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| ### Statement |
| ### Statement |
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| $14.2.13.$ [Insert the problem statement] |
| $14.2.13.$ According to observations from Earth in a spacecraft moving at speed v, |
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| the speed of light did not change, the distances in the direction of movement |
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| decreased by a factor of γ =√1 |
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| , and in the direction perpendicular to |
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| )2 |
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| 1−( |
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| c |
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| the movement — did not change. Simultaneous events in the stationary ship |
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| began to occur at different points in time. The time difference is ∆t = |
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| 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 |
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| from′′′′the Lorentz transformation, which is given in p.a. Show that the resultingtransformation confirms the principle of Galileo’s relativity. |
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| ### Solution |
| ### Solution |
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| We start part by part |
| We start part by part |
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| Part a: Relativistic effects from the Lorentz transformation |
| Part a: Relativistic effects from the Lorentz transformation |
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| Direct Lorentz transformation (S → S'): |
| Direct Lorentz transformation (S → S'): |
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| x' = \gamma (x - vt), \quad y' = y, \quad z' = z, \quad t' = \gamma \left( t - \frac{vx}{c^2} \right) |
| $x' = \gamma (x - vt), \quad y' = y, \quad z' = z, \quad t' = \gamma \left( t - \frac{vx}{c^2} \right)$ |
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| where \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}. |
| $where \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$. |
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| Invariance of the speed of light |
| Invariance of the speed of light |
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| In S: light front according to x = ct. |
| In S: light front according to$ x = ct$. |
| Substitute into x', t': |
| Substitute into $x', t'$: |
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| x' = \gamma (ct - vt) = \gamma t (c - v) |
| $x' = \gamma (ct - vt) = \gamma t (c - v)$ |
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| t' = \gamma \left( t - \frac{v(ct)}{c^2} \right) = \gamma t \left(1 - \frac{v}{c}\right) |
| $t' = \gamma \left( t - \frac{v(ct)}{c^2} \right) = \gamma t \left(1 - \frac{v}{c}\right)$ |
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| Speed in S': |
| Speed in S': |
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| 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 |
| $c' = \frac{x'}{t'} = \frac{\gamma t (c - v)}{\gamma t (1 - v/c)} = \frac{c - v}{1 - v/c} = \frac{c - v}{(c - v)/c} = c$ |
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| We can see that the speed of light is c in both systems. |
| We can see that the speed of light is c in both systems. |
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| Length contraction |
| Length contraction |
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| A rod at rest in S' has fixed ends x_1', x_2'. Its proper length is L_0 = x_2' - x_1'. |
| 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. |
| In S we measure its ends$ x_1, x_2$ at the same instant t. |
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| From the transformation: |
| From the transformation: |
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| x_1' = \gamma (x_1 - vt), \quad x_2' = \gamma (x_2 - vt) |
| $x_1' = \gamma (x_1 - vt), \quad x_2' = \gamma (x_2 - vt)$ |
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| L_0 = x_2' - x_1' = \gamma (x_2 - x_1) = \gamma L |
| $L_0 = x_2' - x_1' = \gamma (x_2 - x_1) = \gamma L$ |
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| \Rightarrow L = L_0 / \gamma. |
| $\Rightarrow L = L_0 / \gamma$. |
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| The length measured in S is smaller. |
| The length measured in S is smaller. |
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| Invariance of perpendicular distances |
| Invariance of perpendicular distances |
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| y' = y, z' = z directly. No change. |
| $y' = y, z' = z $directly. No change. |
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| Relativity of simultaneity |
| Relativity of simultaneity |
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| Events simultaneous in S': t_1' = t_2'. Use the inverse transformation to express t: |
