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How do we know that a photon is massless?

  1. Dec 2, 2013 #1
    How do we know a "photon" is massless?
    Do we have any experimental proof?
    Photon has momentum, it even bends in gravity, doesn't that mean photon has mass?
    A radioactive material will go on emitting radiations until it balances, a way in which it reduces it's mass.
    So, isn't photon the tiniest possible mass in universe?
     
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  3. Dec 2, 2013 #2

    PeterDonis

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    We can't prove that a photon is exactly massless, because there is always some uncertainty in any experiment. But experiments have shown that if the photon does have mass, its mass is too small for us to detect. See this article from the Usenet Physics FAQ:

    http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

    No, it means the photon has energy. There is a PF FAQ on this:

    https://www.physicsforums.com/showthread.php?t=511173

    The term "relativistic mass" is sometimes used as a synonym for energy, but when people say photons are massless, they mean photons have zero invariant mass (some call it "rest mass", but this is a bad term for photons since they can never be at rest), not zero relativistic mass.

    I'm not sure what you mean by "balances": radioactive atoms go on emitting particles of some kind until they have changed into non-radioactive atoms. This process does reduce the mass of the atom, but that's not really why the atoms are radioactive; most radioactive decay chains stop at atoms which are still very massive on the atomic scale (typically lead or somewhere near that on the periodic table), so it can't be that the atoms are trying to lose as much mass as they can. (The decay chains that stop at lighter atoms *start* with lighter atoms, and usually are just one or two steps of emitting beta particles, which involve virtually no change in the mass of the nucleus.)

    Also, radioactive atoms don't just emit photons: they can also emit alpha particles (helium nuclei) and beta particles (electrons).

    Different photons can have different energies; as far as we know, there is no lower limit on the energy a photon can have (which is actually another way of saying that as far as we can tell, photons have zero invariant mass). The same is true for any other particle with zero invariant mass: in the Standard Model of particle physics, the only other such particles are gluons (the particles that mediate the strong nuclear interaction). It is also believed that gravitons, the quantum particles corresponding to gravity, are massless, but we have no way of testing that experimentally.
     
  4. Dec 6, 2013 #3

    Meir Achuz

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    There are many experiments involving decays that emit photons, and the scattering of photons, that all show the photon has zero mass (within the limits of the experiments).
    Also, precision tests of Gauss's law show the photon mass to be zero.
    The verification that photon speed (speed of light) is independent of its energy or frequency also shows that the photon mass is zero.
     
    Last edited: Dec 6, 2013
  5. Dec 6, 2013 #4
    Plus a photon mass would break the electromagnetic gauge symmetry leading to charge conservation violation which has never been observed.
     
  6. Dec 6, 2013 #5
    Place an atom in a metastable state with half life N seconds. Without disturbing the system, 10N seconds later, where apparently the probability amplitude of a radiate wave is large, the mass of the system is unchanged.
     
  7. Dec 6, 2013 #6

    PeterDonis

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    Only it won't be if a photon is emitted: photons have energy. "Mass" in this thread refers to invariant mass, not relativistic mass, i.e., energy.

    What will actually happen in your scenario is that the system will be in a superposition of two states with different energies: the "no emission" state with the original energy, and the "emitted a photon" state with a lower energy (the difference being the energy of the emitted photon). The electromagnetic field will be in a corresponding superposition of a photon present at the emitted frequency, and no photon present at that frequency. As long as no measurement is made to determine whether a photon was emitted, the superposition will continue, with amplitudes for each term evolving according to the appropriate Hamiltonian.

    Note, btw, that measuring the mass of the system counts as a measurement that determines whether a photon was emitted: you will get either the original mass (indicating no emission) or a smaller mass (the original mass minus the energy of the emitted photon), with probabilities determined by the relative amplitudes in the above superposition.
     
  8. Dec 6, 2013 #7
    I will list here some similar observations with apparently contradictory 'conclusions' and then I will show you how these 'contradictions' are resolved.

    1a) A hot brick has more rest mass than a cold brick. As the bricks cools it radiates photons and loses rest mass, yet the photons only carry away energy.
    This suggests energy and momentum contribute to the rest mass.

    2a) The rest mass of brick is independent of its velocity relative to the observer.
    This suggests that momentum does not contribute to the rest mass and appears to contradict (1a).

    3a) A box containing a hot gas with the molecules bouncing around inside the box at high speed has more rest mass than the same box with the same molecules after it has cooled down and the molecules are bouncing around at lower speeds.
    This suggests the momentum of the molecules contributes to the rest mass of the system. This appears to contradict (2a).

    4a) A single atom emits a single photon. The rest mass of the atom is unchanged.
    This appears to contradict (1a).

    All the above 'contradictions' are resolved by noting that the rest mass of a system is the total of all the individual rest masses of the particles plus the total of all the individual momentum energies of the particles, as measured in the zero momentum reference frame. So let's go through the list again.

