# Why does a photon have zero rest mass?

1. Oct 5, 2009

### klij

But according to special relativity, light ALWAYS travels with the light speed c, and is NEVER at rest​

and from this statement i am supposed to gather there is no rest mass. But logic tells me just because light is never at rest, doesn't necessarily mean it would have no mass if it was at rest.

to get its regular mass i used m=hf/c^2

its seems impossible to solve for m0 by using because its undefined.

So am i missing something? or is a more correct description, the rest mass of a photon simply unknowable at this time?

2. Oct 5, 2009

### Bob_for_short

Photon equations are different from the Newton-like ones: there is no force term to accelerate a photon, etc. Its energy is calculated differently: hv. So speaking of photon mass is senseless. What we need to know is how photons are created and how they exchange the energy-momentum with matter. The theory of that does not contain any mass belonging to photon. Moreover, the relativistic relationship between the photon energy and momentum is valid only for m=0.

Last edited: Oct 5, 2009
3. Oct 5, 2009

### klij

so you are saying even the mass equation i used to get its regular mass, is incorrect and its regular mass is undefined to.
so it has no mass even when its in motion.

i just asked this question too.
quote:
the way i understand it is photons excite atoms, causing them to increase in energy. i view this increase in energy as jostling of a atom causing more movement of its proton, neutrons, electrons.

i realize a photon is a particle and a wave(in limited comprehension.)

so should i view the transfer for energy as a vibration(wave form of light) traveling then hits a medium(atom), causing a vibration as it passes through?

or should i view this as a particle passing through passing through a systems(atoms) orbit. much like a rogue planet passing through a solar system effects caused by gravity possible collision?

or neither? ​

i guess the point of these two questions i asked, is to try to make sense of how a mass less particle can interact with matter, when F=MA(but you just said that equation is useless on them)

so how do they interact?

4. Oct 5, 2009

### clem

The more correcct description is that momentum and energy of any object are related by (with c=1) $$E^2-p^2=m^2$$. This means that this combination of E and p is a Lorentz invariant which is the same in any state of motion. The invariant quantity m is then interpreted as the mass of the object. The velocity of the object is related to E and p by
v=p/E. Some algebra then leads to the formula you gave. Looking at the first equation above shows that m=0 for the photon. The equation v=p/E shows that the photon velocity is always 1.

5. Oct 5, 2009

### Staff: Mentor

The general relationship between ("rest") mass, energy and momentum is

$$E^2 = (m_0 c^2)^2 + (pc)^2$$

which works for both "ordinary" particles and for photons (for which $m_0 = 0$).

Note that even in classical electrodynamics, the energy and momentum densities in an electromagnetic wave are related by E = pc, the same as given by the equation above for photons.

6. Oct 5, 2009

### javierR

I responded to your other question before I saw this. I only mentioned there that the electron and photon "fundamentally" interact in such a way that, under appropriate circumstances, a photon and electron can meet, leaving an electron (absorption). This process is "fundamentally" the same as starting with an electron and ending up with an electron and a photon (emission by the electron). This interaction appears in quantum field theory, which is the language for our model of fundamental interactions (see below). Quantum field theory allows us to talk about other particle interactions in a similar way.

Perhaps the confusion you have arises over our first encounter with interactions between objects with mass, in which case they have momenta mv and kinetic energy 1/2 mv^2. In particular, particles would behave like little balls bouncing off each other. The short and cheap answer to your question is that classical mechanics is only an approximate description of nature in which there are little objects with mass and there are electromagnetic fields with propagating waves (light). The true underlying description of nature is quantum mechanics, which becomes clear only when you look carefully at electromagnetic radiation, atomic processes and so on. In that case, the description of interactions changes so that we're not generally talking about little balls bouncing off each other, though there are cases in which you can get away with this picture. In quantum mechanics, as discussed in a previous post here, the energy and momentum of any particle, whether massive or massless, can be expressed in terms of wave properties (frequency and wavelength) associated with the particles. Once you have particles assigned energies and momenta, there is not so much of an impediment to seeing that we should be able to talk about interactions among these particles (mass is no longer so important in this sense).
But even here, one might get confused about the fact that Schrodinger's equation has a mass in it as well. That's because we have to include special relativity to talk about particle near or at the speed of light. The end result is the framework of quantum field theory, in which particles are the smallest excitations of a "quantum field" that can carry energy and momentum. The fact that it is a quantum field means that the particle excitations naturally have wave properties associated to them. The picture of particle interactions in quantum field theory is quite different, as excitations of one type can mix with excitations of another type, rather than picturing balls bouncing off each other or colliding and sticking together. Mass is not relevant in being able to discuss such properties.

7. Oct 5, 2009

### klij

i think i see now why the rest mass must be zero in https://www.physicsforums.com/latex_images/23/2379621-0.png [Broken]

how ever i this other equation https://www.physicsforums.com/latex_images/11/1129330-0.png [Broken] you cant get to https://www.physicsforums.com/latex_images/11/1129330-7.png [Broken]

because what you do is multiply both sides by the demoninator, thus giving you X/X=1. however 0/0=is still undefined. is my math wrong in this

Last edited by a moderator: May 4, 2017
8. Oct 6, 2009

### clem

You are trying to derive something from an incorrect equation.
Where did you get your m=m_0/sqrt from?
The only correct equations come by starting with m^2=E^2-p^2.

