Wikipedia:Reference desk/Archives/Science/2008 November 11

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November 11[edit]

How to move energy faster than the speed of light[edit]

Imagine you have a easily pliable ring millions of miles in diameter. What if one were to stand in the exact center of this ring with a garden hose and release a perfectly even stream of water for one second while doing a 360. The water would then hit the ring causing it to move in a wave that would cover millions and millions of miles (or whatever circumference you would like) in one second. Is this possible?

Thanks--UhOhFeeling (talk) 06:23, 11 November 2008 (UTC)[reply]

Only if you are standing in Philadelphia. CBHA (talk) 07:03, 11 November 2008 (UTC)[reply]
Since a ring with a diameter (2xr) of about 2 million miles (1,911,200 miles) is already at a distance (r) of 4 times the average distance of the Moon from earth (238,900 miles) all sorts of factors come into play, including what keeps your pliable ring taut, at what speed it is rotating, the fact that neither a garden hose as delivery system nor water as a liquid would have much chance of meeting your "perfectly even stream" requirement. And that's just for starters. Having someone hop up and down on the surface of the ring might have a better theoretical chance of producing a wave, but nowhere near any speed of light. 76.97.245.5 (talk) 08:01, 11 November 2008 (UTC)[reply]
Hypothetical example, assume ideal conditions.--UhOhFeeling (talk) 09:28, 11 November 2008 (UTC)[reply]
Doing this with a light beam is much more reliable than with water. Just shine a torch away from you, and quickly rotate so the torch shines in all directions. At long distances from you, the torch beam is moving much faster than light. I thought we had an article mentioning this, and why it is not normally considered to count as FTL (because it doesn't allow superluminal information transmission), but I can't find it. Algebraist 09:54, 11 November 2008 (UTC)[reply]
Indeed. Or replace the water with raido waves and you have a pulsar. Suppose pulsar is at A and sweeps its beam between B and C, many light years apart, in a fraction of a second. Then energy is transferred from A to B (or to C) at the speed of light - but no energy is transferred from B to C. Gandalf61 (talk) 10:01, 11 November 2008 (UTC)[reply]
I guess it's similar to how the group velocity of light can exceed the "speed of light". The key point, as has been said, is that no information is travelling faster that light speed. --Tango (talk) 11:52, 11 November 2008 (UTC)[reply]
That is why you need water (or whatever you want, Hypothetical Example) to impart a force on the pliable strip between B and C creating a physical wave moving faster than the speed of light. Under ideal conditions would this hypothetical work?--UhOhFeeling (talk) 19:58, 11 November 2008 (UTC)[reply]
Why bother? Light can push stuff, so you can probably get a wave going with that if you want to. You still won't have superluminal information transfer. Algebraist 20:00, 11 November 2008 (UTC)[reply]
Sure Light is fine, Hypothetical. Why won't you have superluminal information transfer is my question. Thanks.--UhOhFeeling (talk) 20:04, 11 November 2008 (UTC)[reply]
Because as you stand at A swinging your spotlight (at FTL speeds) from B to C, there's no way that events at B at the time the light hits can influence what you're doing (said events are still way in the future at this point). Thus events at B can't influence C FTL through this mechanism. Algebraist 20:11, 11 November 2008 (UTC)[reply]
The "wave" isn't propagating along the ring, it's being caused separately at each point when the beam of water or light hits it, so there is no information going along the ring (well, there will probably be a wave propagating as well, but that will follow along behind at a slower speed). The only information moving if from the centre of the ring to the ring itself, which travels at sub-light speeds (for water) of light speed (for light) but never faster. --Tango (talk) 20:13, 11 November 2008 (UTC)[reply]
Why would the wave be following along at a slower speed if the energy hit it in such a way that the wall was being pushed in a way that would create a wave faster than light. Couldn't this wave continue after it's initial trip around on it's own? --UhOhFeeling (talk) 01:27, 12 November 2008 (UTC)[reply]
Of course not, any more than dropping a thousand rocks into a pond in a line will create some sort of supersonic ripple that arrives at the other end of the pond immediately. Each rock creates a wave (really a set of waves) that spreads out from its point of impact at a fixed speed, and the presence of other waves is irrelevant. (This isn't very precise for waves on the surface of water, but the idea is correct.) Given that you're trying to make a faster wave, you must be dropping the rocks into the pond with very little time between them: so little time, in fact, that the first rock's waves haven't reached the last rock when it hits! This means that the rest of the pond (in the direction of the line of rocks) will see the wave from the last rock before any of the others have had any effect on it. This is what is meant by the lack of superluminal information: the first rock can't affect the last rock because information about its impact on the water can't get to the last rock's impact.
With the speed of light, the "can't affect" becomes an absolute statement that encompasses all kinds of effects, so it is simply irrelevant to each part of your ring that you have already doused some other part, since the other parts are all too far away to have affected it (yet). At the moment your last bit of water hits the ring, that part of the ring is aware only of that last bit of water, so we can't expect that part of the ring to do anything special like maintain the ludicrous speed of the "wave". (Put differently, there is no single object that we may call "the wave", because it's not causally connected.) Moreover, assuming all the miniwaves propagate at the same speed, no previous part of the wave will ever catch up to the leading edge (where the water hit last), so there won't even be interference at that edge to make anything interesting happen.
In short, the answer is Yes, it's possible, but so what?. The fact that the water was once all in the center with you does not make it one object when it hits the ring, because it too has become causally disconnected. The various events of bits of water striking bits of ring are unrelated, and any "speed" you assign them as a group (by considering the change in position with respect to time of "the most recent place where water has hit") is entirely a construct of your perception.
Finally, note that non-physical speeds such as that of the water-ring contact point are not well defined given the existence of other observers; for an observer moving at an appropriate relativistic velocity, the wetting of the ring may be happening in more than one place at once and be moving in the opposite direction you say it is! --Tardis (talk) 17:11, 12 November 2008 (UTC)[reply]
See also Superluminal Scissors (wot, no article?); here, then. --Sean/76.182.94.172 (talk) 13:36, 11 November 2008 (UTC)[reply]
Cool article, I'm not sure if it applies here as the information would be directly impacting the surface rather than traveling the length of the scissor edge.--UhOhFeeling (talk) 20:02, 11 November 2008 (UTC)[reply]

