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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 [1]. 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 "unanswered 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)

I thought you'd meant to answer my question about "no mass". HOTmag (talk) 10:10, 13 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 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)

One week before you referred me to this article, User:Graeme Bartlett did the same in a previous thread, so I thanked him for the article and also stated I had read it. That was long before you referred me to this article for your first time (now it's your second time). Anyway, after having read it, I asked him some further questions (which you can find in that previous thread). HOTmag (talk) 10:05, 13 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)

A quark inside a proton can't decay. It's only the free one which can. On the other hand, my question to you was about bound quarks, i.e. ones inside a proton, when I asked you why you thought the proton's gravitational mass relied also on any bound gluon's gravitational mass rather than on the mass of bound quarks alone. The experiment you referred me to, was not about bound quarks but rather about quarks detected when they were free, even for less than a milionth of second (before they created new hadrons because of the color confinement). HOTmag (talk) 10:34, 13 March 2024 (UTC)

Gluons contribute to the proton mass per [2], a reliable source. Modocc (talk) 16:00, 13 March 2024 (UTC)

Now you're ignoring my response to your response in which you presented the source. HOTmag (talk) 16:36, 13 March 2024 (UTC)

Certainly "...the contributions to the mass of the proton are..." refers to the proton's bound constituent particles. Your responses simply keep ignoring that. Modocc (talk) 16:50, 13 March 2024 (UTC)

So let's remain in dispute about whether the values of those contributions refer to the values of bound constituent particles (as you claim) or to the values of those particles had they been free (as I claim). HOTmag (talk) 17:35, 13 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:^[3]

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

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)

Thanks. HOTmag (talk) 10:37, 13 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)

I just tried the copilot ai but got nothing useful towards answering this question. Hewer7 (talk) 14:24, 16 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).[4] 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.^[5]

The first of the two studies reported on^[6] 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^[7] 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.^[8]  --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)

"Leaking and uncontained hydrogen led to an explosion that killed four workers and seriously injured a fifth last May at AB Specialty Silicones in Waukegan, Illinois".  --Lambiam 07:13, 13 March 2024 (UTC)

USCSB did a video for that one. DMacks (talk) 07:19, 13 March 2024 (UTC)

Link to USCSB video.  --Lambiam 07:25, 13 March 2024 (UTC)

Also, "Hydrogen is 14 times as flammable as natural gas and can be ignited by static electricity."  --Lambiam 07:22, 13 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 trailing 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)

on your claim that "...everything you learned about Russia was incorrect..." 2601:646:8082:BA0:8014:354A:6A03:A76 (talk) 14:09, 13 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)

Did any other possible alternatives to Hitler (e.g. Drexler, Strasser, Roehm, etc.) have a similar grand vision of global German expansion and/or systematic mass extermination of Jews and other ethnic minorities that he had outlined in Mein Kampf and then seriously attempted to implement, with a large measure of initial success, in real life? 2601:646:8082:BA0:8014:354A:6A03:A76 (talk) 14:09, 13 March 2024 (UTC)

Hitler was pretty good at rising to power. He succeeded after about 12 years, which is pretty good for somebody with such extreme ideas in a democratic state. He was not so good at remaining in power. He lost power (along with his life) after just 12 years, rather short for a dictator. Killing him before he rose to power may very well have been beneficial to the world, doing it after he rose to power maybe not. PiusImpavidus (talk) 09:24, 14 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)

Are we actually certain that nobody turned up to Hawking's time traveller party? :-) NadVolum (talk) 13:00, 13 March 2024 (UTC)

March 13

Lake Kakhovka

What's the situation at the former Lake Kakhovka from a "rewilding" point of view? All satellite images I've been able to find online are close to the moment of the destruction of the dam, but I was curious to know if trees and plants are starting to grow and in general how nature is recoverinf there. Thank :) --LametinoWiki (talk) 00:42, 13 March 2024 (UTC)

Given that the dam was destroyed less than a year ago, we may not have much significant regrowth on "tree" scale yet (especially given that a significant amount of that time period was winter). Additionally, this is a war zone or territory illegally occupied by Russia, so I doubt a ton of scientific teams are able to properly investigate the location. Both the war and occupation conditions could also be an impediment to significant regrowth. --OuroborosCobra (talk) 02:28, 13 March 2024 (UTC)

Two beams of light start moving perpendicular to the initial distance between them. Will it become shorter, because of any gravitational curvature caused by them in spacetime?

I'm asking from an empirical point of view, rather than from a theortical one (as predicted in General Relativity).

I was recently shocked to hear from a local astronomer, that beams of light frequently arriving from pulsars, are known to prove that the distance does not become shorter. As far as you know, is there any evidence that supports - or refutes - what I've heard from that astronomer, emprically speaking? HOTmag (talk) 11:04, 13 March 2024 (UTC)

That's not measurable. I don't know what your local astronomer was talking about but they may have thought of the pulsar timing array. --Wrongfilter (talk) 11:23, 13 March 2024 (UTC)

Thanks. Does the pulsar time array have anything to do with a (constant?) distance between two beams of light arriving from pulsars? HOTmag (talk) 11:43, 13 March 2024 (UTC)

Your astronomer friend is correct from the theoretical point of view though I don't know what they were referring to. Think of having an object going at near the speed of light alongside some light. The light would be deeply red shifted and have very little energy. NadVolum (talk) 12:52, 13 March 2024 (UTC)

As for your first point: Why was the astronomer correct from the theoretical point of view? Doesn't General Relativity predict both beams of light will curve spacetime because of their energy?

