# Special Relativity Explained

## Introduction

Recently I came across Einstein’s Special Relativity again and found I was very confused about the knowledge I once learned in college, so I think I would be better document it in case I forgot again in the future.

In this blog post, I would discuss the Special Relativity and mainly focus on the Lorentz transformation.

## Prerequisites

This is a collection of the definitions of some of the fundamental concepts in physics that are key to understanding Special Relativity.

### Spacetime

In physics, spacetime is any mathematical model which fuses the three dimensions of space and the one dimension of time into a single four-dimensional manifold.

### Event

A position in spacetime is called an event, and requires four numbers to be specified: the three-dimensional location in space, plus the position in time. An event is something that happens instantaneously at a single point in spacetime, represented by a set of coordinates $x$, $y$, $z$ and $t$.

### Frame of Reference

In physics, a frame of reference (or reference frame) consists of an abstract coordinate system and the set of physical reference points that uniquely fix (locate and orient) the coordinate system and standardize measurements within that frame.

### Inertial Frame of Reference

An inertial frame of reference in classical physics and special relativity possesses the property that in this frame of reference a body with zero net force acting upon it does not accelerate; that is, such a body is at rest or moving at a constant velocity.

## Postulates of Special Relativity

Einstein’s Special Relativity is derived based on the following two postulates or assumptions. Although they are postulates, there have not been too many problems for them as suggested by the experimental evidence.

### First Postulate (Principle of Relativity)

The laws of physics take the same form in all inertial frames of reference.

### Second Postulate (Invariance of Light Speed)

As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity $c$ that is independent of the state of motion of the emitting body. It is also equivalent to say, the speed of light in free space has the same value $c$ in all inertial frames of reference.

The experiments we have conducted so far, such as the Michelson-Morley experiment, all support the invariance of light speed.

## Lorentz Transformation

With the two postulates of Special Relativity only, Einstein was able to derive the Lorentz transformation, which is the core of special relativity. The Lorentz transformation was named after Hendrik Lorentz because he and his coworkers happened to find the exact same transformation to explain the invariance of light speed although based on the conjecture of the existence of aether which later turns out not exist.

I would write another article on how to derive Lorentz transformation based on the two postulates for special relativity in the future.

### Definitions

In the spacetime, we have two reference frames, a reference frame $S$ and another reference frame $S’$ moving at a velocity $v$ with respect to it. So the two reference frames in this scenario are inertial reference frame. The coordinate axes in each reference frame are parallel, i.e., the $x$ and $x’$ axes are parallel, the $y$ and $y’$ axes are parallel, and the $z$ and $z’$ axes are parallel, and remain mutually perpendicular. We assume the relative motion is along the coincident $xx’$ axes. At $t = t’ = 0$, the origins of both coordinate systems are the same, $(x,y,z) = (x’,y’,z’) = (0, 0, 0)$.

An event in the time space could be observed and recorded by the observers on the two reference frames using spacetime coordinates $(t,x,y,z)$ in the reference frame $S$ and $(t’,x’,y’,z’)$ in the reference frame $S’$, respectively.

### Lorentz Boost

The Lorentz transformation is linear transformation. It may include a rotation of space; a rotation-free Lorentz transformation is called a Lorentz boost. We are only focusing on the Lorentz boost in this article.

$$

\begin{align}

t^{\prime} &= \gamma (t - \frac{v}{c^2}x)\\

x^{\prime} &= \gamma (x - vt)\\

y^{\prime} &= y\\

z^{\prime} &= z\\

\end{align}

$$

where

$$

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

$$

and it is the Lorentz factor.

The Lorentz boost could also be thought as transforming the event coordinate observed from the rest reference frame to the motion reference frame. It is considered as tensor transformation.

$$

\begin{align}

\begin{bmatrix}

ct^{\prime} \\

x^{\prime} \\

y^{\prime} \\

z^{\prime} \\

\end{bmatrix}

=

\begin{bmatrix}

\gamma & -\beta \gamma & 0 & 0 \\

-\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

\begin{bmatrix}

ct \\

x \\

y \\

z \\

\end{bmatrix}

=

\begin{bmatrix}

\gamma ct - \gamma \beta x \\

\gamma x - \beta \gamma ct \\

y \\

z \\

\end{bmatrix}

\end{align}

$$

where

$$

\beta = \frac{v}{c}

$$

### Inverse Lorentz Boost

It is also straightforward to get the inverse Lorentz boost from the Lorentz boost.

$$

\begin{align}

t &= \gamma (t^{\prime} + \frac{v}{c^2}x^{\prime})\\

x &= \gamma (x^{\prime} + vt^{\prime})\\

y &= y^{\prime}\\

z &= z^{\prime}\\

\end{align}

$$

We could also write the inverse Lorentz boost using tensor transformations.

