| @@ -12,24 +12,38 @@Solution |
| We have: |
| We have: |
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| · System S: attached to the first station (E1), at rest. |
| · System S: attached to the first station (E1), at rest. |
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| · System S': attached to the second station (E2), which moves with velocity v along the x-axis relative to S. |
| · System S': attached to the second station (E2), which moves with velocity v along the x-axis relative to S. |
| We consider$ v > 0$ if E2 moves away from E1. |
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| We consider$ v > 0$ if E2 moves away from E1. |
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| · At $t = 0$ (time of S), E1 is at$ x = 0$ and E2 is at $x = L$. |
| · At $t = 0$ (time of S), E1 is at$ x = 0$ and E2 is at $x = L$. |
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| · From E1, the following are emitted simultaneously: |
| · From E1, the following are emitted simultaneously: |
| · A light pulse (speed c). |
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| · A test probe (constant speed u along x, to be determined). |
| · A light pulse (speed c). |
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| · A test probe (constant speed u along x, to be determined). |
| · The light travels: E1 → E2 (bounce) → E1 (bounce) → E2. |
| · The light travels: E1 → E2 (bounce) → E1 (bounce) → E2. |
| The probe arrives at E2 exactly at the same instant that the light pulse returns to E2 for the third time. |
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| The probe arrives at E2 exactly at the same instant that the light pulse returns to E2 for the third time. |
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| The position of E2 as a function of time:$ x_{\text{E2}}(t) = L + v t$. |
| The position of E2 as a function of time:$ x_{\text{E2}}(t) = L + v t$. |
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| · First leg (E1 → E2): |
| · First leg (E1 → E2): |
| The light catches up to E2 when $c t_1 = L + v t_1 \Rightarrow t_1 = \dfrac{L}{c - v}.$ |
| The light catches up to E2 when $c t_1 = L + v t_1 \Rightarrow t_1 = \dfrac{L}{c - v}.$ |
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| · Second leg (E2 → E1): |
| · Second leg (E2 → E1): |
| At $t_1$, the light reflects and returns to E1. The position of E1 is $x = 0$. |
| At $t_1$, the light reflects and returns to E1. The position of E1 is $x = 0$. |
| $ 0 = c t_1 - c(t_2 - t_1) \Rightarrow t_2 = 2t_1 = \dfrac{2L}{c - v}.$ |
| $ 0 = c t_1 - c(t_2 - t_1) \Rightarrow t_2 = 2t_1 = \dfrac{2L}{c - v}.$ |
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| · Third leg (E1 → E2): |
| · Third leg (E1 → E2): |
| The light leaves E1 again at $t_2 $and reaches E2 at$ t_3$. |
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| The light leaves E1 again at $t_2 $and reaches E2 at$ t_3$. |
| $ c(t_3 - t_2) = L + v t_2 + v(t_3 - t_2)$. |
| $ c(t_3 - t_2) = L + v t_2 + v(t_3 - t_2)$. |
| Solving:$ t_3 - t_2 = \dfrac{L + v t_2}{c - v}$. |
| Solving:$ t_3 - t_2 = \dfrac{L + v t_2}{c - v}$. |
| Substituting $t_2$ and simplifying gives: |
| Substituting $t_2$ and simplifying gives: |
| $ t_3 = \frac{2L}{c - v} + \frac{L(c+v)}{(c - v)^2} = L\,\frac{3c - v}{(c - v)^2}.$ | | $ t_3 = \frac{2L}{c - v} + \frac{L(c+v)}{(c - v)^2} = L\,\frac{3c - v}{(c - v)^2}.$ |
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| This is the total time the light takes to complete its path. | | This is the total time the light takes to complete its path. |
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| The probe moves with speed u and must catch E2 at the same instant $t_3$: | | The probe moves with speed u and must catch E2 at the same instant $t_3$: |
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| $u t_3 = L + v t_3 \quad\Rightarrow\quad t_3 = \frac{L}{u - v} \quad (u > v)$. | | $u t_3 = L + v t_3 \quad\Rightarrow\quad t_3 = \frac{L}{u - v} \quad (u > v)$. |
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| Equating with the expression for $t_3$: | | Equating with the expression for $t_3$: |
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| $\frac{L}{u - v} = L\,\frac{3c - v}{(c - v)^2} \quad\Rightarrow\quad u - v = \frac{(c - v)^2}{3c - v}$. | | $\frac{L}{u - v} = L\,\frac{3c - v}{(c - v)^2} \quad\Rightarrow\quad u - v = \frac{(c - v)^2}{3c - v}$. |