As for your second point: Please notice that according to General Relativity, the impact of a gravitational curvature on a given particle's momentum - does not depend on the particle's energy, whereas a gravitational curvature caused by a given particle - does depend on the particle's energy. That's why my current question is only being asked about the curvature (if any) caused by light, rather than about the impact of that curvature on light. HOTmag (talk) 15:29, 13 March 2024 (UTC)

I don't have the maths for that anywhere near my finger tips but it seems to me that such a prediction would lead to all sorts of problems. Intuitively I would guess that it's difficult to impossible to construct a non-zero curvature field that transforms covariantly. However, in another one of your threads somebody posted a link to a paper demonstrating some sort of attraction between light rays in the lab, I don't recall the details. Be that as it may, any effect is certainly much to small to be measurable in the presence of much larger perturbations that are ubiquitous and unavoidable on Galactic scales (this also precludes an empirical demonstration of a zero effect to any level that would satisfy you). --Wrongfilter (talk) 15:44, 13 March 2024 (UTC)

This is the link you've just mentioned. It mentions no "lab" nor any other empirical clues. All was "theoretically inverstigated" (as clearly stated in the last sentence of the abstarct). That's why my current question is being asked "from an empirical point of view".

As for the maths bothering you, please see our artice pp waves, indicating: The pp-waves solutions, model radiation moving at the speed of light. This radiation may consist of electromagnetic radiation, gravitational radiation, massless radiation...or any combination of these, so long as the radiation is all moving in the same direction....Penrose also pointed out that in a pp-wave spacetime, all the polynomial scalar invariants of the Riemann tensor vanish identically, yet the curvature is almost never zero. [They vanish]...because in four-dimension all pp-waves belong to the class of VSI spacetimes. However, this curvature "is almost never zero" because (as stated in our article about VSI) the solutions of the stress–energy tensor usually have also non-polynomial or non-scalar invariants, which actually do not vanish. HOTmag (talk) 15:59, 13 March 2024 (UTC)

"...light causes no acceleration to a co-propagating test ray." emphasis added. Thus co-propagating rays, like the light arriving from distant stars and pulsars, are unaffected. Modocc (talk) 16:17, 13 March 2024 (UTC)

Not in all cases. See again the last sentence of the abstract. Additionaly, please notice you are the person who presented this link for proving the photons "gravitate" (as you wrote). HOTmag (talk) 16:32, 13 March 2024 (UTC)

Yes not in all cases, nevertheless Wrongfilter and the astronomer are, in theory, both correct. Modocc (talk) 16:54, 13 March 2024 (UTC)

Please notice I asked user:NadVolum "why" the astronomer was correct. Doesn't General Relativity claim every object carrying energy curves spacetime? HOTmag (talk) 17:02, 13 March 2024 (UTC)

Yes and its gravity (GR's or any othr model's) depends on invariant mass. Co-moving photons appear to not have that with respect to each other. Modocc (talk) 17:34, 13 March 2024 (UTC)

Please notice, that General Relativity ascribes - the gravitational curvature - caused by a given particle, to the particle's energy, rather than to the particle's invariant mass. For more details about what properties of the particle curve spacetime, see Stress–energy tensor. HOTmag (talk) 17:44, 13 March 2024 (UTC)

Of course all particles gravitate (I've been consistent on this point) thus each photon contributes to the field (warpinging GR's spacetime). With a photon cloud their relative velocities also contribute to invariant masses which determine to what degree there are interactions between them if any. Modocc (talk) 17:59, 13 March 2024 (UTC)

Since "all particles gravitate" as you emphasize, was the astronomer correct?

I recall the astronomer has claimed - emprically [factually] speaking, that the initial distance between two parallel beams of light arriving from pulsars, does not become shorter, even after finishing the long journey to Earth. According to that apparently empirical [factual] claim, those beams of light don't curve spacetime, do they? HOTmag (talk) 18:13, 13 March 2024 (UTC)

Your astronomer cannot measure that. --Wrongfilter (talk) 18:30, 13 March 2024 (UTC)

I'm not arguing about whether the astronomer can measure it. Yet, the question still remaing is whether the astronomer could've been correct, had the astronomer been able to meausre it. Apparently, what the astronomer claims, from an apparently empirical [factual] point of view, contradicts General relativity, which claims that any given particle having non-zero energy (hence a non-zero stress–energy tensor) is supposed to curve spacetime. I thought, the astronomer's apparently empirical [factual] claim, really contradicted General Relativity, but I wanted to be sure, and that's why I posted this thread. HOTmag (talk) 18:58, 13 March 2024 (UTC)

Apparently you have no idea what "empirical" means. --Wrongfilter (talk) 19:14, 13 March 2024 (UTC)

Sorry for replacing "factual" by "empirical". Anyway, as I have already stated, "I'm not arguing about whether the astronomer can measure it". Whether because I "have no idea what 'empirical' means" (as you claim), or because - even if I do know what it means (me meaning "factual" rather than "empirical") - I don't want to ask about whether the astronomer can measure it, but rather about whether the astronomer could've been correct - had the astronomer been able to meausre it. I thought, the astronomer's claim contradicted General Relativity, but I wanted to be sure, and that's why I posted this thread. HOTmag (talk) 19:26, 13 March 2024 (UTC)

(ec Stop making all these microedits) Maybe the astronomer just wanted to end the conversation... You wanted empirical evidence one way or the other, now don't change that to "factual" (I don't even know what that's supposed to mean). I say the effect would be unobservable, even if there was an effect (I don't know - there are lots of subtleties in GR if one dives in deep), hence the question is empirically undecidable. But you will never be sure, no matter what anyone here says. --Wrongfilter (talk) 19:46, 13 March 2024 (UTC)

Just to be clear:

1. By mistake, I replaced "factual" by "empirical" when I referred to the astronomer's claim about the beams of light arriving from pulsars.

2. However, my original question was about any empirical evidence, any one of you knew of. Yes, I used the word "empirical" in my original question, on purpose, meaning empirical, because I do know what it means.

