In this lecture we will resolve the problem we discovered in Lecture 1 in applying the Galilean transforms to electromagnetic radiation and the speed of light. We will do this by modifying the transformations in such a way that they reproduce the familiar classical results at low, everyday velocities, but also allow for the observation that light appears to travel at the same velocity in every reference frame.

The postulates of special relativity

Einstein proposed two postulates. (Postulates are things that we assume are true as the basis for a theory which we then compare with experiment). The postulates are sometime written in slightly different ways, but in essence are:

Together, these statements are not compatible with the Galilean velocity transformations stated in Lecture 1. Consider the following: I am travelling on a train at \(50\,\rm m/s\) , and you are standing by the side of the track. If I throw a ball forwards along the train at \(10\, \rm m/s\) you will say the speed of the ball is \(50 + 10 = 60\,\rm m/s\). This is simply applying the velocity transformations.

But now let’s say I do the same thing, except rather than throwing a ball, I shine light from a torch along the direction the train is moving in. The second postulate says that, from my point of view, the speed of light is equal to \(c\). If we apply the velocity transforms, the speed you measure would be \((c + 50)\,\rm m/s\). But the second postulate says that you also measure the speed of light to be equal to \(c\). Therefore, if we accept the postulates, then we need to modify the Gailean transformations.

In Chapter 1, we stated the Galilean transformations between two inertial frames of reference, \(S\) and \(S'\), to be:

\[\begin{aligned} x'&=x-ut \iff x = x' + ut \\y'&=y \\z'&=z \\t'&=t \end{aligned}\]

We know that these equations work well for most objects that we interact with, so the modified form of the equations must be such that they still give (approximately) the same answer for situations where we know the Galilean transformations work.

Statement of the Lorentz Transformations

The correct transformations between reference frames are called the Lorentz transformations. They had been discovered before Einstein’s paper by Lorentz as a somewhat ad hoc solution to the problem of the Michelson-Morley experiment, although Lorentz still very much believed in the ether. The transformations are:

The \(\gamma\) term is the ‘Lorentz factor’, often simply called gamma. Sometimes, we also define \(\beta = u/c\) to simplify equations, such that \(\gamma = \sqrt{1/(1-\beta^2)}\).

Notice that to switch between the two sets of equations (primed to non-primed and non-primed to primed) is simply a matter of replacing \(u\) with \(-u\). That is to be expected, after all there is nothing special about which set of co-ordinates we call \(S\) and which we call \(S'\).

We can check that the Lorentz transformation do indeed reduce to the Galilean transformations for most situations within our experience. The key thing to notice is the \(u^2/c^2\) in the gamma term (and also the \(ux/c^2\) in the equations for \(t\) and \(t'\)). \(c\) is a very large number, \(3\times10^8 \,\rm m/s\), whereas most velocities we deal with tend to be quite small.

For example, if we go back to our \(50 \,\rm m/s\) train, \(u^2/c^2\) is of the order of \(10^{-14}\). So if we calculate \(\gamma\): \[\gamma = \sqrt{\frac{1}{1-u^2/c^2}} = \sqrt{\frac{1}{1 -10^{-14}}} \approx 1\]

So if \(\gamma \approx 1\), then \(x' \approx x - ut\), which is the same as the Galilean transformation. More generally we can say that \(\gamma \rightarrow 1\) for \(u \ll c\). Similarly, for the \(t\) transformation, assuming \(ux \ll c^2\) we similarly have \(t \approx t'\). So, at the kinds of speeds we typically experience, the Galilean and Lorentz transformation give essentially the the same result. We therefore only need to use the more complicated Lorentz transformations when we have speeds are some reasonable fraction of \(c\).

Derivation of the Lorentz Transformations

We now show that the Lorentz transformations are the only transformations that are consistent with the two postulates of special relativity.

We will assume that space is isotropic and homogeneous. That is to say that the laws of physics do not depend on the direction we are moving in or where we are in space. If that is the case then we can write the most general possible transformations between two inertial frames as \[\begin{aligned} x' &= \gamma(x-ut) \\x &= \gamma(x'+ut') \end{aligned}\] Note that we are calling the constant \(\gamma\), but at this point in the derivation we don’t know what \(\gamma\) is yet!

