Question

1. doppler ship suppose a photon is emitted in the lab with an angle φ measured with respect to the x - axis and frequency f. the doppler ship is moving at a constant velocity β = tanh θ along the x - axis with respect to a lab. show that the frequency f and angle φ measured with respect to the x - axis of the doppler ship are given by the following f = f cosh θ (1 - tanh θ cos φ) cos φ = {cos φ - tanh θ}/{1 - tanh θ cos φ} solution:

Explanation:

Step1: Use Lorentz - transformation for four - momentum

The four - momentum of a photon is $p^{\mu}=(E/c, p_x,p_y,p_z)$. In terms of frequency $f$ and direction $\varphi$, $E = hf$, $p_x=\frac{hf}{c}\cos\varphi$, $p_y=\frac{hf}{c}\sin\varphi$, $p_z = 0$. The Lorentz transformation along the x - axis is given by $p'^{\mu}=\Lambda^{\mu}_{
u}p^{
u}$, where the Lorentz transformation matrix $\Lambda$ for a boost with rapidity $\theta$ along the x - axis has elements $\Lambda^{0}_{0}=\cosh\theta$, $\Lambda^{0}_{1}=-\sinh\theta$, $\Lambda^{1}_{0}=-\sinh\theta$, $\Lambda^{1}_{1}=\cosh\theta$, and $\Lambda^{i}_{j}=\delta^{i}_{j}$ for $i,j = 2,3$.

Step2: Calculate the new energy (frequency)

The energy of the photon in the new frame $E'=p'^{0}$. We know that $p'^{0}=\Lambda^{0}_{0}p^{0}+\Lambda^{0}_{1}p^{1}$. Since $p^{0}=\frac{hf}{c}$ and $p^{1}=\frac{hf}{c}\cos\varphi$, we have $p'^{0}=\cosh\theta\frac{hf}{c}-\sinh\theta\frac{hf}{c}\cos\varphi$. Since $E' = hf'$ and $\tanh\theta=\frac{\sinh\theta}{\cosh\theta}$, we get $f'=f\cosh\theta(1 - \tanh\theta\cos\varphi)$.

Step3: Calculate the new direction

The x - component of the momentum in the new frame is $p'^{1}=\Lambda^{1}_{0}p^{0}+\Lambda^{1}_{1}p^{1}=-\sinh\theta\frac{hf}{c}+\cosh\theta\frac{hf}{c}\cos\varphi$. The magnitude of the momentum in the new frame $p'=\frac{hf'}{c}$. Then $\cos\varphi'=\frac{p'^{1}}{p'}$. Substituting $f' = f\cosh\theta(1 - \tanh\theta\cos\varphi)$ and simplifying, we get $\cos\varphi'=\frac{\cos\varphi - \tanh\theta}{1 - \tanh\theta\cos\varphi}$.

Answer:

We need to show that $f' = f\cosh\theta(1 - \tanh\theta\cos\varphi)$ and $\cos\varphi'=\frac{\cos\varphi - \tanh\theta}{1 - \tanh\theta\cos\varphi}$.