Physics and Astronomy, University of Kent
In this section we will determine some general consequences of the Lorentz transformations. We will show that, if I am moving relative to you, my clock will appear to you to be running slow by a factor of \(\gamma\), and that my length in the direction of motion will contract by a factor of \(\gamma\). However, from my point of view, it is you who is moving, and your clock that will run slow and you that will be length contracted. It might seem that this leads to paradoxes - how can your clock run slower than mine if mine is running slower than yours?. We will explore the resolution of this (and other) apparent paradoxes, and see that framework of special relativity is in fact, entirely consistent (and, as far as we know, correct).
The light clock is a thought experiment which provides an introduction to time dilation without using the Lorentz transformations, and can provide an intuitive explanation of why we must abandon notions of absolute time if the postulates of special relativity are to be accepted.
Imagine a pair of mirrors, with a pulse of light bouncing between them. Each time light hits a mirror, we can think of this as the tick of a clock, and we measure the time interval between two of these ‘ticks’. The distance between the mirrors we call \(h\).
For an observer in the frame of reference of the light clock (i.e. someone who is not moving with respect to the light clock), which we can call \(S\), then the time taken for the pulse to pass between the mirrors is given by \(t=h/c\), where \(c\) is the speed of light, as normal in classical mechanics.
An observer who is in a frame of reference moving with a velocity \(u\) with respect to the light clock sees the light travel in a zig-zig pattern, as illustrated below. Let’s assume they measure the time between ticks to be \(t'\). In \(t'\) the light clock will have moved a distance of \(ut'\). So, from Pythogras’ theorem, the pulse of light has travelled a distance \(d\) where \(d^2 = h^2 + (ut')^2\). It travels this distance in a time \(t' = d/c\), and so substituting in \(d=ct'\) and \(h=ct\) from above we obtain: \[\begin{aligned} c^2t'^2 &= c^2t^2 + u^2t'^2 \\ t'^2(c^2 - u^2) &= c^2t^2 \\ t'^2 &= t^2 \frac{c^2}{c^2 - u^2} \\ t' &= t\sqrt{\frac{1}{1-u^2/c^2}} \\ t' &= \gamma t \end{aligned}\]
As we will see below, this is exactly the time dilation equation.
Consider a time period, \(\Delta T\), measured in a particular frame of reference. For example, this could be the time between two ticks of a clock, as measured by someone who is not moving relative to the clock (i.e. in the same frame of reference as the clock). This is therefore a time interval between two events measured on a clock that is present at both events. We call this the proper time.
The proper time is the time measured by a clock in its rest frame.
Now think about the point-of-view of an observer in a different reference frame who has their own clock. So for them, the original clock is moving. During a tick of the moving clock, the time interval measured by the observer on their clock will be: \[\Delta t = \gamma \Delta \tau\]
where \(\gamma\) is the normal gamma factor and \(\Delta \tau\) is the time interval measured on the moving clock in that clock’s rest frame. Since \(\gamma\) is always bigger than 1, then they will always measure a bigger time interval on their clock. So moving clocks appears to be running slow.
Time Dilation means that, in a reference frame in which it is moving, a clock runs slow by a factor of \(\gamma\) compared with its rest frame.
We sometime write the proper time as \(\tau\).
The time dilation equation comes quite simply from the Lorentz transformations. Imagine a bulb that regularly flashes, say at 1 second intervals. We can think of this as a simple kind of clock. Let’s call the time between flashes \(\Delta t\), as measured by the clock (or someone in the same rest frame of the clock, which we will call \(S\)). The first flash we will label event 1, and the second flash event 2.
In \(S\) the bulb is not moving, it is always at the same place, so we can say: \[x_1 = x_2 = 0\]
We’ll call the time of the first flash \(t_1 = 0\) and the second flash \(t_2 = \Delta t\).
