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			### Statement

			$14.2.13.$
			According to observations from Earth in a spacecraft moving at speed v,
		@@ -5,13 +5,13 @@Statement
			the speed of light did not change, the distances in the direction of movement
			decreased by a factor of γ =√1
			, and in the direction perpendicular to
		−	v
		−	)2
		−	1−(
		−	c
		+
		+	$v)21−(c$
		+
			the movement — did not change. Simultaneous events in the stationary ship
		−	began to occur at different points in time. The time difference is ∆t =
		−	~~γxv2cx~~ is the difference of coordinates in the direction of the ship’s movement.Prove that all these effects follow from the Lorentz transformation $vx= y, z= z, t)γ,x′= (x − vt)γ, y= (t −2c$ where x, γ, z, and t are coordinates and time describing phenomena in a sta-tionary system; x, y, z, and tare coordinates and time describing phenom~~-′′′′~~ena in the system moving with speed v.b. Get the inverse Lorentz transformation: define x, y, z, t by x, y, z, t
		+	began to occur at different points in time. The time difference is $∆t =
		+	γxv2cx $
		+	is the difference of coordinates in the direction of the ship’s movement.Prove that all these effects follow from the Lorentz transformation $vx= y, z= z, t)γ,x′= (x − vt)γ, y= (t −2c$ where x, γ, z, and t are coordinates and time describing phenomena in a sta-tionary system; x, y, z, and tare coordinates and time describing phenomen in the system moving with speed v.b. Get the inverse Lorentz transformation: define x, y, z, t by x, y, z, t
			from′′′′the Lorentz transformation, which is given in p.a. Show that the resultingtransformation confirms the principle of Galileo’s relativity.


			### 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

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