3. Before I read your response about the immeasurability, I posted this thread asking whether any one of you - knew of any empirical evidence - which could support or refute the astronomer's factual claim about the pulsars, because I wondered whether the astronomer could be correct - assuming the astronomer's factual claim about the pulsars could be resolved by measurement. I thought, this astronomer's factual claim contradicted General Relativity, but I wanted to be sure, and that's why I posted this thread asking about any empirical evidence you knew of - which could support or refute the astronomer's factual claim about the pulsars.

4. In your response to my original question, you've claimed all of that can't be measured. As I have already stated, I'm not going to argue about that. However, I still wonder whether the astronomer could have been correct - had the astronomer's factual claim about the pulsars been resolved by measurement.

Am I clearer now? HOTmag (talk) 20:43, 13 March 2024 (UTC)

The starlight is co-moving like racers in a dead heat so according to the citation the photons gravitate, but not each other. Modocc (talk) 18:27, 13 March 2024 (UTC)

Since "the photons gravitate" (me using your words), so they curve spacetime (don't they?), so how can they avoid gravitating "each other" (me still using your words), as long as the spacetime is being curved by them? HOTmag (talk) 18:58, 13 March 2024 (UTC)

Why they (the comoving photons) don't gravitate each other according to the research paper? Perhaps someone else better versed in GR can answer that question, but co-moving photons do not appear to have invariant mass with respect to each other (although their energies obviously contribute to the invariant masses of larger systems). Note that in reality no photon within our universe exists in isolation from it so the notion that they all contribute mass-energy to the field holds. Modocc (talk) 20:13, 13 March 2024 (UTC)

Plesse notice that also massless photons are influenced by the curvature of spacetime, that actually deflects their trajectory, e.g. when they approach the sun, so the question still remains: Why is neither of them influenced by this curvature when it's caused by the other photon, while you agree "they gravitate" - hence curve the spacetime. HOTmag (talk) 20:51, 13 March 2024 (UTC)

Their path's curvature (or spacetime's curvature) are dependent on the sun's mass. As for their affecting each other, that, will depend, according to the paper I cited, on their relative velocities which are vectors. As for measurement, I assume the strength of their combined field is proportional to their invariant mass and inversely proportional to the distance between them squared: in accordance with other masses, and subject to the same caveats (Newtonian approximation, quantum limits, MOND, etc). Comoving photons have zero invariant mass, but also add mass to larger bodies. Modocc (talk) 23:07, 13 March 2024 (UTC)

Two photons move in parallel paths. There is an electron between them. Do the photons curve spacetime (in addition to the curvature caused by the electron)? If the answer is positive, then do they gravitate each other (in addition to the gravitation caused by the electron)? HOTmag (talk) 01:35, 14 March 2024 (UTC)

At infinity the photons don't gravitate each other, according to the citation, but the electron is a gravitational lens causing their convergence: thus the photons' convergence should have an invariant mass that gravitates them, also in accordance with the citation. Modocc (talk) 02:01, 14 March 2024 (UTC)

What do you mean by "the photons' convergence should have an invariant mass that gravitates them"? I remind, we are talking about two photons co-moving in the same direction, so how can the photons' convergence have an invariant mass? HOTmag (talk) 08:23, 15 March 2024 (UTC)

Convergent rays carry energy and momenta towards a distant focal point (with a component velocity less than c). They are no longer propagating parallel to each other. Modocc (talk) 12:04, 15 March 2024 (UTC)

Can you think of any theoretical situation in which the invariant masses of two given photons will be zero, assuming that the whole world contains massive matter as well? HOTmag (talk) 13:50, 15 March 2024 (UTC)

As long as photons are moving along similar parallel geodesics they have energy, but, of course, it is not invariant. Invariant mass: "The invariant mass, rest mass, intrinsic mass, proper mass, or in the case of bound systems simply mass, is the portion of the total mass of an object or system of objects that is independent of the overall motion of the system." Modocc (talk) 17:37, 15 March 2024 (UTC)

Photons have energy and therefore, according to the Einstein field equations, curve spacetime just like other entities with energy. The predicted effect of two photons on each other is too small to be measured.  --Lambiam 21:06, 13 March 2024 (UTC)

You've responded to a response of mine that was, both addressed to user:Modocc only, and assuming user:Modocc's assumptions only. Do you think you can also respond to my original question under the title of this thread? HOTmag (talk) 22:02, 13 March 2024 (UTC)

You have restored a contribution of mine that I had removed. That is a big no-no. When I originally posted it, I had overlooked the fact that the question had been changed: in the OP they were moving in orthogonal directions, not parallel. The latter is much more tricky, one immediate issue already being that the notion of geodesics being parallel in a non-flat continuum is ill-defined.  --Lambiam 10:37, 14 March 2024 (UTC)

Wow, I'm quite surprised now. Of course I was not aware of the restoration made by me. I still wonder how it was actualy done, maybe because I had not received a warning about an ongoing edit conflict? Anyway it was not done on purpose. I guess you forgive me for this big mistake, for which I apologize from the bottom of my heart, don't you?

