Edits to “Problem”, “Solution”, “Answer”
en/14.2.15.md
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| @@ -1,11 +1,65 @@ | |||
| − | ### | ||
| + | ### Problem | ||
| − | $14.2.15.$ | ||
| + | $14.2.15.$ Using the Lorentz transformation, solve problems $14.2.5^*$ and $14.2.6^*$. | ||
| ### Solution | |||
| − | Studio Cyborg Squad presents | ||
| + | The main idea of the method is that the phase of a plane wave $\Phi = \omega t - kx$ is a Lorentz invariant. It follows that the frequency $\nu$ and the wavenumber $k = \nu/c$ transform during transitions between inertial reference frames according to the same Lorentz transformation formulas as time $t$ and coordinate $x$. | ||
| − | #### Answer | ||
| + | The solution algorithm for both problems is the same: | ||
| + | 1. Transition from the laboratory frame ($K$) to the rest frame of the moving object ($K'$). | ||
| + | 2. Analysis of the process (reflection or refraction) in frame $K'$, where the boundary is stationary. | ||
| + | 3. Inverse transition from $K'$ to the laboratory frame $K$ for the modified wave. | ||
| − | |||
| + | --- | ||
| + | |||
| + | <b>Solution to problem 14.2.5 (Reflection from a mirror)</b> | ||
| + | |||
| + | Let the incident wave travel along the $x$-axis. Its frequency is $\nu$, and the wavenumber is $k = \nu/c$. The mirror moves towards the wave with a velocity $v = \beta c$. The projection of the mirror's velocity in frame $K$ is $V_x = -\beta c$. | ||
| + | |||
| + | <b>Step 1. Transition to the mirror frame $K'$.</b>\ | ||
| + | In the rest frame of the mirror $K'$, the frequency of the incident wave $\nu'$ is determined by the formula of the relativistic Doppler effect for an approaching source (which directly follows from the Lorentz transformations): | ||
| + | $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | ||
| + | |||
| + | <b>Step 2. Reflection in frame $K'$.</b>\ | ||
| + | In frame $K'$, the mirror is stationary. Upon normal reflection from a stationary ideal mirror, the wave frequency is conserved, and the direction of propagation is reversed. | ||
| + | Reflected frequency: $\nu'_{ref} = \nu'$. | ||
| + | Reflected wavenumber (taking into account the direction along the $-x$ axis): $k'_{ref} = -\frac{\nu'}{c}$. | ||
| + | |||
| + | <b>Step 3. Return to the laboratory frame $K$.</b>\ | ||
| + | Frame $K'$ (the mirror) moves relative to $K$ with a velocity $V = -\beta c$. | ||
| + | We apply the inverse Lorentz transformation for the frequency of the reflected wave: | ||
| + | $$ \nu_{ref} = \gamma (\nu'_{ref} + V k'_{ref}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(-\frac{\nu'}{c}\right) \right) $$ | ||
| + | $$ \nu_{ref} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 + \beta) = \nu' \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | ||
| + | Substituting $\nu'$ from Step 1: | ||
| + | $$ \nu_{ref} = \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) \sqrt{\frac{1 + \beta}{1 - \beta}} = \nu \frac{1 + \beta}{1 - \beta} $$ | ||
| + | The change in wave frequency $\Delta \nu$: | ||
| + | $$ \Delta \nu = \nu_{ref} - \nu = \nu \frac{1 + \beta}{1 - \beta} - \nu = \nu \frac{2\beta}{1 - \beta} $$ | ||
| + | |||
| + | --- | ||
| + | |||
| + | <b>Solution to problem 14.2.6 (Wave inside a dielectric)</b> | ||
| + | |||
| + | <b>Step 1. Transition to the dielectric frame $K'$.</b>\ | ||
| + | Similarly to the first problem, in the rest frame of the dielectric $K'$, the frequency of the incident wave is: | ||
| + | $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | ||
| + | |||
| + | <b>Step 2. Entering the dielectric in frame $K'$.</b>\ | ||
| + | In frame $K'$, the boundary of the dielectric is stationary. When passing through a stationary boundary between media, the wave frequency does not change: | ||
| + | $$ \nu'_{in} = \nu' $$ | ||
| + | The wave continues to propagate in the same direction (along $+x$). The phase velocity in the dielectric is $c/n$, so the wavenumber inside is: | ||
| + | $$ k'_{in} = \frac{\nu'_{in}}{c/n} = n \frac{\nu'}{c} $$ | ||
| + | |||
| + | <b>Step 3. Return to the laboratory frame $K$.</b>\ | ||
| + | The velocity of frame $K'$ relative to $K$ is still $V = -\beta c$. | ||
| + | We apply the inverse Lorentz transformation for the frequency of the transmitted wave: | ||
| + | $$ \nu_{in} = \gamma (\nu'_{in} + V k'_{in}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(n \frac{\nu'}{c}\right) \right) $$ | ||
| + | $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 - n\beta) $$ | ||
| + | Substituting the expression for $\nu'$: | ||
| + | $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) (1 - n\beta) = \nu \frac{1 - n\beta}{1 - \beta} $$ | ||
| + | Let us find the difference in frequencies $\Delta \nu$ of the plane wave outside ($\nu$) and inside ($\nu_{in}$) the dielectric: | ||
| + | $$ \Delta \nu = \nu - \nu_{in} = \nu - \nu \frac{1 - n\beta}{1 - \beta} = \nu \frac{1 - \beta - 1 + n\beta}{1 - \beta} = \nu \beta \frac{n - 1}{1 - \beta} $$ | ||
| + | |||
