21

I know this may sound weird. But could a photon be given mass? And if so what would the effects be?

Qmechanic
  • 201,751
LostPecti
  • 402
  • 8
    Possible duplicates: https://physics.stackexchange.com/q/4700/2451 , https://physics.stackexchange.com/q/31994/2451 , https://physics.stackexchange.com/q/191451/2451 and links therein. – Qmechanic Sep 22 '16 at 11:51

5 Answers5

28

A "photon" propagating in an optical medium can be ascribed a nonzero rest mass. I put "photon" in quotes because light in a medium is not, strictly speaking, pure photons but a quantum superposition of excited electromagnetic field and matter quantum states.

In a medium with refractive index $n=1.5$, my calculation here reckons the rest mass of a quantum of this superposition to be:

$$m_0 = \frac{E}{c^2}\sqrt{1-\frac{1}{n^2}}$$

For $n=1.5$ (common glasses like window panes or N-BK7 - microscope slide glass) at $\lambda = 500\rm\,nm$, we get, from $E=h\,c/\lambda$, $m_0=3.3\times 10^{-36}{\rm kg}$ or about 3.6 millionths of an electron mass.

This take on your question is perhaps a little different from what your original question asks for, and is not to be confused with the assignment of a rest mass to the "pure photon" as discussed in the other answers. The assignment of nonzero rest mass to the photon replaces Maxwell's equations with the Proca equation, whose most "startling" characteristic is screening, i.e. photon fields would dwindle exponentially with distance from their sources and light as we know it could not propagate through the universe.

Community
  • 1
Selene Routley
  • 88,112
  • 1
    I hope this is not a stupid question but... is this effect measurable? For example would a matter through which the "photons" pass change it's total mass with the amount of electrons passing? My naive approach was to look for most powerful lasers and compare it to the $c^2$, so given "1.3 PW ($1.3×10^{15}$ W) – world's most powerful laser as of 1998", if one would pass it through the matter that would sustain such energy density, a there should be a few to tens of grams difference in mass, is this right? – luk32 Sep 22 '16 at 11:33
  • Related: https://physics.stackexchange.com/q/1898/2451 and links therein. – Qmechanic Sep 22 '16 at 11:49
  • 1
    @luk32 If that 1.3 PW laser beam is continuous, and the chunk of perfectly transparent matter is 1 light-nanosecond thick (just under 30cm), you'd be increasing its energy by 1.3 MJ, and that would increase its mass by approximately 14.46 nanograms. – PM 2Ring Sep 22 '16 at 12:07
  • 1
    Is this exponential decay (screening) related to classical attenuation inside materials? Is the "screening parameter" the same as the "attenuation coefficient"? – Real Sep 22 '16 at 14:04
  • @Real No. It is non-dissipative. It doesn't really have a classical analogy - a massless field propagates and is always on the move at speed $c$, whereas the fields of influence of a massive particle can "stay put", with the indefinitely lingering (i.e. non dissipating) fields of influence dwindling exponentially from the particle's COM. In this respect, the effect is somewhat analogous to the nonmoving stores of energy that accompany evanescent waves. – Selene Routley Sep 22 '16 at 22:03
  • @WetSavannaAnimalakaRodVance That's fascinating. But intutively it doesn't seem like the photon should be able to "stay put" for very long -- would it in practice scatter after a while? (and perhaps end up propagating as a thermal wave?) – Real Sep 23 '16 at 03:11
  • @Real I think one good way to get your head around this idea is to look at the dispersion relationship from the Klein-Gordon equation and how it behaves as rest mass increases from nought to a significant nonzero value. You have $\omega = \sqrt{k^2+m^2}$ in natural units; the nonzero $m$ allows low frequency wavepacket to "stay put" with $\mathrm{d}\omega/\mathrm{d}k\approx 0$ for $k\ll m$ in contrast with the case with $m=0$, where one gets $\mathrm{d}\omega/\mathrm{d}k=1$ always – Selene Routley Sep 23 '16 at 03:20
  • @WetSavannaAnimalakaRodVance Ah ok, I'll have to study, out of my expertise. Just seems strange that waves can be "confined" indefinitely (this takes thermodynamics into account, right?) Would a better analogy be of a potential confining the waves, sort of like an electron with not enough energy would never escape a hydrogen atom? – Real Sep 23 '16 at 03:27
8

