📂Physics

Lorentz Transformation Derivation

Derivation

• The text might be a bit long, but it’s written very simply, so don’t be afraid to dive in.

Let’s think about light (photon) moving in the plane of $xy$ inertial system (coordinate system) when $t=0$. It starts from the origin and is advancing at an angle of $\theta$ with the $x$ axis.

The new transformation that will replace the Galilean transformation can be said to look like the following.

$$\begin{pmatrix} t^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} & & & \\ & & & \\ & & & \\ & & & \end{pmatrix} \begin{pmatrix} t \\ ct\cos\theta \\ ct\sin\theta \\ 0 \end{pmatrix}$$

Before finding the new transformation, let’s first consider the Galilean transformation.

$$\begin{pmatrix} t^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} \color{red}1 & \color{red}0 & 0 & 0 \\ \color{red}{-v_{0}} & \color{red}1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix}$$

Here, let’s assume that our desired transformation, like the Galilean transformation, does not affect the direction that is not moving. That is, since the $A^{\prime}$ inertial system is moving in the direction of $x$, let’s say that $y$, $z$ components are not affected. Then, the new transformation we want is only the red parts changing from the Galilean transformation. Therefore, let’s state the new transformation as the following.

$$\begin{equation} \begin{pmatrix} t^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} a & b & 0 & 0 \\ e & d & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t \\ ct\cos\theta \\ ct\sin\theta \\ 0 \end{pmatrix} \label{eq1} \end{equation}$$

In alphabetical order, it’s natural for $c$ to come into the place of $e$, but $e$ is used since $c$ is used for the speed of light. Now, from the given conditions, we need to find $a$, $b$, $e$, $d$. Solving the matrix yields the following.

$$\begin{align} t^{\prime} &= at+bct\cos\theta \label{eq2} \\ x^{\prime} &= et+dct\cos\theta \label{eq3} \\ y^{\prime} &= ct\sin\theta \label{eq4} \\ z^{\prime} &= 0 \nonumber \end{align}$$

Rearranging $(2)$ for $t$ yields the following.

$$t = \frac{t^{\prime}}{a+bc\cos\theta}$$

Inserting this into $(3),$ $(4)$ yields the following.

$$\begin{align*} x^{\prime} &= \frac{e+dc\cos\theta}{a+bc\cos\theta}t^{\prime} \\ y^{\prime} &= \frac{c\sin\theta}{a+bc\cos\theta}t^{\prime} \end{align*}$$

According to the theory of relativity, the speed of this light must be $c$ in both $A$ system and in $A^{\prime}$ system. This can be expressed by the following formula.

$$\begin{align} {v^{\prime}_{x}} ^{2}+{v^{\prime}_{y}} ^{2}+{v^{\prime}_{z}} ^{2}=c^2 \label{eq5} \end{align}$$

Using $x^{\prime}$, $y^{\prime}$ obtained above to calculate $v_{x}^{\prime}$, $v_{y}^{\prime}$ yields the following.

$$\begin{align*} v^{\prime}_{x} &= \frac{dx^{\prime}}{dt^{\prime}}=\frac{e+dc\cos\theta}{a+bc\cos\theta} \\ v^{\prime}_{y} &= \frac{dy^{\prime}}{dt^{\prime}}=\frac{c\sin\theta}{a+bc\cos\theta} \\ v^{\prime}_{z} &= 0 \end{align*}$$

Inserting these three equations into $(5)$ yields the following.

$$\begin{align*} && \left( \frac{e+dc\cos\theta}{a+bc\cos\theta} \right)^{2} + \left( \frac{c\sin\theta}{a+bc\cos\theta} \right) ^{2} &= c^{2} \\ \implies && (e+dc\cos\theta)^{2} + c^{2}\sin^{2}\theta &= c^{2}(a+bc\cos\theta)^{2} \\ \implies && e^{2}+2edc\cos\theta + d^2c^{2}\cos^{2}\theta + {\color{blue}c^{2}\sin^{2}\theta} &= c^2a^{2}+2abc^3\cos\theta+b^2c^4\cos^{2}\theta \\ \implies && e^{2}+2edc\cos\theta + d^2c^{2}\cos^{2}\theta + {\color{blue}c^{2}-c^{2}\cos^{2}\theta} &= c^2a^{2} + 2abc^3\cos\theta + b^2c^4\cos^{2}\theta \end{align*}$$

Grouping the above formula for $\cos\theta$ yields the following.

$$c^{2}(d^{2}-1)\cos^{2}\theta + 2edc\cos\theta + (e^{2}+c^{2}) = b^2c^4\cos^{2}\theta + 2abc^3\cos\theta + a^2c^{2}$$