| Events simultaneous in $S': t_1' = t_2'$. Use the inverse transformation to express t: |
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| t = \gamma \left( t' + \frac{v}{c^2} x' \right) |
| $t = \gamma \left( t' + \frac{v}{c^2} x' \right)$ |
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| For two events: |
| For two events: |
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| t_1 - t_2 = \gamma \left[ (t_1' - t_2') + \frac{v}{c^2} (x_1' - x_2') \right] |
| $t_1 - t_2 = \gamma \left[ (t_1' - t_2') + \frac{v}{c^2} (x_1' - x_2') \right]$ |
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| With t_1' = t_2': |
| With$ t_1' = t_2'$: |
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| \Delta t = t_1 - t_2 = \frac{\gamma v}{c^2} (x_1' - x_2') = -\frac{\gamma v}{c^2} \Delta x' |
| $\Delta t = t_1 - t_2 = \frac{\gamma v}{c^2} (x_1' - x_2') = -\frac{\gamma v}{c^2} \Delta x'$ |
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| The magnitude of the time difference is \frac{\gamma v}{c^2} |\Delta x'|, where \Delta x' is the separation in the ship. |
| The magnitude of the time difference is $\frac{\gamma v}{c^2} |\Delta x'|, where \Delta x'$ is the separation in the ship. |
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| Part b: Inverse Lorentz transformation |
| Part b: Inverse Lorentz transformation |
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| From: |
| From: |
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| x' = \gamma(x - vt), \quad t' = \gamma\left(t - \frac{vx}{c^2}\right) |
| $x' = \gamma(x - vt), \quad t' = \gamma\left(t - \frac{vx}{c^2}\right)$ |
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| Solve for x, t in terms of x', t'. |
| Solve for x, t in terms of x', t'. |
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| From the first: x = \frac{x'}{\gamma} + vt. |
| From the first:$ x = \frac{x'}{\gamma} + vt$ |
| From the second: t = \frac{t'}{\gamma} + \frac{vx}{c^2}. |
| From the second:$ t = \frac{t'}{\gamma} + \frac{vx}{c^2}$. |
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| Substitute t into x: |
| Substitute t into x: |
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| x = \frac{x'}{\gamma} + v\left( \frac{t'}{\gamma} + \frac{vx}{c^2} \right) = \frac{x' + vt'}{\gamma} + \frac{v^2}{c^2}x |
| $x = \frac{x'}{\gamma} + v\left( \frac{t'}{\gamma} + \frac{vx}{c^2} \right) = \frac{x' + vt'}{\gamma} + \frac{v^2}{c^2}x$ |
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| 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} |
| $x - \frac{v^2}{c^2}x = \frac{x' + vt'}{\gamma} \;\Rightarrow\; x\left(1 - \frac{v^2}{c^2}\right) = \frac{x' + vt'}{\gamma}$ |
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| Since 1 - v^2/c^2 = 1/\gamma^2: |
| Since$ 1 - v^2/c^2 = 1/\gamma^2$: |
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| \frac{x}{\gamma^2} = \frac{x' + vt'}{\gamma} \;\Rightarrow\; x = \gamma (x' + vt') |
| $\frac{x}{\gamma^2} = \frac{x' + vt'}{\gamma} \;\Rightarrow\; x = \gamma (x' + vt')$ |
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| Now for t: |
| Now for t: |
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| 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' |
| $t = \frac{t'}{\gamma} + \frac{v}{c^2} \gamma (x' + vt') = \frac{t'}{\gamma} + \gamma \frac{v}{c^2}x' + \gamma \frac{v^2}{c^2}t'$ |
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| The coefficient of t' is: |
| The coefficient of t' is: |
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| \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 |
| $\frac{1}{\gamma} + \frac{\gamma v^2}{c^2} = \gamma\left( \frac{1}{\gamma^2} + \frac{v^2}{c^2} \right) = \gamma (1 - \beta^2 + \beta^2) = \gamma$ |
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| Thus: |
| Thus: |
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| t = \gamma \left( t' + \frac{v}{c^2} x' \right) |
| $t = \gamma \left( t' + \frac{v}{c^2} x' \right)$ |
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| Inverse Lorentz transformation: |
| Inverse Lorentz transformation: |
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| \boxed{x = \gamma (x' + vt'), \quad y = y', \quad z = z', \quad t = \gamma \left( t' + \frac{v}{c^2} x' \right)} |
| $\boxed{x = \gamma (x' + vt'), \quad y = y', \quad z = z', \quad t = \gamma \left( t' + \frac{v}{c^2} x' \right)}$ |
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| Confirmation of the principle of relativity |
| Confirmation of the principle of relativity |
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| The inverse transformation is identical to the direct one by replacing v \to -v. |
| 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. |
| 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: |
| In the limit$ c \to \infty (\gamma \to 1, v/c^2 $\to 0), we recover the Galilean transformation: |
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| x = x' + vt', \quad t = t' |
| $x = x' + vt', \quad t = t'$ |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| $\boxed{x = \gamma (x' + vt'), \quad y = y', \quad z = z', \quad t = \gamma \left( t' + \frac{v}{c^2} x' \right)}$ |
