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February 27

Bulletin of the Seismological Society of America in online libraries?

I'm trying to make the cited sources in White Wolf Fault as accessible to readers as possible but I haven't been able to find this journal in the Internet Archive, JSTOR or Wiley. HathiTrust only has the issues from 1911 to 1926. The Seismological Society of America's website has has the issues cited, but it requires library or institutional access. If there are other journal repositories in our Library that have it, I've missed them. I'm stumped. If someone can help out, future readers will be better off...

I do not at all mind having to hunt down the article in the proper repository (keeps me from getting lazy, lol). Just point me in the right direction and off I'll go! Oona Wikiwalker (talk) 07:19, 27 February 2024 (UTC)

The cited articles are also available online for a (hefty) fee at GeoScienceWorld; see here and here. The fee imposed suggests that the publisher enforces its copyright.  --Lambiam 11:38, 27 February 2024 (UTC)

But Library access is within copyright restrictions because they pay for access for their patrons. That's what I'm inquiring after; did I miss a collection within our library that contains this journal?

Oona Wikiwalker (talk) 23:01, 27 February 2024 (UTC)

I've checked the Wikipedia Library and it doesn't provide access. I have added the doi and bibcode to the cite to make it a bit easier for readers to click to the journal. Mike Turnbull (talk) 16:48, 28 February 2024 (UTC)

It's long been an issue for those of us in Wikipedia:WikiProject Earthquakes, particularly for US earthquakes. You either pay the money or take a trip to a library that stocks that journal, which is generally easier said than done. Mikenorton (talk) 17:28, 28 February 2024 (UTC)

A mistake in our article Four momentum

The third paragraph states: p⁰ = E/c², with p⁰ being the momentum in the time dimension.

I'd fixed it: p⁰ = E/c, but somebody reverted. Would you like to give your opinion? HOTmag (talk) 18:06, 27 February 2024 (UTC)

The version with c² is formally correct. Another matter is whether that should be in the lede of the article (and who is "some authors"?). I don't think I've ever seen that convention — whenever I've seen $x^{0}=t$ it was due to setting $c=1$, in which case $p_{0}=E$. --Wrongfilter (talk) 18:13, 27 February 2024 (UTC)

I admit I'm quite surprised. The four momentum is (p⁰, p¹, p², p³), traditionally written also as (E/c, p_x, p_y, p_z), hence: p⁰ = E/c. On the other hand, both sides of the current expression (in the lede): p⁰ = E/c² don't have the same units, so how can it be correct? HOTmag (talk) 18:29, 27 February 2024 (UTC)

I wrote "formally correct", in some weird unit system. I don't think it makes much sense, which is why I removed it from the article. --Wrongfilter (talk) 18:38, 27 February 2024 (UTC)

Which one of the following three statements do you find "weird"?

1. The four momentum is (p⁰, p¹, p², p³).

2. The four momentum is (E/c, p_x, p_y, p_z).

3. Hence: p⁰ = E/c.

HOTmag (talk) 18:51, 27 February 2024 (UTC)

Oh please, give it a rest. The statement in the article had a context that you keep ignoring. --Wrongfilter (talk) 18:55, 27 February 2024 (UTC)

You are allowed to avoid answering my last question, which has nothing to do with the article. Actually, regardless of the article, I was quite curious to know whether you also considered the third statement p⁰ = E/c (which had been my correction before it was reverted): as "weird", while I assumed you accepted the two statements it followed, and that's why I asked you my last question, regardless of the article. But again, nobody forces you to answer my last question. HOTmag (talk) 19:44, 27 February 2024 (UTC)

$p^{0}=E/c$ is standard, nothing weird about that. --Wrongfilter (talk) 20:15, 27 February 2024 (UTC)

So (back to article), my correction p⁰ = E/c in the article was right, and it didn't have to be reverted to p⁰ = E/c². That's what I wanted to know from the very beginning: whether the user who reverted my correction was wrong (as I thought). HOTmag (talk) 22:26, 27 February 2024 (UTC)

No you were wrong in that context. That bit has been removed now from the article because what was there isn't common and as shown here will just cause confusion. It is best to have all the elements using the same units. NadVolum (talk) 23:25, 27 February 2024 (UTC)

I haven't asked about whether the context justified mentioning this formula there, nor about why the whole formula was eventually removed. I've only asked about whether, my replacing the original formulation p⁰ = E/c² by the correct formula p⁰ = E/c, had been a better version than the original version p⁰ = E/c². In other words, I've wanted to know if the user who reverted my correction p⁰ = E/c back to the original version p⁰ = E/c² made the article worse. I think they did, and I've wanted to get your opinion about what I think. All of that is quite clear in my first post, that had actually been posted before the whole formula was eventually removed. HOTmag (talk) 08:07, 28 February 2024 (UTC)

The original statement was a conditional, with an "if... then..." structure. The formula that you changed was in the "then..." part, and was correct under the condition stated in the "if..." part. By changing the "then..." part without taking into account the "if..." part, you introduced an error, and the other user was right in reverting your edit. --Wrongfilter (talk) 08:38, 28 February 2024 (UTC)

I got it now. Thank you. I understand now that the original "then" part (before I changed it) would be correct assuming that the "if" part were correct. However, the "if" part really introduces a definition I've never come across, and should be omitted. HOTmag (talk) 09:52, 28 February 2024 (UTC)

Parachutes and terminal velocity

Hi, I'm currently doing a lab using a small model parachute. As I currently understand it, the terminal velocity can be calculated by equating the weight of the object to the drag force, given by the drag equation. My question is, how can I use this terminal velocity to estimate the time it would take an object to fall a set distance? Do I assume the object starts out at terminal velocity? Do I use a SUVAT equation? Thanks! ARandomName123 (talk)^{Ping me!} 23:05, 27 February 2024 (UTC)

the time to accelerate to the terminal velocity needs to be accounted for. RudolfRed (talk) 00:22, 28 February 2024 (UTC)

So one needs to integrate the varying acceleration. Poor ARandomName123. Zarnivop (talk) 02:18, 28 February 2024 (UTC)

Integration is the bane of my existence. I guess I'll just approximate it then. Thanks for the help! ARandomName123 (talk)^{Ping me!} 03:09, 28 February 2024 (UTC)

Take the height of the fall, divide by the terminal velocity. That gives you a rough answer. So if the falling object gets close to terminal velocity after just a short fraction of the fall (for a human with a parachute, around 50 metres) and the fall is short compared to the scale height of the atmosphere (around 7 kilometres on Earth; else the changing density changes the terminal velocity) and you don't care too much about the second digit of your answer, this may be good enough. Otherwise, you have to integrate the equations of motion. PiusImpavidus (talk) 11:18, 28 February 2024 (UTC)

Yes, that's what I ended up doing. It wasn't actually too far off from the experimental result, only a few fractions of a second. Thanks! ARandomName123 (talk)^{Ping me!} 13:37, 28 February 2024 (UTC)

In the following, we only consider the motion of an idealized point particle along a vertical line. We use the variables $t$ for time, $s$ for the particle's position above $0$, $v={\frac {ds}{dt}}$ for its velocity and $a={\frac {dv}{dt}}$ for its acceleration. Furthermore, $m$ stands for its mass and $g$ for the acceleration of gravity.

The force exerted on the particle is the sum of the forces due to drag and to gravity:

$F=F_{\text{drag}}+F_{\text{grav}}=m\kappa v^{2}-mg=m(\kappa v^{2}-g),$

in which $F_{\text{drag}},$ whose form given by the drag equation, contains a yet unknown constant $\kappa .$ Its value may be determined if the terminal velocity $v_{\infty }$ is known, as follows. Terminal velocity is the limit velocity, approximated arbitrarily closely as the acceleration $a$ tends to $0,$ which means (since then $F=ma$ tends to 0) that the upward drag force cancels the downward gravitational force, or $\kappa v_{\infty }^{2}-g=0.$ It follows that $\kappa =g/v_{\infty }^{2},$ so now we have

${\frac {dv}{dt}}={\frac {F}{m}}=\kappa v^{2}-g=g\left({\frac {v^{2}}{v_{\infty }^{2}}}-1\right).$

This differential equation, combined with the condition that $v=0$ at time $t=0,$ the moment of release, is solved by:

$v=-v_{\infty }\tanh \left({\frac {g\,t}{v_{\infty }}}\right).$

Integrating again, with the condition that at time $t=0$ the particle is released from a given height $h,$ results in:

$s=h-{\frac {v_{\infty }^{2}}{g}}\ln \,\cosh \left({\frac {g\,t}{v_{\infty }}}\right).$

Solving $s=0$ for $t$ gives the time it takes the particle to reach ground level:

$t={\frac {v_{\infty }}{g}}\,{\text{arcosh}}\,\exp \left({\frac {g\,h}{v_{\infty }^{2}}}\right),$

in which ${\text{arcosh}}$ stands for the inverse hyperbolic cosine. Asymptotically, as $h$ tends to infinity, this is equal to

${\frac {h}{v_{\infty }}}+{\frac {v_{\infty }}{g}}\ln 2.$

 --Lambiam 00:36, 1 March 2024 (UTC)

I haven't learned inverse hyperbolic functions yet, but thanks for the detailed answer! ARandomName123 (talk)^{Ping me!} 02:38, 1 March 2024 (UTC)

February 28

Why are airports never an effing big Greek cross?

Two runways in series instead of in parallel (i.e. 30 Back for landing 30 Front for takeoff 12 Back for landing 12F takeoff if wind azimuth near 120) and the paths never have to cross even in the air. And another two in series at right angles that they switch to if crosswind's too strong. In extreme both engines exploded on fire emergencies they clear all runways much easier landing.

Also why'd it take so long to invent faster runway exits with slight angles? That seems like common sense. Sagittarian Milky Way (talk) 02:28, 28 February 2024 (UTC)

Space at existing airports within metropolitan areas is obviously one limiting factor. With completely new builds in rural areas, it could work. Airports are also always looking at ways of minimising noise on the ground. Your proposal would stretch the area affected by aircraft noise to twice the distance. HiLo48 (talk) 02:48, 28 February 2024 (UTC)

I assume you mean noise due to the flight paths of the airplanes. But besides noise, different flight paths from different runway directions may also limit construction height near the airport or simply not be possible due to current constructions. See also [1] [2] and consider whether this works so well under the proposal. Nil Einne (talk) 09:43, 28 February 2024 (UTC)

Well Denver had forced land, forced altitude and enormous property in the middle of nowhere but they made this (one of the most efficient layouts possible).

Denver

Sagittarian Milky Way (talk) 04:21, 1 March 2024 (UTC)

Even away from metropolitan areas flat land suitable for an airport may be at a premium and the shape of the suitable area limits the possibilities. The angles of runway exits were not much of an issue until the introduction of heavy wide-body aircraft, which take longer to reduce speed and don't turn as readily as lighter aircraft. If the space is available, an L-shaped two-runway pattern offers the same crosswind-avoidance advantage as a cross while having a decreased collision risk and giving an easier task in designing the whole airport plan.  --Lambiam 10:03, 28 February 2024 (UTC)

Also consider what happens on two inline runways when there's a departure on one and a go-around on the other: the aircraft going around will end up in the wake vortex of the departing aircraft. So the capacity of two in-line runways isn't really much better than that of a single runway. It's better to put them in a staggered configuration.

