| v | | v |
| )2 | | )2 |
| 1−( | | 1−( |
| c | | c |
| the movement — did not change. Simultaneous events in the stationary ship | | the movement — did not change. Simultaneous events in the stationary ship |
| began to occur at different points in time. The time difference is ∆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 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 | | 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 |
| from′′′′the Lorentz transformation, which is given in p.a. Show that the resultingtransformation confirms the principle of Galileo’s relativity. | | 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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| $\boxed{x = \gamma (x' + vt'), \quad y = y', \quad z = z', \quad t = \gamma \left( t' + \frac{v}{c^2} x' \right)}$ | | $\boxed{x = \gamma (x' + vt'), \quad y = y', \quad z = z', \quad t = \gamma \left( t' + \frac{v}{c^2} x' \right)}$ |
