Physics and Astronomy, University of Kent
In this section we will resolve the problem we discovered in Section 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.
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:
The laws of physics are the same in all inertial frames, one cannot detect one’s absolute velocity by conducting any experiment.
The speed of light is independent of the source, and is constant \(c\) in a vacuum as measured in all inertial reference frames.
Together, these statements are not compatible with the Galilean velocity transformations which we stated to be:
\[\begin{align} x'&=x-ut \iff x = x' + ut \\y'&=y \\z'&=z \\t'&=t \end{align}\]
If I am travelling on a train at \(50\,\mathrm{m/s}\), and you are standing by the side of the track, and I throw a ball forwards along the train at \(10\, \mathrm{m/s}\) then the Galilean transformations you will say the speed of the ball is \(50 + 10 = 60\,\mathrm{m/s}\).
But if instead 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 \((c + 50)\,\mathrm{m/s}\). But the second postulate says that you also measure the speed of light to be equal to \(c\). Therefore, the postulates contradict the Galilean transformations.
We therefore need a modified form of the equations that give (approximately) the same answer for situations where we know the Galilean transformations work, but which are also consistent with the principles of relativity.
The correct transformations between reference frames are called the Lorentz transformations. Equivalent equations had been written down before Einstein’s paper by Hendrik Lorentz, which is why they bear his name, although his physical understanding of them was somewhat different. In any case, the transformations are:
\[\begin{align} x'&=\gamma (x-ut) \quad \Longleftrightarrow \quad x=\gamma(x'+ut') \\y'&=y \\z'&=z \\t'&=\gamma(t-ux/c^2) \quad \Longleftrightarrow \quad t&=\gamma(t'+ux'/c^2) \end{align}\]
where
\[\begin{equation} \gamma = \frac{1}{\sqrt{1-u^2/c^2}} \end{equation}\]
The \(\gamma\) term is the ‘Lorentz factor’, often simply called gamma. Sometimes, we also define \(\beta = u/c\) to simplify equations, such that \(\gamma = 1/\sqrt{1-\beta^2}\). Again, these equations are only valid for the standard configuration of reference frames that we established in Section 1.
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: there is nothing special about which set of co-ordinates we call \(S\) and which we call \(S'\), and we should obtain the same physical results if we swap them.
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, \(3\times10^8 \,\mathrm{m/s}\), whereas most velocities we deal with tend to be quite small.
For example, for the \(50 \,\mathrm{m/s}\) train, \(u^2/c^2\) is of the order of \(10^{-14}\). So if we calculate \(\gamma\): \[\begin{equation} \gamma = \sqrt{\frac{1}{1-u^2/c^2}} = \sqrt{\frac{1}{1 -10^{-14}}} \approx 1 \end{equation}\]
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, if \(\gamma \approx 1\) and \(ux/c^2 \ll t\) then we similarly have \(t \approx t'\).
Therefore, 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 \(c\). What ’reasonable‘ means depends on the level of accuracy you require. For example, for \(u = 0.01c\) the correction factor is approximately \(5 \times 10^{-5}\).
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. Together with some other considerations which are set out in more detail in Rindler, it turns out that the the most general possible transformations between two inertial frames as \[\begin{align} x' &= \gamma(x-ut) \\x &= \gamma(x'+ut') \end{align}\] 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 our chosen standard configuration that at \(t=t'\) we have \(x=x'\). And for similar reasons 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.
Instead, we don’t assume \(t=t'\) and instead apply the second postulate, that light travels at the same speed in both \(S\) and \(S'\). We consider 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: \[\begin{equation} ct' = \gamma(ct-ut) = \gamma t(c-u) \\ct = \gamma(ct'+ut') = \gamma t'(c+u) \end{equation}\] From the second equation we have \[\begin{equation} t = \gamma t' \frac{(c+u)}{c}. \end{equation}\] Substitute this into the first equation and we have \[\begin{equation} ct' = \gamma \frac{[\gamma t' (c+u)](c-u)}{c}. \end{equation}\] The \(t'\) on both sides cancel, and we can multiply both sides by \(c\) to get \[\begin{equation} c^2 = \gamma^2(c+u)(c-u). \end{equation}\] Now, \((c+u)(c-u)\) is equal to \(c^2-u^2\) and so we have \[\begin{equation} \gamma^2 = \frac{c^2}{c^2-u^2}. \end{equation}\] So, dividing through the right hand side by \(c^2\) and taking the square root we have \[\begin{equation} \boxed{\gamma = \sqrt{\frac{1}{1-u^2/c^2}}} \end{equation}\] 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 \[\begin{equation} x = \gamma^2x - u\gamma^2t + u\gamma t'. \end{equation}\] 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 \[\begin{equation} t'= \gamma t - x\frac{\gamma^2 - 1}{u\gamma}. \end{equation}\] We then need to know the following identity, which can be found from the definition of \(\gamma\), \[\begin{equation} \gamma^2 - 1 = \frac{u^2}{c^2}\gamma^2. \end{equation}\] Substituting in this expression for \(\gamma^2 - 1\) leads directly to \[\begin{equation} \boxed{t' = \gamma\bigg(t - \frac{ux}{c^2}\bigg)} \end{equation}\] 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.
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 the standard configuration, 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. Informally, we can think of the Lorentz transformations telling 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.
A more formal way to think about this 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.
Lorentz Transformation Problem A spaceship
departs Earth at a constant speed of \(c/2\). After 1 year (by the pilot’s clock)
the pilot measures an explosion which is 1 light year directly ahead.
According to an observer on Earth, how long after the spacecship left
did the explosion occur, and how far was is from Earth?
Solution:
Let’s define \(S\) = Earth frame and \(S'\) = spaceship frame.
The spaceship leaves the Earth at time \(t_0 = t_0' = 0\).
The event occurs at time \(t\) as measured on Earth and time \(t'\) as measured on the spaceship, at locations \(x\) and \(x'\), respectively.
So in \(S'\): \(x'\) = 1 ly, \(t'\) = 1 year
And in \(S\): \(x\) = ?, \(t\) = ?
And we have \(u = c/2\)
To find the time measured from Earth: \[\begin{align} t &= \gamma(t' + ux'/c^2) \\ &= \frac{1}{\sqrt{1-u^2/c^2}} (t' + ux'/c^2) \\ &= \frac{1}{\sqrt{1-(1/2)^2}} (1 \mathrm{[yr]} + 1/2 \mathrm{[yr]}) \\ &= 2\frac{\sqrt{3}}{3}\frac{3}{2}~\mathrm{yrs}\\ &= \sqrt{3}\approx 1.7~\mathrm{yrs} \end{align}\]
If you’re wondering where the \(c\)s went, one cancels with the \(c\) in the velocity (\(c/2\)). The other converts the distance in light years to a time in years (\(x' = 1\, \mathrm{ly}\) and so \(\frac{x'}{c} = 1 \, \mathrm{yr}\)).
To find the distance measured from Earth, we similarly use: \[\begin{align} x &= \gamma(x' + ut') \\ &= 2\frac{\sqrt{3}}{3} \bigg(1~\mathrm{[ly]} + (c/2)(1~\mathrm{[yrs]}) \bigg) \\ &= 2\frac{\sqrt{3}}{3} (3/2~\mathrm{[ly]}) \\ &= \sqrt{3}\approx 1.7~\mathrm{ly} \end{align}\]