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.