$$

\begin{align}

\begin{bmatrix}

ct \\

x \\

y \\

z \\

\end{bmatrix}

=

\begin{bmatrix}

\gamma & \beta \gamma & 0 & 0 \\

\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

\begin{bmatrix}

ct^{\prime} \\

x^{\prime} \\

y^{\prime} \\

z^{\prime} \\

\end{bmatrix}

=

\begin{bmatrix}

\gamma ct^{\prime} + \gamma \beta x^{\prime} \\

\gamma x^{\prime} + \beta \gamma ct^{\prime} \\

y \\

z \\

\end{bmatrix}

\end{align}

$$

Note that the transformation matrices used in the Lorentz transformation and the inverse Lorentz transformation are inverse to each other.

$$

\begin{align}

\begin{bmatrix}

\gamma & -\beta \gamma & 0 & 0 \\

-\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

\begin{bmatrix}

\gamma & \beta \gamma & 0 & 0 \\

\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

&=

\begin{bmatrix}

\gamma & \beta \gamma & 0 & 0 \\

\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

\begin{bmatrix}

\gamma & -\beta \gamma & 0 & 0 \\

-\beta \gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

\\

&=

\begin{bmatrix}

\gamma^2 - \beta^2 \gamma^2 & 0 & 0 & 0 \\

0 & \gamma^2 - \beta^2 \gamma^2 & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

\\

&=

\begin{bmatrix}

1 & 0 & 0 & 0 \\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \\

\end{bmatrix}

\\

&= I

\end{align}

$$

### Lorentz Coordinate Differences

Suppose we have two events, one event has coordinates $(t_1,x_1,y_1,z_1)$ observed in the reference frame $S$ and coordinates $(t’_1,x’_1,y’_1,z’_1)$ observed in the reference frame $S’$, another one has coordinates $(t_2,x_2,y_2,z_2)$ observed in the reference frame $S$ and coordinates $(t’_2,x’_2,y’_2,z’_2)$ observed in the reference frame $S’$. Note that because reference frame $S’$ is moving along the $xx’$ axes, $y_1 = y’_1$, $z_1 = z’_1$, $y_2 = y’_2$, $z_2 = z’_2$.

We could calculate the coordinate differences between these two events in both the reference frame $S$ and $S’$.

We define the difference between coordinates $t’$ and $x’$ in the reference frame $S’$ are as follows.

$$

\begin{align}

\Delta t^{\prime} &= t_{2}^{\prime} - t_{1}^{\prime}\\

\Delta x^{\prime} &= x_{2}^{\prime} - x_{1}^{\prime}\\

\Delta y^{\prime} &= y_{2}^{\prime} - y_{1}^{\prime}\\

\Delta z^{\prime} &= z_{2}^{\prime} - z_{1}^{\prime}\\

\end{align}

$$

Similarly, the difference between coordinates $t$ and $x$ in the reference frame $S$ are as follows.

$$

\begin{align}

\Delta t &= t_{2} - t_{1}\\

\Delta x &= x_{2} - x_{1}\\

\Delta y &= y_{2} - y_{1}\\

\Delta z &= z_{2} - z_{1}\\

\end{align}

$$

We plug in the Lorentz boost and the inverse Lorentz boost into the coordinate differences in the reference frame $S’$.

$$

\begin{align}

\Delta t^{\prime} &= \gamma(\Delta t - \frac{v}{c^2} \Delta x)\\

\Delta x^{\prime} &= \gamma(\Delta x - v \Delta t)\\

\Delta y^{\prime} &= y_{2} - y_{1}\\

\Delta z^{\prime} &= z_{2} - z_{1}\\

\end{align}

$$

Similarly, the coordinate differences in the reference frame $S$ are

$$

\begin{align}

\Delta t &= \gamma(\Delta t^{\prime} + \frac{v}{c^2} \Delta x^{\prime})\\

\Delta x &= \gamma(\Delta x^{\prime} + v \Delta t^{\prime})\\

\Delta y &= y_{2}^{\prime} - y_{1}^{\prime}\\

\Delta z &= z_{2}^{\prime} - z_{1}^{\prime}\\

\end{align}

$$

## Lorentz Transformation Consequences

The equations for $\Delta y$, $\Delta y^{\prime}$, $\Delta z$, and $\Delta z^{\prime}$ are not interesting because it matches our “conventions”. However, from the equations for $\Delta t$, $\Delta x^{\prime}$, we would find some non-trivial consequences. Relativity of simultaneity, time dilation, and length contraction are the most famous ones.

### Relativity of Simultaneity

Because of the following coordinate difference we have just derived

$$

\begin{align}

\Delta t^{\prime} &= \gamma(\Delta t - \frac{v}{c^2} \Delta x)

\end{align}

$$

If the two events are observed to happen simultaneously in the reference frame $S$, i.e., $\Delta t = 0$, these two events might not be seen simultaneously in the reference frame $S’$, i.e., $\Delta t^{\prime}$ might not be 0.

When $\Delta t = 0$,

$$

\begin{align}

\Delta t^{\prime} &= - \frac{\gamma v}{c^2} \Delta x

\end{align}

$$

In most cases, $\Delta t^{\prime} \neq 0$ unless the two events were observed to happen on the same $x$ coordinates in the reference frame $S$, i.e., $\Delta x = 0$. $\Delta y$ and $\Delta z$ do not have to be zero though.