As for your second remark, I'm surprised again: What do you mean by "the question had been changed: in the OP they were moving in orthogonal directions, not parallel". As far as I remember: from the very beginning, the question indicated in the title (that has never changed) has always been formulated as follows: Two beams of light start moving perpendicular to the initial distance between them. Will it become shorter, because of any gravitational curvature caused by them in spacetime? In other words, from the very beginning the question has always been about two beams of light moving in parallel directions, hasn't it? HOTmag (talk) 08:09, 15 March 2024 (UTC)

Suppose one photon's trajectory is given by its position as a function of time as $(ct,0,0)$ while the other has $(0,ct,\delta ).$ Then, when $t=0,$ each moves perpendicular to the line segment from $(0,0,0)$ to $(0,0,\delta )$ connecting them, and in orthogonal directions with respect to each other. I now think that also in this case they would not gravitationally attract each other. It is difficult to find reliable sources handling the issue, as any effect would be too small to measure, but nevertheless I found this: "This conclusion explains ... why photons do not attract each other during the long trip from the distant galaxies and collapse into a single clump, despite of the fact that photons have gravitational mass." ^[9]  --Lambiam 19:00, 15 March 2024 (UTC)

Careful: that quote is from a paper published in Physics Essays, which looks like a rather dodgy journal. --Wrongfilter (talk) 19:19, 15 March 2024 (UTC)

I was trying to make it simple. There is no reason to invoke General Relativity, Special Relativity is quite enough to see why the rays don't attract each other. If two particles have the same direction and speed as each other then they are at rest relative to each other and only their rest mass counts for the gravitationl attraction between them - even if they are going at near the speed of light relative to us and have a huge mass relative to us. Now think of a photon as the limit of a particle getting a smaller and smaller rest mass but going faster and faster so the relativistic mass stays the same. in the limit we have a particle with zero rest mass but the same energy. Now two of those going along together won't attract each other because their rest mass is zero. NadVolum (talk) 20:57, 13 March 2024 (UTC)

Following your considerations, let's look at the whole picture - taking into account new considerations you didn't take into account, after making a distinction - as you've made - between two cases: A pair of massive particles, and a pair of photons.

1. Two massive particles start moving, in the same direction, perpendicular to the initial distance between them. In their reference frame, the initial distance between them will soon become shorter, because of the gravitational curvature caused by their rest mass (as you have indicated). However [and this is my new consideration you didn't take into account], the distance will become much shorter, in any other reference frame, because the curvature is bigger in that reference frame, because the curvature in that other reference frame is also influenced by their kinetic energy (being their "relativistic mass" in that other reference frame, provided that we allow to use this concept, and I allow, and I assume you allow as well, but even if we don't allow we can still adhere to their kinetic energy instead).

2. Two photons start moving, in the same direction, perpendicular to the initial distance between them. Had they had a reference frame in which they could have been at rest, then: In their reference frame the initial distance between them would have not changed, because the gravitational curvature caused by their zero rest mass - would have been zero as well - as you've indicated. However [and this is my new consideration you didn't take into account], the distance will become shorter, in any other reference frame, because the curvature is bigger in that other reference frame, because the curvature in that other reference frame is also influenced by their kinetic energy (being their "relativistic mass" in that other reference frame, provided that we allow to use this concept, and I allow, and I assume you allow as well, but even if we don't allow we can still adhere to their kinetic energy instead).

HOTmag (talk) 22:02, 13 March 2024 (UTC)

I'm having difficulty with what you said but apparent contraction of lengths only happens in the direction of motion, not at right angles to it. A super speed locomotive may seem shorter but the wheels will remain the same distance apart and stay on the track. Also while going any particular distance the particles will experience less time and a shorter distance than an observer and therefore will not approach each other even the amount one might assume from Newtonian gravity on their rest mass never mind with the extra energy. NadVolum (talk) 00:17, 14 March 2024 (UTC)

I was not referring to the Special relativistic effect of length contraction due to velocity, but rather to the General relativistic effect of difelcting a given photon's trajectory when the photon is in a gravitational field, e.g. when it moves near the sun, so the distance between the photon and the sun becomes shorter because of the effect of gravitational lensing, regardless of the Special relativistic effect of length contraction due to velocity.

That said, do you agree that when an electron and a planet start moving perpendicular to the initial distance between them, the direction will soon become shorter, according to the following three principles:

1. The faster the reference frame moves relative to the planet, the bigger kinetic energy the planet has (hence the bigger relativistic mass the planet has).

2. The bigger kinetic energy (hence relativistic mass) the planet has [in a given reference frame], the shorter the distance between the electron and the planet will become in that reference frame.

Logically, the following principle follows from the combination of the previous ones:

3. The faster the reference frame moves relative to the planet, the shorter the distance between the electron and the planet will become in that reference frame.

If you agree, then try to read again my previous response, but now you should interpret the effect as a General relativistic one, due to gravitational lensing effect that obeys the third principle mentioned above (rather than as a Special relativistic effect of length contraction). HOTmag (talk) 01:22, 14 March 2024 (UTC)

Changing reference frames, in general, does not, I repeat, does not change physical interactions. For example, did you know that the Earth has arbitrary KE! Pick any train, rocket, plane or the like, the Earth has different KE due to the velocity of your chosen reference frame. Instead, g depends simply on the reference frame independent invariant mass of the gravitating body. That said, accelerations of objects indeed changes interactions: such as how Doppler is measured. In other words, physically changing one's reference frame (as opposed to only applying a Galilean transformation) is not trivial. For instance, surfers that ride the crests of ocean waves and then glide off them physically change their reference frames and the frequency of wave interactions increases. Hence, both kinds of change happen at the same time. Modocc (talk) 02:30, 14 March 2024 (UTC)

AFAIK, in General Relativity, gravity is considered to be caused by curvature in spacetime, while this curvature is caused by the stress–energy tensor, while this tensor is defined by the relativistic energy E (and the momentum p along with some other components relating to these energy and momemtum) - of the entity (e.g. a planet) responsible for the gravity, while this relativistic energy E is equivalent to the relativistic mass E/C², rather than to the invariant mass E/C² devided by the gamma factor. HOTmag (talk) 07:52, 15 March 2024 (UTC)

Relativistic masses each contribute to the invariant masses of larger bodies, which in turn, gravitate in accordance with their invariant masses. So yes they gravitate. :-) Modocc (talk) 13:15, 15 March 2024 (UTC)

