B Does my speedy spaceship (.999+C) have a temperature?

[Vanadium 50]

I hesitate to jump in, because the detour suggests that this is really the start of an anti-relativity screed.

However...

Classically, there are several different definitions of temperature. They are equivalent in that they all give the same value for the same system. In relativity, these transform differently so do not agree. They will never agree.

It also doesn't matter. If I have a rocket moving at .99c with respect to my heat bath, it is not in thermal equilibrium with that heat bath. There's no way to say what its temperature "really is" since it doesn't fulfill the conditions to even have a temperature.

 

[pervect]

I gather there are a couple of relativistic treatments of temperature. One of them, as I recall, had inverse temperature as a 4-vector. So $\Delta Q$ became the change in a 4-vector (not just a scalar energy), and the change in inverse temperature also became a 4-vector. And $\Delta S$, which was still a scalar, became the dot-product of these two 4-vectors, the change in energy-momentum, and the inverse temeprature.

However, I was more comfortable with the treatments (which I've also seen) where temperature was always specified in the rest frame of whatever had the temperature.

The reference I recall was https://arxiv.org/abs/physics/0505004. Apologies if I messed up anything in my recollections, it's been a while since I read it. I don't know what the impact factor of this paper was, it looked like a decent place to start to me, but I'm not that familair with thermodynamics.