Lorentz transformation Solutions

Quick note:§

Remembers that in previous articles, we defined the lorentz transformation using

$$\Lambda^T\eta\Lambda = \eta$$

This immidieatly hint us to have $\eta$ in the calculation of invariant, as the definition above is already about invariant of $\Lambda$ no matter what it is it should obey the definition. Let's sandwich it with $X$ and note that $X' = \Lambda X$.

$$\begin{align*} X^T\Lambda^T\eta\Lambda X &= X^T\eta X \\ (\Lambda X)^T\eta(\Lambda X) &= X^T\eta X \\ (X')^T\eta(X') &= X^T\eta X \\ \end{align*}$$

Yeah, we see that indeed $X^T\eta X$ is invariant, write it in index notation it is $\eta_{\mu\nu}X^\nu X^\mu$ being invariant. And it is

$$\dd s^2 := -c^2\dd t^2 + \dd x^2 + \dd y^2 + \dd z^2$$

This invariant, we call it proper length squared, whos actual meaning will be clear later. Although from first principle this is not how we defined lorentz transformation. But since this invariant in universally true, we might as well treat it as an alternative definition of lorentz transformation. A transformation that keep this value invariant.

And note that we can form an invariant of another unit ( of time ) rather than lenght by dividing through $-c^2$. We will get

$$d\tau^2 := dt^2 - \frac{1}{c^2}(\dd x^2 + \dd y^2 + \dd z^2)$$

Again, we will see why is this the proper time later.

Lorentz transformation: general structure§

Let's remind ourselves that we have a generator $G$ for the Lorentz transformation

$$\Lambda = e^{-iG}$$

(using a different convention from previous articles, but the structure stays the same). We have the generator

$$G=i\begin{pmatrix} 0 & \phi_x & \phi_y & \phi_z \\ \phi_x & 0 & -\theta_z & \theta_y \\ \phi_y & \theta_z & 0 & -\theta_x \\ \phi_z & -\theta_y & \theta_x & 0\\ \end{pmatrix}$$

This is the general $\mathfrak{so}(1,3)$, the Lie algebra of $SO^+(1,3)$ — the proper, orthochronous Lorentz group, the part reachable continuously from the identity by exponentiating a generator like $G$. (The full Lorentz group $O(1,3)$ also contains parity and time reversal, sitting in disconnected pieces that no real $G$ can reach this way — we won't need them here.) We split $G$ into six basis generators, exactly matching the six degrees of freedom: three boosts, three rotations.

$$K_x = \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \qquad K_y = \begin{pmatrix} 0 & 0 & i & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \qquad K_z = \begin{pmatrix} 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \end{pmatrix}$$ $$J_x = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \end{pmatrix} \qquad J_y = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \end{pmatrix} \qquad J_z = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

so that

$$G=\phi_xK_x + \phi_yK_y + \phi_zK_z + \theta_xJ_x + \theta_yJ_y + \theta_zJ_z = \boldsymbol\phi\cdot \mathbf K + \boldsymbol\theta\cdot \mathbf J$$

with the meaning of $\boldsymbol\phi$ and $\boldsymbol\theta$ now immediate: one parametrizes boosts, the other rotations.

These six generators don't just sit next to each other — they talk to each other:

$$[J_i,J_j]=i\epsilon_{ijk}J_k, \qquad [J_i,K_j]=i\epsilon_{ijk}K_k, \qquad [K_i,K_j]=-i\epsilon_{ijk}J_k$$

Keep that last relation in your pocket. We won't need it for a while, but it is not decorative.

A fully general six-parameter Lorentz transformation can be written down by exponentiating $G$ directly, but the closed form carries no physical insight on its own. So instead of solving the general case, we'll build up from the physically transparent one. Homogeneity and isotropy of space mean we only need to understand a boost with no rotation — so set $\boldsymbol\theta = 0$ — and by isotropy, any single boost direction is as good as any other. We start with one dimension.

1D boost§

Solution§

Let only $\phi_x$ be nonzero. Call it simply $\phi$, and its generator simply $K$, so that in one dimension

$$K = i\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

Notice

$$K^2 = -I \qquad K^3 = -K \qquad K^4=I \qquad K^5=K \quad\ldots$$

which turns the exponential into a familiar pair of series:

$$\Lambda=e^{-i\phi K}=I-i\phi K + \frac{1}{2!}\phi^2 I - i\frac{1}{3!}\phi^3K + \frac{1}{4!}\phi^4 I+ \dots = I\cosh\phi-iK\sinh\phi$$

so that

$$\Lambda = \begin{pmatrix}\cosh\phi & \sinh\phi \\ \sinh\phi & \cosh\phi\end{pmatrix}$$

Rapidity and velocity§

$\phi$ is just a parameter so far — let's find out what it physically means. Boost into the frame of an object moving as $x=vt$, and demand that in the new frame the object sits still: $x' = 0$.

