12.1: Time Dialation and Space Contraction Revisited

As we've already seen twice, Lorentz transformations do funny things with the measurement of time and space. Here, we'll study these effects once more using spacetime diagrams. First we'll consider time. Suppose you and a friend come together at some point in space and time, which we'll call O (for origin, obviously). You synchronize your (perfect) watches. Your friend next takes off at speed \(u\) in the \(x\)-direction, while you remain stationary. Your inertial frame is thus \(S\) (the black frame in figure \(\PageIndex{1}\)a), whereas your friend's inertial frame is \(S^{\prime}\) (blue frame in figure \(\PageIndex{1}\)). After some time \(t_1\) has passed on your watch, you whip out a telescope and take a look at your friend's watch \({ }^1\) - and see that it is lagging behind yours.

To see why your friend's watch appears to be slow, consider the observation lines in figure \(\PageIndex{1}\)a. You both started at the origin. You have not moved in space since, only in time, so your position in spacetime coincides with the \(c t\) axis. Your friend is moving in both time and space, so in your coordinates your friend's trajectory is a sloped line, but of course in the comoving coordinates of \(S^{\prime}\) your friend is also stationary, so (s) he is moving on the \(c t^{\prime}\) axis. When you look at your friend at \(t=t_1\), you observe him \(/\) her at \(t^{\prime}=t_1^{\prime}\) in \(S^{\prime}\), as shown by the black arrow. From your point of view, \(t_1\) and \(t_1^{\prime}\) are simultaneous, because they lie on a line parallel to your space axis. However, from your friend's point of view, these events are not simultaneous at all! Instead, in \(S^{\prime}, t_1^{\prime}\) is simultaneous with events on the line parallel to the \(x^{\prime}\) axis - so point \(t_0\) (which you already passed) on your time axis. You can find \(t_0\) by projecting from \(t_1^{\prime}\) along the \(x^{\prime}\) axis on the \(c t\) axis - the green arrow in figure \(\PageIndex{1}\)a. By moving away from you, your friend's time thus seems to move slower, a phenomenon we call time dilation. Of course, we can also find out at which point your friend's watch indicates \(t_1\), by backtracking the upper green arrow in figure \(\PageIndex{1}\)a, which shows that that time corresponds to your later time \(t_2\).

Figure \(\PageIndex{1}\): Spacetime diagrams showing (a) time dilation and (b) length contraction of a moving object (blue frame,S′) as observedfrom a stationary point of view (black frame S).

We now have two ways to calculate how much time seems to slow down for your moving friend. On the one hand, the spacetime diagrams in figure \(\PageIndex{1}\) simply represent Lorentz transformations. Since you haven't moved in space, \(x=0\), and equation 11.12 gives \(c t^{\prime}=\gamma(u) c t\), so \(\gamma(u)\) is the dilation factor. Alternatively, we could use that in \(S, t_1\) and \(t_1^{\prime}\) are observed to be simultaneous. We already know that the slope of the \(c t^{\prime}\) line in \(S\) is given by \(\tan (\alpha)=u / c\), where \(\alpha\) is the angle between the \(c t\) and \(c t^{\prime}\) axes. We also know how the units on 

the \(c t^{\prime}\) axis relate to those in \(S\) (equation 12.1). Putting all these together, we have:

\[
\begin{aligned}
t_1 & =\cos (\alpha) \cdot(\text { unit conversion factor }) \cdot t_1^{\prime} \\
& =\frac{1}{\sqrt{1+(u / c)^2}} \cdot \sqrt{\frac{1+(u / c)^2}{1-(u / c)^2}} \cdot t_1^{\prime} \\
& =\gamma(u) t_1^{\prime}
\end{aligned}
\]

A similar thing happens for lengths. Of course, in our new four-dimensional spacetime, we would expect so, as time and space are now intimately linked. Suppose you have an object of length \(x_1\), that you position on your \(x\)-axis at \(t=0\), with one end at the origin and one end at \(x=x_1\) (figure \(\PageIndex{1}\)b). Now consider an identical moving object that passes you by at speed \(u\) at \(t=0\). How long do you measure that object to be? One end is at the origin of the \(S^{\prime}\) frame, which coincides with the origin of your frame; the other end sits at \(x^{\prime}=x_1^{\prime}\) on the \(x^{\prime}\)-axis, which you find by looking along your own time axis (black vertical arrow). To project back to your axis, you have to project from \(x_1^{\prime}\) on your space axis along the time axis of \(S^{\prime}\) - that's the green arrow, which projects \(x_1^{\prime}\) on some point \(x_0<x_1\). From equation (11.12) we find \(x_1^{\prime}=\gamma(u) x_1\), so the length of the object appears shortened by a factor \(\gamma(u)\), or Lorentz contracted.

Note that time dilation and length contraction are really the same thing - parts a and b of figure \(\PageIndex{1}\) are identical, only reflected in the diagonal line with slope 1 that represents the path taken by light. It is therefore also not surprising that the contraction/dilation factor \(\gamma(u)\) is the same in both cases.

A subtle but important point is that time dilation and length contraction are observations you make on an object moving relative to you. As long as you're in an inertial reference frame, you can always define your own frame to be the stationary one - simply pick the one that is moving with you. However, your friend can of course do exactly the same. You see your friend moving along your positive \(x\)-axis at speed \(u\) - and your friend sees you moving along their positive \(x\)-axis at speed \(u\), the only difference being that they define 'positive' the opposite direction you do (but as this is entirely arbitrary, there is no 'right' or 'wrong' choice). Therefore, your friend also sees your watch go slower, and your lengths contracted!