Deriving Lorentz Transformation Using Graphical Method

Spacetime Diagram

I read this derivation in A First Course of General Relativity by Schutz. We are going to take light speed $c=1$, a dimensionaless constant. As before, we start off with the two special relativity postulates

• all laws of physics have the same form in all inertial frames
• light speed is constant for all inertial frames

Let $O$ and $O'$ be two inertial frames in which $O'$ moves at a speed $v$ in positive $x$ direction relative to $O$. Let their origins be the same position initially. The $t'$-axis locates at where $x'=0$ in $O'$ frame. As $O'$ is moving at speed $v$, $t'$-axis should be a slope of $v$ passing through $O$ origin.

In $O'$, a light signal was sent at $t'=-a$ at $x'=0$ to $x'=a$ then reflected back to $x'=0$ at $t'=a$. As light travels at constand speed $c=1$ in all inertial frames, the worldlines have the same slopes in $O$ frame. Their intersection should pass through the $x'$-axis at $x'=a$.

We would like to know the angle between the $x'$-axis and $x$-axis.

For the line $PE$, as we know the speed is $c=1$, slope of $PE$ is 1. Then,

$$y = x +a(\sin\phi-\cos\phi).$$

Similarly, for $RP$,

$$y=-x+a(\cos\phi+\sin\phi).$$

Then, the coordinate of their intersection, $E$, will be

$$\begin{align} P_x + a(\sin\phi-\cos\phi) &= -P_x + a(\cos\phi+\sin\phi) \\ P_x &= a\cos\phi \\ P_y &= a\cos\phi + a(\sin\phi - \cos\phi)\\ &= a\sin\phi. \end{align}$$

Then,

$$\tan\phi'=\frac{a\sin\phi}{a\cos\phi}=\tan\phi.$$ So, $$\phi'=\phi.$$

We now know how to draw a spacetime diagram.

Space-time Interval

Define the spacetime interval as $$\Delta s^2 = -\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2.$$ Similarly, the spacetime interval in $O'$ is $$\Delta s'^2 = -\Delta t'^2 + \Delta x'^2 + \Delta y'^2 + \Delta z'^2.$$ From the graphs, we know that $\Delta x'^{\alpha}$ and $\Delta x^\beta$ are related linearly, so the spaetime interval in $O'$ can be expressed as $$\Delta s'^2 = \sum^3_{\alpha =0}\sum^3_{\beta=0}M_{\alpha\beta}(\Delta x^\alpha)(\Delta x^\beta).$$ For light, $\Delta s^2=0$, $$\Delta t = \Delta r = \left[(\Delta x)^2+(\Delta y)^2+(\Delta z)^2\right]^{1/2}.$$ In $O'$ frame, $$\Delta x'^2 = M_{00}(\Delta r)^2 + 2 \left(\sum^3_{i=1}M_{0i}\Delta x^i\right)\Delta r+\sum^3_{i=1}\sum^3_{j=1}M_{ij}\Delta x^i\Delta x^j.$$ Thus, in order for $\Delta s'^2 = \Delta s^2 = 0$, we have $$\begin{align} M_{0i} &= 0,\, i=1,2,3 \\ M_{ij} &= -M_{00}\delta_{ij}, \, i,j=1,2,3. \end{align}$$ Therefore, $$\Delta s'^2 = M_{00}(\Delta t^2-\Delta x^2-\Delta y^2 -\Delta z^2).$$ Define $$\phi(\vec{v})\equiv M_{00}.$$

Suppose a rod lies on the $y$-axis. Let $A$ be the event of the end of the rod at $t=0$ and $B$ be the event of the start of the rod at $t=0$. Since $y$-axis is perpendicular to the motion of $O'$ frame, $y$-axis and $y'$-axis are parallel to each other. Hence event $A$ and event $B$ lie on the $y'$-axis as well, perpendicular to $t'$-axis. Hence, $$\Delta t'_{AB}=\Delta t_{AB} = 0.$$ The spacetime interval in $O'$ frame is $$\Delta s'_{AB} = (\text{length of rod in }O')^2.$$ So, we have $$(\text{length of rod in }O')^2 = \phi(\vec{v})(\text{length of rod in }O)^2.$$ As the same argument could be made for an inertial frame moving at $-v$ relative to $O$ positive $x$ direction, $$\phi(\vec{v})=\phi(v).$$