9. Oct 6, 2009

### klij

10. Oct 6, 2009

### JustinLevy

First, let's address a slight variation of the title question: Does a photon have zero rest mass?

This alone is problematic, and is what prevents responding to the 'why?' cleanly. Because (as you already mentioned) the photon is always moving at c in an inertial frame, it is more precise to say "The photon has zero invariant mass". The "m" particle physicists use is the invariant mass. For all massive particles, this is the same as its rest mass. So often the invariant mass is also called the rest mass. We know what is meant when someone says "the rest mass of the photon is zero", but I can see how that caused confusion as it is a bit of a misuse.

Now "why?" does the photon have zero invariant mass? Well, a short answer would be that Coulomb's law would be different if the photon had mass. So one could say experiments strongly support such a statement (the current experimental invariant mass upper limit for a photon is amazingly small). A theorist might respond instead by saying the universe apparently has an unbroken U(1) gauge symmetry (am I saying that correctly?).

Now onto your question regarding an atom or similar absorbing a photon and gaining mass. Yes this does seem counter-intuitive based on Newtonian views of mass. The problem is that invariant mass is not additive. For a more concise example, a single photon has zero invariant mass, but a system of just two photons usually has non-zero invariant mass. So do not expect the total (or final) rest mass of a system to be the sum of the constituent (or reactants) rest masses.

EDIT: Also, klij, that equation you mention is using m to refer to the 'relativistic mass', which is another beast entirely, and in my opinion should be avoided. Particle physicists don't really use it, and it is becoming an archaic term like 'transverse mass' and the like. Also, relativistic mass is more precisely defined as m = E/c^2, which indeed would apply to photons as well. What you wrote is what m = E/c^2 simplifies to for a massive (non-zero invariant mass) particle.

Last edited: Oct 6, 2009
11. Oct 6, 2009

### Parlyne

Let me expand on this a little. Maxwell's theory of electromagnetism has this U(1) gauge symmetry, which shows up, for practical purposes, in the freedom to add a total time derivative of a function to the scalar potential and a gradient of the same function the the vector potential (equivalently, adding a 4-gradient to the 4-vector potential) without changing any physical observable.

If we write down a Lagrangian density for the 4-vector potential, we find that including a mass term breaks this gauge symmetry. In other words, the action (the integral of the Lagrangian density over all space and time) is not the same after making a gauge transformation.

By Noether's theorem, we know that any symmetry of nature is related to some kind of conservation law. In the case of the EM gauge symmetry, it's the conservation of electric charge. If the gauge symmetry were broken, electric charge would not be conserved. Since, we've never seen any evidence of the non-conservation of electric charge, we can put extremely strong limits on how much breaking of EM gauge symmetry there can be which, in turn, puts very stringent upper limits on the photon mass.

12. Oct 7, 2009

### jnorman

to be even more simplistic, what if we consider the increase in mass of an atom when it absorbs a photon? the mass (total energy of the system) of the atom is increased when it absorbs a photon, so could you not indicate that difference as the "rest mass" of the photon?

13. Oct 7, 2009

### Count Iblis

This argument is flawed. A massive photon can be described by Proca theory and that theory is invariant under appropriately modified gauge transformations. You get a conserved current in that case as well. From the fact that electric charge is conserved you do not get any limits on the photon mass at all.

Another argument why the photon mass has to be exactly zero was debunked by Bohr in the 1920s. It was argued at that time that since a massive photon has 3 polarization states a black body should radiate much more energy at the same temperature than it actually does. But Bohr pointed out that if you let the photon mass go to zero, the longitudinal component does not interact with matter anymor0, so this does not contribute to black body radiaton.

Also, recently the very strong limits on the photon mass posed by the galactic vector potential were debunked, http://arxiv.org/abs/hep-ph/0306245" [Broken]

I also remember my GR professor in college making the erroneous claim that the GR prediction of the deflection of light by a massive body and its experimental confirmation proves that the photon is exactly massless. He claimed that particles with a mass m>0 in the limit of m to zero would be deflected by an angle that differs by a factor 2 compared to the massless case. But he later retracted his claim. It turns out that there is only a factor of 2 difference if you compare extremely relativistic particles to nonrelativistic particles.

So, I think that the massive photon has been a victim of a lot of Straw Man attacks. :grumpy:

Last edited by a moderator: May 4, 2017
14. Oct 8, 2009

### Haelfix

"A massive photon can be described by Proca theory and that theory is invariant under appropriately modified gauge transformations."

It is not incorrect to say that the Proca mass term for the photon breaks local gauge invariance. I realize that people have played with this a lot over the years, but it is tightly constrained by experiment and the PDG has an empirical upper bound on the mass.

15. Oct 8, 2009

### Count Iblis

The mass term is invariant under gauge transformations, the longitudinal component of the photon enters in the mass term and takes care of that.

The PDG limits are based on the galactic vector potential and are model dependent has pointed out here:

http://arxiv.org/abs/hep-ph/0306245

So, the photon mass could be as large as 10^(-14) eV