The Lorentz transformation equations help make sense of this situation. Given the lengths, time intervals and event coordinates on the platform, we can compute their values as measured on the train (and vice versa) using the equations below: $$ \begin{array}{rl} x'&= \gamma(x - vt)\\[6pt] t'&= \gamma(t - \frac{v}{c^2}x)\\[6pt] y'&= y\\[6pt] z'&= z\\[6pt] \text{where, }\\ \gamma & = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\\ \text{and, }\\ & (x',y',z',t')\text{ and }(x,y,z,t)\text{ are the spacetime coordinates on train and platform.}\\ \end{array} $$
Note that these equations are valid only when relative motion is along the x-axis. The general transformations for motion along all three axes require a deeper discussion and have been left out of scope.
We now proceed to derive these equations with a help of a few mini-gedankens. The procedure is mostly based on Einstein's booklet, with some extras for clarity.
Derivation of Lorentz Transformations
Let us attach coordinate systems named $K$ and $K'$ to the platform and train respectively. As the train is moving at a constant velocity $v$ past the platform, these systems are also called inertial reference frames (or just frames, from here on). Let the origin of $K'$ be at the center of the train, and the origin of $K$ be at the center of the platform.
At the exact instant when the two Bobs pass each other, let their clocks be set to 0.
When Bob's are together at this common origin point, lightning flashes at a distance $x$ and height $y$ from the center of the platform ($z = 0$). Let us say it takes light to travel $t$ ticks on the platform clock to reach Train-Bob. Let us name this event as $E$. So the coordinates of $E$ on the platform are $(x,y,z,t)$. The same event appears as $E'$ on the train with coordinates $(x',y',z',t')$, where $t'$ is the number of ticks on the train clock.
Given this context, we can complete the derivation in four steps:
- Establish linearity. State the equations for each coordinate of $E'$ in the form $x'=f(x,y,z,t)$, $y' = g(x,y,z,t)$ and so on, where $f$, $g$ etc. are simple linear equations.
- Analyze the motion of the origin of the Train frame.
- Analyze how unit lengths are transformed between frames.
- Replace the coefficients in the equations from step 1 with values obtained in steps 2 and 3 to arrive at the final equations.
Step 1 : State the transformations in linear form
The lightning occurs at point $(x,y,z)$ in space and light takes $t$ ticks of a clock to reach T-Bob. The distance travelled by the light is the radial distance $ct$ from event to T-Bob, so: $$ (ct)^2 = x^2 + y^2 + z^2$$
$z$ is known to be 0, so: $$z' = z = 0 \implies (ct)^2 = x^2 + y^2$$
As observed from the train, the lightning occurs about half a train length away at $(x',y')$ and light from it has to travel a distance $ct' > ct$ to reach T-Bob. Note again, that speed of light is constant for both frames. As $ct'$ is the radial distance from $(x',y')$ to T-Bob (as observed from the train), we have: $$(ct')^2 = (x')^2 + (y')^2$$
$ct < ct'$, as, from the platform perspective, T-Bob is moving towards the source of light, whereas from the train perspective, he is stationary.
Motion along the $x$-axis does not cause any changes to lengths on the $y$-axis. Based on this we have: $$ y = y'\\[6pt] (ct)^2 - x^2 = y^2 = y'^2 = (ct')^2 - (x')^2$$
This equality holds true (or is invariant) for any position of lightning and any position of T-Bob along the x-axis on the train, as should be apparent from the mini-gedanken below:
The quantity $c^2t^2 - x^2$ (rather its three dimensional version $c^2t^2 - x^2 - y^2 -z^2$) is called the spacetime interval between events and remains the same in every inertial reference frame.
Given the invariant nature of spacetime interval between reference frames, the story progresses as below:
$$ \begin{array}{lrll} &(ct)^2 - x^2 &= (ct')^2 - (x')^2\\[6pt] &(ct - x)(ct + x) &= (ct' - x')(ct' + x')\\[12pt] \text{When $x$ $\neq$ $\pm$$ct$, }\\[6pt] &(ct - x)/(ct' - x') &= (ct' + x')/(ct+x)& = A \text{ (some constant)}\\[12pt] \text{With algebra,}\\[6pt] &ct' &= \frac{(1 + A^2 )}{2A}ct - \frac{(1 - A^2)}{2A}x&\ldots (1) \\[6pt] &x' &= \frac{(1 + A^2)}{2A}x - \frac{(1 - A^2)}{2A}ct &\ldots (2) \\[6pt] \end{array} \\[6pt] \begin{array}{lrll} \text{(1) and (2) can be restated as,}\\[6pt] &x' &= ax - bct & \ldots (3)\\[6pt] &ct' &= act - bx & \ldots (4)\\[6pt] \text{where,}\\[6pt] &a &= \frac{(1 + A^2)}{2A}\\[6pt] &b &= \frac{(1 - A^2 )}{2A}\\[6pt] \text{and from the givens,}\\[6pt] &y' &= y & \\[6pt] &z' &= z & \end{array} $$ $$ \text{When $x$ $=$ $\pm$$ct$, we can substitute $x$ in (3) and (4) and confirm that linearity holds. } $$Step 2 : Motion at the origin of the Train reference frame
For the origin of $K'$, $x' = 0$. Substituting this in (3) we get:
Note that the origin of $K'$ is the center of the train, which is moving with velocity $v$ relative to the center of the platform. So $x = vt$.
Step 3 : Transformation of unit lengths
Let us take a snapshot of the train from the platform at a moment in time, say $t = 0$. Substituting $t=0$ in (3), we get:
This means that a unit length on the train is measured as $1/a$ units from the platform.
Now, let us take a snapshot of the platform from the train at time $t' = 0$. Substituting $t' = 0$ in (4) we get:
Substituting $t$ from (8) and $b/a$ from (6) into (3), we get:
This means that a unit length on the platform is measured as $a(1 - \frac{v^2}{c^2})$ from the train.
According to the classical principle of relativity, the two reference frames are equivalent. So a unit length on the train as seen from the platform should be the same as a unit length on the platform as seen from the train. So equating (7) and (9) we get:
The symmetry in length contraction is depicted below:
Note that $a$ is the same $\gamma$ that we derived in the length transformation experiment.
Substituting $\gamma$ from (10) into (6):
Step 4 : Replace coefficients
That is all for now.
Watch this spacetime for more gedank-a-dankens on gravity. They could appear in the near future or maybe sometime in the past. 😉