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'