High-speed runway exits appeared in the 1950s, right at the start of the jet age with its high landing speeds and long runways. And at that time, air traffic wasn't so dense that they needed to land 25 planes per hour per runway. PiusImpavidus (talk) 11:38, 28 February 2024 (UTC)

Wouldn't some who could afford to fly then like time-obsessed execs like fast trips? Sagittarian Milky Way (talk) 16:31, 28 February 2024 (UTC)

Corporate execs want bragging rights: "My bizjet can do Mach 0.92," not "I paid the airport $600k for a high-speed taxiway at exactly the right spot, so now I save 30 seconds each time I visit HQ." PiusImpavidus (talk) 23:02, 28 February 2024 (UTC)

Fair enough. Sagittarian Milky Way (talk) 00:14, 29 February 2024 (UTC)

And don't non-inline runways in the same direction cause problems unless very separated? Which would increase taxi distance between runway and terminal. Sagittarian Milky Way (talk) 16:35, 28 February 2024 (UTC)

Depending on the quality of your surveillance system, and ignoring wakes, you need a horizontal separation of 2.5 to 3NM, or a vertical separation of 300m/1000ft. A modern jet climbs about 1000m/30000ft per minute, i.e. it has the necessary vertical separation in 20 seconds. Takeoff speed is 150knots or higher, so at least 2.5NM per minute, and rapidly increasing. So you would have the required horizontal separation in less than a minute. IIRC, the very best airports manage about one takeoff per minute, so separation is not a problem. --Stephan Schulz (talk) 21:14, 28 February 2024 (UTC)

Parallel runways usually have about 500m separation (like 7C/25C and 7R/25L at EDDF). How much you need depends on the operational flexibility you want; parallel departures require less separation than parallel arrivals. 500m additional taxi distance is less than a minute travel time. Or you could put the runways on opposite sides of the terminal (like at EGLL), so there's no additional taxi distance at all. If you have them slightly staggered (like at EDDB) instead of exactly side by side (like at EDDL), you save some taxi distance and you gain some safety during parallel approaches. BTW, LEMD gets close to your solution. It has two parallel departure runways and two parallel arrival runways, pretty much end to end, but with an angle between them. The new terminal 4 is where the runways almost join. PiusImpavidus (talk) 23:40, 28 February 2024 (UTC)

So when I see 9,000 feet on Google that won't help too often vs 500 m? We have 300 yards apart in KEWR slightly staggered and 250 yards in KSFO but those are compromises with limited land. I think KSFO might be the record for big planes in my country. LEMD looks pretty cool though I wonder why they didn't join them, Space Shuttle accidents and Gimli Gliders would have an easier land and maybe they could avoid a tire explosion or gear breakdown from trying to stop so hard. Sagittarian Milky Way (talk) 00:55, 29 February 2024 (UTC)

I suppose KEWR normally uses one runway for arrivals, one for departures, with no parallel arrivals. KSFO does parallel arrivals, but has some restrictions. The aircraft must fly side by side, so that neither can get into the wake of the other (wakes are angled in a crosswind) and the pilots must see the other plane before aligning with the runway, so they can only do parallel arrivals of planes with the same landing speed in good weather. My local airport (EHAM) has 3 parallel runways with 2.1–2.8km separation and 1.5–3km stagger, allowing uncoordinated parallel arrivals on centre and right (left is departures only, right arrivals only) even in poor weather. They didn't need all that separation for that, but the terminal had to fit in the east gap and the west gap was needed for noise reasons.

LEMD was extensively rebuilt about 20 years ago. It used to have 4 crossing runways, 3 of which were closed. The original 18/36 was almost in-line with the new 18L/36R, but was closed. I suppose interference of go-arounds with departures and available space (both land and noise) was the reason. A runway in-line with 14R/32L (the old runway left) to the NW would have led up to La Moraleja, a hill with some expensive housing. PiusImpavidus (talk) 11:41, 29 February 2024 (UTC)

KEWR indeed uses one for arrival and one for departure. Sagittarian Milky Way (talk) 04:24, 1 March 2024 (UTC)

Does Chicago's Midway Airport fit your description? ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:00, 29 February 2024 (UTC)

Close but each direction's only one runway in a row not two. If Denver doubled like this the longest strip would be at least 32,000 feet long (16,000 like the current runway for takeoff and a downwind 16,000 for landing). Sagittarian Milky Way (talk) 21:41, 29 February 2024 (UTC)

Does a given body's total rest energy, have any other meaning, besides the body's ability to totally vanish - becoming energetic radiation?

HOTmag (talk) 10:01, 28 February 2024 (UTC)

A "kilobar" gold bar, with a rest energy of some 90 PJ, has a market value of about €60,000 or $65,000, twice that of a goldbar with half that rest energy. Transformed into "usable energy", the market value of this energy would be much higher, but we do not have the technology for the necessary transformation. The main reason for physicists to use the concept is to avoid conversions in calculating the energy balance of particle interactions, being able to use eV units throughout.  --Lambiam 10:31, 28 February 2024 (UTC)

By a "meaning", I've meant: a physical meaning (rather than an economic one). HOTmag (talk) 10:47, 28 February 2024 (UTC)

The main reason to introduce the concept of rest energy (corresponding to the mass of the body) is that it is necessary to include it in the energy balance to satisfy conservation of energy. This is what enables the conversion into radiation, for example (in practice, other conservation laws may stand in the way of that process). In addition, it determines the inertia and the gravitation of the body (as mass, rather than rest energy). In addition, the internal energy (which is part of the rest energy in the case of composite systems) plays a role in the thermodynamics of the thing (where this is relevant). --Wrongfilter (talk) 10:43, 28 February 2024 (UTC)

Note I asked about total rest energy. I can understand how it's helpful in determining the value of energy of the radiation - the body has turned into, whereas the value of the body's mass is already known. However, does the body's total rest energy - also have any meaning - in determining the inertia or the gravitation (or anything else) of the body whose mass is already known?

To sum up: besides a Gedanken experiment in which a given body having a given value of mass - totally vanishes - becoming energetic radiation, can you think about any other Gedanken experiment, which involves a body having a given value of mass, and which shows how the body's total rest energy can be helpful? HOTmag (talk) 11:28, 28 February 2024 (UTC)

What makes you think that there's a difference between a body's total rest energy and its mass (times c²)? --Wrongfilter (talk) 12:00, 28 February 2024 (UTC)

After the formula E=m^2 was discovered, all of us know there's no difference between a total rest energy and a total mass (times c^2). However, before discovering the formula E=m^2, we couldn't predict the value of energy - of the radiation a given body turned into, even when we did know the value of total mass - the body had had - before it became energetic radiation. This fact proves, that the meaning of a given body's total rest energy, lies (for example) in the body's ability to totally vanish - becoming energetic radiation - whereas the value of the body's total mass is already known. That said, I still wonder if this is the only meaning of the total rest energy, whereas the value of the body's total mass is already known. HOTmag (talk) 12:24, 28 February 2024 (UTC)

The terms "rest energy" and "rest mass" are practically interchangeable, except that some physicists feel compelled to express the latter in units of MeV/c². Others have no such compunction and are happy to write, for example, "a K⁰ particle of a mass of about 650 Mev".^[3]  --Lambiam 16:05, 28 February 2024 (UTC)

Yes, the terms "rest energy" and "rest mass" are practically interchangeable, due to the discovery of the formula E=m^2. However, they are not conceptually interchangeable, as I explained in my previous response you've just responded to. HOTmag (talk) 18:01, 28 February 2024 (UTC)

You keep writing ${\textstyle E=m^{2}}$. The formula is ${\textstyle E=mc^{2}}$. The units don't make any sense the other way.

As Lambiam has pointed out, in some contexts, physicists will adopt a system of units where c = 1 (and sometimes h = 1, too), but you still don't square the value of the mass in that case.

I would argue that the terms "rest energy" and "rest mass" are wholly synonymous. The former indicates nothing other than, "what is the rest mass of this thing, expressed in units of energy?"

PianoDan (talk) 19:31, 28 February 2024 (UTC)

Note I'd written to Wrongfilter: "After the formula E=m^2 was discovered, all of us know there's no difference between a total rest energy and a total mass (times c^2)". So, you can figure out that E=m^2 was of course a repeated typo (in which I only kept omitting the "c" before the "2") in my reciting the best known scientific formula, which I was reciting (correctly) by heart - before I knew to count (and before I knew what it meant)...

As for your second remark: Well, it seems you didn't read all of my response (mentioned above) to Wrongfilter. If I'm wrong and you did read all of it, so let me put it this way: If Einstein thought like you that those terms were synonymous, he wouldn't feel he had to prove the equivalence between those terms (up to c^2). HOTmag (talk) 23:21, 28 February 2024 (UTC)

But he DID prove it, and it's been experimentally verified continuously since then. We don't have to behave as though that hasn't already been thoroughly established. PianoDan (talk) 23:26, 28 February 2024 (UTC)

Yes, he proved it. But this fact doesn't mean those terms are synonynous. Just as the terms "101+10" and "110+1" are not synonymous, even though it's provable that 101+10 = 110+1. Unless we disagree about what "synonymous" means. It seems that you're coming from a practical standpoint, whereas I'm coming from a conceptual standpoint. HOTmag (talk) 23:34, 28 February 2024 (UTC)

If you combine the definition "having the same meaning" of synonymous with the definition "the denotation, referent, or idea connected with a word, expression, or symbol" given for meaning, we can observe that the mathematical expressions "101+10" and "110+1" are two names for the same referent (the integer 111) and so are synonymous in at least some sense. Viewed as strings, formulae, or numerical recipes, they can obviously be distinguished, in which case they have – in another domain than the integers – different referents.  --Lambiam 01:07, 1 March 2024 (UTC)

Yes, the same argument could be made for the apparent synonymy between tg and sin/cos. Anyway the OP (i.e. me) posted this thread while assuming that tg and sin/cos are not synonymous. And if you wonder now whether, then...no, synonymy is not the shortest word containing at least three Ys. syzygy is shorter :) HOTmag (talk) 09:43, 1 March 2024 (UTC)

February 29

Gravitational potential energy

Our article gravitational energy, asserts in the first chapter:

For two pairwise interacting point particles, the gravitational potential energy $U$ is given by $U=-{\frac {GMm}{R}},$

without making any distinction between Classical mechanics and Relativistic mechanics.

I wonder if this definition leads to a vicious circle in Relativistic mechanics, since: According to the formula $E=mc^{2},$ the whole mass $m$ of a given body already contains also the body's gravitational potential energy $U,$ defined above by the whole mass $m,$ whose content contains also $U...,$ so isn't it a vicious circle?

Apparently, it's like when someone mentioning A - who is asked "what do you mean by A" - defines A by B, but when asked "what do you mean by B" - they define B by A. In our case, the whole mass $m$ depends (according to the formula $E=mc^{2})$ also on the energy $U,$ depending (according to the formula $U=-{\frac {GMm}{R}}$) also on $m,$ right? Unless I miss something here.

However, if our article (in the first chapter) did have to make a distinction between Classical mechanics and Relativistic mechanics, then how does the latter avoid the vicious circle mentioned above? HOTmag (talk) 08:42, 29 February 2024 (UTC)

See Mass in general relativity. Graeme Bartlett (talk) 10:55, 29 February 2024 (UTC)

Thank you for the link.

Having read it, let me put my original question this way:

1. Can General Relativity give a clear cut answer to any/both of the following questions?

A. Does a given body's gravitational potential energy contribute to the body's total mass?

B. Does a given body's gravitational potential energy depend on the body's total mass?

HOTmag (talk) 12:13, 29 February 2024 (UTC)

I think the answers are yes and yes. This is what makes General Relativity nonlinear and very hard. This is related to the question whether gravitational waves could exist in GR — Einstein showed that linear waves could exist in the weak-gravity limit (that's easy, and in this regime your questions are quantitatively irrelevant), but had doubts himself whether waves could actually be generated in the strong-gravity regime. This question was only settled in the 1960s. --Wrongfilter (talk) 12:17, 29 February 2024 (UTC)

If (as you say) the answers are yes and yes, i.e. a given body's gravitational potential energy, both contributes to the body's total mass, and also depends on the body's total mass, so it follows logically that the gravitational potential energy, both contributes to itself (it being a part of the total mass), and also depends on itself (for the same reason). Isn't it like Baron Münchhausen who saved himself from drowning by pulling up on his own hair? In other words, isn't it a vicious circle? All of that reminds me of the algebraic equation: x=x+1...

Indeed, sometimes defining an object by itself does not lead to any vicious circle, e.g. in the algebraic equation: 2x+1=x+2, but that's only because we can prove that this equation has a solution (actually a unique one). However, how can we be sure that the interdependence between the total mass and the gravitational potential energy leads to no contradiction, as opposed to the case of the interdependence of the sides of the equation x=x+1? In both cases, an object depends on itself !

