Do Photons Have Mass and Why Does it Matter?

[Evenus1]

hi I was wondering if some one could give me a simpler explanation as to why photons have zero rest mass and of any subject areas I should read up on to better understand this. can we please bear in mind I'm just in 15 so do not have a university grade knowledge. but this is an are of interest that I wish to explore farther.
many thanks
Ewen

 

[BvU]

Hello Ewen,

Difficult to answer without bringing in some Physics (with a capital P). In particular: special relativity. Lots of introductions to that, so look around.
One of the 'results' from special relativity is the Einstein $E = mc^2$ formula. The $m$ in there is not the rest mass $m_0$, but $$m = \gamma m_0$$ with $$ \gamma = {1\over \sqrt{1 - v^2/c^2}} $$ and so this breaks down for something that moves at speed $v = c$, unless $m_0 = 0$.

'A particle with non-zero rest mass that moves with the speed of light would swallow up all the energy in the unverse'

PS

a simpler explanation

simpler than what ?

 

[mathman]

Why anything is almost impossible to answer. As others have answered, the rest mass of a photon is zero because special relativity forces it to be.

 

[ellipsis]

'why' - You can't explain small things with big things, just like you can't explain addition in terms of multiplication (but you can vice-versa). 'Why' is one of those big things.

As far as figuring out facts, though: We recognize patterns (equations) and extrapolate them to the situations we can't directly observe. A lot of this stuff in modern physics is way too big or too small to observe directly - instead we work with things we can play with, figure out models for them, and see what those models output in certain other circumstances.

It turns out the only way you can avoid a division by zero is if the rest mass of a photon is zero. It's fun to think, that's how the math talks to us.

 

[vanhees71]

That the photon is massless is an empirical fact. To be precise, the upper limit of the empirically determined photon mass is very small:

http://pdglive.lbl.gov/Particle.action?node=S000

 

[A. Neumaier]

the upper limit of the empirically determined photon mass is very small:

How can the photon mass be nonzero? It would imply that light moves with less than the speed of light, which sounds contradictory. (At least on the level of ''basic'' questions, such as #1.)

 

[jerromyjon]

How can the photon mass be nonzero? It would imply that light moves with less than the speed of light, which sounds contradictory.

Doesn't sit well with me either... would this be a consistent mass irrelevant of the energy of the photon or would a photon of higher energy have a higher but still negligible mass?

 

[ellipsis]

Doesn't sit well with me either... would this be a consistent mass irrelevant of the energy of the photon or would a photon of higher energy have a higher but still negligible mass?

By "the empirically established upper limit is very small" only means we've proven it is less than a certain size. This says nothing about our established empirical lower limit, which is zero. He is saying:

"Everything else is on the scale of 1, 2, 3. We have proven a photon's mass is less than 0.00000000000000000000001. Based on how the equations work out, we think the mass is in fact zero."

 

[jerromyjon]

we think the mass is in fact zero.

Well energy does have mass which is why I ask if there any proven theories relating the energy of a photon as a real massive effect or if the light speed "outruns" its gravitational effect.

When I think about it sensibly photons travel along side each other for billions of years without "attracting" each other, which I think pretty much rules out "mass" in the gravitational attraction sense.

 

[sheaf]

Some work has been done on these issues: https://en.wikipedia.org/wiki/Bonnor_beam
If the photons are traveling parallel but in opposite directions, they allegedly attract! Of course these calculations are presumably restricted to classical light, so the use of the term "photons" in this context needs some caveating!

 

[vanhees71]

How can the photon mass be nonzero? It would imply that light moves with less than the speed of light, which sounds contradictory. (At least on the level of ''basic'' questions, such as #1.)

It just means that light moves with a speed less than the limiting speed of Minkowski space. There is no fundamental principle in the Standard model that forbids a non-zero photon mass. You even have an Abelian renormalizable gauge theory with a massive gauge boson without Higgs mechanism (Stückelberg model). Only for non-Abelian gauge theories you need the Higgs mechanism to have massive gauge bosons without violating unitarity and causality.

 

[Nugatory]

When I think about it sensibly photons travel along side each other for billions of years without "attracting" each other, which I think pretty much rules out "mass" in the gravitational attraction sense.

Electromagnetic radiation is a form of energy, so it does contribute to the stress energy tensor and have gravitational effects. For example, if I had a perfectly mirrored box such that light could bounce around in it forever, that box would weigh more and produce a stronger gravitational field when it was "full of light" then when it was not.

However, photons are not little objects that can "travel along side each other", so you can't just assign them a mass and a position and expect that they'll gravitate as if that mass was at that position. They don't.

 

[A. Neumaier]

Abelian renormalizable gauge theory with a massive gauge boson

I think you meant ''Abelian renormalizable quantum field theory with a massive vector boson''. A theory without massless vector bosons is never a gauge theory.

 

[Nugatory]

You even have an Abelian renormalizable gauge theory...

I think you meant ''Abelian renormalizable quantum field theory...

And I think neither of you noticed the "B" tag on this thread :)
It's always good to remind people that the easy math-free answers are just a sad and pale shadow of the real thing... but OP is asking for a "simpler" explanation.

 

[dipstik]

perhaps if you looked into why anything has mass and came up with a reason why photons are not subject to such, you could find something like an answer.

something like this: https://en.wikipedia.org/wiki/Higgs_mechanism#The_photon_as_the_part_that_remains_massless

unfortunately its not simple.

 

[BvU]

In the spirit of Nugatory: I have a smart nephew -- same age (15) and sharp enough -- with more or less the same question and I don't know how to bring this without talking nonsense while still helping out ...

 

[andresB]

The mass of the photon is zero because that is what experiment (and the whole way we analyze them) seems to tell us with a high degree of accuracy. IHMO, It is really hard to go beyond that. What can be said is that the current theories made around the experimental results work very well.

