Photon rest mass = 0 wasn't proven experimentally?

[Nick666]

Wasnt it ?

So I stumbled upon this fragment from wikipedia's photon page

"Current commonly accepted physical theories imply or assume the photon to be strictly massless, but this should be also checked experimentally. If the photon is not a strictly massless particle, it would not move at the exact speed of light in vacuum, c "

Of course I was a bit surprised, so I googled and found this https://www.princeton.edu/~romalis/PHYS312/Coulomb Ref/TuCoulomb.pdf

"Up to now, no experiment has
proved the photon rest mass to be nonzero. However, an
experiment that fails to find a finite photon mass does not
prove definitely that the mass is zero. The limits on the photon
mass have approached ever more closely the ultimate limit
determined by the uncertainty principle. So, nobody can assert
that the next experiment will not reveal evidence of a definite,
nonzero mass"

Can someone please explain this more a bit ?

 

[bhobba]

Every measurement has some uncertainly so all you can do is set an upper bound on a photons mass - you can't ever prove its zero.

BTW the uncertainty principle has nothing to do with it. There is nothing in the uncertainly principle that says you can't measure anything with 100% accuracy. Its a technological limitation in that showing something was exactly zero would require an infinite number of digits after the decimal point.

Thanks
Bill

 

[mathman]

Indirect evidence? The speed of light is the same in any inertial system. Doesn't special relativity require this to be true only for massless particles?

 

[Jazzdude]

Every measurement has some uncertainly so all you can do is set an upper bound on a photons mass - you can't ever prove its zero.

That's not generally true. Sometimes only an exact equality allows for certain symmetries that you can test. See neutrino oscillations as an example.

 

[jtbell]

What exact equality leads to neutrino oscillations?

 

[eno31krad]

I was thinking about this aswell- shouldn't a photon always have relativistic mass from E=Mc2? Kinetic energy aside, how can any particle have 0 energy? I always thought of this as analogous to fully cooling something to absolute zero.

 

[TeethWhitener]

shouldn't a photon always have relativistic mass from E=Mc2?

$E=m c^2$ is the first term of the Taylor series expansion of the relativistic energy-momentum relation $E=\sqrt{p^2 c^2 + m^2 c^4}$. The proper equation for a massless particle, derived from the full energy-momentum expression, is therefore $E=pc$.

 

[bhobba]

That's not generally true. Sometimes only an exact equality allows for certain symmetries that you can test. See neutrino oscillations as an example.

Sorry - don't agree.

There is no way you can prove the photons mass is exactly zero - you can only set an upper bound eg
http://usatoday30.usatoday.com/weather/science/wonderquest/photonmass.htm

If you think its possible to measure it as exactly zero I would like to know the exact experiment.

Thanks
Bill

 

[bhobba]

Indirect evidence? The speed of light is the same in any inertial system. Doesn't special relativity require this to be true only for massless particles?

That a velocity exists that is invariant between inertial frames follows from space-time symmetries.

If the photon travels at that invariant speed is an experimental matter.

Gauge invariance, which is the rock bottom essence of QED says it should - but if it really does is another matter.

Thanks
Bill

 

[bhobba]

I was thinking about this aswell- shouldn't a photon always have relativistic mass from E=Mc2?

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

Relativistic mass is an outmoded idea:
http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

You can see its issues if you apply a force to a particle in its direction of motion or at 90% to it.

Thanks
Bill

 

[bhobba]

I always thought of this as analogous to fully cooling something to absolute zero.

It can't - but you are getting confused with what E=MC^2 says. It says mass is a form of energy, not that energy is a form of mass, any more than electrical energy is a form of chemical energy - they are both energy - but different.

Thanks
Bill

 

[vanhees71]

Gauge invariance, which is the rock bottom essence of QED says it should - but if it really does is another matter.

Thanks
Bill

No! There's a loop hole in this argument. The gauge symmetry of QED is an Abelian U(1) symmetry, and there you can have a gauge symmetry with a massive vector boson without invoking the Higgs formalism, i.e., you can well have a theory with electrons, positrons, and massive photons without destroying any of the goodies of standard QED, where the photon mass is assumed to be 0. It can stay even renormalizable in the narrow Dyson sense as QED!

Don't get me wrong: Today, there's no evidence for the mass of the photon not being 0, but as was stressed already in many postings before, whether it's precisely 0 cannot be told from any known symmetry in our models (here the Standard Model of elementary particle physics). So it's a question of measuring the photon mass. In any case, if the photon mass is really not 0, then it's very very tiny. The current upper bound according to the particle data group's Review of Particle physics (2014), it's $m_{\gamma}<10^{-18} \text{eV}$.

That the photon is really described by an Abelian gauge symmetry is also pretty well established (but again of course not proven) from experiment: If the electromagnetic interaction was part of a larger non-Abelian group, the photon must carry a charge of this group. At least the measured electric charge of the photon is also tiny, namely $q_{\gamma}<10^{-35} e$, where $e$ is the charge of a proton.

 

[Nick666]

But if the mass aint exactly zero, that means that they don't move at exactly c but a little slower, in my mind that means that they would have had to be accelerated ... right ?

What energy would be required to accelerate say 10^90 photons with a mass of 10^-70 grams to the speed of c-[(10^-70) *c] ?

 

[vanhees71]

Have fun in calculating this yourself. It's very instructive to do so, showing that sometimes you cannot get a result by typing numbers into a pocket calculator (or any computer with the usual accuracy)!

 

[Nick666]

Yes but what equation do I use ? This ? http://i.imgur.com/YV648.jpg

 

[vanhees71]

Yes!