Therefore, both sides must have the same constant term, coefficient of the first term, and the coefficient of the second term, resulting in the following formula.

$$\begin{align} e^{2}+c^{2} & =a^{2}c^{2} \\ 2edc=2abc^3 \quad &\implies \quad ed = abc^{2} \\ c^{2}(d^{2}-1) = b^{2}c^{4} \quad & \implies \quad d^{2}-1=b^{2}c^{2} \end{align}$$

While we have obtained three conditions, $(6)$, $(7)$, $(8)$, since we have $4$ unknowns to find, we need one more condition. The other equation can be found from the condition when the particle is stationary. The transformation equation for a particle stationary at the origin of the $A$ system is the following.

$$\begin{pmatrix} t^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} a & b & 0 & 0 \\ e & d & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} at \\ et \\ 0 \\ 0 \end{pmatrix} =\begin{pmatrix} t^{\prime} \\ -v_{0}t^{\prime} \\ 0 \\ 0 \end{pmatrix}$$

Therefore, we obtain the following equation.

$$\begin{align*} t^{\prime} &= at \\ et &= -v_{0}t^{\prime}=-v_{0}at \end{align*}$$

Then, we obtain the following equation.

$$\begin{align} e = -v_{0}a \label{eq9} \end{align}$$

If the velocity of the $A^{\prime}$ system is $0$, then because the world line of the $A$ system and the $A^{\prime}$ system are the same, the transformation is as follows.

$$\begin{pmatrix} t^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix}$$

Therefore, when $v_{0}=0$, we obtain the following equation.

$$\begin{align} a=1,\quad b=0,\quad e=0,\quad d=1 \label{eq10} \end{align}$$

Inserting $(9)$ into $(6)$ under the above conditions yields the following.

$$\begin{align*} && \quad{v_{0}}^{2} a^{2}+c^{2} &= c^{2} a^{2} \\ \implies && a^{2} &= \frac{c^{2}}{c^{2}-{v_{0}}^{2}} \\ \implies && a &= \pm\frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} \end{align*}$$

When $v_{0}=0$ by $(10)$, $a=1$, therefore it is as follows.

$$a=\frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}}$$

Inserting the $a$ obtained above into $(9)$ yields the following.

$$e = -v_{0}\frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}}$$

Inserting $(9)$ into $(7)$ yields the following.

$$\begin{align} && ed &= abc^{2} \nonumber \\ \implies && -v_{0} ad &= abc^{2} \nonumber \\ \implies && d &= \frac{bc^{2}}{{-v_{0}}} \end{align}$$

Inserting the $d$ obtained above into $(8)$ yields the following.

$$\begin{align*} && \frac{b^2c^4}{{v_{0}}^{2}}-1 &= b^2c^{2} \\ \implies && \frac{b^2c^4}{{v_{0}}^{2}}-b^2c^{2} &= 1 \\ \implies && \frac{b^2c^{2}}{{v_{0}}^{2}}{(c^{2}-{v_{0}}^{2})} &= 1 \\ \implies && b^{2}&= \frac{{v_{0}}^{2}}{c^{2}(c^{2}-{v_{0}}^{2})} \\ \implies && b &= \pm\frac{v_{0}}{c\sqrt{c2-{v_{0}}^{2}}} \end{align*}$$

Inserting this again into $(11)$ yields the following.

$$d=-\frac{b}{v_{0}}c^{2}=\mp \frac{c^{2}}{v_{0}}\frac{v_{0}}{c\sqrt{c^{2}-{v_{0}}^{2}}}=\mp \frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}}$$

When $v_{0}=0$ by $(10)$, since $d=1$, it is as follows.

$$\begin{align*} d &= \frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} \\ b &= -\frac{v_{0}}{c\sqrt{c^{2}-{v_{0}}^{2}}} \end{align*}$$

Organizing the obtained $a, b, c, d$ yields the following equation.

$$\begin{align*} a &= \frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}}, & b &= -\frac{v_{0}}{c\sqrt{c^{2}-{v_{0}}^{2}}} \\ e &= -v_{0}\frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}}, & d &= \frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} \end{align*}$$

Inserting this into $(1)$ yields the following.

$$\begin{pmatrix} t^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} \dfrac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} & -\dfrac{v_{0}}{c\sqrt{c^{2}-{v_{0}}^{2}}} & 0 & 0 \\ -v_{0} \dfrac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} & \dfrac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix}$$

