Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Manipulating a formula for a relativistic Doppeler shift
Reply to thread
Message
[QUOTE="71GA, post: 4429198, member: 300573"] [h2]Homework Statement [/h2] 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##. [h2]Homework Equations[/h2] 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} [h2]The Attempt at a Solution[/h2] If i draw the picture in black color (image 2) it occurred 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. [B]I wonder how do the Lorentz transformation change? Can anyone tell me?[/B] [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Manipulating a formula for a relativistic Doppeler shift
Back
Top