This is my wonder from a logical viewpoint. But, let's put logic aside, and get back to physics. My practical question is: Can General relativity describe the gravitational energy, in such a way, that this gravitational energy will only depend on the mass's components other than its gravitational potential energy component? Just as we can do the same in the algebraic equation: 2x+1=x+2, by replacing it by a direct equation x=1, i.e. so that x will only depend on a pure number. For simplicity, let's focus on the two object case, assuming that the whole universe only contains two electrons alone (or any two uncharged objects alone). HOTmag (talk) 13:15, 29 February 2024 (UTC)

I've removed the equation from the title of this question, this seems to cause some problems and weird behaviour of the system. --Wrongfilter (talk) 12:19, 29 February 2024 (UTC)

Gravitational potential energy is not associated with just some of the several bodies involved; it is a quantity associated with a system of bodies in gravitational interaction. Inasmuch as one may want to consider an increase in gravitational energy as an increase in mass, it is the mass of the whole multi-body system.  --Lambiam 11:09, 1 March 2024 (UTC)

Of course, but I don't see how all of that has anything to do with how General relativity avoids the vicious circle I asked about. 15:09, 1 March 2024 (UTC) HOTmag (talk) 15:09, 1 March 2024 (UTC)

I'm still not sure whether I fully understand what question B actually means. I'll start with what I think I know for sure (this is essentially weak fields), and then indicate what I think is the case in the general, strong-field case. Taking the equation for the gravitational energy as it is, U is the energy that we need to supply to separate two bodies of mass M and m, initially at distance R from each other, i.e. move both to "infinity". This is a binding energy, it is negative, and causes a mass defect, so that the total mass of the system is less than the sum M + m. If the system (say a binary black hole) is gravitationally bound, the mass defect is ${\tfrac {U}{2c^{2}}}$ — only half of the gravitational binding energy because there is kinetic energy that keeps the system from collapsing, and the balance between gravitational and kinetic energy is given by the virial theorem. Therefore the total mass is $M_{\mathrm {tot} }=M+m-{\tfrac {U}{2c^{2}}}$. This is not a fully general relativistic treatment of the problem; I don't know what the terms "gravitational energy" and "total mass" would mean in GR; they are integrated quantities and integration in GR is a bit of a mystery to me. Having said that, the gravitational field (the metric) carries energy (gravitational waves certainly do transfer energy) so it enters the energy-momentum tensor, and the field/metric consequently appears on both sides of Einstein's equation. There are self-consistent solutions for this, which are hard to find, but they exist and the question is not a logical conundrum or "vicious circle". If you're not happy with this sort of answer (I expect your aren't), you'll have to find an actual expert or dive deep into the GR literature. --Wrongfilter (talk) 16:06, 1 March 2024 (UTC)

I don't get why the mass equivalent of the potential energy is not added to the rest masses of the components, like $M_{\text{tot}}=M+m+{\tfrac {U}{2c^{2}}}$?  --Lambiam 11:11, 2 March 2024 (UTC)

The binding energy is negative, as energy needs to be lost for the system to form a bound state, or conversely you need to supply energy to break a bound system apart. Take a comet that falls in from the Oort cloud ("from infinity"). If unperturbed, it will follow a hyperbolic orbit past the sun and vanish back to infinity. If it passes close to Jupiter, say, it may transfer some of its energy to that planet and be perturbed onto a tighter elliptical orbit. This bound orbit has a lower energy than the previous unbound orbit, hence the binding energy is negative. Similarly for a collapsing gas cloud that forms a star or a galaxy: The collapse leads to a tighter configuration only if the gas cloud loses energy, mostly through radiation (radiative cooling). This loss of gravitational energy is also what powers active galactic nuclei, quasars and such. Gravitational systems that cannot cool radiatively (e.g. dark matter) can cool through other processes like violent relaxation or in a simpler way by accelerating and losing some particles while the remainder are decelerated and become more tightly bound. In black hole mergers, this is quantitatively relevant: The models for the observed gravitational-wave events always have and end state (single black hole) of lower mass than the sum of the original black holes. Long story short, bound systems have lower total energy compared to unbound systems, hence binding energy is negative, hence the total mass is less than the sum of the masses of the constituents. --Wrongfilter (talk) 12:14, 2 March 2024 (UTC)

Using the formula $U=-{\frac {GMm}{R}},$ we have

${\frac {{\text{d}}\,U}{{\text{d}}R}}={\frac {GMm}{R^{2}}},$

which means that the energy increases and decreases with distance. The highest energy is at infinity. Isn't that consistent with a tighter orbit having a lower energy?  --Lambiam 21:43, 2 March 2024 (UTC)

Oh, I didn't see your response until now. Yes, a tighter orbit has a lower energy. But that's just what I was saying in so many words anyway. --Wrongfilter (talk) 12:32, 10 March 2024 (UTC)

Energy of one period for alternating current

In alternating current, as at the output of an amplifier for a dipole antenna, I cannot find a direct formula giving the energy over a period. Is it simply accepted that this value is given by dividing the power by the frequency? Or is there a direct formula and if so which one? Malypaet (talk) 12:10, 29 February 2024 (UTC)

Energy is power per unit time. So "energy over a period" is just power.

The formula for that isn't so much "accepted" as it is simply the definition of the terms. Energy defined to be power per unit time. (Power / unit time) * (time period) = power. Just basic dimensional analysis. PianoDan (talk) 15:32, 29 February 2024 (UTC)

I assume they mean "energy emitted over a period", i.e. power * (time period). For a monochromatic sinusoidal oscillation this would indeed be power / frequency. --Wrongfilter (talk) 15:39, 29 February 2024 (UTC)

If you use "P" for power in watt or joule/s, "E" for energy in joule, $\Delta t$ for the unit of time of 1s and T for a period in second ($T={\frac {\lambda }{c}}$), "Energy is power per unit time" is like $E={\frac {P}{\Delta t}}$ and as what I know is $P={\frac {E}{\Delta t}}$, you write $E={\frac {E}{\Delta t^{2}}}$...People have a great problem in distinguishing Energy (kinetic energy of an object or energy accounted for in an interval of time in an energy flow), and Power that is an energy rate (flow,etc...).

This is why I am looking for an equation directly giving the energy accounted for over a period ($T={\frac {\lambda }{c}}$) of a sinusoidal electric current. Energy in joule equal to W-s (watt⋅second). Malypaet (talk) 21:30, 29 February 2024 (UTC)

OK, mea culpa, I wrote that wrong. Power is energy per unit time, not the other way around. That said, it's STILL just basic dimensional analysis. (Admittedly, using the correct units does help.)

P = E / t. So E = P * t. That's all there is. PianoDan (talk) 23:26, 29 February 2024 (UTC)

March 1

Neanderthal speech

Could Neanderthals and other premodern humans been handicapped by being tongue tied? How could we find out or rule it out if it isn't preserved by fossils? Maybe by DNA?Rich (talk) 06:57, 1 March 2024 (UTC)

The Neanderthal Throat - Did Neanderthals Speak?. Alansplodge (talk) 10:50, 1 March 2024 (UTC)

Neanderthal genomic DNA has successfully been sequenced and analyzed. To use this for testing hypotheses regarding ankyloglossia impacting Neanderthal speech capability, researchers need to identify the genes involved in the development of a human untethered tongue. I do not know whether this has been achieved, but I doubt it has even been attempted. Mutations (possibly epigenetic) connected to ankyloglossia are reported for the MTHFR gene and the TBX22 gene, but these have much more general roles than the development of a normal tongue. (According to sources, most cases of the anomaly (in humans) are sporadic, so I do not understand why the infobox in our article has "Specialty    Medical genetics".) Finally, the handicap impedes the production of coronal consonants, but not of vowels and the labial, dorsal and laryngeal consonants, leaving enough phonemic space for a rich vocabulary.  --Lambiam 10:53, 1 March 2024 (UTC)

Manually igniting a star

In theory, if you were to go to the center of a cloud of hydrogen/helium gas in the middle of space, an area of gas that hasn't undergone nuclear fission, would it be possible to light the gas and start burning it? Assuming you have the oxygen required to actually light the initial flame, could you, in theory, start the process of fission through a simple handheld lighter? Or would the gas not ignite and simply remain how it is until it's compressed and undergoes fission that way? 71.38.197.194 (talk) 23:08, 1 March 2024 (UTC)

You mean fusion, which (like fission, its opposite) is a nuclear reaction. A chemical reaction, such as oxidation, has too little energy density to start fusion; else life would be very different. —Tamfang (talk) 00:31, 2 March 2024 (UTC)

Stars run on fusion which is multiple times more powerful per gram than fission. The scales of either are absolutely insane compared to fire (https://xkcd.com/1162/). By the time fusion starts the star would average about 4 times the density of osmium (the densest material), have a center of 5,400,000 Fahrenheit, 1,000,000 times sea level pressure and thousands and thousands of times the pressure a human could survive & 1,000 times denser than water (522 pounds per 8 fluid ounces), 44 times denser than osmium, 52 times denser than gold, 11,600 times denser than unboiled hydrogen, 816,000 times denser than air and 12,000,000 times denser than the boiled hydrogen in a party balloon. With tens of thousands of Earth masses of hydrogen crushing down with a human-fatal surface gravity 165 times Earth. Only then does the least massive protostar ignite (it's very hard to crush hydrogen to 0.01% the size of mundane maximally compressed hydrogen cause matter takes up space/void-free liquids+solids are hard to compress/electron degeneracy pressure). It's amazing science is even able to fuse thin gas density hydrogen in a medium-size building though that's cause fusion increases exponentially with temperature they just heat the easiest fuel to hundreds of millions of Fahrenheits. There's fusion reactions where if the temperature doubles the burn rate rises 68,719,476,736 times. Whatever the x is watts rise x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times x times (x times itself 35 times). So the lowest mass 0.06-0.07 Sun mass stars won't "use up their fuel" for 380,000,000,000,000 years while the biggest stars are thousands of times more massive but die 190,000,000 times sooner with core temperatures only about a hundred times hotter. When the core of an 8 to 30 Sun mass star "runs out of fuel" the starstuff falls almost as fast as freefall, fusion accelerates exponentially till 11 to 18 billion Fahrenheit which is when large fractions of the energy it'll ever fuse in its life (few million years) fuse in like 2 seconds and even the 100,000,000,000 to 1,000,000,000,000 Earth surface gravity 1.4 to 3 Sun mass center that remains is powerless to stop the core bounce and runway fusion explosion from escaping the gravity forever. With the light of billions of Suns despite 99% of the explosion energy "wasted" not on light but ghost particles that'd average 6,000,000,000,000 miles of solid lead before the slightest interaction. Sagittarian Milky Way (talk) 00:42, 2 March 2024 (UTC)

It is too difficult as the gas is not dense enough. Perhaps a cold fusion technique will work, like Muon-catalyzed fusion. Graeme Bartlett (talk) 09:32, 2 March 2024 (UTC)

To the 3 questions in short: No, No, agreeing No. Less respectfully, I ask the OP to review the wrong assumptions in their question. It's reasonable to suppose that given a handheld lighter, the OP can ignite the gas from a gas cooker. The gas goes on burning with a flame that emits light and heat, and you can assume that the atmosphere supplies sufficient oxygen under pressure to maintain the required stoichiometric ratio of fuel/oxygen. Each of the underlined terms that are relevant to a cooker in your kitchen are irrelevant and as assumptions misleading to understanding the fusion process in stars. That is a subject of intense research, both because it may become a practical power source if researchers solve difficult challenges of initiating fusion (needs high pressure and energy to start) and containing the subsequent reaction, and because science predicts that we can find the origins of chemical elements in star fusion. For that we are indebted to a landmark 1957 paper called B2FH after the initials of its authors. The best use for your hand lighter at this point would be to heat cups of tea for the hard-working fusion researchers. Philvoids (talk) 11:14, 2 March 2024 (UTC)

ITER/DEMO being possibly the most famous horse in the race to the first power plant though if we're lucky one of the others is sooner. Sagittarian Milky Way (talk) 15:04, 2 March 2024 (UTC)

March 2

Why aren't we allowed to use the formula of relativistic momentum, for calculating the positive momentum of slow light not moving in a vacuum?

