Special relativity

Coordinate transformations between inertial reference frames

An event, $s$ , is a location and time coordinate relative to an inertial reference frame, $S$ . For simplicity, we’ll abbreviate the time derivatives as $\overset{s}{˙}$ and $\overset{s}{¨}$ . I’ve written it out below, expanded out into the $x$ , $y$ , and $z$ spatial components to remind us that these are vectors in 3D space.

$position velocity acceleration s = x y z \overset{s}{˙} = \overset{x}{˙} \overset{y}{˙} \overset{z}{˙} = \frac{d s}{d t} (t) \overset{s}{¨} = \overset{x}{¨} \overset{y}{¨} \overset{z}{¨} = \frac{d ^{2} s}{d t ^{2}} (t)$

But what if we want to define the same event relative to a moving reference frame, $S^{'}$ ?

Galilean Transformations

Suppose $S^{'}$ is moving with respect to $S$ at velocity $v$ .

In Newtonian mechanics, the transformation between two inertial reference frames is given by the Galilean transformations. This is the same thing you might have done in high school physics or an engineering course.

$s (t) = s^{'} (t) + v t$

The underlying assumption here is that time $t$ is the same for observers in $S$ and $S^{'}$ . So the we can simply differentiate the position vector to get the velocity and acceleration vectors for the event in the moving reference frame.

$t and d t therefore \overset{s}{˙} \overset{s}{¨} = t^{'} = d t^{'} = \overset{s}{˙}^{'} + v = \overset{s}{¨}^{'}$

Observers in $S$ and $S^{'}$ will measure the same acceleration for the event. Because mass is unchanged by the transformation, $F = ma$ the force measured by the observers will be the same and Newton’s laws hold. All is well and you get an $A +$ in physics class.

Testing against Einstein’s postulates

Let’s see if Einstein’s postulate that the speed of light is the same for all inertial reference frames holds when applying the a Galilean transformation.

We can test this by calculating the speed of a photon in $S$ and $S^{'}$ .

$let \overset{s}{˙} = c, \overset{s}{˙}^{'} = c if then it must be true that \overset{s}{˙} = \overset{s}{˙}^{'} + v c = c + v$

This is not true, so the Galilean transformation violates the speed of light postulate!

The issue comes from the assumption that time is the same for observers in $S$ and $S^{'}$ , which we know from the previous chapter is not true. In most cases, the relative motion $v$ is much smaller than the speed of light, $c$ , so the Galilean transformation is a good approximation. But if $v$ is significant compared to $c$ or we want to be very precise over long distances, we need a more accurate transformation that accounts for the effects of relativity.

Lorentz Boosts

Good news, someone already figured this out! The Lorentz transformations are a set of equations that describe how to transform coordinates between two inertial reference frames that are in relative motion and are consistent with the speed of light postulate.

Shorthand for this operation is to call it a Lorentz boost (source).

Example

Recall our previous example with the astronaut and the astronomer. Let’s say the astronomer’s frame of reference is $S$ and the astronaut’s frame of reference is $S^{'}$ , moving at velocity $v$ relative to $S$ and the $x$ axis is the direction of motion.

The astronomer observes the origin of $S^{'}$ at time $t$ to have a displacement $x$ . The astronomer also observes the displacement of a photon in the astronaut’s light clock from the origin of $S^{'}$ as $x^{'}$ .

moving reference frame

The origin of $S^{'}$ is moving at velocity $v$ relative to $S$ . An event occurs at coordinate $(x^{'}, 0, 0, t^{'})$ in $S^{'}$ and at coordinate $(x, 0, 0, t)$ in $S$ .

The displacement of the origin of $S^{'}$ is $v t$ .
The displacement of the event in $S^{'}$ is $x^{'}$ .
THe displacement of the event in $S$ is the displacement of $S^{'}$ plus the displacement $x^{'}$ after accounting for relativity.

Problem: What is the displacement of the event in $S$ ?

Solution: In previous chapters we derived the time dilation and length contraction equations that relate the time and distance between two inertial reference frames. Let’s apply them here to transform between $S$ and $S^{'}$ .

$t = γ t^{'}, x = \frac{x ^{'}}{γ} where γ = 1/ 1 - (\frac{v}{c})^{2}$ therefore $x = v t + x^{'} 1 - (\frac{v}{c})^{2}, and x^{'} = \frac{x - v t}{1 - ( \frac{v}{c} ) ^{2}}$

Vectorized Lorentz Boosts