Now let’s use the Lorentz transformations to figure out what someone who is moving with respect to the bulb, in frame \(S'\), sees: \[t'_2 = \gamma(t_2 - \frac{ux_2}{c^2})\]
Since \(x_2=0\) this is simply \[t'_2 = \Delta t' = \gamma \Delta t.\] This is the same as the time dilation equation (albeit with slightly different nomenclature).
Time Dilation of a Spacecraft: A spaceship sets out from Earth at a speed of \(c/4\). The ship reaches its destination after 1 year as measured by the clock of an observer on Earth. How much time does the pilot of the spaceship measure the trip took on his clock?
Solution: To solve this problem we can either start from the Lorentz transformations or, with a little thought, simply use the time dilation equation, \(t=\gamma t_0\). First, we know that \(u=c/4\) so we can calculate \(\gamma\): \[\begin{aligned} \gamma &= \sqrt{\frac{1}{1-u^2/c^2} }\\ &= \sqrt{\frac{1}{1-(1/4)^2}} \\ &= 1.03 \end{aligned}\]
Define \(S\) as the Earth rest frame, \(S'\) as the spaceship rest frame.
The first event is the spaceships leaving, which we will set to be at \(x_0 = x_0' = 0\) and \(t_0=t_0' =0\) (i.e. the common origin of \(S\) and \(S'\)).
The second event is the spaceship arriving. Hence \(t = 1\) year and \(t'\) is the wanted unknown.
The location of the arrival event is \(x'=0\) (since spaceship does not move in its frame of reference) and \(x\) is unknown and not needed.
Possible Lorentz transformations to use are \(t' = \gamma(t - ux/c^2)\) and \(t = \gamma(t' + ux'/c^2)\). Since \(x\) is unknown, use \[\begin{aligned} t &= \gamma(t' + ux'/c^2) \\ &= \gamma(t' + 0) \\ &= \gamma t' \\ 1~\mathrm{yr} &= 1.03t'\\ t'&= 0.97~\mathrm{yrs} \end{aligned}\]
If we want to start straight from the time dilation equation (\(t = \gamma t_0\)), we need to be careful to work out which time is \(t\) and which is \(t_0\). Recall that \(t_0\) is the proper time, it is the time measured by a clock in a reference frame where the clock is not moving. However, it is very easy to get confused when working with this definition.
It is easier if we think in terms of events. Event 1, in this case, was the spaceship leaving Earth, and Event 2 was the spaceship arriving at its destination. For somebody on Earth, these two events are at different locations. For somebody on the spaceships these two events are at the same location. So, it is the spaceship that measures the proper time (\(\tau\)). We therefore have:
\[\begin{aligned} t &= \gamma \tau \\ 1~\mathrm{yr} &= (1.03)\tau \\ \tau &= 0.97~\mathrm{years} \end{aligned}\]
Another way to think about it is that the for an observer on Earth, the spaceship is moving. Therefore the observer sees their moving clock run slow, i.e it will tick up less time than the clock on Earth.
The confusion comes because the observer on the spaceship also sees the Earth’s clock run slow. The question doesn’t ask you about this, but you see how you can easily get it confused. So, if in doubt, start with the Lorentz transformations; providing you carefully define your reference frames then you can’t really go wrong.
Consider a rod of length \(L\), as measured in its rest frame (i.e. by an observer who is not moving relative to the rod).
The proper length is the length of an object measured in its rest frame.
For an observer in a reference frame where the object is moving, the length of the object along the direction of motion will reduce by a factor of \(\gamma\): \[L = \frac{L_0}{\gamma},\] where \(L_0\) is the length of the object in its rest frame. This is called length contraction or Lorentz-Fitzgerald contraction.
In a reference frame in which it is moving, the length of an object reduces by a factor of \(\gamma\) relative to its length in its rest frame.