HOTmag (talk) 16:53, 2 March 2024 (UTC)

The momenta that light carries are observed to be largely independent of its speed. For example, when light, which are particle-waves, passes from one medium to another and then back again into the original medium such as through a window pane, unless it gets scattered (see Compton scattering), it emerges with the same speed, frequencies, energies and momenta it started with. This is because the photons' momenta and energies are conserved throughout although their wave velocities differ within the different media. Conversely, the momenta of masses are known to depend on their speed. Modocc (talk) 19:55, 2 March 2024 (UTC)

Yes, I know that the momenta that light carries are observed to be largely independent of its speed. But I was asking about the invalidity of the formula of relativistic momentum: Please notice this formula does not warn: "Don't use me for calculating the positive momentum of slow light not moving in a vacuum", does it? As long as it does not, what makes it void/invalid, for such a calculation, from a formal viewpoint? HOTmag (talk) 20:16, 2 March 2024 (UTC)

On one hand, individual photons don't have rest mass. So there is nothing really to calculate. On the other hand, two or more counterposing photons do have an invariant mass to which the formula can be applied. Modocc (talk) 20:53, 2 March 2024 (UTC)

Isn't is common to regard the photon's rest mass as zero? Note that the formula of relativistic momemtum does not warn: "Don't use me when the rest mass is zero", does it? HOTmag (talk) 21:02, 2 March 2024 (UTC)

The formula + zero mass yields, simply, zero momentum. When we measure lights' momenta, its not zero, unless the system as a whole has a mass. The formula is only used for calculating the momenta of mass. Modocc (talk) 21:18, 2 March 2024 (UTC)

Do you agree, that the formula should have warned: "Don't use me for any zero mass"? HOTmag (talk) 21:30, 2 March 2024 (UTC)

An isolated silvered box containing coherent photons will oscillate with respect to transference of the photons' constant momenta even if these are considerably slowed down. Clearly, the formula is not valid for modeling their momenta. Modocc (talk) 21:47, 2 March 2024 (UTC)

I agree with you that the formula is not valid for modeling those momenta. However, my question is about the formula per se: Why doesn't it warn: "Don't use me for any zero mass"? HOTmag (talk) 21:53, 2 March 2024 (UTC)

No warning is needed because light carries (or has) momentum without that specific formula. Modocc (talk) 23:17, 2 March 2024 (UTC)

A warning is needed for the formula to be valid. Just as the well known algebraic formula "If ab=ac then b=c" must warn: "Don't use me if a=o", for this algebraic formula to be valid. Without such a warning, this algebraic formula may be wrong (e.g. in case a=0), whereas a warning makes this algebraic formula valid because it will never be used in cases mentioned in the warning. The same is true for the relativistic formula of momentum: Without the warning "Don't use me if the rest mass is zero", this physical formula may be wrong (e.g. in case the rest mass is zero), whereas a warning makes this physical formula valid because it will never be used in cases mentioned in the warning. HOTmag (talk) 00:47, 3 March 2024 (UTC)

The fact it's shown to be invalid and said to be invalid, repeatedly by me here, and by others in the literature: suffices. Modocc (talk) 01:25, 3 March 2024 (UTC)

What? Is it really "said to be invalid...by others in the literature"? I will be glad to see any source ascribing "invalidity" to this formula. HOTmag (talk) 01:46, 3 March 2024 (UTC)

It is valid for mass only, but it is not valid for individual photons. Correct? Modocc (talk) 02:01, 3 March 2024 (UTC)

Also, it's because, unlike rest mass, a photon's frequency, energy and momentum are not invariant. Modocc (talk) 02:12, 3 March 2024 (UTC)

However, physicists usually assume the Energy–momentum relation, which is a work-around that is valid for photons. Modocc (talk) 03:38, 3 March 2024 (UTC)

"not valid for individual photons"? I've never seen this formula warn: "Don't use me for individual photons" (What about gluons?), but I've only seen this formula warn: "Don't use me for velocities not slower than C (because arithmetically one is not allowed to devide by zero nor to extract square roots of negative numbers)".

I didn't understand your second remark about the invariance: Note this formula does not demand, of the frequency or of the energy or of the momentum, to be invariant. It only demands of the rest mass (needed for calculating the relativistic momentum) to be invariant. Apparently, every photon's rest mass is invariant, so apparently this formula doesn't warn against using it for slow photons, because no division by zero (nor any extraction of square root of any negative number) is made for them. HOTmag (talk) 09:52, 3 March 2024 (UTC)

For m₀=0, the Energy–momentum relation $E^{2}=(p{\textrm {c}})^{2}+\left(m_{0}{\textrm {c}}^{2}\right)^{2}\,$ is reduced to

$E=pc$ and light's momentum is thus given by E/c in accordance with its energy which is related to its frequency per the Planck relation,

$$E=h\nu .$$

Furthermore, don't use

$$\mathbf {p} =\gamma m_{0}\mathbf {v} \,$$

for light because its momenta are not zero. From what I've read and recall at this point, slow light interaction models typically invoke either group waves, their photons' coupling to quasi-particles and/or lattice calculations involving Quantum electrodynamics. Stuff that is far too complex to expound on. Modocc (talk) 14:17, 3 March 2024 (UTC)

I already know I shouldn't use the formula of relativistic momentum, for any slow photon, because (as you explain well): "its momenta are not zero". That's obvious, and I've known all of that. I remind of you, that my question was only about why this formula doesn't warn: "Don't use me for slow photons", while all other formulas I know - do warn againt using them in forbidden cases. HOTmag (talk) 14:41, 3 March 2024 (UTC)

It is not clear at all how to define the photon momentum in medium. See this. Ruslik_Zero 20:51, 2 March 2024 (UTC)

Thank you for the article, which actually makes a distinction between the kinetic momentum and the canonical momentum.

On the other hand, a given slow photon's momentum could be easily determined (hence defined) - as zero, if we were allowed to use the formula of relativistic momemta, which does not warn: "Don't use me when the rest mass is zero". The question is, why aren't we allowed to use this formula, when it comes to a given slow photon? HOTmag (talk) 21:02, 2 March 2024 (UTC)

(ec) What makes you think that a photon in a medium has mass zero? Please don't forget that a photon is not a simple classical particle but an excitation of a quantum field which interacts with the medium in an a priori rather complicated way. I assume that we can effectively continue to treat it as a particle, certainly with non-zero momentum, possibly with an effective mass due to the interaction with the medium, but I'm out of my depth here. --Wrongfilter (talk) 21:35, 2 March 2024 (UTC)

Don't forget, that by "mass" we actually mean "invariant mass", so if it's zero in a vacuum, it's apparently zero in any medium. Just as my own invariant mass does not change when I pass from a vacuum to any medium. HOTmag (talk) 21:50, 2 March 2024 (UTC)

I wouldn't bet on that. --Wrongfilter (talk) 22:05, 2 March 2024 (UTC)

That's why I was careful when I wrote about the photon's invariant mass: "it's apparently zero in any medium". HOTmag (talk) 22:11, 2 March 2024 (UTC)

That seems like a misinterpretation of "invariant". --Wrongfilter (talk) 22:16, 2 March 2024 (UTC)

Besides photons, about which we can argue whether their invariant mass may change when passing from a given medium to another, do you have in mind any other example of a particle whose invariant mass may change when passing from a given medium to another? HOTmag (talk) 22:28, 2 March 2024 (UTC)

To be clear, both Minkowski and Abraham arguments seem to assume some sort of complicated transference of the photons' momenta to and from the propagation media. Personally, FWIW, I've come to reject both arguments per Occam's razor in favor of any models in which the photons retain their momenta... But that's on me. Modocc (talk) 21:30, 2 March 2024 (UTC)

The photon, with it's rest mass of zero, always travels at the speed of light. Note that even inside a medium most of the space is empty. If "light" travels more slowly in a medium, that is not because of a slower speed of individual photons, but either because the interaction with the medium causes photons to travel a more complicated path ("bouncing between atoms"), or because the photon is absorbed and re-emitted. --Stephan Schulz (talk) 14:30, 3 March 2024 (UTC)

Thanks. This was actually my suspicion.

By the way, do you have any new answer to a close question in one of my previous threads? HOTmag (talk) 14:42, 3 March 2024 (UTC)

Perhaps individual photons do slow down, or if not at least couple with polaritons, and gain effective mass and attract each other due to QED interactions. Upon reading their article they used an effective field theory and I quote: "...where vg is the group velocity inside the medium,..." and a function of vg "...is the effective photon mass...". It is somewhat interesting though. Modocc (talk) 16:17, 3 March 2024 (UTC)

March 3

Gravitational constant in terms of energy

The Gravitational constant is usually written in terms of mass, but it seems to me it would be more naturally written in terms of energy, because after all it's energy that gravitates, not [just] mass.

In those terms the value would be: 8.2627195×10^-45 m/J. (i.e. G / c^4) Which is a much simpler unit. With this formulation of G you can just plug in the energy of a photon, for example, and get results, no need to resort to complicated curvature of space explanations.

The article on the Gravitational constant does not have it written this way - before I add it, I just wanted to see what others thought. Ariel. (talk) 22:50, 3 March 2024 (UTC)

You should only add this if you can cite reliable sources that present the constant in units of this dimension. See WP:OR. Is your unit correct? Shouldn't it be m·kg⁻¹? Given the uncertainty in the value of G, the number of digits in the numerical value you give is unwarranted. I think it should be something more like 8.26271(19) × 10⁻⁴⁵.  --Lambiam 00:18, 4 March 2024 (UTC)

$m/J$ is the correct unit for $G/c^{4}$, although it's not clear to me why G is being divided by $c^{4}$. It seems this is introducing a new constant G', which complicates other things, like Newton's law of gravitation would change from $F=G{\frac {m_{1}m_{2}}{r^{2}}}$ to $F=G'{\frac {m_{1}m_{2}c^{4}}{r^{2}}}$
As to the OP question, it should definitely not be added to the article unless there is a reliable source that presents it in this way; otherwise it would be original research. CodeTalker (talk) 01:08, 4 March 2024 (UTC)

Agreed about the significant digits, I was going to do that if I added it.

It's not a new constant, rather instead of the mass you would put in the energy of the object, and then use this form of G instead of the original. It's more correct that way because it's energy that gravitates, not just mass.

Is it really possible that this is OR, and that no one else thought of defining G in terms of energy instead of mass? That would surprise me. Ariel. (talk) 02:06, 4 March 2024 (UTC)

You can express the gravitational constant in the unit $m^{5}J^{-1}s^{-4}$, but that doesn't change its numerical value. As long as you use SI units, the numerical value remains the same. When you divide the gravitational constant by $c^{4}$, you get a new constant G', and if you want to that instead of G, you have to compensate every formula with another $c^{4}$. For Newton's law of gravity, that makes things more complex. Newton's theory, which is is older and more widely used than General Relativity, suddenly gets a ridiculously small constant and a weird lightspeed to the fourth, despite having nothing to do with the speed of light. In General Relativity, these factors $c^{4}$ may compensate for others, but the theory is still full of powers of $c$, most of which get cleverly absorbed somewhere.

That mass is equivalent to energy doesn't mean that expressing things in energy is more correct. You can just as well express energy as relativistic mass, which is the more common method in GR (not in particle physics, but that doesn't deal with gravity). Physicists tend to use what's most practical and I've never seen anyone use this constant G'.

BTW, have you noticed how the accuracy of G is improved when expressed in parsec per solar mass times (km/s) squared? That's because the uncertainty in the solar mass compensates for the uncertainty in G, the product of which is known much more accurately than either of them separately. PiusImpavidus (talk) 10:48, 4 March 2024 (UTC)

I see now that there is something called the Einstein gravitational constant, which has the same dimension as your energetic constant but differs by a scalar factor:

$\kappa ={\frac {8\pi G}{c^{4}}}\approx 2.076647442844\times 10^{-43}\,{\textrm {N}}^{-1}.$

It is referred to in section Gravitational constant § Definition. (The precision is unwarranted.)  --Lambiam 19:12, 4 March 2024 (UTC)

March 4

Accelerating (massless) gluons by the strong force

Here are two facts, AFAIK:

Fact #1. When it comes to photons, the impact of gravity on them - which actually must change their momentum (according to the very definition of "force" in mechanics), is reflected, either by accelerating them - i.e. by changing their direction (if their motion has just been perpendicular to the direction of the gravitational field), or by their redshifting or blueshifting (if their motion is parallel to the direction of the gravitational field).

Fact #2. When it comes to gluons, they have a color charge; while this color charge - with respect to the strong force, plays the same role, as the electric charge does - with respect to the electric force. Hence, the strong force has an impact on gluons.