$$\begin{align*} ct\sinh\phi + vt\cosh\phi &= 0 \\ \beta := \frac{v}{c} &= -\tanh\phi \end{align*}$$

Set $\gamma := \cosh\phi = (1-\beta^2)^{-1/2}$, so $\sinh\phi = -\gamma\beta$, giving

$$\Lambda = \gamma\begin{pmatrix} 1 & -\beta \\ -\beta & 1 \end{pmatrix}$$

with $\gamma \ge 1$, $|\beta| < 1$. Notice $\gamma$ doesn't care about the sign of $v$ — the transformation matrix has the same structure whether you boost forward or backward, which is just isotropy of space showing up again. That also tells you $\Lambda^{-1}(\beta)=\Lambda(-\beta)$: undoing a boost is the same as boosting the other way.

The parameter $\phi$ itself has a name — rapidity — and unlike velocity, rapidities for collinear boosts simply add. We won't need that fact yet, but it's worth knowing it's there.

Physical consequences§

Everything that follows — time dilation, simultaneity, length contraction — comes out of exactly the same two equations:

$$c\dd t' = \gamma(c\dd t - \beta \dd x), \qquad \dd x' = \gamma(\dd x - \beta c\dd t)$$

What changes from one effect to the next is only which pair of events we plug in, and which simultaneity condition we impose. Watch how much mileage we get out of the same formula.

Time dilation§

Take a clock sitting still in $S$, so its worldline has $\dd x = 0$. Then

$$\begin{align*} c\dd t' &= c\Lambda^t_\mu\,\dd X^\mu \\ &= \gamma(c\dd t-\beta\dd x) \\ &= \gamma c \dd t \end{align*}$$

Since $S$ is the clock's own rest frame, $\dd t$ here is its proper time $\dd\tau$ ( defined as the time measured in the rest frame ), so really

$$\dd t' = \gamma\, \dd\tau, \qquad\text{i.e.}\qquad \Delta t = \gamma\,\Delta\tau$$

The moving observer watches the stationary clock accumulate less proper time than their own coordinate time ticks off — a moving clock runs slow. It's worth stressing that $\dd t' = \gamma \dd t$ is not the universal statement of time dilation; it only holds because we chose $\dd x = 0$. The frame-independent statement is the one with $\Delta\tau$ in it.

And let's we call we call $-\dd s^2/c^2 = \dd\tau^2$ and we can see it why here, since it is reference frame independant ( invariant ), we use the rest reference frame ($(\dd x^i)^2 = 0$). Then we find

$$-\dd s^2/c^2 = \dd t^2 = \dd\tau^2$$

Length contraction (the wrong way to see the concept)§

Now for length — and here we need to slow down, because length is a trickier thing to define than time. Time can be read off along a single worldline. Length can't: it needs the positions of two endpoints, at the same time, in whichever frame is doing the measuring.

If you note down where the front of a train is in the morning, and where its back is at night after it's crossed an entire country, that difference is not the train's length — you've just measured the gap between two unrelated events. So the measurement has to be simultaneous. Let's see what happens if we're careless about whose simultaneity we mean.

Take two endpoint events simultaneous in $S$, so $\dd t = 0$, and push them through the spatial transformation:

$$\begin{align*} \dd x' &= \Lambda^x_\mu \dd X^\mu \\ &= \gamma(\dd x-\beta c\dd t) \\ &= \gamma \dd x \end{align*}$$

So the moving object comes out longer by a factor of $\gamma$. That should make you suspicious — length contraction is supposed to shrink things, not stretch them. Something in this calculation isn't measuring what we think it's measuring.

Relativity of simultaneity§

Here's the catch. We fixed $\dd t = 0$ — simultaneous in $S$ — but we never checked whether those same two events are simultaneous in $S'$. Let's check:

$$\begin{align*} c\dd t' &= c\Lambda^t_\mu\,\dd X^\mu \\ &= \gamma(c\dd t - \beta \dd x)\\ &= -\gamma\beta\dd x \end{align*}$$

Since $\dd x \neq 0$ for two distinct endpoints, $\dd t' \neq 0$. The two events we used — simultaneous in $S$ — are not simultaneous in $S'$. That's exactly the mistake in the train story above: we measured one end and the other end at different times, just dressed up in Lorentz-transformation language instead of morning-and-night language. What we computed as "$\gamma \dd x$" was never the train's length in $S'$ at all.

The same logic runs in reverse. Force $\dd t' = 0$ instead — simultaneous in the moving frame — and see what that requires of $\dd t$:

$$\begin{align*} c\dd t' = 0 &= c\Lambda^t_\mu\,\dd X^\mu \\ &= \gamma(c\dd t-\beta\dd x) \\ c\dd t &= \beta\dd x \end{align*}$$

So events simultaneous in $S'$ are generally not simultaneous in $S$ either. Simultaneity isn't something the two frames agree on — it's frame-dependent, full stop, and it was hiding inside the Lorentz transformation the whole time.

