22 May 2013 (UTC)[reply]

Again, the answers above are good. I take issue though with Nimur's statement "So this is more a matter of whether a mathematical model is applicable to a physical phenomenon." because it may give the false impression that the butterfly effect is due to some limitation of the models. That's not the correct interpretation of the effect. The models do display the butterfly effect not because they aren't good models. They display the effect because they ARE GOOD models and (correctly) capture a feature of real weather systems - namely, the butterfly effect. Dauto (talk) 17:17, 22 May 2013 (UTC)[reply]

It's useful to explain how this came about. In the early 1960's Edward Lorenz was working on computer simulations of weather systems tracking across the North Atlantic over a simulated 2 week period. His simulations produced convincing results and he was happy. One day he had to re-run a previous simulation. One particular parameter of the simulation should have been set to 0.506127 - and had been set to that on that first run. But on the second run, he only bothered to type 0.506...meh, close enough, right? Wrong! That tiny change of one or two parts in ten thousand was enough to produce UTTERLY different results. That's what "chaos theory" is all about. It's perhaps better called "sensitivity to initial conditions". If a teeny-tiny change at the start of some process is magnified to a GIGANTIC change in the end result, then the process is said to be "chaotic". If one part in 10,000 can change the weather over a couple of weeks of simulated time - then one part in 100,000,000 will create the same amount of change over a month, one part in 10,000,000,000,000,000 over two months. Over six months, the displacement of a single atom over the diameter of an atom is enough to cause drastic changes in the way the atmosphere ends up. Hence, the flapping of a butterfly's wing is more than enough to cause violent weather anywhere on the planet - given enough time...or to cause violent weather not to happen...or to have no effect whatever. Either way! SteveBaker (talk) 18:58, 22 May 2013 (UTC)[reply]

But, why do you get stable overall patterns, given that a "butterfly" can bring the whole system out of sync? Some places are mostly rainy, and some are mostly sunny. OsmanRF34 (talk) 19:05, 22 May 2013 (UTC)[reply]

Damp proof course

Is a damp proof course part of the building code in the usa for a concrete block house? --Jason1267 (talk) 13:53, 22 May 2013 (UTC)[reply]

In the United States, building codes are set at the state, county, and municipal levels. You'd need to be more specific about location. Nimur (talk) 13:58, 22 May 2013 (UTC)[reply]

I am interested in Columbus, Ohio and Miami, Florida--Jason1267 (talk) 14:02, 22 May 2013 (UTC)[reply]

For example, Columbus' Department of Building and Zoning links to the 2011 Ohio Building Code. You can telephone their office for guidance. Often, the city will refer you to a building contractor to answer these sorts of questions, because the answers almost always depend on lots of details; but you can survey the code yourself to get an idea. Nimur (talk) 14:08, 22 May 2013 (UTC)[reply]

Which is bigger?

Which is the bigger ratio, one atom vs one human being, or one human being vs the entire galaxy?114.75.53.116 (talk) 16:10, 22 May 2013 (UTC)[reply]

Ratio of what? Size? Mass? Volume? uhhlive (talk) 16:12, 22 May 2013 (UTC)[reply]

Do you mean by length, by mass, or something else? You can get a rough answer from Orders_of_magnitude_(length) or Orders_of_magnitude_(mass) by calculating the ratios directly. SemanticMantis (talk) 16:15, 22 May 2013 (UTC)[reply]

As pointed out above, the question could've been more precise. Turns out that Galaxy-to-human ratio is much larger than the human-to-atom ratio no matter the criteria used. Dauto (talk) 17:42, 22 May 2013 (UTC)[reply]

Buying pure elements

Where can I easily get:

List of pure and metallic elements

The following discussion has been closed. Please do not modify it.

metallic lithium
metallic sodium
metallic potassium
metallic rubidium
metallic caesium
metallic beryllium
metallic calcium
metallic strontium
metallic barium
metallic radium
metallic uranium
metallic plutonium
metallic tantalum
metallic cadmium
metallic gallium
metallic indium
metallic thallium
metallic scandium
metallic yttrium
pure polonium
pure fluorine
pure chlorine
pure bromine
pure iodine
pure astatine
pure boron
pure arsenic
pure argon
pure neon
pure krypton
pure xenon
pure radon?