Please notice the current discussion is not about whether they gravitate but rather about your previous response in which you've emphasized: "Changing reference frames, in general, does not, I repeat, does not change physical interactions... g depends simply on the reference frame independent invariant mass of the gravitating body". In this previous response, you responded to a response of mine to user:NadVolum, in which I claimed "The bigger kinetic energy (hence relativistic mass) the planet has [in a given reference frame], the shorter the distance between the electron and the planet will become in that reference frame", so it seems your previous response meant to claim that gravity did not depend on relativistic masses. However, if you're claiming now that it does depend on them, then we agree, as far as our current discussion (about my response to user:NadVolum and about your previous response to my response) is concerned. HOTmag (talk) 13:50, 15 March 2024 (UTC)

I was careful to provide enough context with each of my points to avoid contradiction. Note that I said relativistic masses contribute to proper masses which constrains the problem of determining what curvature(s) are present within and outside the bodies under consideration. My first point was that unlike picking reference frames, any curvature present due to g is not arbitrary, and my second point was that for any fixed reference frame differences in invariant masses correspond to differences in curvature so yes we have to apply relativistic mass when calculating them. Modocc (talk) 21:59, 15 March 2024 (UTC)

There will be a gravitational wave associated with the two photons, it is generated when they are emitted. Like a single wave on water it has no effect if you're going along with it like a surfer but if you go across it has a larger effect the faster you cross it. NadVolum (talk) 13:23, 15 March 2024 (UTC)

Considering the indent of your current response, are you reponding now to my respose to user:Modocc? HOTmag (talk) 13:50, 15 March 2024 (UTC)

Irrelevant to what I said. You were not satisfied with a simple explanation using Special Relativity that fully explains what happens. As far as I can make out you want a General Relativity solution about why others would feel a gravitational attraction to the photons but the photons wouldn't feel any to each other. My description using a surfer is just descriptive, a fuller explanation would need you to read the papers and delve through the maths of General Relativity I believe. NadVolum (talk) 15:50, 15 March 2024 (UTC)

Hump

In a freight yard, is there any advantage to putting three lead tracks on the hump, as opposed to the more usual multiples of two? 2601:646:8082:BA0:8014:354A:6A03:A76 (talk) 13:34, 13 March 2024 (UTC)

I don't see anything about what you are saying at that article. Have you something describing what you are talking about? NadVolum (talk) 21:12, 13 March 2024 (UTC)

If you look at the photos in the article Classification yard, the hump in these photos always has two leads, never three -- my question is, is there any particular reason why this is the case? 2601:646:8082:BA0:8014:354A:6A03:A76 (talk) 02:13, 14 March 2024 (UTC)

I don't know how to see which tracks are leads, but in any case, a larger number requires a more extensive hump. If the hump is a constructed hill, as suggested in Rail yard § Freight yards, the extra cost and complexity may not be outweighed by an operational advantage.  --Lambiam 10:19, 14 March 2024 (UTC)

I looked at several examples of hump yards and found three layouts for the leads:

• A single lead going up the hump, after the downhill splitting multiple times into the bowl. For example in Cincinnati, Kansas City, Houston (all US), Odesa (Ukraine). This appears to be the dominant layout in the US.
• Two leads, merging into one on the downhill before splitting multiple times into the bowl. For example in Vienna (Austria), Hamburg, Mannheim (south side) (both Germany), Riga (Latvia), Antwerp (Belgium). This appears to be the dominant layout in Western Europe.
• Two leads with a scissors crossover on the downhill before splitting multiple times into the bowl. For example in Metz (France), Rotterdam (Netherlands), Mannheim (north side) (Germany), Shanghai (China).

All yards named after the nearby large city, not the name of the yard itself.

Now keep in mind the difference between Western European couplings and those used in other areas. Western Europe still uses screw couplings. Before humping a series of wagons, the train is first pushed slightly onto the hump, to compress the couplings. Then the couplings are loosened a bit, so that they can be unhooked. Then the wagons are pushed over the hump, whilst a worker uses a club to knock the link of the coupling off the hook, without stopping the train or getting between the wagons. This isn't needed elsewhere, as trains there use claw couplings. So having two leads is useful in Western Europe, as one train is pushed over the hump whilst the next is prepared.

Having a scissors crossover instead of a merge followed by a split allows sorting two trains at the same time, each into half the bowl tracks. In Metz, one can sort one train onto all 48 tracks or two trains onto 24 tracks each. You gain some flexibility for small jobs, at the cost of two sets of points and a diamond. I suppose that small jobs are more common in Europe, with shorter and more frequent trains, than in the US.

Three or more lead tracks gives diminishing returns, whilst rapidly making the required pointwork more complex. PiusImpavidus (talk) 11:02, 14 March 2024 (UTC)

Thanks! So, just as I thought -- adding a third lead gives you more than 50% higher construction costs (as well as operational costs, and per-capita aspirin consumption by the switchmen) in exchange for less than 50% higher throughput (and it gets worse from that point on)! (Incidentally, the only hump I've ever seen with three leads was a fictional one at Knapford Yards -- or was it Tidmouth Yards? -- in Thomas & Friends (one of the old classic episodes, "Old Iron" "Pop Goes the Diesel" I think) -- and I have to remark, although these yards as a whole look impressive on the screen, they are actually laid out pretty poorly, and the hump seems to have been added pretty much as an afterthought!) 2601:646:8082:BA0:0:0:0:2EB (talk) 13:55, 14 March 2024 (UTC)

And two lead tracks with a scissors crossover gives redundancy. Any of the sets of points can be taken out of service for maintenance, still allowing the yard to operate at half capacity. PiusImpavidus (talk) 10:48, 16 March 2024 (UTC)

