You have said:
If,for instance,the relative motion observed between two frames of reference is that of uniform acceleration, how can we determine which frame is the unaccelerated system? It is obviously not possible.
and

Another part of this very question is also: How can we call the occupied frame of reference as being inertial regardless of whether other frames of reference are accelerating with respect to the occupied frame of reference?

Both these questions have been answered below.

Why would it not be possible? If you are in a reference frame which is accelerating at all, then you will experience pseudo-forces(forces whose source is not determined in that frame). That will tell you that your frame is accelerating. Moreover,if the relative motion between two frames is that of uniform acceleration,then both are accelerating! You do not have to determine WHICH is accelerating! The presence of acceleration(uniform or not) for any reference frame, guarantees that you will experience pseudo-force if you are in it.
for example, if you throw a ball from a height,it seems to hit the ground after travelling a path perpendicular to ground. but the actual trajectory is not so. as the ball falls it is deflected due to Coriolis force,which is a pseudo-force. so technically the earth is not an inertial frame of reference in any way since we can never point to a source who caused this Coriolis force!

You have said:
Resnick states that the frame of reference he occupies is an unaccelerated one. With respect to what? If accelerated motion were to be observed with respect to other frames of reference, how are we to determine that we occupy an inertial frame of reference at all?

According to Resnick he occupies an inertial frame that means, in his frame, Newton's first law holds true. obviously you need a reference object.
when we say a car travels at 75m/s then we actualy mean it travels 75m/s with respect to, say,a stationary tree. but it would travel at 50m/s with respect to another car travelling with 25m/s. so you need a reference object.

From what I understand, the Schrödinger equation describes how the wave function of a quantum system evolves in space over a given time (I am referring to a relativistic version of the Schrödinger equation).

First, there is no relativistic Schrodinger equation. The correct relativitic generalisation is Diracs equation, but even that is a kind of approximation to the true theory and one must work with the whole apparatus of QFT = QM + SR.

My understanding is that the equation essentially describes the evolution of the probability of a quantum measurement as a classical system.

The equation *directly* decribes the evolution of the probability amplitude and not probability per se; the evolution of probabilities is derived. The amplitude in some sense, is the 'square root' of probability, and is one of the basic concepts that distinguishes Quantum Mechanics from Classical Mechanics.

What I'm ultimately wondering is if the probabilities calculated from the wave function whose evolution is described by the Schrödinger equation depend on the reference frame of the observer

Yes it does. The basic problem in canonical quantisation is that we cannot make a *covariant* choice of creation and annihilation operators.

## Best Answer

There is a classic treatise on "Relativity, Thermodynamics and Cosmology" from R. Tolmann from the 1930s - it is still referenced in papers today. This generalises Thermodynamics to Special Relativity and then General Relativity. As a simple example the transformation law for Temperature is stated as: $T=\sqrt(1-v^2/c^2)T_0$ when changing to a Lorentz moving frame.

Another example is that "entropy density" $\phi$ is introduced, which is also subject to a Lorentz transformation. Finally this becomes a scalar with an associated "entropy 4-vector" in GR. The Second Law is expressed using these constructs by Tolmann.

There is some discussion in Misner, Thorne and Wheeler too.

Of course both these texts also include lots of regular General Relativity Theory which you may not need.