 

[HomogenousCow]

The mass of the photon is zero because that is what experiment (and the whole way we analyze them) seems to tell us with a high degree of accuracy. IHMO, It is really hard to go beyond that. What can be said is that the current theories made around the experimental results work very well.

There's more to it than that, you can't write down a massive Lagrangian with the properties you want for the electromagnetic interaction. (I think, could be wrong)

 

[jerromyjon]

There's more to it than that, you can't write down a massive Lagrangian with the properties you want for the electromagnetic interaction. (I think, could be wrong)

Some work has been done on these issues: https://en.wikipedia.org/wiki/Bonnor_beam

This was indicating that in parallel light beams the electric and magnetic forces cancel out, but in antiparallel beams there is an attractive element since the fields don't cancel.
I guess that does make simple sense to me that a photon coming towards you could have some very slight amount of mass, but one moving away from you is just "gone", never to be seen by your eyes or sensed as massive, unless it gets reflected! And then that is another whole "can of worms"...

 

[andresB]

There's more to it than that, you can't write down a massive Lagrangian with the properties you want for the electromagnetic interaction. (I think, could be wrong)

Yes, but the properties we want from the lagrangian are the ones that are ok with the experiments.

For example, at classical level, you can write a lagrangian with a combination of any power of the relativistic scalar and even powers of the relativistic pseudoscalars and that lagrangian will be relativistic invariant and gauge invariant (for example, Euler-Heisenber or the Born-infled ones). But for classical electrodynamics we choose a lagrangian linear in the scalar. The reasons is because that is the one that gives Maxwell's equation, and we want Maxwell's equation because that's the one that works .

 

[HomogenousCow]

Yes, but the properties we want from the lagrangian are the ones that are ok with the experiments.

For example, at classical level, you can write a lagrangian with a combination of any power of the relativistic scalar and even powers of the relativistic pseudoscalars and that lagrangian will be relativistic invariant and gauge invariant (for example, Euler-Heisenber or the Born-infled ones). But for classical electrodynamics we choose a lagrangian linear in the scalar. The reasons is because that is the one that gives Maxwell's equation, and we want Maxwell's equation because that's the one that works .

What I meant is, if you start with a local gauge symmetry,there's no quadratic mass term. If you "believe" that the electromagnetic interaction should be a gauge interaction, then the theory implies that the photon has to be massless.
Again, my exposure to this stuff is minimal and I could be very wrong.

 

[jerromyjon]

What I meant is, if you start with a local gauge symmetry, you can't write down a quadratic mass term.

Search Axriv for a 1998 paper titled "The gravitational interaction of light: from weak to strong fields" , I don't mean to cause a "hit" tied to a "B" thread... I'm not even sure the right way to link it.

 

[andresB]

What I meant is, if you start with a local gauge symmetry, you can't write down a quadratic mass term. If you "believe" that the electromagnetic interaction should be a gauge interaction, then the theory implies that the photon has to be massless.
Again, my exposure to this stuff is minimal and I could be very wrong.

I don't know at the moment, but assuming you are correct my point still stands. The whole reason to do the gauge thing is because it have proved to have usefulness to build a theory that is ok with the observed facts.

 

[vanhees71]

I think you meant ''Abelian renormalizable quantum field theory with a massive vector boson''. A theory without massless vector bosons is never a gauge theory.

No! In the Abelian case you can have massive vector mesons without the Higgs mechanism and without violating the physics which are even renormalizable in the strict Dyson sense. See, e.g.,

N. M. Kroll, T. D. Lee, and B. Zumino. Neutral Vector Mesons and the Hadronic Electromagnetic Current. Phys. Rev., 157:1376, 1967.
http://link.aps.org/abstract/PR/v157/i5/p1376

J. C. Collins. Renormalization. Cambridge University Press, Cambridge, New York, Melbourne, 1986.

 

[vanhees71]

And I think neither of you noticed the "B" tag on this thread :)
It's always good to remind people that the easy math-free answers are just a sad and pale shadow of the real thing... but OP is asking for a "simpler" explanation.

Some questions cannot be answered on the B-level. On B-level I'd simply state that the masslessness of the photon is an assumption in the Standard Model of elmentary particle physics, based on the very low upper bound of $m_{\gamma} < 10^{-18} \text{eV}$ without a deeper theoretical reason like a symmetry principle.

 

[A. Neumaier]

In the Abelian case you can have massive vector mesons

Yes, but massive vector bosons are never gauge invariant, since gauge invariance is a consequence of the masslessness of an irreducible vector representation only.

 

[Vanadium 50]

Um, everyone, this thread is marked "B". Do you think these answers are at a level that the OP understands?

 

[A. Neumaier]

Um, everyone, this thread is marked "B". Do you think these answers are at a level that the OP understands?

Maybe you can split it into a second thread with more advanced mark?

 

[vanhees71]

Yes, but massive vector bosons are never gauge invariant, since gauge invariance is a consequence of the masslessness of an irreducible vector representation only.

Well, that's simply not true. It's known as the Stückelberg formalism (see the references, I gave before).

For massless vector bosons you have necessarily a gauge theory for massive vector bosons you can have an Abelian gauge theory without Higgs mechanism.

 

[DarioC]

Here comes the reductionist.

Hey the kid just wants something he can handle in his head. Try this: Yes mass and energy are "always" equivalent, but---I personally like to think of a light photon as a quantum bundle of energy. I picture it as a perturbation of spacetime that is created by a shift in energy levels of a mass type particle. It propagates through space "like" a radio wave, but only in discrete units.

It is not a super-subminature ping pong ball, however convenient it is to call it a particle. What happens when matter gets in it's way I will leave up to the someone else, it is not a phenomena that I have dug into.