Should photons be considered massless?

[WannabeNewton]

What does mass even mean in a general relativistic sense?

For test particles it's the exact same thing as in SR. For extended bodies (such as black holes, stars, etc.) the situation is much more complicated: http://en.wikipedia.org/wiki/Komar_mass

 

[Jonnyb302]

If you want to nitpick, I can easily point out that that is a scalar equation, i.e. it signifies the amplitude only. So the negative sign is meaningless in this case.

Sorry, it was just meant as a joke. Honest, no offense intended.

But that's my point. You only care about this from ONE point of view, i.e. the mass-energy equivalence. Instead of looking at it as a conversation factor, you are assigning a physical meaning of one being the same as the other. This is absurd if we apply that same logic to other well-known equality!

I agree with you. In a gravitational setting it seems difficult to distinguish the two. But I realize now that there are other fields where mass and energy really are separate in a meaningful way.For example this:

As I've stated, if photons have mass, a number of consequences should occur, including the fact that it must also decay.

Thanks for the help guys, this has been thought provoking and informative. The more advanced physics I learn, the more I have to question the original stuff I think.

 

[BruceW]

I'd say (in relativity), there is no exact analogue to non-relativistic 'mass'. This is the answer I think maybe you were looking for.

edit: and this is not surprising. New theory, new concepts, right?

 

[vanhees71]

The notion of mass is indeed quite different in Newtonian compared to special-relativistic theory. It's more simple in the relativistic case ;-).

Anyway, the argument starts with the formulation of quantum theory in Minkowski or Galilei space-time, which leads you to investigate the unitary ray representations of the corresponding parts of the groups that are continuously connected to the identity, i.e., the proper orthochronous Poincare and (inhomogeneous) Galilei groups, respectively.

In the case of the Poincare group you'll figure out that mass is a Casimir operator of the corresponding Lie algebra given by the relation, $m^2=p_{\mu} p^{\mu}$, where $p^{\mu}$ are the generators of the space-time translations. The further analysis turns out that any ray representation is induced by a unitary representation of the covering group of the Poincare group, i.e., instead of the proper orthochronous Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$ you use its covering $\mathrm{SL}(2,\mathbb{C})$. In this way you come to massive, massless and tachyonic representations. The latter seem not to lead to a sensible physical theory (except for non-interacting tachyons, but these are useless because not observable).

In the case of the Galilei group, it turns out that there are non-trivial central extensions of the group, and mass is the central charge. There is no physically sensible unitary representation of the Galilei group or it's covering group, but only the central extension with the non-zero mass as the central charge.

The subtle difference is that this implies a mass-superselection rule, i.e., there cannot be superpositions of states with different mass, which you don't have in relativistic quantum theory. The latter possibility is realized in nature on the elementary-particle level by the neutrinos, which always are produced in flavor eigenstates which are mixtures of mass eigenstates with different masses, leading to the well-established neutrino oscillations.