Is concussion or neck injury common in horned animals?[edit]

Or, at least, the ones attacked to their heads?

I suppose the answer is kind of straightforward - the ones who would suffer them were sorted out long ago because they didn't survive - but it's one of those questions I have if I awaken in the middle of the night.

Reading about sheep (redirected from rams), I didn't even find much about their horns, though considering that the article says injuries are one of the main causes of death for them, maybe it does still happen.

I would also think that perhaps said horns are so incredibly thick that the constant head butting never gets through to their skulls. However, I also wonder if that doesn't increase the chance of neck injuries.

Thanks.Somebody or his brother (talk) 13:06, 11 November 2008 (UTC)[reply]

See the comments here for some thoughts on the relative resistance of head-butting animals to traumatic brain injury. - Nunh-huh 16:04, 11 November 2008 (UTC)[reply]
Which begs the question, if a sheep is concussed, or suffers cognitive impairment, how does one know? It isn't like the sheep was writing novels or working out differential calculus before the injury... --Jayron32.talk.contribs 19:31, 11 November 2008 (UTC)[reply]
You could devise some simple tests for the sheep to compare cognition before and after. Even measuring pupil reactivity or something. Plasticup T/C 23:48, 11 November 2008 (UTC)[reply]
The force of the collision can't "stay" in the horns (unless they deform but that is besides the point...) The force has to dissipate somewhere, which is down the horns, across the head, down the neck, into the body and to whatever is anchoring the animal in place ( In this case the hooves). Here is some info as it related to woodpeckers (http://web.cornell.edu/blogs/theessentials/?p=395) I bet some of it carries over... 152.16.15.23 (talk) 19:48, 11 November 2008 (UTC)[reply]

How many cells are in an embryo when it splits into twins?[edit]

I understand that identical twins are formed when an embryo splits very early on. How many cells are in the embryo when it splits? Is it always the very first split that results in two embryos? —Preceding unsigned comment added by JeremyStein (talkcontribs) 15:07, 11 November 2008 (UTC)[reply]