Here is a concrete example. If you are traveling on an extremely fast rocket, and there are two events (on two locations along the direction you are traveling) that are seen simultaneously by the people on earth. However, in your eyes, these two events did not happen simultaneously.

### Time Dilation

Because of the following coordinate difference we have just derived

$$

\begin{align}

\Delta t^{\prime} &= \gamma(\Delta t - \frac{v}{c^2} \Delta x)

\end{align}

$$

If the two events are observed to happen on the same $x$ coordinates from the reference frame $S$, the time between the two events observed from the reference frame $S’$ is longer than what is observed from the reference frame $S$.

When $\Delta x = 0$,

$$

\begin{align}

\Delta t^{\prime} &= \gamma \Delta t

\end{align}

$$

Note that because the velocity of everything which has mass could not be greater or equal than $c$, for things that have mass, $\gamma > 1$. Then we have

$$

\begin{align}

\Delta t^{\prime} \geq \Delta t

\end{align}

$$

Here is a concrete example. If you are traveling on an extremely fast rocket, and there is a guy jumping on the early (no change in $x$ before he jumped and after he landed). It is measured 0.1 seconds for this jump on earth. However, you would measure probably 1 year for this jump. Then you would think, wow this jump is so slow.

In addition, because of the following coordinate difference we have just derived

$$

\begin{align}

\Delta t &= \gamma(\Delta t^{\prime} + \frac{v}{c^2} \Delta x^{\prime})

\end{align}

$$

If the two events are observed to happen on the same $x’$ coordinates from the reference frame $S’$, the time between the two events observed from the reference frame $S$ is longer than what is observed from the reference frame $S’$.

When $\Delta x^{\prime} = 0$,

$$

\begin{align}

\Delta t &= \gamma \Delta t^{\prime}

\end{align}

$$

Then

$$

\begin{align}

\Delta t \geq \Delta t^{\prime}

\end{align}

$$

Here is a concrete example. If you are traveling on an extremely fast rocket, and you are jumping in the spaceship (no change in $x’$ before you jumped and after he landed). It is measured 0.1 seconds for this jump in your spaceship. However, the people on earth would measure probably 1 year for this jump. Then the people on the earth would think, wow this jump is so slow.

This is the interesting part. You on the spaceship see the people on the earth are moving extremely slow and their clock moving is slower than yours, whereas the people on the earth see you on the spaceship are moving extremely slow and your clock moving is slower than theirs. You on the spaceship would think you would die much earlier than the people on the earth, whereas the people on the earth would think they would die much earlier than you on the spaceship.

### Length Contraction

We can use a measuring rod to determine the distance between two events along the $xx’$ axes ($\Delta x$ or $\Delta x^{\prime}$) simultaneously ($\Delta t = 0$ or $\Delta t^{\prime} = 0$) from any reference frame.

Suppose a measuring rod is at rest and aligned along the $x$-axis in the reference frame $S$. The length of the rod is $\Delta x$. To measure $\Delta x^{\prime}$ in the reference frame $S’$, the events of the two rod ends have to be measured simultaneously in the reference frame $S’$, which means $\Delta t^{\prime} = 0$.

We plug in $\Delta t^{\prime} = 0$ into

$$

\begin{align}

\Delta x &= \gamma(\Delta x^{\prime} + v \Delta t^{\prime})\\

\end{align}

$$

We get

$$

\begin{align}

\Delta x &= \gamma \Delta x^{\prime}\\

\end{align}

$$

Therefore,

$$

\begin{align}

\Delta x^{\prime} &= \frac{\Delta x}{\gamma}\\

\end{align}

$$

and

$$

\begin{align}

\Delta x^{\prime} \leq \Delta x\\

\end{align}

$$

The length of the rod measured in the reference frame $S’$ is shorter than the length of the rod measured in it resting reference frame $S$.

Suppose a measuring rod is at rest and aligned along the $x’$-axis in the reference frame $S’$. The length of the rod is $\Delta x’$. To measure $\Delta x$ in the reference frame $S$, the events of the two rod ends have to be measured simultaneously in the reference frame $S$, which means $\Delta t = 0$.

We plug in $\Delta t = 0$ into

$$

\begin{align}

\Delta x^{\prime} &= \gamma(\Delta x - v \Delta t)\\

\end{align}

$$

We get

$$

\begin{align}

\Delta x^{\prime} &= \gamma \Delta x\\

\end{align}

$$

Therefore,

$$

\begin{align}

\Delta x &= \frac{\Delta x^{\prime}}{\gamma}\\

\end{align}

$$

and

$$

\begin{align}

\Delta x \leq \Delta x^{\prime}\\

\end{align}

$$

The length of the rod measured in the reference frame $S$ is shorter than the length of the rod measured in its resting reference frame $S’$.

The length contraction does share logic similarities to time dilation.

## Conclusions

In my opinion, Lorentz transformation is simply a mathematical coordinate transformation that transforms the event in the spacetime from one rest reference frame to another motion reference frame moving at constant velocity respect to the rest reference frame.

## References

Special Relativity Explained