1
$\begingroup$

In my textbook, the time dilation equation is presented as $\Delta t' = \gamma \Delta t,$ where $ \gamma = 1/ \sqrt{1 - v^2/c^2} $. My understanding of this equation is as follows (correct me if I'm wrong): An observer is moving in a system $K'$ relative to a clock placed in a stationary system $K$. For every $\Delta t$ seconds that pass in the $K$ system, the moving observer in $K'$ only measures a passage of $\Delta t'$ seconds. Is that a correct interpretation?

A few pages later my textbooks states the following:

The proper time $\Delta t'$ measured on a clock in the $K'$ system is related to the time $\Delta t$ measured on a clock fixed in the $K$ system by $\Delta t' = \frac{\Delta t}{\gamma}$. The clock moving in the $K'$ system measures the proper time because it is present at both events.

I don't exactly understand what that second equation means. What does it tell us about the passage of time in the two systems? It almost seems like it is saying the opposite of what the first equation states. Any help is greatly appreciated.

$\endgroup$

2 Answers 2

3
$\begingroup$

The notation is confusing. Typically, a $\Delta t$ and $\Delta t'$ denote the coordinate time difference of two events.

For example, consider two events with coordinates in the unprimed system $(t_1, x_2)$ and $(t_2, x_2)$. The coordinate time difference for these two events is

$$\Delta t = t_2 - t_1$$

For the same two events (this is crucial), the coordinates in the prime frame are $(t'_1, x'_2)$ and $(t'_2, x'_2)$ thus

$$\Delta t' = t'_2 - t'_1$$

If the primed coordinate system has velocity $v$ in the unprimed system then, by the Lorentz transformation

$$\Delta t' = \gamma \left(\Delta t - \frac{v\Delta x}{c^2} \right)$$

Now, for the special case that $\Delta x = 0$, the two events are co-located in the unprimed system. Thus, $\Delta t$ equals the elapsed time according to a clock (at rest) located there.

Since all inertial observers agree on the elapsed time according to one clock, this elapsed time is invariant and is thus a proper time.

So, when $\Delta x = 0$, $\Delta t$ is a proper time and the above Lorentz transformation gives

$$\Delta t' = \gamma \Delta t$$

One crucial property of a proper time is that it is the smallest coordinate time difference between the events; in any other relatively moving inertial system, the coordinate time difference will be larger since $\gamma \ge 1$. This is why it is often said "moving clocks run slowly".


I don't exactly understand what that second equation means.

I've read that passage several times and I'm still uncertain what the author is trying to say.

However, if it is the case that $\Delta t'$ is a proper time, it follows from the above that $\Delta x' = 0$ so, from the Lorentz transformation we have

$$\Delta t = \gamma \left(\Delta t' + \frac{v\Delta x'}{c^2} \right) = \gamma \Delta t'$$

or

$$\Delta t' = \frac{\Delta t}{\gamma} $$

$\endgroup$
0
$\begingroup$

So the clock which measures proper time is the one for which the two events occur at the same location. So lets say you are travelling on a train at constant velocity and you get hungry and eat a bar of chocolate. In your reference frame on the train you start eating the bar (event 1) and finish eating it (event 2) at the same spatial coordinate. So the time difference between these two events is the proper time $\Delta t'$. To an external observer (on a platform or something) the time difference will be $\gamma$ times longer.

I think what your book means by "present" is that you measure proper time by the clock which is stationary with respect to the two events. Although this is a pretty term to use as really any frame has clocks present for both events they just aren't at the same location.

The first case is kind of confusing as there are no clear events to tell which frame is measuring proper time. But the second formula written is definitely correct. (Proper time should always be shorter)

All observers would measure proper time in their own frame in which they are stationary. You need some physical thing to happen to differentiate one frame from another and say that a particular frame is the one which measures proper time for your system.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.