Determining the rest frame Sorry if this has been asked before in some way, I did try looking through the special relativity tag and didn't really see a good answer to my question.
I am having trouble determining which frame is the proper-time frame outside of the simplest cases. The definition I was given is where the event starts and ends in the same place, but that doesn't really seem to define it well in most cases, like the following example:

An observer on a spacecraft moving at $0.700c$ relative to the Earth finds that a car takes 40.0 min to make a trip. How long does the trip take to the driver of the car?

Although I now know the proper frame is the car frame, to me, it would seem that the proper frame is the spacecraft, because someone standing on the craft doesn't have to move to see the car at the start or end point, while the person in the car starts the clock in one location and stops it at another, the destination.
Is there a more rigorous definition for the proper frame or perhaps another way to think about how the clock in each frame "moves"?
 A: An observer in Special Relativity never moves himself, because if he moves, SR doesn’t say how he must take measurements, or better to say that SR is silent about that.
An observer conducts measurements in his „own rest frame “. In simple words, the rest frame is a lattice of synchronized clocks (or at least two synchronized clocks). An observer can place these clocks at certain distance from each other or even in every point of space. When an even takes place at certain point of space, an observer can assign space and time coordinate to this event. He just takes readings of a synchronized clock in  immediate vicinity to the event.
https://en.wikipedia.org/wiki/Observer_(special_relativity)
https://en.wikipedia.org/wiki/Length_contraction#/media/File:Observer_in_special_relativity.svg
https://arxiv.org/abs/physics/0512013
A stationary observer in reference frame $S$ conducts measurements this way:
the observer places clock $C_1$ at coordinate $x_1$ of his frame and light clock $C_2$ at coordinate $x_2$ of his frame.
Then this observer sends a beam of light from clock $C_1$ towards clock $C_2$.  He assumes, that one - way speed of light is c (Einstein synchrony convention). Since he knows distance and speed of light, he synchronizes these clocks, so as they show “the same time” in reference frame $S$.
https://en.wikipedia.org/wiki/Einstein_synchronisation
Then this observer can measure rate of moving clock.
Imagine that clock $C'$ passes by clock $C_1$  at moment of time $t_1$ first and clock $C_2$ at moment of time $t_2$ some later. At these moments, readings of the moving clock and the corresponding fixed clock of reference frame $S$ next to it are compared.


Let the counters of moving clock measure the time interval $\tau _ {0}$ during the movement from the point $x_ {1}$ to the point $x_ {2}$ and the counters of clocks $C_1$ and $C_2$ of the fixed or “rest” frame $S$, will measure the time interval $\tau$. This way,
$$\tau '=\tau _{0} =t'_{2} -t'_{1},$$
$$\tau =t_{2} -t_{1} \quad (1)$$
But according to the inverse Lorentz transformations we have
$$t_{2} -t_{1} ={(t'_{2} -t'_{1} )+{v\over c^{2} } (x'_{2} -x'_{1} )\over \sqrt{1-v^{2} /c^{2} } } \quad (2)$$
Substituting (1) into (2) and noting that the moving clock is always at the same point in the moving reference frame  $S'$, that is,
$$x'_{1} =x'_{2} \quad (3)$$
We obtain
$$\tau ={\tau _{0} \over \sqrt{1-v^{2} /c^{2} } } ,\qquad (t_{0} =\tau ') \quad (4) $$
This formula means that the time interval measured by the fixed clocks is greater than the time interval measured by the single moving clock. Time in reference frame S is running $\gamma$ times faster from the point of view of moving clock $C'$. This means that the moving clock lags behind the fixed ones, that is, it slows down.
The animation below depicts a rest frame (a row of synchronized clocks) and moving clock (single clock).
Each observer in SR repeats the same procedure and attaches this lattice of Einstein - synchronized clocks to himself. This way we can see, that in SR observers never agree on a certain “mutual” frame, but each of them thinks that he is in the center of the universe and the others are jumping around him.
 
A: There's no such thing as "the rest frame" (or "proper frame"). Instead, you'll hear about things like "the rest frame of the spaceship" or "the rest frame of the car" or "the rest frame of the Earth". Sometimes people do just say "the rest frame", but in those cases it's supposed to be clear from context which object's rest frame is intended. (If it isn't, that's bad communication.)
The rest frame of an object is one in which the object is at rest. That's the definition. In your example, the spaceship has a rest frame, the car has a rest frame, and the Earth has a rest frame. All of these frames are in motion relative to one another.
It might be helpful for you to consider how the situation appears in each frame. For example, in the rest frame of the car:


*

*there is a spaceship moving at approximately $0.700c$

*there is a planet (Earth) moving at whatever speed is indicated on the car's odometer, maybe $60\ \mathrm{mi./hr.}$

*the car is not moving, i.e. at rest

