# Does light have no mass?

1. Feb 12, 2005

### godzilla7

Do photons really have 0 mass and how can we be sure, when we slow light to a standstill using condensates, it does not dissappear so e=mc^2 E=0 x c^2, I know about the various equations to explain waves Dirac et al, but how do we equate lights masslessness? how do we get that value? More a question of quantum mechanics theory than of lights mass; also if I send a single photon through a difraction gratin why is it still bent by the grating when it does not interact with anything is this gravity or is it something quantum?

2. Feb 12, 2005

### mathman

Light has no REST mass. However light travels at the speed of light! In the equation E=mc2, m is the rest mass divided by the Lorentz transformation factor, which is 0 for anything going at the speed of light. Therefore the m is now 0/0, so the equation is meaningless for light. Light does have energy using the equation E=hf (f=frequency) and is affected by gravity.

3. Feb 12, 2005

### dextercioby

Yes,no experiment so far indicated otherwise.

Trust me,it hasn't anything to do with QM...

Why would you say that...?
As for the second question,it's not gravity and light diffraction is much more easy-to-do,comprehendable & intuitive assuming light being a wave...

Daniel.

P.S.I answered only what i considered not fully clarified by the previous poster.

4. Feb 12, 2005

### Crosson

Mathman, it is not true that E = mc^2 is meaningless for light and you cannot apply the gamma factor to things moving at the speed of light.

The real equation is E = sqrt (m^2 C^4 + p^2c^2) where p is the relativistic momentum, and m is the rest mass. This equation is the one discovered by einstien, and it applies to everything in the universe. Notice it reduces to E - mc^2 for a particle at rest.

And look, a particle with no rest mass (light) can have a legitamate energy E = pc . If you use the debroglie wave relation wavelength = h/p you can combine this with E = pc to get E = hf for light (photons).

5. Feb 14, 2005

### godzilla7

Re:indeed

Yeah I think I'm looking at light as a discreet partical rather than a wave, thinking of the water in a tank experiments is an easier way to comprehend it.

And I think that's the point we still can't say wether light has mass or not for sure, all we can say is experimental evidence tends to suggest it does not, but then if light has an almost indescribably small mass say 1x10^-37Kg then there would doubtless be no experimental evidence.

Mind you light's being affected by gravity is something I find relatively( excuse the pun) easy to understand, warping of space time is easier to get to grips with than quantum mechanics equations.

Thanks guys

Last edited by a moderator: Feb 14, 2005
6. Feb 14, 2005

### einstone

Putting velocity =c in the equation (mass)( sqrt(1 - (velocity/c)^2)) = The rest mass, we can conclude that photons are massless. ( I always avoid division ).
But imagine a cosmic ray which( in presence of a potential) splits into a couple of particles! We must resort to the 'second quantisation'
as the term goes.Since the effect of gravity has been mentioned, I think it pertinent to ask whether the curvature of spacetime is taken into acount
while dealing with the states of the photon & the particles.(Even if these particles are virtual, the calculation should, I think, involve the curvature of space time).
I'm, with great respect,
Einstone.

7. Feb 14, 2005

### dextercioby

We cannot include the "curvature of space-time",for 2 reasons:
1.If we assume the background of QED to be curved (instead of flat),the theory,even by absurd assumed renormalizable,would not serve us anything.QED is VERY good as it is now & has been over the last 60 years.
2.We cannot include Einstein gravitons (and associated,Rarita-Schwinger gravitinos).The Sugra which would result would be nonrenormalizable.

Daniel.

8. Feb 15, 2005

### godzilla7

Re??? pardon my ignorance

Surely in any equation all significant variables have to be taken into account; sorry if I'm being ignorant but I've not yet studied the subject that long? But, in a practical experiment if we don't take note of gravitations lensing effect would we not get an innacurate result, or are theese equations only valid in theoretical maths? So if QED does not take account of other variables and assumes flat space then QED is palpably unsuitable for application in any real environment be it space or a lab in a lab of course we could ignore many of the smaller variables but not in space? Am I missing the point here? Probably am but Like e=mc^2 is not the whole picture of general relativity, without taking note of gravity in QED then the Lorenz Dirac Schrodinger et al equations are nonsense surely when applied to a practical experiment?

9. Feb 15, 2005

### dextercioby

No,everything we have so far works magnificiently at microscopical level,simply because the gravitational effects can be neglected...And we don't have yet a theory which can include them,without genereting infinities which have no physical relevance.

Daniel.

10. Feb 15, 2005

### cen2y

I must say that there doesn't seem to generate any paradoxes giving light certain masses, such as gravitational affect (though not generating gravitation), the m (not the rest mass) in the equation above, mass to be accounted for in system (such as center of mass), and so on. Anyone willing to refute the later, please check https://www.physicsforums.com/showthread.php?t=61002, I even added a more detailed derivation at the end (as I forgot to replace the v in the initial and the edit buttons' hiding).