B Relativistic Photon Mass: Can It Exist?

[virgil1612]

TL;DR Summary

Photons do not have rest mass, but do they have "relativistic", or "effective", or "dynamic" mass?

Photons have 0 rest mass. But could I talk about relativistic, or dynamic photon mass, that would be the solution of
hf = mc^2 ? The relativistic mass would be m = m0/sqrt(1-v^2/c^2), where m0 is the rest mass, so 0, and v = c, so the denominator is also 0. The previous equations would give 0/0, so indeterminate, and could have a finite value. Am I allowed to think in these terms?
Thanks, Virgil.

 

[Vanadium 50]

Am I allowed to think in these terms?

Sure, you're allowed, but it isn't correct.

Photons are massless.

 

[virgil1612]

Massless as per rest mass, but why couldn't I talk about that "relativistic" mass?

 

[PeroK]

Massless as per rest mass, but why couldn't I talk about that "relativistic" mass?

Rest mass, by definition, is the mass of a particle in its rest frame. As photons do not have a rest frame, they do not have a well-defined rest mass.

A better approach is to take a particle's invariant mass, encapsulated in the most important formula in SR: $$E^2 = p^2c^2 + m^2c^4$$ where $E$ and $p$ are the particle's energy and magnitude of momentum. You can reaarange this formual to definine the invariant mass of a particle as: $$m = \frac 1 {c^2}\sqrt{E^2 - p^2c^2}$$ This formula holds for massive particle and photons, with $E = pc$ and $m = 0$ for photons.

 

[PeroK]

Massless as per rest mass, but why couldn't I talk about that "relativistic" mass?

PS relativistic mass is a largely obsolete term (although it lives on in populat science sources and still has a few adherents). The convention on this forum is not to use it:

https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[virgil1612]

Rest mass, by definition, is the mass of a particle in its rest frame. As photons do not have a rest frame, they do not have a well-defined rest mass.

A better approach is to take a particle's invariant mass, encapsulated in the most important formula in SR: $$E^2 = p^2c^2 + m^2c^4$$ where $E$ and $p$ are the particle's energy and magnitude of momentum. You can reaarange this formual to define the invariant mass of a particle as: $$m = \frac 1 {c^2}\sqrt{E^2 - p^2c^2}$$ This formula holds for massive particle and photons, with $E = pc$ and $m = 0$ for photons.

Thanks Perok, that's definitely useful. So we shouldn't even talk about relativistic mass, it's a wrong concept that leads us astray.

 

[PeroK]

Thanks Perok, that's definitely useful. So we shouldn't even talk about relativistic mass, it's a wrong concept that leads us astray.

Yes, if you want to study SR seriously and, especially, if you move on to General Relativity, it's a dead-end.

 

[virgil1612]

Yes, if you want to study SR seriously and, especially, if you move on to General Relativity, it's a dead-end.

Thanks.

 

[mitochan]

Though tricky, the system of a pair of photons generated by particle, anti-particle annihilation has mass because energy in CoM system is not zero. More generally the system of photons of different momentum has mass.

 

[vanhees71]

Massless as per rest mass, but why couldn't I talk about that "relativistic" mass?

Photons don't have a rest mass, because as massless particles they are never at rest in any inertial reference frame. There is no such thing as "relativistic mass". It's an outdated idea for about 112 years now!

 

[Nugatory]

Though tricky, the system of a pair of photons generated by particle, anti-particle annihilation has mass because energy in CoM system is not zero. More generally the system of photons of different momentum has mass.

This is correct, but doesn't help much with the (non)question of photon mass. The rest mass of a system is not, in general, equal to the sum of the rest masses of its components and there is no internally consistent and generaly applicable way of attaching the discrepancy to any single part of the system.

 

[pervect]

Summary:: Photons do not have rest mass, but do they have "relativistic", or "effective", or "dynamic" mass?

Photons have 0 rest mass. But could I talk about relativistic, or dynamic photon mass, that would be the solution of
hf = mc^2 ? The relativistic mass would be m = m0/sqrt(1-v^2/c^2), where m0 is the rest mass, so 0, and v = c, so the denominator is also 0. The previous equations would give 0/0, so indeterminate, and could have a finite value. Am I allowed to think in these terms?
Thanks, Virgil.

What you're trying to talk about is energy of the photon. You'll create no controversy and you'll be able to communicate clearly if you stick with E = hf.

Why would you think you needed another name for energy using (adjective) mass? Usually the desire to do that is motivated by something, and a discussion on what that something is can occasionally be prodocutive, though not necessarily short.

But the way your question was phrased was to just ask about terminology, and the hands-down winner on terminology is to call h*f "energy".

 

[andresB]

Photons have 0 rest mass. But could I talk about relativistic, or dynamic photon mass, that would be the solution of
hf = mc^2 ?

I don't see why you can't make such a definition. However, I also fail to see what would be gained in doing so.

 

[vanhees71]

You can make such a definition, but it's not useful. In relativity, both in SR (and even more importantly in GR), it is important to stick to quantities that are covariant under Poincare (in GR even under general coordinate) transformations.

From the mathematical structure of the Poincare group which determines the mathematical structure of special-relativistic spacetime (Minkowski space) it is clear that energy and momentum together build a four-vector and that one of the Casimir operators of the group, which is $p_{\mu} p^{\mu}=m^2 c^2$ is an invariant (a scalar). The Casimir operators are part of the specification of a physical system, and as it turns out from experiment, the photon has an invariant mass of zero (the empirical boundary is $10^{-18} \text{eV}/c^2$).

The introduction of a socalled "relativistic mass" is misleading and superfluos, because it is just the energy of the system divided by $c^2$, which for itself is not a covariant object but together with momentum transforms as the temporal component of a four-vector. So by introducing a "relativistic mass" you don't gain anything but introduce a lot of confusion. That's why in modern physics nobody introduces such a quantity anymore.