Special Topic: Lorentz Transformation

Layout of the physical world can be described in two equivalent ways:

Lorentz transformation translates event description from lattice of one inertial frame to that of another inertial frame.

L.2 Faster than light?

If I fly at 4/5 the speed of light and fire a bullet that I observe to fly forward at 4/5 the speed of light, isn’t the bullet flying at 8/5 the speed of light relative to the observer on the ground?

No. Velocities don’t simply add. The speed the observer will measure will be 40/41 the speed of light.

How did we get this result? Using Lorentz transformation.

Space and time separations between two events typically don’t have the same values in different inertial frames.

Here the events are:

Using the invariance of the spacetime interval, we can calculate the separations in the frame of a passenger on the bullet. But to get the separations in the frame of the observer on the ground, we need Lorentz coordinate transformation equations.

L.3 First Steps

Goal. Find $t$ and $x$ given $t'$, when $x' = 0$.

Assume we have a rocket moving at speed $v_\text{rel}$ relative to the observer in the lab. Take the spatial origins of the two frames to coincide at time $0$.

Let a spark occur at the origin of the rocket frame ($x' = 0$) at time $t$ of the lab frame.

Since the rocket is moving at speed $v_\text{rel}$ relative to the lab frame, the spark will occur at position $x = v_\text{rel} t$.

Now, using the invariance of the spacetime interval, we can calculate $t'$,

$$ t' = t \left( 1 - v_\text{rel}^2 \right)^{1/2} $$

Or

$$ t = \frac{t'}{\left( 1 - v_\text{rel}^2 \right)^{1/2}} $$

We define the time stretch factor as

$$ \gamma = \frac{1}{\left( 1 - v_\text{rel}^2 \right)^{1/2}} $$

So we have $t = \gamma t'$ (when $x' = 0$).

Then, we have $x = v_\text{rel} \gamma t'$ (when $x' = 0$).

L.4 Form of the Lorentz Transformation

In this section the book argues that the general form of the transformation must be linear.

That is, the general form should be:

$$ \begin{aligned} t &= B x' + D t' \\ x &= G x' + H t' \end{aligned} $$

Why? since we are free to choose any event as the reference event. See the book for more details.

L.5 Completing the Derivation

Using the special case in section L.3, we can find $D$ and $H$:

$$ \begin{aligned} t &= B x' + \gamma t' \\ x &= G x' + v_\text{rel} \gamma t' \end{aligned} $$

Demanding the invariance of interval between any pair of events whatsoever, leads to completed form of the Lorentz transformation:

$$ \begin{aligned} t &= \gamma \left( t' + v_\text{rel} x' \right) \\ x &= \gamma \left( x' + v_\text{rel} t' \right) \end{aligned} $$

or

$$ \begin{aligned} t &= \frac{v_\text{rel} x' + t'}{\left( 1 - v_\text{rel}^2 \right)^{1/2}} \\[0.5em] x &= \frac{x' + v_\text{rel} t'}{\left( 1 - v_\text{rel}^2 \right)^{1/2}} \end{aligned} $$

L.6 Inverse Lorentz Transformation

$$ \begin{aligned} t' &= \frac{-v_\text{rel} x + t}{\left( 1 - v_\text{rel}^2 \right)^{1/2}} \\[0.5em] x' &= \frac{x - v_\text{rel} t}{\left( 1 - v_\text{rel}^2 \right)^{1/2}} \end{aligned} $$

L.7 Addition of Velocities

Add light velocity to light velocity: gives light velocity!

Law of addition of velocities:

$$ v = \frac{v' + v_\text{rel}}{1 + v' v_\text{rel}} $$

Example 1. A rocket moves at $4/5$ the speed of light relative to the lab observer. A bullet is fired from the rocket at $4/5$ the speed of light relative to the rocket. What is the speed of the bullet relative to the lab observer?

$$ v = \frac{4/5 + 4/5}{1 + (4/5)(4/5)} = \frac{8/5}{1 + 16/25} = \frac{8/5}{41/25} = \frac{40}{41} $$

Example 2. Suppose that bullet that is fired is in fact a beam of light. What is the speed of the light beam relative to the lab observer?

$$ v = \frac{1 + 4/5}{1 + (1)(4/5)} = \frac{9/5}{9/5} = 1 $$