Doppler Shift

  • #1
stunner5000pt
1,447
2
uestion is How does the second order term in teh relativistic doppler shift (v/c)^2 compare to the total classical Doppler Shift for the observer receeding away from the source?


now the doppler shift (relativistic) is [tex] \Delta f = |f_{0} (1-\sqrt{\frac{1-\beta}{1+\beta}})|[/tex]

for the classical shift it is

[tex] \Delta f = |f_{0} (1-\frac{v_{rel}}{v}-1)| = | f_{0} \frac{v_{rel}}{v}|[/tex]

but i dont see a (v/c)^2 term anywhere? How am i supposed to do this??
refers to part E opf this thread
https://www.physicsforums.com/showthread.php?t=64390
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
45,425
1,876
Do a binomial expansion of:
[tex] \sqrt{\frac{1-\beta}{1+\beta}} = \frac{\sqrt{1-\beta^2}}{1+\beta}[/tex]
 

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