| ### Statement | | ### Statement |
| | | |
| $14.2.18.$ | | $14.2.18.$ |
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| a. According to cosmonauts ’ observations, the body inside the spacecraft per- | | a. According to cosmonauts ’ observations, the body inside the spacecraft per- |
| forms harmonic motion with a frequencyω | | forms harmonic motion with a frequencyω |
| and an amplitude A along the | | and an amplitude A along the |
| 2π | | 2π |
| spacecraft axis z = Asinωt. How will the axial coordinate of this body be re- | | spacecraft axis z = Asinωt. How will the axial coordinate of this body be re- |
| lated to time according to observations from the Earth, if the ship is moving | | lated to time according to observations from the Earth, if the ship is moving |
| away from the Earth at a speed βc? | | away from the Earth at a speed βc? |
| b. Solve the problem of point a if the body inside the ship, according to the | | b. Solve the problem of point a if the body inside the ship, according to the |
| observations of astronauts, made the same harmonic motion across the axis | | observations of astronauts, made the same harmonic motion across the axis |
| of the ship, y = A sin ωt. | | of the ship, y = A sin ωt. |
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| ### Solution | | ### Solution |
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| Lorentz transformation (spaceship S' → Earth S) | | Lorentz transformation (spaceship S' → Earth S) |
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| The spaceship moves in the +z direction with velocity $v = \beta c$ | | The spaceship moves in the +z direction with velocity $v = \beta c$ |
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| $z = \gamma (z' + vt'), \qquad t = \gamma \left(t' + \frac{v}{c^2}z'\right), \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$ | | $z = \gamma (z' + vt'), \qquad t = \gamma \left(t' + \frac{v}{c^2}z'\right), \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$ |
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| The inverse transformations (spaceship from Earth) are: | | The inverse transformations (spaceship from Earth) are: |
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| $z' = \gamma (z - vt), \qquad t' = \gamma \left(t - \frac{v}{c^2}z\right)$ | | $z' = \gamma (z - vt), \qquad t' = \gamma \left(t - \frac{v}{c^2}z\right)$ |
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| We start with (a) Longitudinal motion ($z' = A \sin \omega t'$) | | We start with (a) Longitudinal motion ($z' = A \sin \omega t'$) |
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| In the spaceship frame, the body oscillates about the origin (z' = 0). From Earth, the center of oscillation moves with velocity$ \beta c$, so the coordinate z of the body will be: | | In the spaceship frame, the body oscillates about the origin (z' = 0). From Earth, the center of oscillation moves with velocity$ \beta c$, so the coordinate z of the body will be: |
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| $z = \underbrace{\beta c t}_{\text{center}} + \Delta z$ | | $z = \underbrace{\beta c t}_{\text{center}} + \Delta z$ |
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| where$ \Delta z$ is the position relative to the center. To find the relationship between z' and t' observed from Earth, we use the inverse transformations: | | where$ \Delta z$ is the position relative to the center. To find the relationship between z' and t' observed from Earth, we use the inverse transformations: |
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| $z' = \gamma(z - \beta c t), \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ | | $z' = \gamma(z - \beta c t), \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ |
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| Substituting into the equation of motion $z' = A \sin \omega t'$ | | Substituting into the equation of motion $z' = A \sin \omega t'$ |
| we obtain: | | we obtain: |
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| $\gamma(z - \beta c t) = A \sin\left[\omega \gamma \left(t - \frac{\beta}{c}z\right)\right]$ | | $\gamma(z - \beta c t) = A \sin\left[\omega \gamma \left(t - \frac{\beta}{c}z\right)\right]$ |
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| Solving approximately gives: | | Solving approximately gives: |
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| $\boxed{z' = \frac{A}{\gamma} \sin\!\left(\frac{\omega t'}{\gamma}\right)\left(1 + \frac{\beta z'}{\omega c}\right)}$ | | $\boxed{z' = \frac{A}{\gamma} \sin\!\left(\frac{\omega t'}{\gamma}\right)\left(1 + \frac{\beta z'}{\omega c}\right)}$ |
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| We have in this case | | We have in this case |
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| The amplitude is reduced by a factor of $\gamma$ (length contraction). | | The amplitude is reduced by a factor of $\gamma$ (length contraction). |
| The frequency is reduced by $\gamma$ (time dilation). | | The frequency is reduced by $\gamma$ (time dilation). |
| The phase depends on position. | | The phase depends on position. |
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| b) Transverse motion | | b) Transverse motion |
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| The motion is perpendicular to the direction of relative motion. Transverse coordinates do not contract (y = y'), and the time t' is uniformly dilated: | | The motion is perpendicular to the direction of relative motion. Transverse coordinates do not contract (y = y'), and the time t' is uniformly dilated: |
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| $y = y', \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ | | $y = y', \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ |
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| However, since the motion is transverse, we can directly use$ y = A \sin \omega t' $ | | However, since the motion is transverse, we can directly use$ y = A \sin \omega t' $ |
| and substitute t' as a function of t: | | and substitute t' as a function of t: |
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| $\boxed{y' = A \sin\!\left(\frac{\omega t'}{\gamma}\right)}$ | | $\boxed{y' = A \sin\!\left(\frac{\omega t'}{\gamma}\right)}$ |
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| In this case | | In this case |
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| The amplitude A does not change (there is no contraction in the transverse direction). | | The amplitude A does not change (there is no contraction in the transverse direction). |
| The frequency is reduced by $\gamma$ (time dilation). | | The frequency is reduced by $\gamma$ (time dilation). |
| The oscillation is perfectly harmonic in S', and upon passing to S an additional dependence appears because t' varies with z if the motion is not purely transverse. Nevertheless, for a fixed point in S, the observed frequency is $\omega/\gamma$ | | The oscillation is perfectly harmonic in S', and upon passing to S an additional dependence appears because t' varies with z if the motion is not purely transverse. Nevertheless, for a fixed point in S, the observed frequency is $\omega/\gamma$ |
