Do photons, phonons and electrons have mass?

  • #51
vanhees71
Science Advisor
Insights Author
Gold Member
17,820
8,783
Wasn't the original question, whether photons, phonons, and electrons have mass?

Well, I tried to answer that in #46 assuming that the OP is familiar with what these (quasi-)particles are and that we are taking about QFT (relativistic for photons of course, but for quasiparticles like photons it also makes sense for non-relativistic QFT).
 
  • #52
Demystifier
Science Advisor
Insights Author
Gold Member
11,761
4,193
These articles have energy but do they have mass?
If you print them then they obviously have, but if you read them as pdf's on a computer screen, then it's a very nontrivial physical question. :wink: :oldlaugh:
 
  • #53
TeethWhitener
Science Advisor
Gold Member
1,980
1,415
(NB—Thread had no prefix, so possibly answering this at the wrong expertise level.)

Some phonons (specifically optical phonons) have a non-zero energy at the center of the Brillouin zone (at zero momentum), so I suppose you could call this the “rest energy” of those phonons—and associate it with a rest mass. I’m not sure what that gains you, though.
 
  • #54
8
2
Electrons don't come at rest. Electron rest mass is the mass of an electron as measured when its speed is zero relative to an observer. A photon never comes at rest thus its rest mass is 0. But why can't be the rest mass of a photon be measured when its speed is zero relative to the observer?
Photons have zero mass and are not at rest in any reference frame. If you are trying to determine how photons are affected by gravity, the equivalent property for mass would be ## \frac {hf} {c^2}## .
 
  • #55
vanhees71
Science Advisor
Insights Author
Gold Member
17,820
8,783
And this leads to wrong conclusions. Gravity should be treated within GR, and a naive photon model, i.e., treating the em. wave as a massless point particle has its justification in the eikonal approximation of the Maxwell equation in a spacetime background. E.g., you get the famous deflection of light on the Sun, which lead to Einstein's fame in the public in 1919, when Eddington et al confirmed the prediction of this deflection from GR. The naive idea to just take ##hf/c^2## as the mass of the photon in a Newtonian gravitational field leads to only half the value, but of course the value from GR turned out to be right.

The reason, why for "photons" the Newtonian approximation gets wrong by a factor 2 precisely is the fact that it is always a relativistic object. That's why the Newtonian solution for the Kepler motion in a Newtonian (weak) gravitational field is correct for slow massive objects as testparticles in a gravitational field is a (very) good approximation to the full general-relativistic solution but not for relativistic "objects" like photons.
 
  • #56
8
2
Not always i.e. inside glass they are moving slower,and inside heavy water of a nuclear reactor very slow.
That's because the photon is being absorbed and re-emitted by the atoms it runs into, however in-between the atoms, it travels a c.
 
  • Skeptical
Likes weirdoguy and Motore
  • #57
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
26,971
10,791
Why do people jump on to old threads only to post incorrect things? Especially when this was put to rest on the previous page?

If this were true, the refractive indices in gasses would depend on pressure, but not temperature. They depend on both. If this were true, the refractive index of graphite would be much lower than diamond (because of densities). In fact, it's higher.
 
  • #58
8
2
And this leads to wrong conclusions. Gravity should be treated within GR, and a naive photon model, i.e., treating the em. wave as a massless point particle has its justification in the eikonal approximation of the Maxwell equation in a spacetime background. E.g., you get the famous deflection of light on the Sun, which lead to Einstein's fame in the public in 1919, when Eddington et al confirmed the prediction of this deflection from GR. The naive idea to just take ##hf/c^2## as the mass of the photon in a Newtonian gravitational field leads to only half the value, but of course the value from GR turned out to be right.

The reason, why for "photons" the Newtonian approximation gets wrong by a factor 2 precisely is the fact that it is always a relativistic object. That's why the Newtonian solution for the Kepler motion in a Newtonian (weak) gravitational field is correct for slow massive objects as testparticles in a gravitational field is a (very) good approximation to the full general-relativistic solution but not for relativistic "objects" like photons.
There is really nothing Newtonian about light; in fact, it is the basis of relativity (all of us are moving at the speed of light, just not through space as much as through time as illustrated by a space-time light cone diagram).

