# Why photons have no mass

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1. Jan 13, 2016

### 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

2. Jan 13, 2016

### 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
simpler than what ?

3. Jan 13, 2016

### 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.

4. Jan 13, 2016

### 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.

5. Jan 14, 2016

### 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

6. Jan 14, 2016

### A. Neumaier

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.)

Last edited: Jan 14, 2016
7. Jan 14, 2016

### jerromyjon

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?

8. Jan 14, 2016

### ellipsis

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."

9. Jan 14, 2016

### jerromyjon

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.

Last edited: Jan 14, 2016
10. Jan 14, 2016

### sheaf

Some work has been done on these issues: https://en.wikipedia.org/wiki/Bonnor_beam
If the photons are travelling 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!

11. Jan 14, 2016

### vanhees71

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.

12. Jan 14, 2016

### Staff: Mentor

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.

13. Jan 14, 2016

### A. Neumaier

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.

14. Jan 14, 2016

### Staff: Mentor

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.

15. Jan 14, 2016

### dipstik

16. Jan 14, 2016

### 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 ...

17. Jan 14, 2016

### 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.

18. Jan 14, 2016

### HomogenousCow

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)

19. Jan 14, 2016

### jerromyjon

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"...

20. Jan 14, 2016

### andresB

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 .

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