    1b) A hot brick has more rest mass than a cold brick. As the bricks cools it radiates photons and loses rest mass, yet the photons only carry away energy.
    A single photon has no zero momentum reference frame and no rest mass. However, a group of photons moving in opposite directions has a definable zero momentum reference frame and so as a group they have a definable rest mass. The total rest mass of the system (brick plus photons) does not change, but since the photons as group have rest mass, the brick now has less rest mass than before the photons were emitted.

    2b) The rest mass of brick is independent of its velocity relative to the observer.
    This is because the rest mass of the brick is defined in its zero momentum reference frame.

    3b) A box containing a hot gas with the molecules bouncing around inside the box at high speed has more rest mass than the same box with the same molecules after it has cooled down and the molecules are bouncing around at lower speeds.
    Since a zero zero momentum reference frame can be defined for the system of moving molecules, their individual momentum energies contribute to the rest mass of the system (box + molecules).

    4b) A single atom emits a single photon. The rest mass of the atom is unchanged.
    The total rest mass of the system remains unchanged before and after the emission of the photon. Since the single photon has no rest mass, the rest mass of the atom must remain unchanged. It does however gain momentum relative to its original rest frame.

    None of the above prove that a photon has zero rest mass, but hopefully they demonstrate that a photon having non zero momentum and zero rest mass does not cause any contradictions.
     
  9. Dec 6, 2013 #8

    PeterDonis

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    This is not correct; the rest mass of the atom is reduced. Work out the math:

    In the original rest frame of the atom, we start out with an atom of mass ##m##; we end with an atom of mass ##m'## moving to the left at velocity ##v##, and a photon with momentum (and energy--we're using units where ##c = 1##) ##p## moving to the right. Conservation of momentum gives ##\gamma m' v = p## and conservation of energy gives ##m = \gamma m' + p##. Substituting the first equation into the second, we get ##m = m' \gamma (1 + v)##, which makes it clear that ##m > m'##.

    Note that the total rest mass of the system does remain unchanged; it's still ##m## (since the original rest frame of the atom is the zero momentum frame). This example illustrates the counterintuitive fact that rest mass is not additive.
     
    Last edited: Dec 6, 2013
  10. Dec 7, 2013 #9
    Yes of course. My bad! The momentum of the photon contributes to the total rest mass of the system, so the rest mass of the atom must be less after the emission of the photon.
    Agree.
     
    Last edited: Dec 7, 2013
  11. Dec 8, 2013 #10
    You miss the point. You probably think there are spatial points having various physical properties. I say your 'photon' has no velocity associated momentum until it is measured. So who is right? It doesn't matter. No one has made sense of it.

    In the mean time, before this historic event heralding new physics, you must admit that the electromagnetic field could have zero momentum. The common perception is a tiny little thing that has objective existence dashing out in some objective direction. I think this is nonsense--my opinion. However,given that the actual nature is unknown, your particular claim is not validated.
     
    Last edited: Dec 8, 2013
  12. Dec 8, 2013 #11

    PeterDonis

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    I'm not sure how you are getting this from what I said; in fact I'm not even sure what you think it means.

    That's not what quantum mechanics says. Quantum mechanics says what I said: until there is an actual measurement, the electromagnetic field is in a superposition of a state with a photon (and nonzero momentum) and a state without a photon (and zero momentum). Since the EM field's state is correlated with the state of the atom, the atom must also be in a superposition of a state with the original mass and a state with mass reduced by the energy of the emitted photon. That is not consistent with your statement that "the mass is unchanged"; the latter statement requires that the atom is in a mass eigenstate.

    I'm not sure what this means either.

    What new physics? Standard quantum mechanics already makes correct predictions about this scenario.

    Sure, standard quantum mechanics agrees with this: the EM field has a momentum eigenstate with eigenvalue zero, the state with zero photon number. But in your scenario, if there has not been a measurement, the EM field is not in this state: it's not in *any* momentum eigenstate. It's in a superposition of momentum states.

    This may be the "common perception", but it's not what standard quantum mechanics says, and it wasn't the basis for my claims, so I'm not sure why you bother bringing it up. Indeed, it seems that you are making a similar mistake about the EM field momentum and the atom's mass: see below.

    If "the actual nature is unknown", then your claim that "the electromagnetic field could have zero momentum" is not "validated" either, since it assumes that we at least know that "the actual nature" is an electromagnetic field that has a property called "momentum". But if we allow that assumption, then it is your claim, not mine, that is questionable, because your claim assumes that the momentum of the EM field must always have a definite value (and the same for the mass of the atom): in other words, you assume that we have a tiny little thing with a definite mass and a field filling all space with a definite momentum. That is just as much an erroneous "common perception" as the idea that the photon is a tiny little thing dashing out in some objective direction.
     
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