Many physicists define m as strictly rest mass, but for undergrad physics students, m is taught as representing effective mass for simplicity as relativity is a new concept for most of the students. Defining m as the rest mass, for light, m=0.

It's probably more accurate to think of massless particles in terms of momentum. BTW, the famous equation E=mc2 is actually not the complete equation. The complete equation is E2=(pc)2 + (mc2)2 . For a photon, the rest mass is zero so this equation reduces to E2=(pc)2 + 0 or simply E=pc where p=h/λ. Basically for photons, it's not the mass; it's the momentum.
 
  • #59
vanhees71
Science Advisor
Insights Author
Gold Member
17,820
8,783
Where does this strange claim about "we'd all move with the speed of light" come from? I guess, it's paraphrasing the mathematical identity that the four-velocity of a particle obeys always ##u_{\mu} u^{\mu} = c^2##, but it is misleading to say everything moves with the speed of light.

It's not many physicists but all physicists that define ##m## as strictly the invariant mass. For massive particles it's the rest mass. For photons the mass is 0 and thus never a rest mass, because a photon moves always with the speed ##c## (in the vacuum). Of course photons are no massless particles in a literal sense. They cannot even localized, because there's no position operator for them in the strict sense. That's why I said that the naive photon picture as used in GR has to be understood in the sense of the iconal approximation of the Maxwell equations (see, e.g., Landau&Lifshitz vol. 2).

I think at no level of teaching one should tell students of the 21st century about several different sorts of "relativistic masses". Einstein comitted this sin only in the very beginning of relativity around 1905 but then stated one should use mass only as the invariant mass.
 
  • #60
PeterDonis
Mentor
Insights Author
2020 Award
34,303
12,548
it is the basis of relativity (all of us are moving at the speed of light

This is not the "basis of relativity". It is a particular viewpoint which, while it can be said to have some basis in the actual math (see below), is never found in actual textbooks or peer-reviewed papers, but only in pop science books, articles, and TV shows.

Even the basis in the actual math is questionable. As @vanhees71 notes, one can observe that the invariant length of the 4-velocity ##u^\mu## of an object with nonzero rest mass is ##\sqrt{u^\mu u_\mu} = c##, and the square root of this could be thought of as the "speed through spacetime" of the object. However, the main point of doing this, according to the pop science sources where this viewpoint appears, is so that we can say that, as the velocity of the object relative to some chosen frame increases, more of the object's "speed through spacetime" becomes "speed through space" instead of "speed through time". And the limiting case of this is claimed to be a light ray, which moves at ##c## "through space" and therefore doesn't move "through time" at all.

However, this nice-seeming viewpoint conceals a critical flaw: the invariant length of a light ray's 4-momentum is zero, not ##c##. There is no continuity between the case of nonzero rest mass (4-velocity with invariant length ##c##) and zero rest mass (4-momentum with invariant length ##0##) of the kind that is claimed by this viewpoint.

There have been plenty of previous PF threads on this (often prompted by one of Brian Greene's TV specials where he pushes this viewpoint), but it's been a while since the last one.
 
  • Like
  • Informative
Likes etotheipi and vanhees71
  • #61
vanhees71
Science Advisor
Insights Author
Gold Member
17,820
8,783
That's a very important point. As the analysis of the representation theory of the proper orthochronous Poincare group in the context of relativistic QFT reveals, the massless case is special, and the limit "##m \rightarrow 0##" is anything but trivial. That's also the mathematical reason for the fact that one must not think about photons as pointlike objects traveling with the speed of light wrt. any (inertial) reference frame.

The correct semiclassical point of view of the "photon", as usually treated in GR textbooks, is that this is in fact the eikonal approximation of Maxwellian electrodynamics. It describes the behaviour of wave vectors in the sense of geometric optics. The point-particle-photon picture can sometimes be a shortcut in deriving interesting things about em.-wave propagation (e.g., in the GR context the gravitational bending of light) but it must not be mistaken as a point-particle interpretation of photons. This was an erroneous point of view in the early days of the "old quantum theory", which is out of date for at least 97 years!
 

Related Threads on Do photons, phonons and electrons have mass?

  • Last Post
Replies
22
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
13
Views
7K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
11
Views
3K
Replies
5
Views
906
Replies
11
Views
6K
  • Last Post
2
Replies
34
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
8K
Top