My question is: What's the impact of the strong force on gluons, bearing in mind that gluons have no mass, so apparently the strong force can't accelerate them by changing - their speed - i.e. the absulote value of their velocity.

In other words: Is the way gluons are influenced by the strong force, identical to the way photons are influneced by gravity, even though the explanation of the impact of gravity is usually ascribed to the curved spacetime - which is not the case when it comes to the strong force?

But maybe I'm wrong, so although the gluon is massless, its speed (i.e the absolute value of the gluon's velocity) is not as "constant" as the photon's speed is, so also the gluon's speed can be changed by the strong force. Can it? HOTmag (talk) 05:07, 4 March 2024 (UTC)

Basically you are asking: Since Gluons are massless they always travel at the speed of light, so how can they accelerate/change energy? But so far as I know Gluons are virtual only, there are no free Gluons, and virtual particals do not need to conserve momentum or energy in isolation. Ariel. (talk) 06:06, 4 March 2024 (UTC)

Gluons were first detected in 1978, and I guess they were detected as free. Quark–gluon plasma, in which gluons are free, was first detected in the year 2000. Additionally, glueballs are assumed to have their gluons free.

By the way, I won't be surprised if a given (massless) gluon's speed turns out to be non-constant (hence different from the speed of light). See my thread here. HOTmag (talk) 06:36, 4 March 2024 (UTC)

The observations that could best be explained as involving gluons, made starting in 1978, did not "detect" gluons. One can say they were "discovered". The researchers studied the hadronic decay of the Y boson ϒ meson and identified that this decay was probably mediated by gluons. Only in 1979 had enough evidence been collected to assign a sufficiently high statistical significance to the gluon explanation to confirm the physical existence of this thus far theoretical particle.  --Lambiam 12:49, 4 March 2024 (UTC)

It was the ϒ meson (Greek upsilon), not the (hypothetical) Y boson. --Wrongfilter (talk) 13:02, 4 March 2024 (UTC)

Fact 1 doesn't work. A photon has no mass, so the gravitational force acting on it is zero. The resulting acceleration is zero divided by zero, which is undefined. In other words, classical mechanics doesn't apply to photons. Newton's laws can be bent in such a way that they predict deflection of photons in a gravitational field, but the predicted deflection is only half that predicted by General Relativity. Observations are consistent with GR. PiusImpavidus (talk) 11:18, 4 March 2024 (UTC)

Due to your comment, I realized I had to strike out the content of parentheses containing the word "force" in my original post, and that's what I've just done. However, I also realized that your comment had no impact on my question, being about gluons, rather than about photons. HOTmag (talk) 12:11, 4 March 2024 (UTC)

On a scale less than about 0.8 fm gluons are the force carriers of the strong force; see Strong interaction § Behavior of the strong interaction. It is not particularly meaningful to assert that the strong force has an impact on gluons; in some sense they are the strong force.  --Lambiam 13:01, 4 March 2024 (UTC)

Gravitational lensing

Before asking my question, let me begin with three facts regarding it:

1. By the classical definitions in Mechnaics, a mass of a given particle, reflects the particle's resistance to accelaration: The bigger mass the particle carries, the bigger its resistance to acceleration is.

2. By the classical definitions in Mechnaics, the angle of deflecting the light by gravitational lensing, can be regarded as a kind of acceleration, actually a normal/radial one: Indeed, it does not change the speed (the absolute value of the veloicity), yet it does change the velocity's direction.

3. By General relativity, the angle of deflecting the light by gravitational lensing, only depends on properties of the gravitational lens (actually its mass), and on the distance between the lens and the light, yet not on any propery of the light itself (e.g. its momentum/color).

Question: Am I allowed to infer - from these three facts alone (regardless of special relativity), that light has a constant mass (e.g. zero mass), independent of the light's momentum/color? For, if the light's mass varied - e.g. depended on the light's momentum/color - so that blue photons were "heavier" than red ones, then by #1 - a red photon's resistance to this photon's normal/radial accelaration - would be smaller than a blue photon's resistance to this photon's normal/radial accelaration, and so by #2 - the angle of deflecting red photons by a given gravitational lens - would be wider than the angle of deflecting blue photons by that very gravitational lens, as opposed to #3. Hence (apparently), light has a constant mass, independent of the light's momentum/color. Is my deduction correct, from a logical point of view (regardless of special relativity)? HOTmag (talk) 10:19, 4 March 2024 (UTC)

Light has no mass. But any masses going in the same direction and speed as each other would be deflected the same. The orbit of a satellite around the earth does not depend on its mass - well not unless we had one that had a mass comparable to the earth! NadVolum (talk) 10:54, 4 March 2024 (UTC)

Oops, I forgot that a given body's gravitational acceleration does not depend on the body's mass. Sorry for this huge (contemporary) mistake... HOTmag (talk) 11:05, 4 March 2024 (UTC)

Do you have in mind examples, of an exceedingly weak gravitational field (if not the weakest one), ever measured?

HOTmag (talk) 12:48, 4 March 2024 (UTC)

Well there's [4]. Or maybe Modified Newtonian dynamics is the sort of thing that you're interested in. NadVolum (talk) 13:30, 4 March 2024 (UTC)

Thx. HOTmag (talk) 15:28, 4 March 2024 (UTC)

Regardless of the following four formulas, is there any other way (whether an empirical one or a theoretical one) to prove that light has no mass?

Denoting: the speed of light by $c,$ the velocity by $v,$ the gamma factor by $\gamma ,$ the mass by $m,$ the momentum by $P,$ and the energy by $E,$ the four formulas I'm currently ignoring are:

1. $P=\gamma mv$

2. $E=\gamma mc^{2}$

3. $E^{2}=P^{2}c^{2}+m^{2}c^{4}$ (bearing in mind that $E_{light}=cP_{light})$

4. $M_{rel}=\gamma m$ (while $M_{rel}$ denotes the so-called "relativistic mass").

I'm asking the question in the title, because the formulas mentioned above were originally developed/deduced for bodies slower than light, so these formulas won't be reliable (in my eyes) for determining whether light has any mass - e.g. a mass for creating a gravitational field.

Additionally, for defining force, I only use the formula $F=dp/dt$ (which actually defines the force as the change in momentum with respect to time), rather than the formula $F=ma$ (which actually defines the force as the product of mass and acceleration).

I would also like to ignore the theory of elementary particles (including gauge theory), according to which photons have no mass. HOTmag (talk) 15:33, 4 March 2024 (UTC)

It cannot be proved on theoretical grounds, but in general: Rest mass is invariant (it's measured to be the same quantity in all inertial reference frames) and massless energy is not. A cloud of photons has both gravity and rest mass. Taken individually a photon's frequency is not invariant, it's arbitrary; its blueness/redness depends on the observer. There are empirical constraints on its mass, see Photon#Experimental_checks_on_photon_mass. Modocc (talk) 17:52, 4 March 2024 (UTC)

Has it ever been empirically proved that a cloud of photons has gravity/mass? I emphasize I'm asking from an empirical point of view, because I guess it would be impossible to theoretically prove this - without relying on any of the formulas mentioned above which I currently ignore. HOTmag (talk) 18:22, 4 March 2024 (UTC) HOTmag (talk) 18:22, 4 March 2024 (UTC)

Does conservation of mass-energy and Electron-positron annihilation suffice? Modocc (talk) 18:32, 4 March 2024 (UTC)

Are you referring now to my new question about the cloud of photons? If you are, then - using the conservation law you've mentioned - along with the annihilation you've mentioned, how do you prove that a cloud of photons has gravity - without relying on any of the formulas I've mentioned in my first post? Note that Relativity theory allows mass to disappear and to re-appear, as long as the total energy is conserved. HOTmag (talk) 18:38, 4 March 2024 (UTC)

I didn't notice your question(s) earlier. Certainly, the photons' gravity can and should be empirically demonstrated: for example, by measuring the deflection of ultra-thin beams of energetic light which cross paths in extremely close proximity to each other. The gravitational deflection(s) will be small and these can be measured by redirecting the beams into an interferometer. In the confines of a small Earth-bound setting a ring of mirrors can be added such there are a multitude of repeated crossings of the beams (in a vacuum) that would accumulate their deflections if necessary. One would need to calibrate the beams independently and then turn them on together to see if there is an observable effect. Ideally one would repeat the experiment and vary its parameters to get a more complete picture of their interactions if any. Modocc (talk) 22:31, 4 March 2024 (UTC)

Do you think such an experiment (or any similar one) has ever been carried out? HOTmag (talk) 22:48, 4 March 2024 (UTC)

Not to my knowledge. Perhaps we simply need to instigorate it. :-) Modocc (talk) 23:09, 4 March 2024 (UTC)

The Eddington experiment measured the gravitational deflection of starlight passing near the Sun. Philvoids (talk) 01:34, 5 March 2024 (UTC)

That showed that the sun gravitates light. The unanswered question is whether or not photons reciprocate and gravitate. The difficulty with that is photons are not known to interact with each other via gravity, and if they do their angular deflections would be very very small anyway, perhaps too small to detect. Modocc (talk) 02:45, 5 March 2024 (UTC)

The two words "reciprocate and" in your recent response, are needless, in my opinion. Even without them, the "unswered question" (as you name it) basically remains the same question. HOTmag (talk) 08:55, 5 March 2024 (UTC)

You cannot use relativity to derive the mass of the photon or anything else; relativity per se doesn't know what a friggin photon is. You tell relativity: "Here's a particle with mass $m=0$ and energy $E=h\nu$" (info comes from other theories, e.g. de Broglie for the energy), and then you ask "How does it behave?", and relativity will tell you that. For instance, it tells you that the momentum of the photon is $p=E=h\nu$, using equation (3), which is the only relevant equation here. --Wrongfilter (talk) 12:40, 5 March 2024 (UTC)

1. Do you think that a given frigging photon has no mass?

If you do, then my original question was:

2. How do you know that? Do you only know that by (some of) the equations I've indicated in my first post?

HOTmag (talk) 13:19, 5 March 2024 (UTC)

I can only repeat myself. All these equations are part of the theory of relativity. Relativity doesn't know what a photon is, therefore these equations cannot tell you what the mass of a photon is. You need to look elsewhere for that. --Wrongfilter (talk) 13:25, 5 March 2024 (UTC)

Let's leave Relativity theory. Personally, do you think (or estimate) that light has no mass? HOTmag (talk) 13:32, 5 March 2024 (UTC)

This is a reference desk and not a place for personal opinions. --Wrongfilter (talk) 13:36, 5 March 2024 (UTC)

Let's leave personal opinions. Can science determine, that a given body - whose speed does not depend on any observer - has no mass? HOTmag (talk) 13:40, 5 March 2024 (UTC)

See Photon#Experimental_checks_on_photon_mass. Modocc (talk) 15:49, 5 March 2024 (UTC)

I've read this chapter many times over the recent years. Yet, I don't see how it answers my question, about "no mass", rather than about a mass smaller than a given tiny limit. HOTmag (talk) 15:58, 5 March 2024 (UTC)

See Invariant mass for the details on how physicists calculate it. Modocc (talk) 17:46, 5 March 2024 (UTC)

The article only states:

1. Systems whose four-momentum is a null vector (for example, a single photon or many photons moving in exactly the same direction) have zero invariant mass and are referred to as massless. But unfortunately, the article doesn't explain why "a single photon" mentioned in the parentheses is an "example" of that. In other words, the article doesn't explain why a photon's four momentum is a nul vector. It wouldn't have been a null vector, had we assumed that a photon does have a (positive) mass.

2. Thus, the mass of a system of several photons moving in different directions is positive, which means that an invariant mass exists for this system even though it does not exist for each photon...For example, rest mass and invariant mass are zero for individual photons. But unfortunately, the article doesn't explain why this is true.

