Do photons have mass? Why not?

[Dale]

Could you clarify? There's a few different possible things you could mean with this, and I'm not sure which one (or all of them) you're talking about.

I mean every vacuum spacetime ($T^{\mu\nu}=0$) with curvature ($R^\mu{}_{\nu\eta\xi}\ne 0$). That includes the various gravitational waves and also Schwarzschild and Kerr spacetimes.

 

[Matterwave]

I mean every vacuum spacetime ($T^{\mu\nu}=0$) with curvature ($R^\mu{}_{\nu\eta\xi}\ne 0$). That includes the various gravitational waves and also Schwarzschild and Kerr spacetimes.

Gotcha, thanks for clarifying!

So curvature does not require a gravitational source.

It depends a bit on what you mean by "source". If you mean just your criterion $T^{\mu\nu}=0$ then I fully agree. Indeed, the Schwarzschild and Kerr solutions are everywhere vacuum. The variable $M$ (and $J$ for Kerr) which appears in the metric though could be considered a boundary condition which "sources" (now differently defined) the curvature and in real life would correspond to the mass or angular momentum of the astrophysical object under considerarion.

For the gravitational wave case, free plane wave solutions are mathematically possible, just as they are for Maxwell equations. But in real life we still expect "sources" (in the second sense) for that radiation (e.g. merging Neutron stars or black holes).

 

[Dale]

It depends a bit on what you mean by "source". If you mean just your criterion Tμν=0 then I fully agree.

Yes. That is what I mean.

$T^{\mu\nu}\ne 0$ implies $R^\mu{}_{\nu\eta\xi}\ne 0$. But $R^\mu{}_{\nu\eta\xi}\ne 0$ does not imply $T^{\mu\nu}\ne 0$

in real life would correspond to the mass or angular momentum of the astrophysical object

Real life may include things like primordial black holes.

 

[Motore]

If I'm traveling away from the source approaching C will the gamma radiation still not travel towards me at C?

It will, but because energy is frame dependent the "gamma" ray will not be in the gamma energy range in this reference frame. It could be in the radio energy range.

 

[Mike_bb]

Photons have mass. The mass is 0. "Massless" is somewhat unfortunate term.

 

[Cyosis]

Photons have mass. The mass is 0. "Massless" is somewhat unfortunate term.

I am not sure if I agree. If we take the Proca Lagrangian as an example and work out the the three polarization vectors and then naively set m=0 we run into problems, the longitudinal one explodes. The longitudinal polarization vector doesn't exist in the massless theory. So the physical degrees of freedom differ as well.

 

[javisot]

For the gravitational wave case, free plane wave solutions are mathematically possible, just as they are for Maxwell equations. But in real life we still expect "sources" (in the second sense) for that radiation (e.g. merging Neutron stars or black holes).

If the sun were to disappear instantly, you would continue to experience its gravity for some time, even though the source no longer exists.

AdS, for example, is another case of a vacuum with curvature; in this instance, the concept of the "source of the curvature" is even more complex.

 

[bob012345]

Photons have mass. The mass is 0. "Massless" is somewhat unfortunate term.

That’s like saying I’m a billionaire but I just happen to have zero billion dollars at the moment.

 

[Mike_bb]

That’s like saying I’m a billionaire but I just happen to have zero billion dollars at the moment.

The term "massless" means "mass is undefined". But mass is fundamental property then term "massless" isn't correct in that case.

 

[Dale]

The term "massless" means "mass is undefined".

The term “massless” can also mean “has 0 mass” or even “has negligible mass”. Words are messy and can have multiple distinct meanings.

 

[Matterwave]

Real life may include things like primordial black holes.

Could you clarify or elaborate what you are trying say? This is a true statement, I feel like I'm just missing your point.

If the sun were to disappear instantly, you would continue to experience its gravity for some time, even though the source no longer exists.

The Sun can't disappear instantly without violating GR and the EFEs. Local conservation of stress energy is enforced by the Bianchi identities.

AdS, for example, is another case of a vacuum with curvature; in this instance, the concept of the "source of the curvature" is even more complex.

Fair point, though our universe is not ADS. But you do highlight that in my conception I group the cosmological constant $\Lambda$ with the source terms $T^\Lambda_{ab}=-\frac{\Lambda}{8\pi}g_{ab}$