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+### Statement
+
+$14.2.13.$ [Insert the problem statement]
+
+### 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'
+
+#### Answer
+
+[Insert a concise answer or boxed result]