HOTmag (talk) 18:40, 5 March 2024 (UTC)

Consider two red photons a and b from stationary sources approaching you with velocities +|v| and -|v| from opposite directions. Now switch to another reference frame such that the source of a is moving towards you and the source of b is receding. a is bluer and b is redder. Doppler shift changes what energy is measured for each photon, but it doesn't change their total energy. That way the invariant mass of the proton and my newspapers didn't change if they were on my bike, I was walking with them, or being read by the numerous customers I delivered them to. Physicists are simply tallying (empirically) an invariant mass-energy for each particle. Modocc (talk) 19:18, 5 March 2024 (UTC)

I understand your thought experiment (a well known one), but I don't understand why you think it proves that the mass is zero. Why zero? HOTmag (talk) 20:50, 5 March 2024 (UTC)

I don't think that at all. Your 1st quote about zero mass is simply empirical, and I explained the 2nd quote to give it some context. Modocc (talk) 21:53, 5 March 2024 (UTC)

When an electron and a positron collide - annihilating each other, does the gravitational field - having been created by them - disappear?

Assuming that light has no gravity. HOTmag (talk) 18:33, 4 March 2024 (UTC)

The photons carry away the mass-energy from the collision and redistributes it to other parts of the system. So the field itself has to change (in accordance with its mass-energy distribution). Modocc (talk) 18:53, 4 March 2024 (UTC)

Their combined energy and momentum do not change but are now carried by photons. In general relativity, the gravitational field is not ascribed to individual bodies but a universal field entirely determined by the stress–energy tensor, which only depends on the distribution of energy and momentum.  --Lambiam 18:56, 4 March 2024 (UTC)

Is it a response to me or to Modocc?

Anyway, bearing in mind that the two original massive particles no longer exist after the light was emitted, is it carrying now the original gravitational field? If this light is not, then which entity is, in this case? Is this gravitational field carried now by no entity? In other words: does spacetime get curved, yet by no entity? I'm quite surprised... HOTmag (talk) 19:01, 4 March 2024 (UTC)

It was meant to be a response to the original question; I had not noticed Modocc's response. Your question above ignores the nature of the gravitational field iin general relativity, which is not ascribed to individual bodies. So the photons do not carry a gravitational field, but they influence the gravitational field.  --Lambiam 19:44, 4 March 2024 (UTC)

I thought your first response had been meant to be a response to Modocc, because of its indent. Anyway, back to your main response: Assuming theoretically - that the only gravitational field that existed in the whole universe before the collision of the two massive particles - was the gravitational field created by them, you actually claim it still exists after the collision - although the whole universe (in our theoretical case) currently contains light only - actually being the light emitted after the collision, so you actually claim we've got a curved spacetime - without any mass - we assuming that light has no mass. That's why I was so surprised, having read your first response. I'd always thought, that spacetime could only be curved by a presence of mass. Had I been wrong? HOTmag (talk) 19:56, 4 March 2024 (UTC)

Yes, if you only consider only matter to have mass and do not equate other forms of energy with mass. The concept of mass in general relativity is notoriously difficult to define; see the section Mass in general relativity § Defining mass in general relativity: concepts and obstacles. Spacetime is curved by the presence of energy and momentum; see the second paragraph of Einstein field equations in combination with Stress–energy tensor. Photons have energy and momentum.  --Lambiam 07:08, 5 March 2024 (UTC)

I'm surprised again. If (as you actually claim) light curves spacetime, i.e. creates a gravitational field, i.e. has (by defintion) an active gravitational mass, so what's the physical meaning of a given photon having "no mass" - as commonly accepted to claim? On the second hand, if - by saying that light has "no mass" - one actually means light has no passive gravitational mass, then one actually says nothing, because the change in a given body's momentum due to the gravitatoinal force (while defining any force as $F=dp/dt$ rather than as $F=ma$) - as well as the body's gravitational normal/radial acceleration, are not influenced by the body's passive gravitational mass - nor even by whether it's a positive one or a zero one. On the third hand, if - by saying that light has "no mass" - one actually means light has no inertial mass, then what does a given photon's inertial mass mean, bearing in mind that light is not influenced by any force other than the gravitational one (e.g. when a given photon's path is deflected when it approaches the sun)...

This surprise, as described in the previous paragraph, takes me back to the title of my previous thread.

By the way: regarding your first ten words: "Yes, if you only consider only matter to have mass", please see the first paragraph in our article Einstein field equations: "In the general theory of relativity, the Einstein field equations (EFE; also known as Einstein's equations) relate the geometry of spacetime to the distribution of matter within it". This first claim in the lede, is sourced by a reliable source. So, would you suggest that the lede in this article be fixed somehow, e.g. by replacing "matter" by "mass" (or by "mass-energy"), along with replacing the source (ibid.) by another one? HOTmag (talk) 08:49, 5 March 2024 (UTC)

A photon has a rest mass equal to zero.

Have you considered getting a proper textbook on relativity? I just pulled my Rindler from the shelf. PiusImpavidus (talk) 11:08, 5 March 2024 (UTC)

I wonder how your remark has anything to do with my question in the current thread. Have I ever claimed the photon has a positive mass?

As for your question: Have you read my previous thread? HOTmag (talk) 11:16, 5 March 2024 (UTC)

In your post above my last post, you asked about the physical meaning of "a photon has no mass", followed by some discussion on active or passive gravitational mass and inertial mass. I answered that it's the rest mass that's zero. In other words, I gave a direct answer to the question you asked just above. If that's irrelevant, you asked the wrong question.

Yes, I read your previous thread. It sounds like you don't care too much about actually understanding physics, so I assume the answer to my question is no. PiusImpavidus (talk) 18:43, 5 March 2024 (UTC)

Yes, I know that the photon's rest mass must be zero (because the photon can't be at rest). That's said, my question you've quoted, which should have been quoted in its full version: "what's the physical meaning of a given photon having 'no mass' - as commonly accepted to claim?", was actually meant to be addressed to all those physicists (not me but for example Okun) who claim that there's only one kind of mass, which I and you call "rest mass", and which they call "the mass" - i.e. "the only one mass possible". They don't make any distinction between, what I and you call "rest mass", and any other kind of mass. If you think (like me) that there is more than one kind of mass, then we can become good friends. However, as far as they are concerned, they actually claim that "a photon has no mass", without making any distinction between a photon's rest mass and a photon's relativistic mass. In their opinion, if a photon's mass is zero (e.g. when considering the "rest mass"), then any mass of the photon must be zero, simply because there's only one mass possible.

As for your second remark ("It sounds like you don't care too much about actually understanding physics"). Unfortunately, you didn't explain why you thought so (about a person who thought she could become your friend), so I can't respond to this remark with the same elaboration appearing in my previous thread. HOTmag (talk) 19:08, 5 March 2024 (UTC)

Before general relativity, mass was thought to be a well-defined and reasonably understood physical concept, unambiguous with respect to a preferred inertial frame of reference. That is no longer the case. I referred you to the section Mass in general relativity § Defining mass in general relativity: concepts and obstacles. Did you read this? Do you expect us, instead of referring you to sources, to come up with solutions to these obstacles?  --Lambiam 21:21, 5 March 2024 (UTC)

What do you mean by "the system"? The only system has been the original pair of electron-positron, which turned out to be light. HOTmag (talk) 19:00, 4 March 2024 (UTC)

Although you didn't constrain them to a closed system or to a hypothetical one they still gravitate and affect each other (at least in theory).Modocc (talk) 19:06, 4 March 2024 (UTC)

After the collision, they no longer exist, do they? HOTmag (talk) 19:13, 4 March 2024 (UTC)

The photons which are created contribute to the stress–energy tensor of the gravitational field of spacetime. They gravitate. Modocc (talk) 19:38, 4 March 2024 (UTC)

Thank you for this link. I was not aware of this theoretical effect. I still wonder, though, if it has ever been empirically detected. I also wonder if it's mentioned in Wikipedia. HOTmag (talk) 19:43, 4 March 2024 (UTC)

And the destructive collision will make some tiny gravitational waves. So not all energy will end up as photons. Graeme Bartlett (talk) 07:16, 5 March 2024 (UTC)

Gravitational waves can only be created, if the curvature in spacetime is already given - the waves characterizing the manner of how this given curvature propogates in spacetime.

That said, I'm asking if the very curvature really still exists after the collision, assuming that the whole universe had only contained the electron and the positron before they collided and annihilated each other while emitting light only, about which I'm actually asking whether it can curve spacetime (along with creating gravitational waves with respect to this curvature), assuming that the whole universe no longer contains matter after the collision. HOTmag (talk) 09:50, 5 March 2024 (UTC)

In theory, all massless particles gravitate. Consider protons: Their gluons are massless but these have enough energy to account for the proton's mass which gravitates. Modocc (talk) 11:45, 5 March 2024 (UTC)

A given proton is composed of quarks and gluons. Every quark "inside" the proton (i.e. contrary to a "free" quark) carries, not only what would be its mass if the quark were free, but also another component of mass which reflects the potential energy associated with the strong force carried by the gluons. So why do you think the proton gravitates due to the gluons, rather than due to those two components of mass of the quarks contained in the proton? HOTmag (talk) 12:40, 5 March 2024 (UTC)

Oops. It was in my thoughts but I didn't type "most of" when I said they have "...enough energy to account for the proton's mass...". Note, I didn't specify why or how. Modocc (talk) 15:24, 5 March 2024 (UTC)

Also after you add "most of", the question I asked you in my previous response remains the same. HOTmag (talk) 15:29, 5 March 2024 (UTC)

See Proton: "Using lattice QCD calculations, the contributions to the mass of the proton are the quark condensate (~9%, comprising the up and down quarks and a sea of virtual strange quarks), the quark kinetic energy (~32%), the gluon kinetic energy (~37%), and the anomalous gluonic contribution (~23%, comprising contributions from condensates of all quark flavors)". Modocc (talk) 16:35, 5 March 2024 (UTC)

Note, that the quarks this experiment refers to, have become "free" quarks, after the stationary proton split into several moving particles carrying "kinetic energy", like: "up and down quarks", a sea of "virtual strange quarks", and moving gluons. On the other hand, my question to you was about the quarks "inside" the stationary proton, rather than about "free" quarks outside it. The question also explains why there's a difference between, the mass of a quark "inside" the proton, and the mass of a "free" quark. HOTmag (talk) 17:53, 5 March 2024 (UTC)

The mass (invariant) of a proton is what is measured when it is stationary. They calculated the gluons contributions per the invariant mass article. Modocc (talk) 18:22, 5 March 2024 (UTC)

Yes, the mass of the proton is the mass of the stationary proton, but the masses of the quarks and gluons are masses of those components after they became free, i.e. after the proton split into them while they (including "sea of virtual strange quarks") became particles carrying "kinetic energy". HOTmag (talk) 18:49, 5 March 2024 (UTC)

My sandwich contains numerous bound and free particles. That fact doesn't change their contributions to its mass. Modocc (talk) 19:38, 5 March 2024 (UTC)

The difference between a quark inside a proton (i.e. a bound quark) and a quark outside a proton (i.e. a free quark) is as follows: All agree that a given proton is composed of quarks and gluons. However, contrary to a "free" quark, every quark "inside" the proton carries, not only what would be its mass if the quark were free, but also another component of mass which reflects the potential energy associated with the strong force carried by the gluons.

Note that the article equates the mass of the stationary proton (i.e. before it split into plenty of particles) with the sum of the masses/energies of those many particles (including a "see of strange" ones) into which the proton split. My question to you was: Why do you rule out the following option: If the calculation had only referred to the quarks "inside" the proton before it split, the mass of the proton would have turned out to be exactly the sum of all the bound quarks inside it alone, so that its gravity would have only depended on this sum alone, without relying on any gluon. HOTmag (talk) 20:00, 5 March 2024 (UTC)

See Color confinement. Quarks are always in a bound state with each other. Modocc (talk) 20:30, 5 March 2024 (UTC)

Yes, but after a proton splits into moving quarks (including "a sea of strange" ones), they become free - even for a millionth of second - until they create new hadrons. Without their being free - even for a millionth of second, they couldn't have been detected, but they have! Further, the calculation of the mass of a proton - as the sum of the masses/energies of those many particles (including a "see of strange" ones) into which the proton split, actually relies on this millionth of second - during which those particles were free (untill they created new hadrons because of the color confinement). HOTmag (talk) 20:37, 5 March 2024 (UTC)

Pumped with enough energy we have observed Quark matter that decays. That doesn't change their contributions to the proton mass. BTW, I've other things to tend to so I will be disappearing soon. Modocc (talk) 21:16, 5 March 2024 (UTC)

When you disappear, what will happen to your gravitational field?  --Lambiam 21:24, 5 March 2024 (UTC)

March 5

Gravitational lens

Let a given photon, move perpendicular to a given gravitational field created by a given star. According to General relativity, the photon's trajectory will be deflected by the angle $\theta ={\frac {4GM}{v^{2}r}}$ toward the star, whereas: $v$ denotes the photon's velocity (i.e. $v=c$), $G$ denotes the universal constant of gravitation, $M$ denotes the star's mass, and $r$ denotes the distance between the star and the photon.