The article you link to answers yoru question in the first paragraph and the start of the second: "If the zygote splits very early (in the first two days after fertilization), each cell may develop separately its own placenta (chorion) and its own sac (amnion). These are called dichorionic diamniotic (di/di) twins, which occurs 18–36% of the time.[15] Most of the time in MZ twins the zygote will split after two days, resulting in a shared placenta, but two separate sacs. These are called monochorionic diamniotic (mono/di) twins, occuring 60-70% of the time[15]. In about 1-2% of MZ twinning the splitting occurs late enough to result in both a shared placenta and a shared sac called monochorionic monoamniotic (mono/mono) twins.[15] Finally, the zygote may split extremely late, resulting in conjoined twins."
I'm not sure how many cells the earliest splitting equates to, but it's clear that splitting is possible at almost any point in the pregnancy. 137.108.145.10 (talk) 15:35, 11 November 2008 (UTC)[reply]
well, i'm not sure that conjoined twins are split later on, so much as twins which have fused later on. the requirement for splitting the embryo is that it must happen early enough so that each cell is totipotent i.e. can take off and start an embryo on its own. this is a handy property for lab tinkering; you take an early embryo, gently take the cells apart, and you have a bunch of identical embryos. repeat as desired. according to our article, this stops at about 4 days. i'm not expert, so i'll accept that. doesn't sound unreasonable. at this point the cells begin to become increasingly specialized; for instance, the first thing is that cells become to think of themselves as one end of the embryo or the other. from fruitfly work, if you then transplant cells from one end to the other things go haywire, but if you move the cells at each end around on the same end, no problem. the specialization then gets more specific from there downhill, like at one point you can swap the future leg and antenna and each will switch to the correct item for where it now is, but the next day it won't. this implies that it's the neighboring cells that somehow tell the developing cells what they're supposed to be, and as each part develops the cells within tell each other what they're supposed to be doing. it's fascinating stuff. caveat: i haven't read anything about it in a decade or so, so this represents state of the art for the end of the last millenium. Gzuckier (talk) 15:53, 11 November 2008 (UTC)[reply]

Totipotent cells are present in the morula up to the 16-cell stage. At 32 cells, the morula begins to differentiate. So it isn't necessarily the first split that leads to identical twins. A split anywhere up to 16 cells can lead to identical twins. Axl ¤ [Talk] 11:30, 12 November 2008 (UTC)[reply]

basic anatomy[edit]

why do you taste eyedrops after applying them to your eyeballs? —Preceding unsigned comment added by 69.246.63.171 (talk) 18:53, 11 November 2008 (UTC)[reply]

It's because the drops travel down the nasolacrimal duct into your nasal cavity and from there down onto the back of your tongue. Fribbler (talk) 19:01, 11 November 2008 (UTC)[reply]
That's awesome!! Mac Davis (talk) 22:55, 11 November 2008 (UTC)[reply]
I wonder if it goes in reverse: i.e. spicy foods causing one's eyes to burn. I still figure the majority of the cause of that is airborn molecules but now I wonder... 152.16.15.23 (talk) 23:58, 11 November 2008 (UTC)[reply]
A dentist once injected novocaine into the inside of my cheek and it went up into my eye, which went numb. The tears that came out and ran down my cheek from the stinging tasted of novocaine. She said "oops". --Sean 01:03, 12 November 2008 (UTC)[reply]

How does the exchange of virtual photons mediate attractive vs. repulsive coulomb force?[edit]

I understand that, in QED, it is the exchange of virtual photons that mediates that coulomb force via the transfer of momentum. However, what I am having difficulty tracking down is how the charge information is carried. (I.e. whether the transfer should function as attractive or repulsive force.) I'm willing to work my way through the technical details if I can just be pointed toward a good online source, such as some scientific journal articles. (The virtual particle FAQ didn't prove useful for me, unfortunately.) Any suggestions? Thank you.—RJH (talk) 20:11, 11 November 2008 (UTC)[reply]

Hmm, no help here then.—RJH (talk) 16:54, 14 November 2008 (UTC)[reply]

pain[edit]

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.
This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page. --~~~~

glass recycling projects third world[edit]

We live in the Namib desert in Namibia and require information about recycling glass for a community project. Is there a way of reusing the glass bottles to manufacture glass tiles or beads or any other product to either sell to tourists or reuse in a sustainable way? Your advise will be much appreciated.

