homework and exercises - Proper time for an accelerating object

As far as I have read so far, proper time is the time measured on the clock of an inertial frame moving uniformly with respect to another inertial frame. The concept and the mathematical expression for proper time is originated from the concepts of relativity of simultaneity and time dilation, both of which are evident from the fact that the quantity "interval" between two events remains constant in all the inertial frames. The conclusion is that the quantity proper time has a meaning only when we are talking about an inertial frame of reference.

I encountered a question in my exam: $$ x(t) = \sqrt{(b^2)+((ct)^2)} $$ The equation of motion of a particle in the ground frame of reference is given by the above equation. Calculate the expression for proper time. (This question is taken by Griffiths, Electrodynamics book).

I have two doubts about this question:

Does it make sense to define proper time for an accelerating object?

Assuming that the answer for Q1 is yes, then is it calculated by transforming coordinates into a new reference frame moving with velocity v for every small time dt? i.e., for every small change in dt there is a change in velocity of the particle as seen from ground frame. So, do I have to change my frame for every dt time, and sum up the dT? dT - infinitesimal proper time.

Answer

In special relativity, you have to choose as frame of reference which is an inertial frame. In this inertial frame, you may consider the movement of any object, whatever this movement is (accelerated or not).

Let the coordinates of the moving object, relatively to an inertial frame $F$, be $x$ and $t$. We can consider an other initial frame $F'$, which coordinates of the moving object, relatively to $F'$, are $x'$ and $t'$

The heart of special relativity is that exists an invariant which is $c^2 dt^2 - \vec dx^2 = ds^2$. This means that : $c^2 dt^2 - \vec {dx}^2 =ds^2= ds'^2 = c^2 dt'^2 - \vec {dx'}^2$. All inertial frames, when looking at the moving object, agree on the same value $ds^2$

Now, at some instant $t_0$, you may always consider a inertial frame $F'(t_0)$ which has, at this instant, the same speed as the moving object, relatively to $F$. Of course, you will have a different inertial frame $F'(t)$ for each instant. However, the key point is, that the instantaneous speed of the moving object relatively to $F'(t)$ is zero, that is, you have $dx' =0$, so you may write : $ds^2 = c^2 dt^2 - \vec {dx}^2 = ds'(t)^2=c^2dt'^2$

The time $t'$ defined in this manner is called the proper time of the moving object, and is noted $\tau$ ($c^2 dt^2 - \vec {dx}^2 = c^2d\tau^2$). It represents the time elapsed for a clock moving with the moving object.

With your problem, note that if you take the parametrization :

$$\left\{ \begin{array}{l l} ct= b ~sh (\frac{c \tau}{b}) \tag{1}\\ x= b ~ch (\frac{c \tau}{b}) \end{array}\right.$$ you will find, with a little algebra, that, first, $x(t) = \sqrt{(b^2)+((ct)^2)}$, and secondly, that $c^2 dt^2 - \vec {dx}^2 = c^2d\tau^2$ (We suppose here $dy=dz=0$).

So, $\tau$ is the proper time, that you are looking for, and you may find a expression of $\tau$ relatively to $t$, by inversing the first equation of the parametrization $(1)$ :

$$ \tau = \frac{b}{c}~Argsh (\frac{c t}{b}) \tag{2}$$