Why must they take this form? They must be linear, i.e. they cannot contain some power of \(x\) or \(t\). If they contained a power of \(x\) then this would violate homogeneity. This means that \(\gamma\) cannot be a function of \(x\) (although it may be a function of \(u\)). There cannot be a constant term since this would violate isotropy. And for the same reason the two transforms (i.e. for \(x\) and \(x'\)) must be identical except for exchanging \(u\) with \(-u\).

Prior to the development of special relativity we would have said that time is measured the same in both frames of reference, and hence \(t=t'\). If we substitute that into the equations above, the only way to make it work is if \(\gamma=1\). This gives us the Galilean transformations.

But now let’s abandon the constraint that \(t=t'\) and see what happens. Moving to the second postulate, we have that light travels at the same speed in both \(S\) and \(S'\). So if we imagine a pulse of light moving in \(S\) at \(c\) the distance it will travel in time \(t\) is simply given by \(x=ct\). In reference frame \(S'\) we similarly have that \(x'=ct'\). Notice that there is no \(c'\), the second postulate says that light travels at \(c\) in all reference frames (i.e. \(c'=c\)).

Now, if we substitute \(x=ct\) and \(x'=ct'\) into our general transformations, we have: \[ct' = \gamma(ct-ut) = \gamma t(c-u) \\ct = \gamma(ct'+ut') = \gamma t'(c+u)\] From the second equation we have \[t = \gamma t' \frac{(c+u)}{c}.\] Substitute this into the first equation and we have \[ct' = \gamma \frac{[\gamma t' (c+u)](c-u)}{c}.\] The \(t'\) on both sides cancel, and we can multiply both sides by \(c\) to get \[c^2 = \gamma^2(c+u)(c-u).\] Now, \((c+u)(c-u)\) is equal to \(c^2-u^2\) and so we have \[\gamma^2 = \frac{c^2}{c^2-u^2}.\] So, dividing through the right hand side by \(c^2\) and taking the square root we have \[\boxed{\gamma = \sqrt{\frac{1}{1-u^2/c^2}}}\] which is exactly the \(\gamma\) factor in the Lorentz transformations.

We can then obtain an expression for \(t'\) by substituting \(x' = \gamma(x-ut)\) into \(x = \gamma(x'+ut')\) giving \[x = \gamma^2x - u\gamma^2t + u\gamma t'.\] From this point you could solve this directly for \(t'\) or \(t\) by substituting in the expression for \(\gamma\). Alternatively, we can rearrange to obtain \[t'= \gamma t - x\frac{\gamma^2 - 1}{u\gamma}.\] We then need to know the following identity, which can be found from the definition of \(\gamma\), \[\gamma^2 - 1 = \frac{u^2}{c^2}\gamma^2.\] Substituting in this expression for \(\gamma^2 - 1\) leads directly to \[\boxed{t' = \gamma\bigg(t - \frac{ux}{c^2}\bigg)}\] which again is what we want. We have therefore shown that the Lorentz transformations are the only possible transformations consistent with the postulates of special relativity.

Working with the Lorentz Transformations

If we know the space and time co-ordinates of an event in one reference frame \(S\) (\(x\) and \(t\)) then we can use the Lorentz transformations to calculate its space and time co-ordinates (\(x'\) and \(t'\)) in another reference frame, \(S'\), providing we know the velocity, \(u\) which \(S'\) is moving with relative to \(S\). Bear in mind that, when not otherwise stated, we are assuming that we are working in our standard co-ordinate system, where the two observers agree that \(t = t'\) at the instant when the zero of the \(S'\) frame moves past the zero of the \(S\) frame.

Another factor to keep in mind is the light travel time. When we talk about the \(t\) co-ordinates of an event being measured by an observer, we don’t mean the time at which an observer would see the event if they were looking at it through a telescope, This is because it would take some amount of time for the light to travel from the event to the observer and her telescope. This is not what the Lorentz transformations are telling us about. The Lorentz transformations tell us what the observer would measure even if they corrected their measurement to allow for the travel time of the light. To put it another way, special relativity is not telling us about some apparent change in time due to the time it takes light to travel, but a real physical effect.

One way this is sometimes expressed is to imagine a reference frame, \(S\) or \(S'\), as being an infinite lattice of clocks, moving rigidly together, with all the clocks perfectly synchronised. The Lorentz transformations tell us what would be measured by a clock in the lattice that is located at the event, not what an observer looking at that clock through a telescope would see.