Let $O''$ be another inertial frame moving at $-v$ with respect to $x'$ direction of $O'$. So, in fact, $O''$ is essentially identical to $O$ frame. $$\begin{align} \Delta s''^2 &= \phi(v)\Delta s'^2 \\ \Delta s'^2 &= \phi(v)\Delta s^2. \end{align}$$ Substituting $\Delta s'$, we have $$\Delta s''^2 = \phi(v)^2\Delta s^2 = \Delta s^2.$$ Then, $$\phi(v) = \pm 1.$$ As shown before, $$\Delta s'_{AB} = (\text{length of rod in }O')^2 = \phi(\vec{v})(\text{length of rod in }O)^2.$$ and length cannot be negative, $$\phi(v) = 1.$$ We come to the conclusion that $$\bbox[5px,border:2px solid #666] { \Delta s'^2 = \Delta s^2 }$$

Time Dilation

Consider events only on the $tx$ plane. All events lie on the hyperbola $$-t^2+x^2 = -1$$ have the same spacetime interval with respect to origin. In $O'$ frame, 1 unit of time stationary at $x'=0$ in $O'$ frame will have the spacetime interval $$-t'^2 = -1.$$ As spacetime interval is the same in all inertail frames, as proven above, this event also lies on the hyperbola $-t^2+x^2=-1$. Thus, this event lies on the intersection of the hyperbola with the $t'$-axis (i.e. $x'=0$)

On that intersection, $t$ is $$-t^2+x^2=-1,$$ in which $$v = \frac{x}{t}.$$ So, 1 unit of time in $O'$ frame will be observed as $$t = \frac{1}{\sqrt{1-v^2}}$$ units of time in $O$ frame. Thus, the duration in $O'$ observed in $O$ will be $$\bbox[5px,border:2px solid #666] { \Delta t = \frac{\Delta t'}{\sqrt{1-v^2}}. }$$ This is the time dilation.

Length Contraction

At $C$, in $O'$, $$(x',t')=(l,0).$$ The spacetime interval is $$-t^2+x^2=x'^2 = l^2.$$ The $x'$-axis is $$vx=t.$$ So, we know at point $C$ $$\begin{align} x_C &= \frac{l}{\sqrt{1-v^2}} \\ t_C &= \frac{vl}{\sqrt{1-v^2}}. \end{align}$$ As we know the slope of $BC$ is $$\begin{align} \frac{x_C-x_B}{t_c-0} &= v \\ x_B &= x_C - vt_C \\ &= \frac{l}{\sqrt{1-v^2}} - \frac{v^2l}{\sqrt{1-v^2}}\\ &= l\sqrt{1-v^2}. \end{align}$$ This is known as length contraction.

Lorentz Transformation

We know that $x'^\alpha$ relates with $x^\beta$ linearly. Let $$\begin{align} t' &= \alpha t + \beta x \\ x' &= \gamma t + \sigma x \\ y' &= y\\ z' &= z, \end{align}$$ where $\alpha$, $\beta$, $\gamma$ and $sigma$ are constant. The $t'$-axis is a line with equation $$vt - x = 0$$ and the $x'$-axis is a line with equation $$vx - t = 0.$$ At $t'$-axis (i.e. $x'=0$), $$t' = \alpha\left(t + \frac{\beta}{\alpha}x\right) = 0.$$ At $x'$-axis (i.e. $t'=0$, $$t' &= \sigma\left(\frac{\gamma}{\sigma}t + x\right) = 0.$$ So, we know that $$-\frac{\beta}{\alpha} = -\frac{\gamma}{\sigma} = v.$$ Substitute into the equations, we have $$\begin{align} t' &= \alpha(t-vx) \\ x' &= \sigma(x-vt). \end{align}$$ By the invariance of the spacetime interval, $$\begin{align} -\Delta t'^2 + \Delta x^2 &= -\Delta t^2 + \Delta x^2\\ \alpha &= \pm \frac{1}{\sqrt{1-v^2}}. \end{align}$$ When $v=0$, $x'$ was supposed to equal $x$, so the minus is rejected. Then we have the relation $$\begin{align} t' &= \frac{1}{\sqrt{1-v^2}}(t-vx) \\ x' &= \frac{1}{\sqrt{1-v^2}}(x-vt) \\ y' &= y\\ z' &= z. \end{align}$$ This is known as the Lorentz transformation.