Whoop whoop pull up ^{Bitching Betty | Averted crashes} 16:59, 22 May 2013 (UTC)[reply]

Fisher Scientific and McMaster-Carr. A few of these (Polonium?) cannot be easily gotten anywhere. Nimur (talk) 17:01, 22 May 2013 (UTC)[reply]

Polonium can be bought at United Nuclear. In the same way as many other radioactive isotopes. OsmanRF34 (talk) 17:55, 22 May 2013 (UTC)[reply]

What is your geographical location? Most of these things can be obtained in pure analytical form, from local laboratory suppliers. However, since the Home Guard (or whatever they are called where you are) are looking for anybody that can possibly be sourcing materials for terrorist devices, expect a visit from the men in black or the FBI. P.S. Why, Oh Why, are you asking? Why do you need this stuff?Aspro (talk) 17:16, 22 May 2013 (UTC)[reply]

To be fair, Fisher sells these under their "Science Education" program, typically marketing to high school and college-level chemistry students and teachers. Items that are more hazardous - like polonium and radon - are not usually available without raising a few hackles; but with lots of paperwork, and a reasonable degree of oversight and accountability, a credible educational institution can often acquire these sorts of things; they do have uses other than causing havoc. But just because you're an enthusiast with no ill intention, that doesn't mean you won't get a little extra attention at the airport for the rest of your life! Nimur (talk) 17:21, 22 May 2013 (UTC)[reply]

Tiny amounts of polonium are available in such things as anti-static brushes, with no special red tape needed. --Trovatore (talk) 18:54, 22 May 2013 (UTC)[reply]

You can get an entire element collection here, from the Red Green & Blue Company, for around 500 pounds. There are also occasional element collections on eBay, usually in the range of a few hundred dollars. --Bowlhover (talk) 18:48, 22 May 2013 (UTC)[reply]

Not really credible — how do you sell a sample of francium? My guess is that the "francium" tube contains a sample of natural uranium ore, which at any given time will contain a few atoms of francium. Probably the same for several of the other short-lived elements in the U-238 decay chain. --Trovatore (talk) 18:52, 22 May 2013 (UTC)[reply]

Nothing in ebay seems genuine anymore. Anyway, a serious company could ship radioactive material that have a short half-life. It only would need to ship it shortly after obtaining it, and not keep it on stock. OsmanRF34 (talk) 19:01, 22 May 2013 (UTC)[reply]

Using small LCDs for other purposes

The very small and simple LCD screens we find in cheap watches, which display only black numbers (each made by four vertical and three horizontal sticks), are just grey coated glass slides. Theoretically I know they have "liquid crystals" in them which turn black from transparent when current is applied. I don't find any visible inlets which could take electricity inside them. Yes they do have soft a rubber line on their base which contacts the numerous metal contacts held in a long line along the circuit board that clearly seems to be the appropriate number of "sticks" that form the digits. My problem is that how can one directly manipulate them ? Is the usual 1.5 direct current from a small battery appropriate ? 124.253.173.16 (talk) 19:24, 22 May 2013 (UTC)[reply]

"Non-Newtonian Play-Doh"

For a business assignment, we are to make Play-Doh (salt dough), create a brand for it, market it, etc. To be creative, I thought back to my days of elementary school when we would mix cornstarch and water to make "oobleck," and I was considering adding the oobleck substance into the play-doh mixture, which we will need to make ourselves (using 1 cup of water, 1/2 cup salt, 1 cup flour, 1 tbsp oil, 1 tbsp cream of tartar). If we added cornstarch and water to the mixture, would it make the play-doh non-Newtonian in any way? Would it ruin the mixture? Would it just not do anything? Thanks so much! 174.93.65.84 (talk) 19:52, 22 May 2013 (UTC)[reply]

Wikipedia:Reference desk/Science