Johan Raubenheimer <email redacted> —Preceding unsigned comment added by 41.198.48.196 (talk) 21:53, 11 November 2008 (UTC)[reply]

Glass is a pretty recyclable material - you just need a sufficiently hot furnace - and people with the right skills to reform the liquid glass into new shapes. I guess in the Namib, the problem will be finding fuel for the furnace. I wonder if it would be possible to make a solar furnace with enough power to do the job? Making glass animals is a pretty easy skill to learn - you can do that with a small propane torch - you don't even need a furnace. If you have someone with the imagination to produce something unique, then you would have something to sell to tourists. The trick will be in finding a compelling and unique art style. If you can find the talent - you could sell the product on the Internet too. Small, high value product that don't have high shipping costs are a great choice for Internet marketting. SteveBaker (talk) 00:06, 12 November 2008 (UTC)[reply]
Also see Bottle wall for one use not requiring melting/furnaces. There are others. We used to make vases by coating and/or cutting glass bottles. Get a bunch of people together and brainstorm. 76.97.245.5 (talk) 07:33, 12 November 2008 (UTC)[reply]

surface of last scattering[edit]

Cosmic_background_radiation#Features says:

Accordingly, the radiation from the sky we measure today comes from a spherical surface, called the surface of last scattering. This represents the collection of points in space (currently around 46 billion light-years from the Earth—see observable universe) at which the decoupling event happened long enough ago (less than 400,000 years after the Big Bang, (13.7 billion years ago)) that the light from that part of space is just reaching observers.

If the event occurred 14 billion years ago in that space, shouldn't it only be 14 billion light-years away instead of 46 billion ly? At first I was thinking that we're moving away so fast that it takes 4 times as long for the light to reach us, but that violates special relativity, right? So how can both those numbers be right? Thanks 72.236.192.238 (talk) 22:44, 11 November 2008 (UTC)[reply]

It doesn't violate special relativity because there isn't actually any movement involved, it's just the space inbetween expanding. When we talk about the universe expanding that doesn't mean everything in it is moving outwards it means the actual spacetime itself is expanding. The standard analogy is to think of a balloon. Before you blow up the balloon, draw some dots on it. Now as you blow it up those dots get further and further apart, but none of them has actually moved relative to the balloon. --Tango (talk) 22:56, 11 November 2008 (UTC)[reply]
See Metric expansion of space#Understanding the expansion of space. The pictures illustrate a nearly identical situation (light from 12–13 billion years ago, instead of 13.7, from an object that's now 28 billion light years away, instead of 46). -- BenRG (talk) 23:10, 11 November 2008 (UTC)[reply]
It is really quite simple (simpler than the above answers make it sound). The thing we see as the surface of last scattering is moving rapidly away from us (Hubble expansion). Today we can see light from it that was emitted 13 billion years ago, and we infer from where it was then and how fast it has been moving that it should now be 46 billion light years away. In other words, it is simply a matter that the separation is believed to have increased a lot during the time it took the initial light to reach us. Dragons flight (talk) 23:40, 11 November 2008 (UTC)[reply]

Black knees[edit]

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.
This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page. --~~~~
If you are concerned about unusual symptoms you experience after physical exertion, you should speak to a medical expert. TenOfAllTrades(talk) 23:24, 11 November 2008 (UTC)[reply]

I'm not concerned about the symptoms; it's a physiological question. Not worth a wasting the Doctors time. So I'll ask agin, why do my knees look like they are bruised when I come back from a long run, but it isn't a bruise as it goes away a couple of hours later. --TrogWoolley (talk) 00:21, 14 November 2008 (UTC)[reply]

any way to guess how long it took someone to write something?[edit]