Length contraction, done properly§

Armed with that, let's redo the measurement honestly. If we want the train's proper length, we need to measure both ends at the same time in the train's own rest frame $S'$ — that means $\dd t' = 0$, not $\dd t = 0$. Using the relation we just derived, $c\dd t = \beta \dd x$, substitute into the spatial transformation:

$$\begin{align*} \dd x' &= \Lambda^x_\mu \dd X^\mu \\ &= \gamma(\dd x-\beta c\dd t) \\ &= \gamma(\dd x - \beta^2 \dd x)\\ &= \gamma(1-\beta^2 )\dd x \\ &= \frac{1}{\gamma}\dd x \end{align*}$$

And we write $\dd x$ as the proper length $L_0$ and $\dd x'$ as the measured length. Where the proper length is deifned as the length measreud in the rest frame.

$$L = \frac{L_0}{\gamma}$$

This is length contraction. The earlier factor of $\gamma$ wasn't wrong arithmetic — it was the right arithmetic applied to the wrong pair of events. The deeper lesson: a Lorentz transformation doesn't just rescale numbers, it also reshuffles which events count as simultaneous, and length contraction is really a simultaneity effect wearing a geometry costume.

And since, now proper length is defined, let's see why the invariant $\dd s^2$ is the proper length ( squared ). Well, since it is invariant ( it's value doesn't depend on the reference frame ), we can evaluate it's value in any reference frame that makes the meaning trivial. We chose the frame that makes $dt = 0$, Which measure the length of the object in the rest frame, which by definition of proper length.

$$\dd s^2 = -c^2\dd t^2 + \dd x^2 + \dd y^2 + \dd z^2 = \dd x^2 + \dd y^2 + \dd z^2 = L_0^2$$

Boost of a moving object§

Now let's keep the velocity explicit instead of setting it to zero. Suppose an object moves at velocity $u$ in $S$, so $\dd x = u\dd t$, and we boost by $v$:

$$\begin{align*} \dd x' &= \Lambda^x_\mu \dd X^\mu \\ &= \gamma(\dd x-\beta c\dd t) \\ &= \gamma(u-v)\dd t \\ \frac{\dd x'}{\dd t'} &= \gamma(u-v)\frac{\dd t}{\dd t'} \\ u' &= \gamma (u-v) \frac{\dd t}{\gamma\left(\dd t - \dfrac{v}{c^2}\dd x\right)} \\ u' &= \frac{u-v}{1 - \dfrac{uv}{c^2}} \end{align*}$$

If $u,v \ll c$, the denominator collapses to $1$ and we recover the Galilean $u' \approx u - v$, exactly as it should.

Boost of light§

Now push it to the extreme: set $u = c$.

$$u'=\frac{c - v}{1-\dfrac{v}{c}}=c$$

Light stays light speed no matter how you boost. That's not an extra rule bolted on afterward — it falls straight out of the same transformation we've been using all along.

Doppler effect§

A source at rest in $S$ emits with period $T_0$ — its proper period. Time dilation alone tells us the observed period lengthens to $T = \gamma T_0$. But that's not the whole story: while the wave takes time to cross the widening gap, the observer (moving away, say) keeps retreating further, so each successive crest has a little farther to travel than the last. That extra travel time is $t = s/c = (v/c)T$, and folding it in gives

$$T_\text{obs}=(1 + \beta)\gamma T_0 = T_0\sqrt\frac{1+\beta}{1-\beta}$$

so the observed frequency is

$$f_\text{obs}=\sqrt\frac{1-\beta}{1+\beta}\,f_0$$

This is the longitudinal Doppler effect — receding along the line of sight gives you both the time-dilation piece and the widening-gap piece together. If instead the motion is purely transverse — perpendicular to the line of sight, so the separation isn't changing — that second piece vanishes and only the dilation survives:

$$f_\text{obs}=\frac{1}{\gamma}f_0$$

Aberration§

Doppler changes a wave's frequency. The same boost, applied to a different object, also changes the direction it appears to arrive from — that's aberration.

