| ### Problem | | ### Problem |
| | | |
| $14.2.15.$ Using the Lorentz transformation, solve problems $14.2.5^*$ and $14.2.6^*$. | | $14.2.15.$ Using the Lorentz transformation, solve problems $14.2.5^*$ and $14.2.6^*$. |
| | | |
| ### Solution | | ### Solution |
| | | |
| 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 $x$. | | 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 $x$. |
| | | |
| The solution algorithm for both problems is the same: | | The solution algorithm for both problems is the same: |
| 1. Transition from the laboratory frame ($K$) to the rest frame of the moving object ($K'$). | | 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. | | 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. | | 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> | | <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 $V_x = -\beta c$. | | 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>\ | | <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): | | 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): |
| $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | | $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ |
| | | |
| <b>Step 2. Reflection in frame $K'$.</b>\ | | <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. | | 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 frequency: $\nu'_{ref} = \nu'$. |
| Reflected wavenumber (taking into account the direction along the $-x$ axis): $k'_{ref} = -\frac{\nu'}{c}$. | | 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>\ | | <b>Step 3. Return to the laboratory frame $K$.</b>\ |
| Frame $K'$ (the mirror) moves relative to $K$ with a velocity $V = -\beta c$. | | 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: | | We apply the inverse Lorentz transformation for the frequency of the reflected wave: |
| $$ \nu_{ref} = \gamma (\nu'_{ref} + V k'_{ref}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(-\frac{\nu'}{c}\right) \right) $$ | | $$ \nu_{ref} = \gamma (\nu'_{ref} + V k'_{ref}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(-\frac{\nu'}{c}\right) \right) $$ |
| $$ \nu_{ref} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 + \beta) = \nu' \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | | $$ \nu_{ref} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 + \beta) = \nu' \sqrt{\frac{1 + \beta}{1 - \beta}} $$ |
| Substituting $\nu'$ from Step 1: | | Substituting $\nu'$ from Step 1: |
| $$ \nu_{ref} = \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) \sqrt{\frac{1 + \beta}{1 - \beta}} = \nu \frac{1 + \beta}{1 - \beta} $$ | | $$ \nu_{ref} = \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) \sqrt{\frac{1 + \beta}{1 - \beta}} = \nu \frac{1 + \beta}{1 - \beta} $$ |
| The change in wave frequency $\Delta \nu$: | | The change in wave frequency $\Delta \nu$: |
| $$ \Delta \nu = \nu_{ref} - \nu = \nu \frac{1 + \beta}{1 - \beta} - \nu = \nu \frac{2\beta}{1 - \beta} $$ | | $$ \Delta \nu = \nu_{ref} - \nu = \nu \frac{1 + \beta}{1 - \beta} - \nu = \nu \frac{2\beta}{1 - \beta} $$ |
| Similarly to the first problem, in the rest frame of the dielectric $K'$, the frequency of the incident wave is: | | Similarly to the first problem, in the rest frame of the dielectric $K'$, the frequency of the incident wave is: |
| $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ | | $$ \nu' = \nu \sqrt{\frac{1 + \beta}{1 - \beta}} $$ |
| | | |
| <b>Step 2. Entering the dielectric in frame $K'$.</b>\ | | <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: | | In frame $K'$, the boundary of the dielectric is stationary. When passing through a stationary boundary between media, the wave frequency does not change: |
| $$ \nu'_{in} = \nu' $$ | | $$ \nu'_{in} = \nu' $$ |
| 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: | | 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: |
| $$ k'_{in} = \frac{\nu'_{in}}{c/n} = n \frac{\nu'}{c} $$ | | $$ k'_{in} = \frac{\nu'_{in}}{c/n} = n \frac{\nu'}{c} $$ |
| | | |
| <b>Step 3. Return to the laboratory frame $K$.</b>\ | | <b>Step 3. Return to the laboratory frame $K$.</b>\ |
| The velocity of frame $K'$ relative to $K$ is still $V = -\beta c$. | | 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: | | We apply the inverse Lorentz transformation for the frequency of the transmitted wave: |
| $$ \nu_{in} = \gamma (\nu'_{in} + V k'_{in}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(n \frac{\nu'}{c}\right) \right) $$ | | $$ \nu_{in} = \gamma (\nu'_{in} + V k'_{in}) = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu' + (-\beta c) \left(n \frac{\nu'}{c}\right) \right) $$ |
| $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 - n\beta) $$ | | $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \nu' (1 - n\beta) $$ |
| Substituting the expression for $\nu'$: | | Substituting the expression for $\nu'$: |
| $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) (1 - n\beta) = \nu \frac{1 - n\beta}{1 - \beta} $$ | | $$ \nu_{in} = \frac{1}{\sqrt{1 - \beta^2}} \left( \nu \sqrt{\frac{1 + \beta}{1 - \beta}} \right) (1 - n\beta) = \nu \frac{1 - n\beta}{1 - \beta} $$ |
| Let us find the difference in frequencies $\Delta \nu$ of the plane wave outside ($\nu$) and inside ($\nu_{in}$) the dielectric: | | Let us find the difference in frequencies $\Delta \nu$ of the plane wave outside ($\nu$) and inside ($\nu_{in}$) the dielectric: |
| $$ \Delta \nu = \nu - \nu_{in} = \nu - \nu \frac{1 - n\beta}{1 - \beta} = \nu \frac{1 - \beta - 1 + n\beta}{1 - \beta} = \nu \beta \frac{n - 1}{1 - \beta} $$ | | $$ \Delta \nu = \nu - \nu_{in} = \nu - \nu \frac{1 - n\beta}{1 - \beta} = \nu \frac{1 - \beta - 1 + n\beta}{1 - \beta} = \nu \beta \frac{n - 1}{1 - \beta} $$ |
| | | |
| #### Answer | | #### Answer |
| $14.2.5$. Will increase by $\Delta \nu = \nu \frac{2\beta}{1-\beta}$. | | $14.2.5$. Will increase by $\Delta \nu = \nu \frac{2\beta}{1-\beta}$. |
| $14.2.6$. Frequency difference is $\Delta \nu = \nu \beta \frac{n - 1}{1 - \beta}$. | | $14.2.6$. Frequency difference is $\Delta \nu = \nu \beta \frac{n - 1}{1 - \beta}$. |
