# 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).

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.

• Thank you for asking a well-formulated homework question. I added the "homework" tag in, because even though it isn't real homework, the tag seems to apply. Click on the tag for more details about the tag. – Abhimanyu Pallavi Sudhir Sep 22 '13 at 14:53
• I mentioned the question just to make it more clear. I mentioned griffiths because it is a very well established book, and people will not doubt the legitimacy of the question. But, this is a problem in general. The issue is with the understanding of the concept. Not the above mentioned problem. – Rajath S Sep 22 '13 at 18:16
• I know that, however, due to the homework policy How do I ask homework questions on Physics Stack Exchange? , any question arising in an educational context or from the context of solving a problem is considered to be tagged as homework. – Abhimanyu Pallavi Sudhir Sep 23 '13 at 1:58

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 $(1)$ :

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

• Thanks for the reply. I have a small doubt. We can always choose an inertial frame of reference which is moving at the same speed as the particle with respect to the ground frame and sum up the change in time. But, how do we know that acceleration will not introduce any complication into concept of time? How can we be sure that the time in accelerated frame is same as the one found using the above method? – Rajath S Sep 23 '13 at 12:17
• Because we are working with infinitesimal variations $dt$ and $dx$. Any hypothetic correction would be at the second order ($(dx)^2, (dt)^2, dx dt$) and is neglectible relatively to $dx$ and $dt$. See also this very interesting post about proper acceleration – Trimok Sep 23 '13 at 15:07
• can you lead me to a source such as a book, where I can read this in depth? thanks. – Rajath S Sep 23 '13 at 18:44
• @RajathS : I have not a book very detailed on this problems, my advice is to look on the web with the keyword "proper acceleration", I found some interesting papers, Susskind lectures, Teaching relativity, or much more complex, movement in a non-inertial frame – Trimok Sep 24 '13 at 8:16
• I'm learning General Relativity now. May be I'll understanding this better after that. Thanks. – Rajath S Sep 24 '13 at 12:27

The proper time is well defined in SR, SR + acceleration and indeed GR, and is invarient in all three. In this case the invarience of the proper time means you can just use the elapsed time for the observer at rest.

The equation you've been given is a thinly disguised version of the relativistic rocket equation:

$$d(t) = \frac{c^2}{a} \left(\sqrt{1 + \left(\frac{at}{c}\right)^2} - 1 \right)$$

($$c$$ is the speed of light here - it isn't clear if $$c$$ means the speed of light or just a constant in the equation you've been given). The derivation of the equation for the relativistic rocket is given in Gravitation by Misner, Thorne and Wheeler, chapter 6.

• could you explain how it is valid for SR+acceleration and GR? It is pretty obvious that it is valid because it is a problem given in a standard book such as Griffiths. But, how it is correct is the question. – Rajath S Sep 22 '13 at 20:37
• my question is not about understanding the equation or solving the problem. I want to understand how proper time is defined for an accelerating object. May be I should rephrase my question. – Rajath S Sep 22 '13 at 20:43
• The line interval (proper time times $-c^2$) is defined by $ds^2 = \eta_{ab}x_a x_b$. For a freely moving observer this simplifies to $\Delta s^2 = -\Delta t^2$. For a non-inertial observer like the one in your question you have to integrate $ds$. – John Rennie Sep 23 '13 at 5:59

Proper time is defined for an arbitrary time-like trajectory $x^\mu(t)$ and an arbitrary spacetime metric $g_{\mu\nu}(x)$ by the formula $$\tau=\int\sqrt{g_{\mu\nu}(x)\dot{x}^\mu\dot{x}^\nu} dt.$$ I would recommend to compare this with the general formula for the length of a curved path to get a feeling that this is something invariant, and does not depend on the choice of coordinates if the metric is computed correctly in the new coordinates.