Let a photon travel in $S$ at angle $\theta$ to the $x$-axis, so its momentum components are $p_x=p\cos\theta$, $p_y=p\sin\theta$, with $E=pc$. Photon momentum transforms as a four-vector, exactly the way $(ct,x)$ did — we haven't shown why $(E/c,\,p_x,\,p_y,\,p_z)$ forms a four-vector yet, that's a later article, so take it on credit for now:

$$p_x' = \gamma(p_x - \beta E/c), \qquad E'/c = \gamma(E/c - \beta p_x)$$

Divide the first by the second, and use $p_x/(E/c)=\cos\theta$:

$$\cos\theta' = \frac{\cos\theta - \beta}{1-\beta\cos\theta}$$

A source moving toward you doesn't just blueshift — its light also gets funneled forward, toward $\theta'=0$, no matter which direction it was actually emitted in. Push $\beta\to1$ and nearly everything a fast-moving source radiates ends up crammed into a narrow forward cone. That's the headlight effect, and it's the same $\Lambda$ we've had since the first page.

2D boost§

By isotropy, a boost pointed in some other single direction is no more interesting than the 1D case — the math gets uglier, the physics doesn't. The only reason 2D boosts deserve their own section is what happens when you chain two boosts in different, non-collinear directions. (Chain two boosts in the same direction and you just get one bigger boost — nothing new there.)

So look back at the generator, and consider composing two boosts:

$$\Lambda = e^{-i\phi_x K_x}e^{-i\phi_y K_y}$$

For small $\phi_x, \phi_y$, expand to leading order:

$$\Lambda = (I-i\phi_xK_x)(I-i\phi_yK_y) = I - i(\phi_xK_x+\phi_yK_y) - \phi_x\phi_y K_xK_y + O(\phi^3)$$

This is where that commutation relation we set aside earlier finally gets used. Recall $[K_x,K_y] = -iJ_z$, so $K_xK_y = K_yK_x - iJ_z$, and substituting,

$$\Lambda = I - i(\phi_xK_x+\phi_yK_y) - \phi_x\phi_y K_yK_x + i\phi_x\phi_y J_z + O(\phi^3)$$

There it is: a $J_z$ term. Two pure boosts, composed, have produced a piece of a rotation.

(That's only the leading term. For finite rapidities, the exact rotation angle $\theta_W$ between two perpendicular boosts of rapidity $\phi$ and $\chi$ works out to $\tan(\theta_W/2)=\tanh(\phi/2)\tanh(\chi/2)$, which reduces to our $\phi_x\phi_y$ result once both are small. We won't derive the finite case here, but it's worth knowing the infinitesimal calculation isn't hiding anything qualitatively different from the exact one.)

Here's the physical picture behind that algebra. Take a rod already moving along $x$, and in its own instantaneous rest frame, give both of its ends a simultaneous sideways kick in $y$ — a "rigid" perpendicular push. Simultaneous in that frame, that is. That frame is reached from the lab by an $x$-boost, and an $x$-boost's relativity of simultaneity depends on $x$-position — so two events that share a $y$ or $z$ coordinate but differ in $x$, simultaneous in the boosted frame, are not simultaneous back in the lab. Since the rod's two ends are separated along $x$ — the same direction as the boost it's already carrying — the "simultaneous" kick lands on one end before the other, as clocked in the lab. By the time you look at the whole rod at a single lab instant, one end has had a head start accelerating sideways and the other hasn't caught up — so the rod appears to have tilted, not because anything physically twisted it, but because the two kicks were smeared apart in time. And this only shows up when the boost you add is genuinely non-collinear with the one already there — a further kick purely along $x$, collinear with the first boost, never desynchronizes anything.

This composition of non-collinear boosts producing a rotation is called a Wigner rotation. Let an object's velocity change continuously — think of it as an unbroken chain of these infinitesimal non-collinear boosts — and the accumulated rotation of its comoving frame is what we call Thomas precession.

$$\text{non-commuting boosts} \;\longrightarrow\; \text{Wigner rotation} \;\longrightarrow\; \text{Thomas precession (continuous acceleration)}$$

It's worth stating the group-theoretic version of what just happened: pure boosts do not close under composition. Compose two non-collinear ones and you land outside the set of pure boosts entirely, in something with a rotation baked in. Boosts alone aren't a subgroup of the Lorentz group — which is exactly why $\mathfrak{so}(1,3)$ needed both $\mathbf K$ and $\mathbf J$ from the very first page. $\mathbf K$ by itself doesn't close under commutation; $[K_i,K_j]\propto J_k$ kicks you straight back out into rotation territory.

So that commutation relation we asked you to remember, $[K_i, K_j] = -i\epsilon_{ijk}J_k$, was never just bookkeeping. At the infinitesimal level it's already saying: change the direction you're boosting in, and a rotation is unavoidable.

This isn't just a geometric curiosity. The textbook non-relativistic calculation of spin-orbit coupling in hydrogen — the electron's spin interacting with the effective magnetic field produced by its own orbit — comes out a factor of 2 too large compared to experiment, until you account for the fact that the electron's rest frame is itself undergoing exactly this kind of precession as it orbits the nucleus. Thomas precession supplies the missing factor of $1/2$. A purely kinematic fact about composing boosts turns out to matter for the fine structure of every atom you're made of.