Question: What will the angle be, if we replace the photon by a massive particle, its properties being the same as before (except its velocity $v$ which will be slower than $c$ of course). Will the angle be a half of the angle mentioned above? HOTmag (talk) 15:11, 5 March 2024 (UTC)

There's no distinction - energy and mass are equivalent. The deflection of the light is just a much greater version of the same effect as the prcession of Mercury. NadVolum (talk) 18:54, 5 March 2024 (UTC)

Thx.

Here you've written "there's no distinction", but you've also written "No big difference" in the edit summary (see the history page), so I wonder what's more exact.

Additionally, what do you mean by "a much greater version"? HOTmag (talk) 19:43, 5 March 2024 (UTC)

Massive particles are deflected by half that angle. Ruslik_Zero 19:56, 5 March 2024 (UTC)

Now I'm a bit confused, because of the contradiction between your reply and the previous reply above yours. Of course, if anyone of you could supply a source (or any argument analogous to a source), I would be much less confused (if at all). HOTmag (talk) 20:23, 5 March 2024 (UTC)

The general formula, for massless and massive particles, is:^[5]

$\theta ={\frac {2GM}{v^{2}r}}\left(1+{\frac {v^{2}}{c^{2}}}\right),$

where $v$ is the velocity at a large distance, before any acceleration due to the gravitational attraction. When $v=c,$ the factor in parenthesis equals $2.$ When $v\ll c,$ it approaches $1.$  --Lambiam 10:47, 6 March 2024 (UTC)

March 6

Pathogens that increase crop yields

Is there any plant virus/bacterium/whatever that has been associated with increased crop yields? I remember seeing a paper a long time ago that a few plant pathogens (I think it was viruses) increase the fruit/seed yield of their hosts, along with the usual disease symptoms, and discussed reasons why this might be so. Note that I am asking about "pathogens"; symbiotic or commensal relationships or endophytes and mycorrhiza are not what I am asking about. Jo-Jo Eumerus (talk) 08:14, 6 March 2024 (UTC)

There's this possible example; Ustilago esculenta on Zizania latifolia. Abductive (reasoning) 20:44, 7 March 2024 (UTC)

So, active symbiosis? If so, a widely known symbiosis is the one legumes and rhizobia bacteria formed. Zarnivop (talk) 21:58, 8 March 2024 (UTC)

Not symbiosis, precisely not symbiosis. Pathogens, viruses and diseases. I saw years ago a paper discussing that sometimes, plants produce more/bigger fruits/seeds after an infection with viruses or phytoplasms, but I can't find it again. Jo-Jo Eumerus (talk) 10:00, 9 March 2024 (UTC)

Perhaps some sort of last-ditch reproduction? If so, the crop quality will suffer, and be lower in nutrient value, maybe by adding too much endosperm or by making more smaller seeds or fewer larger ones. Or the plant may switch from perennial to annual. Its overall reproductive success will be impaired. Abductive (reasoning) 11:52, 10 March 2024 (UTC)

"Symbiosis" merely means a systematic interaction between species, which can be mutualistic, commensalistic, or parasitic. The rhizobia–legume symbiosis is mutualistic. If you exclude all forms of symbiosis you exclude parasitic symbiosis and thereby all tolerated (non-lethal) infections by pathogens. The Ustilago esculenta wild rice smut destroys the flowering structures of the infected plant, which therefore cannot make seed. It is not to the natural reproductive advantage of the host, but is artificially advantageous through human cultivation, which makes it difficult to assess its placement in the categorization mutualism—commensalism—parasitism.  --Lambiam 12:26, 9 March 2024 (UTC)

Correct, the only reason Ustilago esculenta increases the crop yield of Zizania latifolia is because it prevent the plant from making high-protein seeds. The plant then sequesters carbon—that it would otherwise consume to produce those seeds—in its stems. This process increases the mass of the crop. Abductive (reasoning) 11:44, 10 March 2024 (UTC)

Thing is, "symbiosis" is ofteb restricted to examples of mutually beneficial interactions. I wss specifically not asking about these. JoJo Eumerus mobile (main talk) 12:00, 10 March 2024 (UTC)

Words have no biological reality. You asked for a pathogen that increased crop yield, and I gave one. Abductive (reasoning) 23:25, 10 March 2024 (UTC)

Has the cubic correction to redshift versus luminosity been tested, or is it likely to be?

A 2017 paper mentions this (details are in the draft article under the heading An alternative to Dark Energy) but the value did not seem to be available then. This is for the draft article: Draft:Shockwave Cosmology Hewer7 (talk) 16:49, 6 March 2024 (UTC)

All I could find were posts of people asking whether this cubic correction had been measured.  --Lambiam 21:17, 6 March 2024 (UTC)

March 9

Scientists Thought Only Humans Learn Complex Behaviors from Others. They Were Wrong

I'm reading an article from Scientific American titled Scientists Thought Only Humans Learn Complex Behaviors from Others. They Were Wrong about chimps and bumblebees learning behaviors from others. It says, "New studies in bees and chimps challenge the long-held assumption that only humans can learn from innovative peers". I thought it was already established that whales have culture (i.e. learned behavior).[6] A Quest For Knowledge (talk) 11:46, 9 March 2024 (UTC)

Many of the assumptions that humans are the only species that can do X, Y or Z are zombie memes, which keep resurfacing long after they should have been put to rest. (On the other hand, reports that also species U can do X tend to oversimplify the issues and exaggerate the achievements). I can report, though, as a definitive fact, that Homo sapiens is the only species to wonder what makes it unique.  --Lambiam 12:34, 9 March 2024 (UTC)

The headline writers at Nature were more constrained:

Bees and chimpanzees learn from others what they cannot learn alone

It has been argued that human culture rests on a unique ability to learn from others more than we could possibly learn alone in a lifetime. Two studies show that we share this ability with bumblebees and chimpanzees.^[7]

The first of the two studies reported on^[8] writes, in its abstract,

Increasing evidence suggests that animal culture can, like human culture, be cumulative: characterized by sequential innovations that build on previous ones.

The qualification "increasing" already contradicts the bold SciAm pronouncement about what scientists "thought". What the team conducting the study showed was, as they state, a new finding for invertebrates, to wit that bumblebees socially learn behaviours so complex that they lie beyond the capacity of any individual to independently discover during their lifetime. Personally, I'm more inclined to describe this as having shown the capability of collectively finding solutions to complex problems whose complexity poses intractable problems to mere individuals – which obviously involves learning from each other, but not for example intergenerational transmission. The second study reported on^[9] only established that chimpanzees can acquire a skill by social learning that they cannot independently master. One group was trained to master a task, and then an untrained group could learn from the trained chimpanzees, or from others who had learned the skill by observing already skilled individuals. This is IMO hardly unexpected and does not merit the breathlessness of the headlines.  --Lambiam 13:14, 9 March 2024 (UTC)

My response to statements such as that in the title here is "Which scientists?" HiLo48 (talk) 22:36, 9 March 2024 (UTC)

The scientists that are baffled.  --Lambiam 06:12, 10 March 2024 (UTC)

March 10

Could humans and any kind of farm herbivore live on the same diet for a year?

What about with added vitamins and minerals but not added essential amino or fatty acids? Sagittarian Milky Way (talk) 20:22, 10 March 2024 (UTC) Same percent of each food but obviously the big species will eat a lot more kg per month. Sagittarian Milky Way (talk) 20:26, 10 March 2024 (UTC)

Humans and Pigs could probably live on the same foods for extended periods, even it it would not necessarily be optimum for both. {The poster formerly known as 87.81.230.195} 51.198.186.221 (talk) 20:32, 10 March 2024 (UTC)

What about non-omnivores like cows, sheep, goats and horses? Could they live on like grains and the less disgusting fruits and vegetables like legumes with a little oil and sea salt or would they need lots of sucky food like spinach or raw salad? A surprising number of specific vegetables seem to be bad for horses. As is too much bread or horse treats like apples. Sagittarian Milky Way (talk) 21:55, 10 March 2024 (UTC)

Is there some human staple that's less bad than others? Maybe undried corn kernels cause it's kind of vegetably? There's so many that aren't common in the West like millet and quinoa. Sagittarian Milky Way (talk) 21:59, 10 March 2024 (UTC)

". . . less disgusting fruits and vegetables . . .", ". . . sucky food like spinach or raw salad . . ." – you seem to have some peculiar attitudes towards conventional foodstuffs (he says, having just enjoyed a salad.) {The poster formerly known as 87.81.230.195} 51.198.186.221 (talk) 02:28, 11 March 2024 (UTC)

I wonder if he's strong to the finach. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:15, 11 March 2024 (UTC)

Blargh salad! According to supertaster ~25% of humans are supertasters and ~25% are insensitive. Thus the bitter vegetables and sour fruits suck, the rest are okay to delicious. Orange juice: the lite version of the few times I threw a vinegar capful down my throat as fast as I could in 5th grade. Sweet, salty or moderately spicy usually good though, and bitter food can be worse than "glue" liquor, cause evolution wants humans to die? Also Brits say Hershey's® milk chocolate has a strong vomit note not in European chocolate and I don't know what they're talking about, it's barely worse than a delicious imported Icelandic chocolate. Sagittarian Milky Way (talk) 07:45, 11 March 2024 (UTC)

Icelanders also eat fermented Greenland shark, so their chocolate may not be typical of European chocolate generally. As a Brit who has taken an advanced tasting course (for Real ale, as it happens, including taste traces caused by things not right, like stale hops), I agree that US chocolate has a noticable vomit note. {The poster formerly known as 87.81.230.195} 51.198.186.221 (talk) 09:10, 11 March 2024 (UTC)

The rancid note in Hershey bars is supposed to be due to butyric acid. catslash (talk) 16:51, 11 March 2024 (UTC)

In Britain I think they don't add butyric acid to Hershey's® or Hershey's® With Almonds. In America they must add it on purpose to increase profit or they wouldn't keep doing it. Cilantro tastes like soap to 20% of humans. The bitter vegetables are some of the most healthy foods and the tasty junk foods some of the worst. These things don't always make sense. Sagittarian Milky Way (talk) 17:45, 11 March 2024 (UTC)

I'm not recommending this nothing but-potatoes-for-a-year diet, but it seemed to work for him. -- Jack of Oz ^{[pleasantries]} 06:38, 11 March 2024 (UTC)

I suspect that a strictly adhered-to nothing-but-beer diet would also work.  --Lambiam 16:52, 11 March 2024 (UTC)

3650 bottles of beer on the wall, 3650 bottles of beer ... Clarityfiend (talk) 23:11, 11 March 2024 (UTC)

Back in the day, there was a popular idea (I suppose we would now call it a meme) that one could live sustainably on a diet of two (I think) crates of Guinness (i.e. 24 pint bottles) and a loaf of bread per week. {The poster formerly known as 87.81.230.195} 51.198.186.221 (talk) 23:32, 11 March 2024 (UTC)

The Sumerian staple only diet modern version? Sounds a bit fat-deficient? Sagittarian Milky Way (talk) 17:39, 12 March 2024 (UTC)

Such a diet would be deficient in Vitamin B12, but if you ate some insects on your plant matter, you may get some. Eating most kinds of grass would have too much fibre and not enough nutrition, but if you stuck to grass seeds and the juicy part of the stalk, not so bad. Graeme Bartlett (talk) 03:30, 12 March 2024 (UTC)

Unicode CLDR for filtering mixed scripts

The last time I checked the English Wikipedia's regex filters for page titles, a lot of them were designed to prevent mixing of more than one non-Latin script. Could this be done more efficiently by a MediaWiki function that would look up each character's script in the Unicode CLDR and reject the title if more than one non-Latin language-specific script was found this way, or if a Latin and a non-Latin letter were in the same word (delimited by a normal space (\x20) or a printable punctuation mark)? NeonMerlin 21:57, 10 March 2024 (UTC)

It can probably be done more efficiently than it is done now, which is true for a lot of functionality. Whether it is worth adding this to Wikipedia's annual wish list for new MediaWiki functionality is better discussed at WP:VPR. Mixing Latin-alphabet and non-Latin-alphabet letters in, for example, a user name (I am not User:Lambiаm) is IMO much more suspect than using two different language-specific non-Latin scripts.  --Lambiam 10:01, 11 March 2024 (UTC)

March 12

Hydrogen balloon safety

My mom's birthday was recent. We had some people over, brought a few presents, and got a helium happy birthday balloon which I thought was ridiculously expensive for what it was (why? inflation! Heh). It turns out helium itself is quite expensive even in such small quantities. Plus it's a limited and somewhat scarce resource, yada yada.

