Manipulating a formula for a relativistic Doppeler shift

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Homework Statement


The spaceship is approaching Earth with a speed ##\scriptsize 0.6c## under an angle
of ##\scriptsize 30^\circ##. What frequency does an observer on Earth measure if
spaceship is sending frequency ##\scriptsize 1.00\cdot10^9Hz##.

Homework Equations


[Broken]
Lets say we take the standard configuration when ##\scriptsize x'y'## is moving away from system ##\scriptsize xy## (image 1). By knowing that the phase is constant in all frames ##\scriptsize \phi=\phi'## we can derive the Lorenz transformations for a standard configuration.

Derivation (using the parametrization):
\begin{align}
\phi &= \phi'\\
-\phi &= -\phi'\\
k \Delta r - \omega \Delta t &= k' \Delta r'- \omega'\Delta t'\\
[k_x , k_y , k_z][\Delta x , \Delta y , \Delta z] - \omega \Delta t &= [{k_x}'\! , {k_y}'\! , {k_z}'][\Delta x'\! , \Delta y'\! , \Delta z']\! - \!\omega'\Delta t'\\
k_x \Delta x + k_y \Delta y + k_z \Delta z - \omega \Delta t&= {k_x}'\Delta x' + {k_y}' \Delta y' + {k_z}' \Delta z'\! - \!\omega' \Delta t'\\
{k_x} \gamma \Bigl(\!\Delta x' + u\Delta t' \!\Bigl) + {k_y} \Delta y' + {k_z} \Delta z' - \omega \gamma \left(\Delta t' + \Delta x' \frac{u}{c^2}\right)&= ...\\
{k_x} \gamma \Delta x' + k_x \gamma u\Delta t' + {k_y} \Delta y' + {k_z} \Delta z' - \omega \gamma \Delta t' - \omega \gamma \Delta x' \frac{u}{c^2}&= ...\\
\gamma \Bigl(\!k_x - \omega \frac{u}{c^2}\! \Bigl) \Delta x' + k_y \Delta y' + k_z \Delta z' - \gamma \Bigl(\omega - {k_x} u \Bigl) \Delta t' &= k_x' \Delta x' + k_y' \Delta y' + k_z' \Delta z' - \omega' \Delta t'\\
\end{align}
Lorentz transformations and their inverses (are derived similarly):
\begin{align}
&\boxed{\omega' = \gamma\Bigl(\omega - {k_x} u \Bigl)} & &\boxed{\omega = \gamma\Bigl(\omega' + {k_x}' u \Bigl)}\\
&\boxed{k_x' = \gamma \Bigl(k_x - \omega \frac{u}{c^2} \Bigl)} & &\boxed{k_x = \gamma \Bigl(k_x' + \omega' \frac{u}{c^2} \Bigl)}\\
&\boxed{k_y' = k_y} & &\boxed{k_y = {k_y}'}\\
&\boxed{k_z' = k_z} & &\boxed{k_z = {k_z}'}
\end{align}
We can express Lorentz transformations and their inverse using some trigonometry (##\scriptsize k_x = k \cos{\xi} = \frac{\omega}{c} \cos{\xi}##, ##\scriptsize k_y = k \sin{\xi} = \frac{\omega}{c} \sin{\xi}## and ##\scriptsize k_z = 0##) as:
\begin{align}
&\boxed{\omega' = \gamma \, \omega \! \Bigl(1 - \cos{\xi} \frac{u}{c} \Bigl)}& &\boxed{\omega = \gamma \, \omega' \! \Bigl(1 + \cos{\xi'}\frac{u}{c} \Bigl)}\\
&\boxed{k_x' = \gamma \, \frac{\omega}{c} \! \Bigl(\cos{\xi} - \frac{u}{c} \Bigl)}& &\boxed{k_x = \gamma \, \frac{\omega'}{c} \! \Bigl(\cos{\xi'} + \frac{u}{c} \Bigl)}\\
&\boxed{k_y' = \frac{\omega}{c} \sin{\xi}} & &\boxed{k_y = \frac{\omega'}{c} \sin{\xi'}}\\
&\boxed{k_z' = k_z} & &\boxed{k_z = {k_z}'}
\end{align}

The Attempt at a Solution



[Broken]

If i draw the picture in black color (image 2) it occured to me that solving this case could be possible by simply using a relativistic Doppeler effect shift equation for 2 bodies which are closing in (in which i would use the ##\scriptsize u_x = u \cdot \cos 30^\circ##).

$$\nu = \nu' \sqrt{\frac{c+u_x}{c-u_x}} \approx 1.78\cdot 10^8Hz$$

Am i allowed to solve this case like this?

I wasnt so sure about the above solution, so i tried to get the similar situation to the one i had in image 1. I noticed that if i rotate coordinate systems (image 2 - systems which are colored in red) i get fairly similar configuration, with the ##\scriptsize \xi## and ##\scriptsize u## a bit different than the ones in image 1. I wonder how do the Lorentz transformation change? Can anyone tell me?
 
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Answers and Replies

  • #2
Im not sure which images you are talking about...you sure they were uploaded correctly?
 
  • #3
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Im not sure which images you are talking about...you sure they were uploaded correctly?

Yes.
 

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