| + | #### Answer | ||
| + | $14.2.5$. Will increase by $\Delta \nu = \nu \frac{2\beta}{1-\beta}$. | ||
| + | $14.2.6$. Frequency difference is $\Delta \nu = \nu \beta \frac{n - 1}{1 - \beta}$. | ||
| @@ -1,11 +1,65 @@ | |||
| ### |
### Problem | ||
| $14.2.15.$ |
$14.2.15.$ Using the Lorentz transformation, solve problems $14.2.5^*$ and $14.2.6^*$. | ||
| ### Solution | ### Solution | ||
| Studio Cyborg Squad presents | The main idea of the method is that the phase of a plane wave $\Phi = \omega t - kx$ is a Lorentz invariant. It follows that the frequency $\nu$ and the wavenumber $k = \nu/c$ transform during transitions between inertial reference frames according to the same Lorentz transformation formulas as time $t$ and coordinate $x$. | ||
| #### Answer | The solution algorithm for both problems is the same: | ||
| 1. Transition from the laboratory frame ($K$) to the rest frame of the moving object ($K'$). | |||
| 2. Analysis of the process (reflection or refraction) in frame $K'$, where the boundary is stationary. | |||
| 3. Inverse transition from $K'$ to the laboratory frame $K$ for the modified wave. | |||
| --- | |||
| <b>Solution to problem 14.2.5 (Reflection from a mirror)</b> | |||
| Let the incident wave travel along the $x$-axis. Its frequency is $\nu$, and the wavenumber is $k = \nu/c$. The mirror moves towards the wave with a velocity $v = \beta c$. The projection of the mirror's velocity in frame $K$ is $V_x = -\beta c$. | |||
| <b>Step 1. Transition to the mirror frame $K'$.</b>\ | |||
| In the rest frame of the mirror $K'$, the frequency of the incident wave $\nu'$ is determined by the formula of the relativistic Doppler effect for an approaching source (which directly follows from the Lorentz transformations): | |||
| $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | |||
| <b>Step 2. Reflection in frame $K'$.</b>\ | |||
| In frame $K'$, the mirror is stationary. Upon normal reflection from a stationary ideal mirror, the wave frequency is conserved, and the direction of propagation is reversed. | |||
| Reflected frequency: $\nu'_{ref} = \nu'$. | |||
| Reflected wavenumber (taking into account the direction along the $-x$ axis): $k'_{ref} = -\frac{\nu'}{c}$. | |||
| <b>Step 3. Return to the laboratory frame $K$.</b>\ | |||
| Frame $K'$ (the mirror) moves relative to $K$ with a velocity $V = -\beta c$. | |||
| We apply the inverse Lorentz transformation for the frequency of the reflected wave: | |||
| $$ \nu_{ref} = \gamma (\nu'_{ref} + V k'_{ref}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(-\frac{\nu'}{c}\right) \right) $$ | |||
| $$ \nu_{ref} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 + \beta) = \nu' \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | |||
| Substituting $\nu'$ from Step 1: | |||
| $$ \nu_{ref} = \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) \sqrt{\frac{1 + \beta}{1 - \beta}} = \nu \frac{1 + \beta}{1 - \beta} $$ | |||
| The change in wave frequency $\Delta \nu$: | |||
| $$ \Delta \nu = \nu_{ref} - \nu = \nu \frac{1 + \beta}{1 - \beta} - \nu = \nu \frac{2\beta}{1 - \beta} $$ | |||
| --- | |||
| <b>Solution to problem 14.2.6 (Wave inside a dielectric)</b> | |||
| <b>Step 1. Transition to the dielectric frame $K'$.</b>\ | |||
| Similarly to the first problem, in the rest frame of the dielectric $K'$, the frequency of the incident wave is: | |||
| $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | |||
| <b>Step 2. Entering the dielectric in frame $K'$.</b>\ | |||
| In frame $K'$, the boundary of the dielectric is stationary. When passing through a stationary boundary between media, the wave frequency does not change: | |||
| $$ \nu'_{in} = \nu' $$ | |||
| The wave continues to propagate in the same direction (along $+x$). The phase velocity in the dielectric is $c/n$, so the wavenumber inside is: | |||
| $$ k'_{in} = \frac{\nu'_{in}}{c/n} = n \frac{\nu'}{c} $$ | |||
| <b>Step 3. Return to the laboratory frame $K$.</b>\ | |||
| The velocity of frame $K'$ relative to $K$ is still $V = -\beta c$. | |||
| We apply the inverse Lorentz transformation for the frequency of the transmitted wave: | |||
| $$ \nu_{in} = \gamma (\nu'_{in} + V k'_{in}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(n \frac{\nu'}{c}\right) \right) $$ | |||
| $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 - n\beta) $$ | |||
| Substituting the expression for $\nu'$: | |||
| $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) (1 - n\beta) = \nu \frac{1 - n\beta}{1 - \beta} $$ | |||
| Let us find the difference in frequencies $\Delta \nu$ of the plane wave outside ($\nu$) and inside ($\nu_{in}$) the dielectric: | |||
| $$ \Delta \nu = \nu - \nu_{in} = \nu - \nu \frac{1 - n\beta}{1 - \beta} = \nu \frac{1 - \beta - 1 + n\beta}{1 - \beta} = \nu \beta \frac{n - 1}{1 - \beta} $$ | |||
| #### Answer | |||
| $14.2.5$. Will increase by $\Delta \nu = \nu \frac{2\beta}{1-\beta}$. | |||
| $14.2.6$. Frequency difference is $\Delta \nu = \nu \beta \frac{n - 1}{1 - \beta}$. | |||