Would some like to point out where the leads are in this diagram just so I know. It never occurred to me before that you need a sorting algorithm to build trains despite it being obvious and despite having probably gone past these kind of places numerous times. Sean.hoyland (talk) 11:19, 14 March 2024 (UTC)

It's the track from the approach yard to the classification yard, leading up to the hump. The hump is located just before the first retarder. PiusImpavidus (talk) 11:48, 14 March 2024 (UTC)

Thanks. So the red rectangles are retarders I guess. Sean.hoyland (talk) 11:54, 14 March 2024 (UTC)

Just saw the legend which says in clear large font Retarders...Sean.hoyland (talk) 11:55, 14 March 2024 (UTC)

arrest, dearrest / de-arrest, arrested

What's the meaning of "arrest and dearrest" on page 684 of https://archive.org/details/proteintransfero0000unse/page/684/mode/2up?q=dearrest and "de-arrest of arrested" on page 3 of https://archive.org/details/arxiv-1206.2024/page/n1/mode/2up?q=%22de-arrest%22? Mcljlm (talk) 21:42, 13 March 2024 (UTC)

The verb to arrest means, "to keep (something) from moving". What is kept in arrest in these cases are molecules, in the first case specifically proteins; in the second case, those of which a glass is comprised. The verb to dearrest means, "to release (something) from an existing arrest". The form de-arrest is a spelling variant.  --Lambiam 10:06, 14 March 2024 (UTC)

Thanks Lambiam. Until a few days ago I'd not come across dearrest/de-arrest at all {originally I was only intending to ask about that word}, and then only relating to people. Later I found it used relating to ships and goods in historical documents (the OED doesn't mention any use later than 1791). Is its use recent and common in a scientific context? Is the hyphen's use more or less frequent in scientific publications? Mcljlm (talk) 16:49, 14 March 2024 (UTC)

Regarding ships and boats: arresting a boat (in a legal sense) was certainly a thing up to the 1990s, and I would guess still is. In the early '90's my father, who worked at a legal firm, passed on to me an obsolete word processor (an IBM Displaywriter System) that had just been replaced at his firm; on the data disk were several form documents, including one for arresting a boat. 51.198.186.221 (talk) 00:43, 15 March 2024 (UTC)

It's dearrest/de-arrest which the OED doesn't have any mention for later than 1791.. Could it be that another word was used? Perhaps I should ask at the Humanities and/or Language reference desks. Mcljlm (talk) 04:52, 15 March 2024 (UTC)

The 1791 use actually uses the unhyphenated spelling dearrest. Here is a use of de-arrest, with a hyphen, from 1898.  --Lambiam 17:46, 15 March 2024 (UTC)

Since it's an edition of a 1334 document it would be interesting to know if the hyphen is in the original and if not why it is in this edition. Mcljlm (talk) 10:05, 16 March 2024 (UTC)

Until 1905, the so-called atomic theory – the notion that matter was composed of molecules which in turn were composed of elemental atoms – was a hypothetical concept. There was in fact considerable opposition to atomic theory. This only changed with the publication of Einstein's article on Brownian motion. So most of the use of the term in this scientific sense of a molecule being arrested is indeed relatively recent. Proponents of the atomic theory did use the term arrest, though, already in the 19th century (e.g, in 1874, "the heat of arrested motion"^[10]). Nevertheless, this is not a term of art but the use of a word in an accepted common sense applied to a specific context. The term dearrest is rather rare, and it is hard to tell which spelling variant is more common.  --Lambiam 20:58, 14 March 2024 (UTC)

Explained by the BBC (which appears to uses 'de-arrest' consistently): Who, what, why: What does it mean to be de-arrested? . -- Verbarson  ^talk_edits 13:49, 15 March 2024 (UTC)

I became aware of the use of de-arrest/dearrest relating to contemporary police action about a week ago Verbarson when someone quoted from a Metropolitan Police statement. I then noticed its use in the other contexts I mentioned. Mcljlm (talk) 15:38, 15 March 2024 (UTC)

References

March 15

Can't remember the name of this organic chemistry phenomenon where conjugation "extends" a functional group

I think I remember reading a Wikipedia article about this, but I can't find the name of this phenomenon at all. Pinging @Smokefoot: maybe you might know since you are an active editor of organic chemistry articles. Here I have drawn the concept:

This is the phenomenon in which a double bond "extends" a functional group by conjugation, so that it behaves similarly to the parent functional group. The bottom compound, an "extended" carboxylic acid, has similar chemistry as the top carboxylic acid, except the two parts of the functionality (C=O and -OH) are separated by an additional double bond. The reason for this similarity is because conjugation can transfer charges by resonance, e.g. the bottom compound is acidic like a carboxylic acid because the negative charge on the deprotonated -OH oxygen can move to the carbonyl oxygen by resonance. Michael7604 (talk) 05:50, 15 March 2024 (UTC)

See or search dicarbonyl or keto-enol equilibrium. @Michael D. Turnbull and DMacks:. --Smokefoot (talk) 14:37, 15 March 2024 (UTC)

@Michael7604 You already had the key concept in your title: see conjugated system and the articles linked from it. My favourite related topic is, of course, the Michael addition reaction. Mike Turnbull (talk) 16:28, 15 March 2024 (UTC)

A vinylogous group. DMacks (talk) 18:26, 15 March 2024 (UTC)

This is it, thank you Michael7604 (talk) 21:15, 15 March 2024 (UTC)

Fictitious force. Is the opposite phenomenon called a: "Real" force? "Physical" force? "External" force? "Natural" force?

