### Relativity

# 8. Length contraction and time dilation

Physics Capsule received the following question from a reader:

`Irene N.S. asks, ‘We observed in [the film] ‘Interstellar’ a group of astronauts travel through a wormhole and reach another galaxy. In the movie they stay the same age and time seems to be slower in space than on Earth. How does this happen?’`

First of all, Irene, thank you for the great question. To explain exactly how the characters in *Interstellar* stay the same age we would have to either introduce a considerable amount of mathematics or hope that you would to be satisfied with vague descriptions. Since neither of those sound interesting we will have to seek an indirect route.

We can begin with the special theory of relativity, which is simpler and much easier to understand, and use that to examine how time can actually run at different rates. Once we are convinced of this flexibility of time, a verbal explanation of gravitational time dilation (which is really what you are looking for in your question) will be much more convincing.

# Transformations

A transformation equation is one that helps us describe an event (such as the position or motion of an object or perhaps a lightning strike) consistently across multiple frames of reference. Although deriving these transformations can be incredibly complicated we will try to simplify them as far as possible.

## Galilean transformations

We talked about Inertial frames of reference back in chapter 1 and we will use the same idea here to describe what are called Lorentz transformations. To arrive at those we need a more primitive transformation called Galilean transformations.

Let us say, with reference to the figure above, that some event takes place at $x'$. Say a bulb bursts there. The position $x'$ is at rest with respect to the frame on the right, shown to be moving with some velocity $\mathbf{v}$. This is called the primed frame because we use primes $x'$ and $y'$ to represent it. There is of course a $z$ in the so-called rest frame and a $z'$ in the primed frame but we will restrict ourselves to two dimensions for now.

Right off the bat we can state that, after some time period $t$, the position of the event is related to $x$ from the perspective of an observer in the rest frame as the origin $x'$ of the primed frame plus the distance the primed frame moved in time $t$. This distance is simply the product of velocity and time $-vt$ or $vt'$ depending on which frame we view it from (also, these have not been expressed vectorially yet). Mathematically, $$\mathbf{x} = \mathbf{x'} + \mathbf{v}t'$$and the vice versa is also true: $\mathbf{x'} = \mathbf{x} - \mathbf{v}t$. These are known as **Galilean transformations**.

## Lorentz transformations

Let us now work on the assumption that these Galilean transformations are incorrect at relativistic speeds, i.e. speeds comparable to that of light. After all we have no indication whatsoever that they are universal. Let us say, therefore, that there exists some correction factor that still needs to be applied; call this $\gamma$. Our equations must then become

$$\begin{equation}\mathbf{x} = \gamma ( \mathbf{x'} + \mathbf{v}t' ) \label{eq:x}\end{equation}$$

and

$$\begin{equation}\mathbf{x'} = \gamma ( \mathbf{x} - \mathbf{v}t ) \label{eq:x'}\end{equation}$$

That the same gamma factor applies to both is a crucial requirement. What is driving us at this point is the fact that all observers in physics are equivalent which means the laws of physics must not change for different observers. That is to say, what physics you deduce must not depend on where you observe from. It is with this intention that we seek a universality for gamma and for our transformation equations. Our task now is to determine exactly what this gamma is quantitatively.

We have been careful so far in not making any assumptions about time. Galilean transformations put the time $t$ measured by an observer in the unprimed frame as equal to the time $t'$ measured by an observer in the primed frame. Lorentz transformations, which is what we are formulating now, do not make this assumption. Indeed let us say that the unprimed observer measures $x$ at time $t$ and the primed observer measures $x'$ at $t'$ thereby calling to the fore Einstein's key idea: that $c$ in a vacuum is the speed limit of the universe.

From the simple velocity–time–displacement relationship we have $x = ct$ and $x' = ct'$ for our two frames of reference. Let us now multiply the left- and right-hand sides of eq. $\eqref{eq:x}$ and $\eqref{eq:x'}$ to arrive at $\gamma$.

$$\begin{align*}\mathbf{x} \mathbf{x'} &= \gamma^2 ( \mathbf{x'} \mathbf{x} + \mathbf{v} \mathbf{x} t' - \mathbf{x'} \mathbf{v} t - v^2 tt') \\[0.5em] c^2tt' &= \gamma^2 ( c^2tt' - v^2 tt' )\end{align*}$$

$$\begin{equation}\therefore \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \label{eq:gamma}\end{equation}$$

This factor $\gamma$ is called the Lorentz factor and plays an extremely important role in many relativistic calculations. What it tells us, in effect, is that if two observers in a primed and an unprimed frame, as shown above, measured an event and located it at $(x,t)$ and $(x',t')$ respectively in their frames of reference then the coördinates are related by eq. \eqref{eq:x}, \eqref{eq:x'} and \eqref{eq:gamma}.

