Question

Relativistic Doppler Effect

Answer 1

Does anyone have a simple proof for this, ideally just a Lorentz transform or a few lines of calculus, and not a whole load of algebra....

\(f_{obs \ receding} = \sqrt{\frac{1-v/c}{1+v/c}}f_{source}\)

Answer 2

here we have to keep in mind that the wave quadrivector:
\[\Large \left( {{k_0},{\mathbf{k}}} \right)\]
will transfor itself like the position quadrivector:
\[\Large \left( {{x_0},{\mathbf{x}}} \right) = \left( {ct,{\mathbf{x}}} \right)\]

Answer 3

transform*

Answer 4

so we can write:
\[\Large \left\{ \begin{gathered}
{k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\
k{'_{||}} = \gamma \left( {{k_{||}} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\
{\mathbf{k}}{'_ \bot } = {\mathbf{k}} \bot \hfill \\
\end{gathered} \right.\]

Answer 5

now if we consider the case in which
\[\Large {\mathbf{\beta }},\;{\mathbf{k}}\]
have the same direction, we can rewrite the first equation, as below::
\[\Large \omega ' = \gamma \omega \left( {1 - \beta } \right)\]
being:
\[\Large {k_0} = \frac{\omega }{c}\]
Now keeping mind that:
\[\Large \beta = \frac{V}{c},\;\gamma = \frac{1}{{\sqrt {1 - \frac{{{V^2}}}{{{c^2}}}} }}\]
where V is the relative speed between those inertial system to which our Lorentz transforms are referring to, we got your relationship

Answer 6

oops.. the second equation, is wrong, here is the right one:

Answer 7

\[\Large \left\{ \begin{gathered}
{k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\
k{'_{||}} = \gamma \left( {{k_{||}} - \beta {k_0}} \right) \hfill \\
{\mathbf{k}}{'_ \bot } = {{\mathbf{k}}_ \bot } \hfill \\
\end{gathered} \right.\]

Answer 8

nevertheless the Doppler effect comes from the first equation

Answer 9

that looks wonderful and exactly what i was looking for. i only need to do it along x so i can probably simplify even more, eg draw it on some frames.

i've been going round in circles on this - i tried just transforming the wavelength in the hope it would "just work out" but it doesn't - so thank you very much!

Answer 10

:)

Answer 11

.