Lorentz transformations continued YouTube

Directions perpendicular to the motion

So far, we have only considered transformations in the direction parallel to the axis of motion, but what about the other two spatial directions perpendicular to the motion.

Assuming we carry on restricting motion to the $x$-axis, then the two perpendicular coordinates are unaffected by the motion, $$\begin{align*}y' &= y\\z' &= z\end{align*}$$

Figure 4.1 - Directions perpendicular to the motion

Figure 4.1 - Directions perpendicular to the motion

To see why this is true, suppose we have one frame moving at speed $v$ towards a second, stationary frame. Identical meter sticks have been laid out along the $y$-axes of the two frames, as in figure 2.1.

We want to show that the lengths of the two sticks are the same in both frames.

Figure 4.2 - The view from the centre-of-mass frame

Figure 4.2 - The view from the centre-of-mass frame

Consider the centre-of-mass frame, labelled with subscript $c$ in figure 2.2, which remains exactly midway between the two existing frames.

An observer at rest in this frame will see a completely symmetrical situation. That is, two frames are moving towards the centre-of-mass at speed $\tfrac{v}{2}$.

Figure 4.3 - Endpoints coincide in centre-of-mass frame

Figure 4.3 - Endpoints coincide in centre-of-mass frame

Eventually the two frames will coincide at the origin of the centre-of-mass frame.

Due to the symmetry of the situation, the endpoints of the identical meter sticks must coincide at the events, $E_O, E_s$.

To be sure, suppose that the primed stick was shorter than the unprimed stick as measured in the centre-of-mass frame. The observer could rotate his coordinates $180^{\circ}$ so that his $x$-axis was pointing in the opposite direction to the two moving frames and their roles would be reversed, implying that the primed stick would now be longer than the unprimed stick. This would mean that the original $x$-direction is a special direction of space, something which is not accepted in physics.

For completeness, we note that since the endpoints coincide in the centre-of-mass frame, they must coincide in all frames, due to the invariance of events, and so the length of the sticks must be measured to be the same in all frames. That is, lengths in the $y$- direction are unaffected by motion and we can make the same argument for the $z$-direction.

Hence, the full set of Lorentz transformations, where the primed frame is moving at speed $v$ relative to a stationary frame, along the common $x$-direction is $$\begin{align*}t' &= \frac{t - vx}{\sqrt{1 - v^2}}\\x' &= \frac{x - vt}{\sqrt{1 - v^2}}\\y' &= y\\z' &= z\end{align*}$$

Invariant hyperbolae

The Lorentz transformations stretch and squeeze space-time in certain directions. To see this we can consider how various points and lines are shifted by the transformations. For example, we can consider what happens to light-paths, which have the form $$t \pm x = k$$ in the stationary frame and $$t' \pm x' = k'$$ in the moving frame, for constants $k, k' \in \mathbb{R}$.

Using the Lorentz transformations, $$\begin{align*}t' - x' &= \frac{\left(t - vx\right) - \left(x - vt\right)}{\sqrt{1 - v^2}}\\&= \frac{\left(t - x\right)\left(1 - v\right)}{\sqrt{1 - v^2}}\\&= \left(t - x\right) \sqrt{\frac{1 - v}{1 + v}}\end{align*}$$ and $$\begin{align*}t' + x' &= \frac{\left(t - vx\right) + \left(x - vt\right)}{\sqrt{1 - v^2}}\\&= \frac{\left(t + x\right)\left(1 - v\right)}{\sqrt{1 - v^2}}\\&= \left(t + x\right) \sqrt{\frac{1 + v}{1 - v}}\end{align*}$$

If we write $\alpha(v) = \sqrt{\frac{1 - v}{1 + v}} \lt 1$, then $$\begin{align*}t' - x' &= \alpha(v) \left(t - x\right)\\t' + x' &= \frac{1}{\alpha(v)} \left(t + x\right)\end{align*}$$ which means that the $+x$-direction paths are squeezed together and the $-x$-direction paths are stretched apart.

However, the square of the space-time separation is invariant, that is $$\begin{align*}t'^2 - x'^2 &= \left(t' - x'\right) \left(t' + x'\right)\\&= \alpha(v) \left(t - x\right) \frac{1}{\alpha(v)} \left(t + x\right)\\&= \left(t - x\right) \left(t + x\right)\\&= t^2 - x^2\end{align*}$$

The curves, $$t^2 - x^2 = k^2$$ are called the invariant hyperbolae of Minkowski space-time and are shown in the javascript widget below.

Notice that, as the relative velocity increases, frames (and everything else) are skewed along the hyperbolae, whilst the hyperbolae themselves remain constant.