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David d.
PhD in Physics
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Calculus
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Question:

The following differential equations describe the angle of the pointers of two different clocks: $$\frac{\text{d}\theta_1}{\text{d}t} = \omega_1 - k(\theta_1 - \theta_2)$$ $$\frac{\text{d}\theta_2}{\text{d}t} = \omega_2 + k(\theta_1 - \theta_2)$$ Can these two clocks synchronize?

David d.
Answer:

The two clocks will synchronize when they beat at the same rate $$\left(\frac{\text{d}\theta_1}{\text{d}t} - \frac{\text{d}\theta_2}{\text{d}t} = 0\right)$$. One can subtract the first equation from the second to obtain: $$\frac{\text{d}(\theta_1-\theta_2)}{\text{d}t} = \omega_1-\omega_2 - 2k(\theta_1 - \theta_2)$$ or in a more compact notation $$\frac{\text{d}\Delta\theta}{\text{d}t} = - 2k\Delta\theta + \Delta\omega$$. For positive $$k$$, the difference between the two clocks will decrease until it reaches its minimum value when $$\frac{\text{d}\Delta\theta}{\text{d}t} \approx 0$$. The minimum value is $$0 = - 2k\Delta\theta + \Delta\omega$$, i.e., $$\Delta\theta = \Delta\omega/2k$$.

Advanced Physics (Special Relativity)
TutorMe
Question:

Describe a stationary electromagnetic wave inside a 1D well moving at constant speed close to the speed of light. Tip: write the stationary wave as the sum of two plane waves moving in opposite directions and apply the relativistic Doppler effect.

David d.
Answer:

A stationary wave is of the form $$\sin\left(\frac{n\pi}{L} x\right)\sin\left(\frac{n\pi}{L} c t\right) = \sin\left[\frac{n\pi}{L} (x-ct)\right] + \sin\left[\frac{n\pi}{L} (x+ct)\right]$$ for a stationary well. For a well in motion at velocity $$v$$, the forward propagating wave is shifted to a higher frequency by the relativistic Doppler effect while the backward propagating wave is shifted to a lower frequency. In such a well, the stationary wave is: $$\sin\left[\frac{n\pi}{L} \frac{1-\frac{v}{c}}{\sqrt{1-v^2/c^2}}(x-ct)\right] + \sin\left[\frac{n\pi}{L} \frac{1+\frac{v}{c}}{\sqrt{1-v^2/c^2}}(x+ct)\right] = \sin\left[\frac{n\pi}{L\sqrt{1-v^2/c^2}} (x-vt)\right]\sin\left[\frac{n\pi}{L\sqrt{1-v^2/c^2}} c (t-vx/c^2)\right]$$ This is a stationary wave in a box of length $$L\sqrt{1-v^2/c^2}$$, but centered around $$x-vt$$ and with a time variable of $$t-vx/c^2$$. One can recognize the Lorentz transformation: $$x' = \frac{x-vt}{\sqrt{1-v^2/c^2}}$$ $$t' \,= \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}$$

Physics
TutorMe
Question:

Describe the stationary wave inside a one-dimensional well that is moving with constant speed relative to the wave's medium. Tip: write the stationary wave as a sum of two plane waves propagating in opposite directions.

David d.
Answer:

A stationary wave is of the form $$\sin\left(\frac{n\pi}{L} x\right)\sin\left(\frac{n\pi}{L} c t\right) = \sin\left[\frac{n\pi}{L} (x-ct)\right] + \sin\left[\frac{n\pi}{L} (x+ct)\right]$$ for a stationary well. For a well in motion at velocity $$v$$, the forward propagating wave is shifted to a higher frequency by the Doppler effect while the backward propagating wave is shifted to a lower frequency. In such a well, the stationary wave is: $$\sin\left[\frac{n\pi}{L} (1-\frac{v}{c})(x-ct)\right] + \sin\left[\frac{n\pi}{L} (1+\frac{v}{c})(x+ct)\right] = \sin\left[\frac{n\pi}{L} (x-vt)\right]\sin\left[\frac{n\pi}{L} c (t-vx/c^2)\right]$$ This is a wave with the same wavelength and frequency as the stationary wave, but centered around $$x-vt$$ and with a time variable of $$t-vx/c^2$$. One can recognize the Lorentz transformation without the Lorentz factor: $$x' = x-vt$$ $$t' \,= t\,-vx/c^2$$

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