I'm looking for the most useful term, intended to exclude fictitious forces. HOTmag (talk) 08:38, 15 March 2024 (UTC)

Each of the suggested terms is usable if you take away the ornamentation of Scare quotes. Philvoids (talk) 10:31, 15 March 2024 (UTC)

They should only be considered to be quotation marks. That's because I was looking for "the most useful term" (Btw now I'm quoting myself), i.e. a term that could be quoted from sources using useful terminology. HOTmag (talk) 10:45, 15 March 2024 (UTC)

A commonly used contrasting term is true force.^[11][12][13]  --Lambiam 17:25, 15 March 2024 (UTC)

Meh. Yes, true force has some popularity, sometimes even in quotes, but don't expect everybody to understand you when you mention true forces. And despite Newton's ideas on absolute space and preferred reference frames that only have constant velocity relative to it, in real life we deal with the equivalence principle, under which those fictitious forces are as real as gravity. Which is quite real in classical mechanics. In fact, they are gravity. In a free-falling lift, we just declare gravity zero. On a merry-go-round, we just declare the centrifugal force and Coriolis force part of gravity. Good luck describing orbital manoeuvring or accretion in compact binaries in an inertial frame. General Relativity (I think you'll love that) even makes it its basic principle. Turns out that the entire universe spinning around our stationary merry-go-round causes frame dragging, which exactly matches those fictitious forces. PiusImpavidus (talk) 19:57, 15 March 2024 (UTC)

That was Einstein's original view, but I think the matter has become somewhat controversial. We have an article on Mach's principle. --Trovatore (talk) 21:06, 15 March 2024 (UTC)

Just as a slight extension, in dynamics we invent a fictitious force m*a, called a D'Alembert force, and can then treat it as a statics problem. It isn't always useful. Greglocock (talk) 23:36, 15 March 2024 (UTC)

March 16

Weld; Syren

This 1878 report makes a passing reference to "Weld's Sound Experiment". What was that, and who was Weld?

It also mentiones "Syren and Galton's Whistle"; the latter is Francis Galton, but who was Syren? Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 21:28, 16 March 2024 (UTC)

My guess is a siren whistle. --Wrongfilter (talk) 21:49, 16 March 2024 (UTC)

Galton's whistle is similar to a boy scout's whistle . It differs in that it has a piston provided with a screw and a milled head... (it's an ultrasonic dog whistle).

From Oscillations and Waves p. 297. Alansplodge (talk) 21:56, 16 March 2024 (UTC)

The most promising Weld is Alfred Weld, who was a bit of a scientist, but I fail to find any indication that he may have dealt with sound. --Wrongfilter (talk) 22:02, 16 March 2024 (UTC)

And but nay. In duo together with Ernst Chladni, Alfred will make too much of a youngster, for those classicals. --Askedonty (talk) 01:33, 17 March 2024 (UTC)

What classicals? This is about the first annual meeting by the Midland Union of Natural History Societies. Also, they had no scruples looking at a toddler like Galton's whistle, which had been invented a mere two years before. But don't hesitate to suggest another Weld if you've got one. --Wrongfilter (talk) 07:29, 17 March 2024 (UTC)

Wiktionary:syren#Noun: "Obsolete form of siren". Alansplodge (talk) 22:03, 16 March 2024 (UTC)

For the meaning of "Galton's Whistle", see Dog whistle.  --Lambiam 00:53, 17 March 2024 (UTC)

Isaac Weld.^[14]  --Lambiam 10:24, 17 March 2024 (UTC)

Resolved

 – Thank you. all. Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 10:50, 17 March 2024 (UTC)

Length of a photon

A couple of questions about photons.

Firstly, am I right to feel slightly uncomfortable about the following passage? From the article photon:

However, experiments confirm that the photon is not a short pulse of electromagnetic radiation; a photon's Maxwell waves will diffract, but photon energy does not spread out as it propagates, nor does this energy divide when it encounters a beam splitter. Rather, the received photon acts like a point-like particle since it is absorbed or emitted as a whole by arbitrarily small systems, including systems much smaller than its wavelength, such as an atomic nucleus (≈10⁻¹⁵ m across) or even the point-like electron.

It's trying, I think, to make a point that a quantum-mechanical wavefunction is different from a classical wave, because it can suffer wavefunction collapse; and because it always represents a single whole particle (and also that a photon is a point particle -- which I understand to mean that its overall (extended) wavepacket can be evolved by integrating it forward via point to point propagators).

But am I right to feel uncomfortable with the argument about it not being distributed because it can interact with objects that have a very small scale? I seem to remember that in the classical treatment of radio waves and antennas, an antenna might be essentially just a single vertical pole; but may interact with a radio wave that may have a wavelength of hundreds of metres, if the antenna is part of a circuit that has the right resonant frequency. If I remember correctly, the radio wave induces a current in the circuit, which in turn can be thought of as exciting an induced radio wave of its own that has an opposite phase in the far field, thus effectively partially cancelling the original wave; and so, despite what may be its small physical size, the antenna can have an effective area ("aperture") that can be very considerably larger. (Did I remember that right? It's a long time ago since those lectures on EM ...)

Now I'm not saying that that is directly equivalent to quantum measurement, because measuring the location of a photon absorbs all of it, not just part of it (give or take whatever discussion we might have about decoherence...) But am I right that the antenna does maybe suggest that we should perhaps be just a little more careful before making a blanket statement that an object cannot be distributed if it can interact with a system much smaller than its own wavelength?

The remark that "the photon energy does not spread out as it propagates" also makes me feel a bit uncomfortable. If one thinks of gravitational lensing, surely it's reasonable to imagine a photon travelling both sides of the gravity well while still retaining self-coherence, with a corresponding energy flow associated with both sides?