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| Therefore, the speed of the probe as measured from the first station is: | | Therefore, the speed of the probe as measured from the first station is: |
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| $\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c - v}}$. | | $\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c - v}}$. |
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| For an observer at E2 (system S'), the probe moves with speed u' given by the relativistic velocity addition formula. | | For an observer at E2 (system S'), the probe moves with speed u' given by the relativistic velocity addition formula. |
| Since S' moves at speed v relative to S: | | Since S' moves at speed v relative to S: |
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| $u' = \frac{u - v}{1 - \dfrac{u v}{c^2}}$. | | $u' = \frac{u - v}{1 - \dfrac{u v}{c^2}}$. |
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| Substituting u and simplifying: | | Substituting u and simplifying: |
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| $u - v = \frac{(c - v)^2}{3c - v}, \qquad | | $u - v = \frac{(c - v)^2}{3c - v}, \qquad |
| 1 - \frac{u v}{c^2} = 1 - \frac{v\,c(c+v)}{c^2(3c - v)} = \frac{(3c - v)c - v(c+v)}{c(3c - v)} = \frac{(3c + v)(c - v)}{c(3c - v)}$. | | 1 - \frac{u v}{c^2} = 1 - \frac{v\,c(c+v)}{c^2(3c - v)} = \frac{(3c - v)c - v(c+v)}{c(3c - v)} = \frac{(3c + v)(c - v)}{c(3c - v)}$. |
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| $u' = \frac{\dfrac{(c - v)^2}{3c - v}}{\dfrac{(3c + v)(c - v)}{c(3c - v)}} = \frac{c(c - v)}{3c + v}$. | | $u' = \frac{\dfrac{(c - v)^2}{3c - v}}{\dfrac{(3c + v)(c - v)}{c(3c - v)}} = \frac{c(c - v)}{3c + v}$. |
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| $\boxed{u' = \frac{c(c - v)}{3c + v}}$. | | $\boxed{u' = \frac{c(c - v)}{3c + v}}$. |
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| This is the speed at which the probe approaches the second station according to E2's instruments. | | This is the speed at which the probe approaches the second station according to E2's instruments. |
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| By reciprocity between inertial systems, if E2 moves with speed v relative to E1, then E1 moves with speed -v relative to E2. Hence: | | By reciprocity between inertial systems, if E2 moves with speed v relative to E1, then E1 moves with speed -v relative to E2. Hence: |
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| $\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}$. | | $\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}$. |
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| The probe constitutes another inertial system (S''). For the probe: | | The probe constitutes another inertial system (S''). For the probe: |
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| · The first station moves with speed -u (magnitude u). | | · The first station moves with speed -u (magnitude u). |
| · The second station approaches with speed u' calculated above. | | · The second station approaches with speed u' calculated above. |
| Thus, the probe's team records: | | Thus, the probe's team records: |
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| $\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad | | $\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad |
| v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}$. | | v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}$. |
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| #### Answers | | #### Answers |
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| $\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c -v}}$ | | $\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c -v}}$ |
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| $\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}$. | | $\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}$. |
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| $\boxed{u' = \frac{c(c - v)}{3c + v}}$ | | $\boxed{u' = \frac{c(c - v)}{3c + v}}$ |
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| $\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad | | $\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad |
| v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}$. | | v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}$. |