Here, $\begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix}$ is referred to as a time-space four-vector. However, since $x, y, z$ is a unit of length and only $t$ is a unit of time, to align the units, $ct$ is used instead of $t$ (since time*velocity=distance and $c$ is the speed of light).

$$\begin{array}{ccc} t^{\prime} = \dfrac{c}{\sqrt{\ \ }}t -\dfrac{v_{0}}{c} \dfrac{1}{\sqrt{\ \ }} x && ct^{\prime} =\dfrac{c}{\sqrt{\ \ }}ct - \dfrac{v_{0}}{\sqrt{\ \ }}x \\ & \implies & \\ x^{\prime} =\dfrac{-v_{0}c}{\sqrt{\ \ }}t +\dfrac{c}{\sqrt{\ \ }}x && x^{\prime}= \dfrac{-v_{0}}{\sqrt{\ \ }}ct + \dfrac{c}{\sqrt{ \ \ }}x \end{array}$$

The 4-vector with aligned units is as follows.

$$\begin{pmatrix} ct^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} \dfrac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} & -\dfrac{v_{0}}{\sqrt{c^{2}-{v_{0}}^{2}}} & 0 & 0 \\ \dfrac{-v_{0}}{\sqrt{c^{2}-{v_{0}}^{2}}} & \dfrac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}$$

This transformation is precisely the new transformation that fits into Maxwell’s equations (the speed of light being $c$ in any coordinate system). This transformation is called the Lorentz transformation. However, because it’s too complicated to write in the form above, it’s easier to make the common parts constants. The following $\gamma_{0}$ is called the Lorentz factor.

$$\gamma_{0} \equiv \frac{c}{\sqrt{c^{2}-{v_{0}}^{2}}}=\frac{1}{\sqrt{1-\frac{{v_{0}}^{2}}{c^{2}}}}$$

And let’s state $\beta_{0}$ as follows.

$$\beta_{0} \equiv \frac{v_{0}}{c}$$

Then, the Lorentz factor can be represented more simply.

$$\begin{align*} \gamma_{0} &= \frac{1}{\sqrt{1-{\beta_{0}}^{2}}} \\ \frac{-v_{0}}{\sqrt{c^{2}-{v_{0}}^{2}}} &= \frac{-v_{0}}{c}\frac{1}{\sqrt{1-\frac{{v_{0}}^{2}}{c^{2}}}}=-\gamma_{0}\beta_{0} \end{align*}$$ Inserting it into the Lorentz transformation yields the following.

$$\begin{pmatrix} ct^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix} = \begin{pmatrix} \gamma_{0} & -\gamma_{0}\beta_{0} & 0 & 0 \\ -\gamma_{0}\beta_{0} & \gamma_{0} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}$$

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Explanation

Although the Galilean transformation led to the derivation of the Lorentz transformation due to the parts it couldn’t explain, it isn’t entirely incorrect. If $v_{0}$ is negligibly small compared to the speed of light $c$ in the Lorentz transformation, i.e., if $\frac{v_{0}}{c}=0$, then the Lorentz transformation takes the same form as the Galilean transformation.

$$\begin{align*} \begin{pmatrix} t^{\prime} \\ x^{\prime} \\ y^{\prime} \\ z^{\prime} \end{pmatrix}&=\begin{pmatrix} \dfrac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} & -\dfrac{v_{0}}{c\sqrt{c^{2}-{v_{0}}^{2}}} & 0 & 0 \\ \dfrac{-v_{0}c}{\sqrt{c^{2}-{v_{0}}^{2}}} & \dfrac{c}{\sqrt{c^{2}-{v_{0}}^{2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} \\ &= \begin{pmatrix} \dfrac{1}{\sqrt{1-{(v_{0}/c)}^{2}}} & -\dfrac{v_{0}}{c^{2}}\dfrac{1}{\sqrt{1-{(v_{0}/c)}^{2}}} & 0 & 0 \\ \dfrac{-v_{0}}{\sqrt{1-{(v_{0}/c)}^{2}}} & \dfrac{1}{\sqrt{1-{(v_{0}/c)}^{2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} \\ &= \begin{pmatrix} \dfrac{1}{\sqrt{1-{0}^{2}}} & -0\dfrac{1}{\sqrt{1-{0}^{2}}} & 0 & 0 \\ \dfrac{-v_{0}}{\sqrt{1-{0}^{2}}} & \dfrac{1}{\sqrt{1-{0}^{2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} \\ &=\begin{pmatrix} 1& 0 & 0 & 0 \\ -v_{0} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix} \end{align*}$$