Solution

We start part by part

Part a: Relativistic effects from the Lorentz transformation

Direct Lorentz transformation (S → S'):

$x' = \gamma (x - vt), \quad y' = y, \quad z' = z, \quad t' = \gamma \left( t - \frac{vx}{c^2} \right)$

$where \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$.

Invariance of the speed of light

In S: light front according to$x = ct$.
Substitute into $x', t'$:

$x' = \gamma (ct - vt) = \gamma t (c - v)$

$t' = \gamma \left( t - \frac{v(ct)}{c^2} \right) = \gamma t \left(1 - \frac{v}{c}\right)$

Speed in S':

$c' = \frac{x'}{t'} = \frac{\gamma t (c - v)}{\gamma t (1 - v/c)} = \frac{c - v}{1 - v/c} = \frac{c - v}{(c - v)/c} = c$

We can see that the speed of light is c in both systems.

Length contraction

A rod at rest in S' has fixed ends $x_1', x_2'$. Its proper length is $L_0 = x_2' - x_1'$.
In S we measure its ends$x_1, x_2$ at the same instant t.

From the transformation:

$x_1' = \gamma (x_1 - vt), \quad x_2' = \gamma (x_2 - vt)$

$L_0 = x_2' - x_1' = \gamma (x_2 - x_1) = \gamma L$

$\Rightarrow L = L_0 / \gamma$.

The length measured in S is smaller.

Invariance of perpendicular distances

$y' = y, z' = z$directly. No change.

Relativity of simultaneity

Events simultaneous in $S': t_1' = t_2'$. Use the inverse transformation to express t:

$t = \gamma \left( t' + \frac{v}{c^2} x' \right)$

For two events:

$t_1 - t_2 = \gamma \left[ (t_1' - t_2') + \frac{v}{c^2} (x_1' - x_2') \right]$

With$t_1' = t_2'$:

$\Delta t = t_1 - t_2 = \frac{\gamma v}{c^2} (x_1' - x_2') = -\frac{\gamma v}{c^2} \Delta x'$

The magnitude of the time difference is $\frac{\gamma v}{c^2} |\Delta x'|, where \Delta x'$ is the separation in the ship.

Part b: Inverse Lorentz transformation

From:

$x' = \gamma(x - vt), \quad t' = \gamma\left(t - \frac{vx}{c^2}\right)$

Solve for x, t in terms of x', t'.

From the first:$x = \frac{x'}{\gamma} + vt$
From the second:$t = \frac{t'}{\gamma} + \frac{vx}{c^2}$.

Substitute t into x:

$x = \frac{x'}{\gamma} + v\left( \frac{t'}{\gamma} + \frac{vx}{c^2} \right) = \frac{x' + vt'}{\gamma} + \frac{v^2}{c^2}x$

$x - \frac{v^2}{c^2}x = \frac{x' + vt'}{\gamma} \;\Rightarrow\; x\left(1 - \frac{v^2}{c^2}\right) = \frac{x' + vt'}{\gamma}$

Since$1 - v^2/c^2 = 1/\gamma^2$:

$\frac{x}{\gamma^2} = \frac{x' + vt'}{\gamma} \;\Rightarrow\; x = \gamma (x' + vt')$

Now for t:

$t = \frac{t'}{\gamma} + \frac{v}{c^2} \gamma (x' + vt') = \frac{t'}{\gamma} + \gamma \frac{v}{c^2}x' + \gamma \frac{v^2}{c^2}t'$

The coefficient of t' is:

$\frac{1}{\gamma} + \frac{\gamma v^2}{c^2} = \gamma\left( \frac{1}{\gamma^2} + \frac{v^2}{c^2} \right) = \gamma (1 - \beta^2 + \beta^2) = \gamma$

Thus:

$t = \gamma \left( t' + \frac{v}{c^2} x' \right)$

Inverse Lorentz transformation:

$\boxed{x = \gamma (x' + vt'), \quad y = y', \quad z = z', \quad t = \gamma \left( t' + \frac{v}{c^2} x' \right)}$

Confirmation of the principle of relativity

The inverse transformation is identical to the direct one by replacing v \to -v.
This means that if S' moves with velocity v relative to S, then S moves with velocity -v relative to S'. No system is privileged; the physical laws have the same form in both.
In the limit$c \to \infty (\gamma \to 1, v/c^2$\to 0), we recover the Galilean transformation:

$x = x' + vt', \quad t = t'$