# Effects of relativistic transformations

## Length contraction

The length of an object, which we will keep at one dimension for simplicity, is given by two points instead of one: say $x_0'$ and $x_1'$ marking the two ends of an object of length $l' = x_1' - x_0'$ lying along the *x*-axis in the primed frame.

The length $l$ of the object as measured by an observer in the unprimed frame is then related to $l'$ as follows: $$l' = x_1' - x_0' = \gamma ( x_1 - vt_1 - x_0 + vt_0 )$$

Assuming that the object is small enough that the two time measurements $t_1'$ and $t_0'$ of the two ends of the object are made simultaneously (at $t_1' = t_0'$) we can cancel out the second and fourth terms on the right-hand side of the equation above to get a simpler form:

$$\begin{equation}l' = \gamma l \label{eq:lc}\end{equation}$$

where $l = x_1 - x_0$ maintaining the same notation but without primes. Note that because $v**length contraction**.

## Time dilation

The effects of varying length can be seen in terms of time as well but as a reverse phenomenon. On the surface, think of this as another case of the velocity—time—displacement relationship. If the velocity is to be maintained at a constant comparable to $c$ then for contracting lengths the times must expand or **dilate**. Of course this is verbal and somewhat a less accurate picture than we can draw using elementary mathematics.

From eq. \eqref{eq:x} we can write the time $t'$ as $$t' = \frac{x - \gamma x'}{v \gamma}$$ which, on substituting from eq. \eqref{eq:x'} gives us $$\begin{align*}t' &= \frac{x}{v \gamma} - \frac{\gamma x}{v} + t \\[.5em] &= t - \left[ \gamma^2 \right] \frac{x}{v}\end{align*}$$

It is easy to see that $$\gamma^2 - 1 = \gamma \frac{v^2}{c^2}$$ as a result of which we end up with

$$\begin{equation}t' = \gamma \left( t - vx/c^2 \right)\end{equation}$$

Now suppose an arbitrary time interval $t'$ was measured in the primed frame as $t' = t_1' - t_0'$ (say). We can calculate the same interval as measured in the unprimed frame as $t = t_1 - t_0$ which is determined as follows:

$$\begin{align*}t' &= t_1' - t_0' \\[.5em] &= \gamma \left( t_1 - vx_1/c^2 - t_0 + vx_0/c^2 \right)\end{align*}$$

which, since the times $t_0$ and $t_1$ are made at the same spatial coördinate, $x_0 = x_1$ and we have

$$\begin{equation}t' = \gamma t\end{equation}$$

as the time as measured in the primed frame. Since $\gamma$, as we stated previously, is an increasing factor, the time measured in the primed frame, i.e. the frame moving at speeds comparable to $c$ appears to be longer than in the unprimed frame.

The takeaway here is that time appears to move at different rates for two observers even if all they are doing is moving relative to each other. All of these are special relativistic effects.

# Gravitational time dilation

The time dilation in *Interstellar* is not exactly special relativistic, however. Our purpose up to this point was to see that time can run at different rates. Whereas in our previous example it was relative motion, general relativity, along similar lines, tells us that the rate of time can vary even with gravity.

For an object moving close to a massive body,

$$t = t' \sqrt{1-2GM/Rc^2}$$

where $t$ is the time measured by the object itself and $t'$ is that measured by an observer at a great distance away from the massive body. $M$ and $R$ are the mass and radius of the massive object (the black hole from *Interstellar* for example) while $G$ is the universal gravitational constant. Clearly, for extremely massive objects the time interval $t$ is sufficiently smaller than $t'$, i.e. time slows down in a strong gravitational field.

So it is not that time runs slower in space, rather that it runs slower for the main characters in the film because they are under the influence of a strong gravitational field from a black hole; others further from the black hole experience a much faster rate of time and age longer before the two groups of people meet again. One last point: you may be interested to know that this is really an exaggerated form of a type of relativistic puzzles among which the most famous is the *twin paradox*.