Beyond that, the second thing I wanted to ask about (which was what led me to look at the photon article in the first place, to see if it anywhere touched on the subject) was this: I seem to remember, way back at school level, being told that a typical wavepacket of a photon of visible light (eg perhaps from a sodium vapour lamp) had a longitudinal length of the order of about a metre (maybe give or take an order of magnitude either way). Does that seem plausible / viable / about right ?

One constraint would seem to be the time <-> frequency uncertainty relation. The overall linewidth of the source would seem to set a limit to the maximum range of frequencies that could be associated with such a photon, and therefore a minimum to the length of time (and so therefore spatial length) that the wavepacket must persist. Presumably there is also a maximum length of time one pictures the emission event taking -- perhaps even measurable by trying to measure the coherence time of such a source using an interferometer at lower and lower light levels ? Is this sort of thinking at all on the right track?

Thanks in advance, Jheald (talk) 22:05, 16 March 2024 (UTC)

The following, unfortunately unsourced, passage is copied from Point particle § In quantum mechanics:

Nevertheless, there is good reason that an elementary particle is often called a point particle. Even if an elementary particle has a delocalized wavepacket, the wavepacket can be represented as a quantum superposition of quantum states wherein the particle is exactly localized. Moreover, the interactions of the particle can be represented as a superposition of interactions of individual states which are localized.

Does this help to ease your uncomfortability?  --Lambiam 00:48, 17 March 2024 (UTC)

Thanks @Lambiam:. No problem with what you've quoted, so far as it goes.

But I do think it can be useful (and meaningful) to try to give a sense of what the whole overall superposition - the whole wavefunction - may look like, when this is possible (for a typical environment or observational set-up).

So eg for an electron, it's a point particle, but it's also useful to be able to talk about the whole atomic or molecular orbital that may be a good representation of the whole wavefunction in particular circumstances.

And similarly, it seems to me, for a photon, it's useful to have an idea of what a whole photon wavefunction might be like in a particular circumstance; and in respect of an interaction, what shape the whole overall superposition of point interactions (as referred to in your quote) might take, to represent eg a whole molecular orbital interacting in typical way with the entirety of a photon wavepacket, and what basic picture that may give for a complete scattering (elastic or inelastic) or emission event. That's something I'd quite appreciate some input on, at least in respect of the second part of my question.

In respect of the first part of my question, while I feel comfortable with what you've just quoted from point particle, it doesn't seem to me to address the things I suggested gave me a bit of discomfort with the bit of text from photon. Do you think it should make me feel more comfortable with that text? Thanks, Jheald (talk) 14:00, 17 March 2024 (UTC)

I think you're quite right not to feel comfortable about its wavefunction. It is possible to get pictures that represent it. But thinking a particle was actually at any of the point when it has not been observed to be there - well ... have a look at Quantum Cheshire cat and marvel at its ... well whatever. NadVolum (talk) 15:51, 18 March 2024 (UTC)

March 18

Special Relativity. Is it possible to calculate the velocity of a system, composed of two bodies, we given their different inertial masses and velocities?

An equivalent question: What's the system's (relativistic) inertial mass, we bearing in mind that the conservation of (relativistic) mass, as a general rule, does not hold in Relativity Theory.

2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 16:58, 18 March 2024 (UTC)

There is an article on the general problem Two-body problem in general relativity. Good luck with it! NadVolum (talk) 19:08, 18 March 2024 (UTC)

Oh, I forgot to point out that my question was only about Special Relativity, the given masses being the inertial ones (rather than the gravitational ones). Due to your comment, I've just added above, that I'm only interested in finding the system's velocity (or inertial mass) in Special Relativity. 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 19:34, 18 March 2024 (UTC)

Given the four-momenta of the two bodies, $p_{i}=(E_{i},\mathbf {p} _{i})=(\gamma _{i}m_{i},m_{i}\mathbf {v} _{i})$ ($i=1,2$; the last expression only for bodies with finite mass), the combined four-momentum is $p_{\mathrm {tot} }=p_{1}+p_{2}$. The mass of the two-body system is given by $m_{\mathrm {tot} }^{2}=p_{\mathrm {tot} }\cdot p_{\mathrm {tot} }=(E_{1}+E_{2})^{2}-(\mathbf {p} _{1}+\mathbf {p} _{2})^{2}$. The centre-of-mass velocity is given by $(\mathbf {p} _{1}+\mathbf {p} _{2})/m_{\mathrm {tot} }$. (I've set $c=1$; spatial (three-)momentum is denoted by $\mathbf {p}$). For photons use $E=|\mathbf {p} |=h\nu$. --Wrongfilter (talk) 09:58, 19 March 2024 (UTC)

For being more explicit, let's assume that the bodies: have masses m,n, and move at velocities u,v (respectively) on the same line. What's the system's velocity, using m,n,u,v only? 2A06:C701:427F:A900:B552:C10D:40A2:7B31 (talk) 11:13, 19 March 2024 (UTC)

Aliens: from cavemen to NASA

I have been reading many books about extraterrestrial life lately. Although there is no actual example to study, the usual approach is to study which things are required to happen or be there, which ones can be likely, which ones not so much, and the feasibility of achiving similar results in other ways (such as life being made of an element other than carbon, or with a solvent other than water). I have seen this approach used to see the chances of a given planet to turn into a Earth-like planet, and for life, once formed, to survive and evolve.

But there is a missing step: once we get full ecosystems with diverse animals, how likely would it be for some animal to evolve into a human-like intelligence? Which would be the requirements and obstacles? And if one actually does, how likely would it be to go all the way from rocks and sticks to space programs? I'm not asking to discuss that ourselves, but to point me some author that did. The article Extraterrestrial intelligence is almost devoid of content and does not help with this search. Cambalachero (talk) 19:44, 18 March 2024 (UTC)

Are you familiar with the Fermi paradox? 41.23.55.195 (talk) 05:39, 19 March 2024 (UTC)