1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Manipulating a formula for a relativistic Doppeler shift

  1. Jun 28, 2013 #1
    1. The problem statement, all variables and given/known data
    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##.

    2. Relevant equations
    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):
    \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'\\
    Lorentz transformations and their inverses (are derived similarly):
    &\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}'}
    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:
    &\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}'}

    3. The attempt at a solution


    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?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 28, 2013 #2
    Im not sure which images you are talking about...you sure they were uploaded correctly?
  4. Jun 28, 2013 #3
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Manipulating a formula for a relativistic Doppeler shift