12: Spacetime Diagrams

To study the effect of Lorentz transformations on the observations done by people in different inertial reference frames, we will make use of a handy tool: the spacetime diagram (figure \(\PageIndex{1}\), sometimes also called the Minkowski diagram). At first glance, these diagrams are very similar to (time, space) plots you're probably familiar with, except with the axes swapped (figure \(\PageIndex{1}\)a). We plot distance horizontally, and time (or rather, \(c\) times time) vertically. A stationary object then has a vertical 'trajectory' (known as its worldline), and a moving object a trajectory with slope \(c / v\), with \(v\) the object's speed. We can go to a comoving frame through a Galilean coordinate transformation (figure \(\PageIndex{1}\)b, equation 11.1). The comoving frame \(S^{\prime}\) has an x -axis \(x^{\prime}\) that coincides with that of the stationary frame, and a time axis \(c t^{\prime}\) that has the same slope as the speed of the moving object. Consequently, all point's on the object's trajectory have the same space coordinate, whereas time keeps on running at the same pace as before. (This might seem counterintuitive - didn't we change time? No, we rotated the time axis, but that doesn't change time itself, as time coordinates are determined by projecting on the time axis parallel to the space axis. Space changed: points are not projected on the same spot on the \(x\) and \(x^{\prime}\) axis, even though the axes themselves coincide).

What about the Lorentz transformations? To study them, we need to consider what happens to the speed of light, which by the light postulate must remain unchanged. Because of our choice of axes (which includes a choice of units), light travels along a line of slope 1 in our spacetime diagram (figure \(\PageIndex{1}\)c). Anything slower than light has a trajectory with a slope larger than 1 . To have a trajectory with slope less than 1 , you'd need to go faster than light. Therefore, you cannot travel from the origin into the region closer to the space than to the time axis. Objects can be there of course - but if you start at the origin, you can't travel there, or even see these objects, as not even light can travel fast enough to connect you. We'll come back to this point in section 12.2. For now, we note that the Lorentz transform must map the line of slope 1 on a line of slope 1 . The only way to do so is to rotate the space and time axes both, and both by the same amount, as is done in figure \(\PageIndex{1}\)d. Here time and space have both changed: the stationary (black, \((x, t)\) ) and moving (blue, \(\left(x^{\prime}, t^{\prime}\right)\) ) observer measure different values for the time and space coordinates, as given by the Lorentz transformations (11.12). To find the angle \(\alpha\) over which the axes must rotate, consider a stationary object in \(S^{\prime}\) : its worldline must coincide with the \(c t^{\prime}\) axis, while it moves with velocity \(u\) in frame \(S\), so the slope of the \(c t^{\prime}\) axis in \(S\) must be \(c / u\), and we have \(\tan (\alpha)=u / c\).

A subtle but important point is that the Lorentz transformations do not only change the orientation of the time and space axis, but also their units (which is what you might have expected, as you know about time dilation and length contraction already). To see how this works, consider the point \(\left(x^{\prime}, c t^{\prime}\right)=(0,1)\) in \(S^{\prime}\) (so a unit distance away from the origin on the time axis). The Lorentz transformations give us the coordinates of this point in \(S:(x, c t)=(\gamma(u)(u / c), \gamma(u))=\gamma(u)((u / c), 1)\). The length of this interval in \(S^{\prime}\) is 1 ; the length of the same interval in \(S\) is \(\gamma(u) \sqrt{1+(u / c)^2}\). We conclude that the units of \(c t^{\prime}\) and \(c t\) are related through this factor. A completely analogous calculation shows that the same holds for the units of \(x^{\prime}\) and \(x\) (as again we might have guessed, due to the symmetry of time and space now present in our system), so we have:

\[
\frac{c t^{\prime} \text { unit }}{c t \text { unit }}=\sqrt{\frac{1+(u / c)^2}{1-(u / c)^2}}=\frac{x^{\prime} \text { unit }}{x \text { unit }}
\]

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\({ }^1\) Well, not really - it takes time for the light from your friends watch to reach you. We'll assume you're smart, and have corrected for this already.