Does Lorentz transformation affect mass density, pressure or temperature of a physical object?

Question

A cylinder filled with gas and moving along its axis would be Lorentz contracted, diminishing its volume. Viewed from stationary frame, would the pressure of the gas be higher than in the co-moving frame?

Answer 1

To tackle the body of your question first: pressure is just part of the stress-energy tensor, which is the object one usually considers in relativistic contexts. The individual components of this are not invariant under Lorentz transformations.

Whether temperature is Lorentz invariant is hotly (excuse the pun) debated, as seen in the answers to Is temperature a Lorentz invariant in relativity?, but the consensus seems to be that temperature alone is not invariant.

As for mass density, what do you even mean by that? "Mass" is generally defined to be equivalent to "rest mass" and is hence invariant. The volume clearly varies due to length contraction, so mass divided by volume must also vary.

Answer 2

For density the answer can be given easily. For pressure a little more thought is needed. For temperature I don't know much about it but will comment.

For density there are two kinds you might be interested in: the density of a scalar quantity which is itself invariant (e.g. electric charge, rest mass) or the density of a scalar quantity which is not (e.g. energy). The term 'mass density' is ordinarily interpreted to mean 'the amount of rest mass per unit volume'. Rest mass is an invariant.

For a fluid we can construct a quantity whose components are density and 3-current of a quantity which is itself a scalar invariant (e.g. rest-mass density and current density of rest mass). Such a quantity is a 4-current. It can be written as a column vector like this: $$ J = \left(\rho,\; {\bf j} \right)^T $$ where $^T$ is a transpose. Any given fluid element has a rest frame (we do not consider turbulent flow). Let $\rho_0$ be the mass density in the rest frame of a fluid element. In this frame the 4-current will take the form $$ J_{\rm rest} = \left(\rho_0,\; 0,0,0\right)^T $$ In other frames you will then find $$ J = \Lambda J_{\rm rest} $$ where $\Lambda$ is the Lorentz transformation. This tells you how $\rho$ transforms (and also $\rm j$).

To get the transformation of pressure, note that pressure is a component of the stress-energy tensor, and that tensor transforms like any 2nd-rank tensor. In some simple cases one finds the pressure to be the same between one frame and another, but in general one does not. However the interpretation of precisely what quantity is indicated by the word 'pressure' may not be straightforward if the relevant fluid element is moving (relative to some given inertial frame).

Finally for temperature you can, of course, observe that velocities are readily transformed and thus one can get a velocity distribution in whatever frame one wants. But there is room for debate about what quantity shall be given the name 'temperature' when the energy distribution is not the Boltzmann distribution. In the case of thermal radiation the standard practice is, I think, to go to the rest frame of the radiation (i.e. the frame where there is no net energy flux). This is how the temperature of the CMB radiation is assigned, for example.