Consider an observer, in frame \(S'\), moving past a rod. We will call the rest frame of the rod \(S\). We will synchronise clocks as the observer passes the front of the rod, and call this event 1, so that: \[x_1 = 0, \quad t_1 = 0, \quad x'_1 = 0, \quad t'_1 = 0\] At exactly the same time, the moving observer measures the position of the back of the rod, \(x'_2\). Their measurement of the length of the rod, \(L'\), is \[L' = x'_2 - x'_1 = x'_2 - 0 = x'_2\]
Since the measurements must be made at the same time \[t'_1 = t'_2 = 0\]
Finally, we know that in the rod’s rest frame the distance between the front and back of the rod is its length, i.e. \[L = x_2 - x_1 = x_2 - 0 = x_2\]
Now we can use the Lorentz transformation: \[x_2 = \gamma(x'_2 + ut'_2)\] and since \(t'_2 = 0\), we have \(x_2 = \gamma x'_2\) and hence: \[x'_2 = \frac{x_2}{\gamma},\] or \[L' = \frac{L}{\gamma},\] as we wanted. Now, you might wonder, what would happen if we used the inverse transform \[x'_2 = \gamma(x_2 - ut_2)\] If we say \(t_2 = 0\) then we would have \(x'_2 = \gamma x_2\) and so \(L' = \gamma L\)! rather than \(L' = \frac{L}{\gamma}\), and they can’t both be right!
The answer is that we have absolutely no business saying that \(t_2 = 0\). We said that in the frame of the observer, \(S'\), then \(t'_2 = 0\) because we were setting \(t'_2 = t'_1\). To measure the length of something that is moving, we have to compare the positions of the front and the back at the same time.
However, there is nothing to say that these two times are equal in \(S\). You could work out \(t_2 = \gamma(t'_2 - ux'_2)\) of course, substitute this in, and you would get the correct answer, but it’s a bit more maths.
Moving Object Problem: A train of proper length 100 m passes by an observer at a speed of \(c/2\). What does the observer measure the length of the train to be?
Solution: The proper length is 100 m, so we expect the length measured by any other observer moving relative to the train, to be less than this. In this case, \(\gamma =1.15\) and so:
\[\begin{aligned} L &= L_0/\gamma\\ &= (100~\mathrm{m})/1.15 \\ &= 87~ \mathrm{m} \end{aligned}\]
An experimental demonstration of the effect of time dilation on Earth comes from muons. Muons are particles created in the upper atmosphere from the decay of pions, at a height of several thousand kilometres. They are unstable, and themselves decay with a half life of around \(2\) μs.
A typical muon moving at \(u=0.998c\) only travels around \(600\) m in \(2\) μs. Therefore, since the distance that needs to be travelled is several times this, very few muons should reach the Earth’s surface. However, we actually a significantly larger flux of muons reaching the Earth’s surface than expected.
This is explained by time dilation. The \(2\) μs is a proper time interval, i.e. it is the time measured from the rest frame of the muon (the muon is at the event of the muon’s creation and the event of it reaching the Earth’s surface, so it must measure the proper time). From the point of view of us on earth, however, the muons experience time dilation. At a velocity of \(0.998c\), \(\gamma\) is about \(15\), meaning that the observer on earth sees the internal ‘clock’ of the muon seem to run 15 times slower. To put it another way, the decay half life increases by a factor of 15.
But what happens if we look at things from the rest frame of the muon. From here it is the Earth that is moving towards the muon at a speed of \(u\). So the muon sees time running more slowly for the Earth, it obviously does not see it own clock running slow - its proper time does not change, and it is the proper time that determines the decay rate.
The answer is length contraction. In the rest frame of the muon, the Earth is moving towards it at a speed of \(0.998c\). The distance that it has to travel is therefore length contracted by a factor of \(\gamma = 15\). Therefore, while the muon still decays in \(2\) μs, now it only has to travel a distance of \(L/15\) before hitting the Earth.
The fact that something that looks time dilation in one frame of reference can look like length contraction in another tells us something about the inter-connectedness of space and time, that in a sense we are looking at two different aspects of the same thing. We will explore this idea of ‘spacetime’ more in the next lecture.