As far as I understand, if you gave a photon mass, the immediate consequence would be that photons wouldn't travel at speed $c$, and instead we would need to consider a new limiting speed for light, less than $c$. Even if your question seems a bit odd, it is pretty interesting, since the consideration of massive photons could give an explanation to dark energy.

gradStudent
  • 276
  • So the photon could be given mass but would lose speed? So if I gave it a small amount if mass, it would still effect the speed significantly? – LostPecti Sep 22 '16 at 07:02
  • 7
    You're right that a massive photon wouldn't travel at speed $c$, but depending on how exactly the photon acquired a mass, it might still be that the limiting speed for the photon (just as for other massive objects) remains $c$. – gj255 Sep 22 '16 at 08:22
  • @J..., you are right. However, it's not necessary that the limiting speed of the massive photon be strictly less than $c$, as suggested in this answer. – gj255 Sep 22 '16 at 14:32
  • @gj255 Indeed - I misread the answer, I see what you meant now. – J... Sep 22 '16 at 14:34
6

If the photon mass arises from a Higgs mechanism then the photon mass can be switched on and off using an external magnetic field, as explained in this article. Above a critical magnetic field strength the Higgs vacuum expectation value will vanish and the photon becomes massless.

Count Iblis
  • 10,114
6

Define mass. If you mean "rest mass", then the answer is no, the rest mass of a photon is zero, otherwise it couldn't propagate at light speed.

If you however mean "gravitational mass", then the answer is: It already has "mass" due to the infamous mass-energy equivalence $E=mc^2$ since a photon's energy equals $E=hf$ (where $h$ is Plack's constant and $f$ is the photon frequency). So a photon's gravitational mass is

$$m = \frac{hf}{c^2}.$$

But as the other answers already point out, the frequency and therefore "mass" depends on the reference frame! Also I am not aware of any experiments that had enough energy in a photon field to actually generate an observable gravitational influence...

Tobias Kienzler
  • 6,768
  • 3
  • 41
  • 60
  • https://physics.stackexchange.com/q/248616/ – UKH Sep 22 '16 at 14:34
  • @Unnikrishnan Can you clarify your point? If you follow to the dupe, you'll find http://physics.stackexchange.com/a/34356/97 "gravity affects anything with energy" (and of course vice versa) – Tobias Kienzler Sep 22 '16 at 17:58
  • But the answer to the above link (my own question) says that it is not right to tell that light has a mass of $h\nu/c^2$ – UKH Sep 23 '16 at 00:32
  • @Unnikrishnan That's oversimplified. It is just as "not right" as claiming a high velocity particle has "mass" $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ - the modification to the mass of non-relativist physics formulae to still "work", but yes, in reality rather "wrong". – Tobias Kienzler Sep 23 '16 at 05:48
  • Okay I understand where the fault creep in. Thank-you for sharing your time. – UKH Sep 23 '16 at 07:55
3

If you give mass to a photon, one major consequence would be the loss of gauge invariance.

The Lagrangian for electromagnetism, $$\mathcal{L}\ =\ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ is invariant under the gauge transformation $$A_\mu\ \rightarrow\ \ A_\mu \ +\ \partial_\mu \Lambda $$ But, if you add a mass term, $m^2A_\mu A^\mu$ to the Lagrangian, it won't respect this symmetry anymore.

Sucheta
  • 437
  • 3
  • 10
  • 5
    It is wrong to say that a photon mass will break gauge invariance. There are gauge invariant ways in which photon can acquire mass. For example Higgs mechanism, Chern simmons theory in 3d, 2d QED etc. – Tuhin Subhra Mukherjee Sep 22 '16 at 07:13
  • Yeah, you are right @Tuhin Subhra Mukherjee. I just considered the simplest case. In fact, there are theories for massive photons as well. But, I am not an expert. Thanks for pointing this out! – Sucheta Sep 22 '16 at 08:14