Resolved

I'm wondering if there is any way to come up with an approximate guess as to how long it took someone to write something. in particular, how could I go about guessing how long it may have taken to write these paragraphs in particular? What about a more general way to guess? —Preceding unsigned comment added by 83.199.126.76 (talk) 23:07, 11 November 2008 (UTC)[reply]

Probably the wrong place. Touch typists can manage up to 1k per minute, although 400 characters per minute is more typical. --Stephan Schulz (talk) 23:21, 11 November 2008 (UTC)[reply]
Wikipedia on this article cites it at 50-70 words per minute. I guess I'd use that as a baseline but obviously it's going to depend on the individual. —Cyclonenim (talk · contribs · email) 23:25, 11 November 2008 (UTC)[reply]
Two things: one, what does "probably the wrong place" mean? Two: why do you both talk about typing speed, when I mean writing speed. We're not talking about chatting (or how I'm writing now), but composition as well....surely it's less than blind typing, especially when we're talking about references as well... Can anyone give me a minute estimate for my above link specifically, and how you made that guess? Thank you!
You're in the wrong place because this problem is not well enough defined to have a useful scientific answer. I gave you a lower limit. Composition speed is often considered half of typing speed, but that of course depends on how well the writer is familiar with the topic and if they know what they actually want to say. You might get some anecdotal evidence at the language or humanities desk. --Stephan Schulz (talk) 00:10, 12 November 2008 (UTC)[reply]
Okay, so I cut and paste the paragraphs into a character counter and divided by five, because hte typing speed article says every five characters counts as a word (regardless of actual space locations), and got 1022. So if the guy composed it continuously, and was able to type all that at 60 words per minute, then if composition is half of typing speed (30 wpm) it took him 34 minutes. If he was very thoughtful and typed half as slowly, it took him 68 minutes. And the upper limit is about 15 hours, because that's how long there was between when what he's replying to appeared, and his reply.... So... anywhere from 34 minutes to 15 hours. Can we do ANY better than that?


I think it's pretty clear that the answer is "No". We know the time when the edit was committed - but we don't know when the person started to type it. It's certainly not just a matter of typing speed - thinking speed matters too. Also, if they proof-read and then go back and fix up errors or decide to expand upon a particular thought...there is truly no way to put a number on it. SteveBaker (talk) 23:49, 11 November 2008 (UTC)[reply]
Do you agree with my reasoning of between 34 minutes and 15 hours? Can you find a trick to improve it by a TAD? (such as a way to figure out that it's almost certainly less than ten hours -- ALL of us know this intuitively!)
No - I certainly don't agree. We have no idea whether this was bulk cut/pasted from other sources - or whether this was a synthesis of some other document written in advance - or whether this is a very thoughtful piece composed by someone with many interruptions who goes through many drafts before hitting that 'Send' button. Heck, he may have been thinking about this for a year and only just now decided to put his words into print. There is quite simply no way to know - and any kind of speculation is just that: "speculation". SteveBaker (talk) 04:49, 12 November 2008 (UTC)[reply]
Thank you. You have me convinced. I realize I can't even have an upper limit of even the lifetime of the poster, since the paragraphs could include cut-and-paste from deceased people as well! :( I guess the conclusion is exactly what you first said, no way to know even approximately. You can put a green checkmak, resolved, next to this question (I don't know how to do it). Thanks!
FYI: You add {{resolved}} just under the title of the question. SteveBaker (talk) 18:11, 12 November 2008 (UTC)[reply]
May I ask why it has to be less than 10 hours? And on a second note, sometimes people leave their computer in the middle of typing, so really even 10 hours is possible. —Preceding unsigned comment added by 76.69.241.185 (talk) 00:56, 12 November 2008 (UTC)[reply]
I understand that Flaubert wrote Madame Bovary at a rate of about 1 sentence per day. --Sean 01:12, 12 November 2008 (UTC)[reply]
At the other end of the spectrum, Anthony Trollope wrote 1000 words an hour, day in, day out for years, according to Brittanica Concise Encyclopedia. Clarityfiend (talk) 04:13, 12 November 2008 (UTC)[reply]
Leonhard Euler likely more than that. Mac Davis (talk) 22:22, 12 November 2008 (UTC)[reply]