Hydrogen went out of fashion as a lifting gas after the 1932 Hindenberg explosion, but that was an airship. A 1 foot diameter party balloon would have roughly 1 gram of H2 inside if my math is right. If that gets ignited, is it more of a danger than, say, dropping a lit match by accident? I can imagine a rather loud pop, but would it be enough to damage hearing, blow out windows, or anything like that? There is a standard chemistry class experiment where you have H2 bubbling up from a liquid and put a lit wooden splinter into it and there's a pop, but I guess that would be more like milligram amounts. Thanks. 2601:644:8501:AAF0:4043:7961:893C:EC1 (talk) 02:03, 12 March 2024 (UTC)

(Side comment: That was actually 1937.) --142.112.220.50 (talk) 03:03, 12 March 2024 (UTC)

It makes a fairly loud pop from when I last did that. I doubt it would shatter windows but it can make your ears ring. Cheers! 🇺🇲JayCubby✡ plz edit my user pg! Talk 02:16, 12 March 2024 (UTC) If you want to use hydrogen, I'd be more concerned about how to safely make it, so don't try it unless you already have most of the materials and are willing to spend a good bit.

Electrolysis of water is a good place to start. 🇺🇲JayCubby✡ plz edit my user pg! Talk 02:19, 12 March 2024 (UTC)

If the gas is pure H₂, igniting it is not more dangerous than igniting the gas of a gas stove. But oxyhydrogen, hydrogen gas mixed with enough oxygen, like one part O₂ to two parts H₂, will explode with a loud bang and can permanently damage the hearing of bystanders.^[10]  --Lambiam 09:35, 12 March 2024 (UTC)

Classroom demo.  --Lambiam 09:45, 12 March 2024 (UTC)

...or burns from the flash fire (Fred Grandy on the set of The Love Boat for example). DMacks (talk) 14:10, 12 March 2024 (UTC)

That was a whole collection of balloons, in a confined space, the back of a cab. Literally "in your face".  --Lambiam 15:10, 12 March 2024 (UTC)

There was no hydrogen explosion in the 1937 Hindenburg disaster; it just burned rapidly, but no bang. Some of the crew up in the envelope were killed by the hydrogen fire, others by lithobraking after the rapid decent or by burning spilled diesel fuel, ignited by the burning envelope. And the flames were the result of the burning envelope and impurities in the hydrogen, as hydrogen itself has an invisible flame. This made hydrogen balloons less popular, but, as our German article explains (de:Gasballon), hydrogen balloons are still used for humans and weather balloons, at least in some countries, like Germany. With proper safety measures and in the hands of professionals, it's safe enough. Amateurs normally use helium, as they might be stupid enough to fill the balloon indoors.

So, if this tiny bit of hydrogen in the party balloon is ignited, not much will happen. The pop won't be louder than that of an air filled balloon. If, whilst filling the balloons, you get a leak, then somebody unplugs a phone charger, you may get a big boom, destroying the house. PiusImpavidus (talk) 12:08, 12 March 2024 (UTC)

How do you know? Have you tried it? ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:54, 12 March 2024 (UTC)

Fail-safety of the air brakes in the Lac-Mégantic rail disaster

I have been reading the Lac-Mégantic rail disaster article, and what happened with the air brakes doesn't seem to be consistent with what is stated in the railway air brake article.

The Lac-Mégantic rail disaster article says that the air brakes failed because the locomotive was shut down:

"Air brakes on the train are supplied with air from a compressor on each locomotive. When a locomotive is shut off, the compressor no longer supplies the brake system with air. An air brake pipe connects to each car and locomotive on the train. When air leaks from the various components, the air pressure drops. If the system is not recharged with air, the locomotive air brakes will become ineffective and provide no braking force."

"With all the locomotives shut down, the air compressor no longer supplied air to the air brake system. As air leaked from the brake system, the main air reservoirs were slowly depleted, gradually reducing the effectiveness of the locomotive air brakes. At 00:56, the air pressure had dropped to a point at which the combination of locomotive air brakes and hand brakes could no longer hold the train, and it began to roll downhill toward Lac-Mégantic [...]"

As described above, the air brakes are thus fail-deadly: if power is not available to supply air, the brakes fail. This seems to me to be an atrocious, terrible design which makes no sense, so I started reading the railway air brake article. That article instead says:

"Modern trains rely upon a fail-safe air brake system that is based upon a design patented by George Westinghouse on April 13, 1869. [...] In various forms, it has been nearly universally adopted."

"The Westinghouse system uses air pressure to charge air reservoirs (tanks) on each car. Full air pressure causes each car to release the brakes. A subsequent reduction or loss of air pressure causes each car to apply its brakes, using the compressed air stored in its reservoirs."

"Unlike the straight air system, the Westinghouse system uses a reduction in air pressure in the train line to indirectly apply the brakes."

"The Westinghouse system is thus fail-safe -- any failure in the train line, including a separation ("break-in-two") of the train, will cause a loss of train line pressure, causing the brakes to be applied and bringing the train to a stop, thus preventing a runaway train."

This seems to contradict the other article by claiming that Westinghouse air brakes are "nearly universally adopted" and fail-safe. So, what is going on here? Did the train in the Lac-Mégantic rail disaster use a fail-deadly straight air system instead of the fail-safe Westinghouse system? If so, that puts into question the claim that the Westinghouse system is nearly universally adopted and raises the question of why that train did not use the Westinghouse system. If the train did use the Westinghouse system, and that system really is fail-safe, then the loss of power and air should have applied the brakes and held the train in place, unless the Westinghouse system is not really fail-safe.

—SeekingAnswers (reply) 04:29, 12 March 2024 (UTC)

It's a difference of short-term vs long-term. Short-term, loss of pressure in the pipe (such as car becoming separated from the engine, or the engine compressor being shut off and the pipe leaking) causes automatic braking using the stored pressure in each car's reservoir. Long-term, there could be enough slow leaks that the system on each car no longer has enough pressure to keep the brakes applied. The air pressure is needed for the actual braking process, not just signaling to apply the brake. Part of the intact system involves replenishment of each car's supply from the compressor. I assume there is some standard for how long a disconnected rail car needs to maintain its braking. DMacks (talk) 08:22, 12 March 2024 (UTC)

The referenced TSB investigation summary explains this a bit better. There were automatic (fail-safe) and independent (fail-deadly) brakes. Both are described in Railway air brake. Leakage in the automatic brakes was too slow and they had not (yet) applied. The WP article could maybe be improved here. (edit: fix spelling) Alien878 (talk) 08:35, 12 March 2024 (UTC)

There were three braking systems on the train: the automatic Westinghouse-type train brake, the independent direct air brakes on the locomotives and handbrakes on all vehicles. The Westinghouse brake relies on stored compressed air to actuate the brakes and the pressure in the continuous pipe (the "train line") to keep them off. The stored air will eventually leak, but in normal operation this is topped up from the train line via a valve. A slow leak in the train pipe will cause an ineffective operation of the brakes until the stored pressure has reduced to that of the train pipe; as the report has it: "Due to the slow decrease in brake pipe pressure, no automatic brake application occurred". The direct brakes use compressed air from the locomotive to apply and hold on the brakes; this is independent of the Westinghouse system. Finally the hand brakes are applied by (typically) turning a wheel which screws the brakes on.

The engineer brought the train to a stand using the automatic brakes. He then applied the direct brakes. He applied the hand brakes on the locomotive "consist". The four training locomotives were shut down. He then performed a hand brake effectiveness test which should involve releasing all air brakes and gently nudging the loco to ensure that the hand brakes are holding. However he forgot to release the direct brakes, and so the test merely confirmed that direct brakes and hand brakes held the train. The train was then left with the lead loco running, the direct brakes applied and some handbrakes, and the train pipe energised so that the Westinghouse automatic brakes were held off. Following the fire and subsequent shutdown of the lead locomotive the air started to leak out from the direct brakes until eventually the train started to roll downhill. The train did not part until the derailment, so there was no full application of the Westinghouse system, the slow release mentioned above ensured that the Westinghouse brakes remained off.

The full report can be read at Railway Investigation Report R13D0054, but be aware that it is a technical report running to 191 pages! Martin of Sheffield (talk) 11:00, 12 March 2024 (UTC)

Penetrating the time-space continuum

Will it ever become possible to penetrate the time-space continuum? I frequently fantasize about having some national government send a drone back in time to 1916 in order to kill Lenin in the hope that doing this would have prevented the 1917 Bolshevik Revolution and thus secured much better 20th and 21st centuries for Russia in a parallel universe (our real universe would remain unchanged due to the grandfather paradox).

Lenin was quite literally the worst thing to happen to Russia, with his creation of a totalitarian one-party making Stalin's subsequent rise to power much easier (does one think that Stalin would have won any free and fair multiparty elections in Russia in the 1920s?) and with the Soviet Union making Tsarist Russia seem like an extraordinarily mild pussycat in comparison. Tsarist Russia at least nominally allowed opposition parties, even though it tried to weaken them and spy on them, and certainly allowed for free emigration, unlike the Bolshevik regime. And it executed a couple of orders of magnitude less people than the Bolsheviks did. The Bolshevik takeover of Russia wasn't even a good thing for most Bolsheviks themselves, who subsequently got purged and killed by Stalin. How ironic! Creating a monster-state and then personally being devoured by that very same monster-state! 172.56.186.104 (talk) 05:06, 12 March 2024 (UTC)

I think you should read our article time travel which talks on the topic covering possibility and paradoxes. Also see Arrow of time, which you are tying to reverse. Using information perhaps it is possible to go into time reversal for very small systems. But the amount required is too colossal to send humans back in time. And changing history is unlikely. It is more likely that you would find out everything you learned about Russia was incorrect, and you would have to make it so that it was incorrect for everyone. Graeme Bartlett (talk) 06:02, 12 March 2024 (UTC)

What information is needed for time reversal? And what about sending an extremely tiny AI-powered drone back in time, the size of a small flea? 172.56.186.104 (talk) 07:55, 12 March 2024 (UTC)

Supposing you achieved that, how would you know? And what makes you think things would turn out any better than they did? ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:27, 12 March 2024 (UTC)

I've heard it suggested that Trotsky would not have been capable of launching a mass Bolshevik uprising in Russia outside of Petrograd in late 1917 like Lenin was, which in turn would have made it much easier for any Bolshevik uprising that Trotsky would have launched in Petrograd to get crushed. Unless you're suggesting that a non-Bolshevik-led Russia would have become just as bad in due time. 172.56.186.104 (talk) 07:55, 12 March 2024 (UTC)

I'm saying (not just suggesting) that there's no way to know what might happen. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:51, 12 March 2024 (UTC)

A running joke in the Time Travel sub-genre is that the Time Patrol (or whoever) have to keep foiling attempts by well-meaning temponauts to assassinate Hitler because the outcome would be much worse.

It's inherently difficult if not impossible to predict how future history will play out or how an alternative history might have gone, because there are so many interacting factors, but historians (I believe) broadly agree that Hitler was an incompetent military strategist and industrial organiser (see Führerprinzip), and any replacement would likely have been more successful. {The poster formerly known as 87.81.230.195} 51.198.186.221 (talk) 18:18, 12 March 2024 (UTC)

National governments keep sending killer robots back in time to kill inventors of time travel before they make their inventions. We just happen to be on a timeline in which all these missions have been successful.  --Lambiam 09:06, 12 March 2024 (UTC)