Новое решение
en/14.2.18.md
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| + | ### Statement | ||
| + | |||
| + | $14.2.18.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Lorentz transformation (spaceship S' → Earth S) | ||
| + | |||
| + | The spaceship moves in the +z direction with velocity $v = \beta c$ | ||
| + | |||
| + | $z = \gamma (z' + vt'), \qquad t = \gamma \left(t' + \frac{v}{c^2}z'\right), \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$ | ||
| + | |||
| + | The inverse transformations (spaceship from Earth) are: | ||
| + | |||
| + | $z' = \gamma (z - vt), \qquad t' = \gamma \left(t - \frac{v}{c^2}z\right)$ | ||
| + | |||
| + | We start with (a) Longitudinal motion ($z' = A \sin \omega t'$) | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $z = \underbrace{\beta c t}_{\text{center}} + \Delta z$ | ||
| + | |||
| + | 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: | ||
| + | |||
| + | $z' = \gamma(z - \beta c t), \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ | ||
| + | |||
| + | Substituting into the equation of motion $z' = A \sin \omega t'$ | ||
| + | we obtain: | ||
| + | |||
| + | $\gamma(z - \beta c t) = A \sin\left[\omega \gamma \left(t - \frac{\beta}{c}z\right)\right]$ | ||
| + | |||
| + | Solving approximately gives: | ||
| + | |||
| + | $\boxed{z' = \frac{A}{\gamma} \sin\!\left(\frac{\omega t'}{\gamma}\right)\left(1 + \frac{\beta z'}{\omega c}\right)}$ | ||
| + | |||
| + | We have in this case | ||
| + | |||
| + | The amplitude is reduced by a factor of $\gamma$ (length contraction). | ||
| + | The frequency is reduced by $\gamma$ (time dilation). | ||
| + | The phase depends on position. | ||
| + | |||
| + | b) Transverse motion | ||
| + | |||
| + | The motion is perpendicular to the direction of relative motion. Transverse coordinates do not contract (y = y'), and the time t' is uniformly dilated: | ||
| + | |||
| + | $y = y', \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ | ||
| + | |||
| + | However, since the motion is transverse, we can directly use$ y = A \sin \omega t' $ | ||
| + | and substitute t' as a function of t: | ||
| + | |||
| + | $\boxed{y' = A \sin\!\left(\frac{\omega t'}{\gamma}\right)}$ | ||
| + | |||
| + | In this case | ||
| + | |||
| + | The amplitude A does not change (there is no contraction in the transverse direction). | ||
| + | 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 $\omega/\gamma$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $14.2.18.$ [Insert the problem statement] | |||
| ### Solution | |||
| Lorentz transformation (spaceship S' → Earth S) | |||
| The spaceship moves in the +z direction with velocity $v = \beta c$ | |||
| $z = \gamma (z' + vt'), \qquad t = \gamma \left(t' + \frac{v}{c^2}z'\right), \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$ | |||
| The inverse transformations (spaceship from Earth) are: | |||
| $z' = \gamma (z - vt), \qquad t' = \gamma \left(t - \frac{v}{c^2}z\right)$ | |||
| We start with (a) Longitudinal motion ($z' = A \sin \omega t'$) | |||
| 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: | |||
| $z = \underbrace{\beta c t}_{\text{center}} + \Delta z$ | |||
| 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: | |||
| $z' = \gamma(z - \beta c t), \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ | |||
| Substituting into the equation of motion $z' = A \sin \omega t'$ | |||
| we obtain: | |||
| $\gamma(z - \beta c t) = A \sin\left[\omega \gamma \left(t - \frac{\beta}{c}z\right)\right]$ | |||
| Solving approximately gives: | |||
| $\boxed{z' = \frac{A}{\gamma} \sin\!\left(\frac{\omega t'}{\gamma}\right)\left(1 + \frac{\beta z'}{\omega c}\right)}$ | |||
| We have in this case | |||
| The amplitude is reduced by a factor of $\gamma$ (length contraction). | |||
| The frequency is reduced by $\gamma$ (time dilation). | |||
| The phase depends on position. | |||
| b) Transverse motion | |||
| The motion is perpendicular to the direction of relative motion. Transverse coordinates do not contract (y = y'), and the time t' is uniformly dilated: | |||
| $y = y', \qquad t' = \gamma\left(t - \frac{\beta}{c}z\right)$ | |||
| However, since the motion is transverse, we can directly use$ y = A \sin \omega t' $ | |||
| and substitute t' as a function of t: | |||
| $\boxed{y' = A \sin\!\left(\frac{\omega t'}{\gamma}\right)}$ | |||
| In this case | |||
| The amplitude A does not change (there is no contraction in the transverse direction). | |||
| 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 $\omega/\gamma$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||