Say a spaceship is traveling at a certain velocity v (>>c) and it emits light from the nose of my spaceship in the direction of travel. The speed of light is finite and hence there should be a change of wavelength of light. How does the speed affect the momentum transfer does my light have lower momentum thanks to the direction in which it has been emitted?

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    $\begingroup$ I think you mean $v \ll c$ ([dollar][backslash]ll c[dollar] $\endgroup$
    – JEB
    Sep 17 at 21:49

1 Answer 1


So you have a reference frame, $S$, in which your spaceship is moving at $v$. There's also the spaceship's rest frame $S'$.

The photon has wavenumber (in $S'$):

$$ k' =\frac{2\pi}{\lambda'} $$

with a 4-vector:

$$ k'_{\mu} \equiv (\omega'/c, k', 0, 0)=(k', k') $$

where I have dropped the transverse dimensions.

The 4-momentum:

$$ p'_{\mu} \equiv (E'/c,\vec p') = (p',p') = \hbar k'_{\mu}$$

To find the momentum in $S$, use the inverse Lorentz transformation:

$$k_{\mu} = \Lambda_{\mu}^{\ \nu}k'_{\nu} $$

I'll do the $x$ component:

$$k_1 = \gamma(k'_1+ \frac v c k'_1) $$

(drop the subscripts since $k_0 = k_1$):

$$k = \gamma(1+\frac v c)k' $$

so that:

$$p = \hbar k = \gamma(1+\frac v c)p'$$

So the momentum is larger in $S$.

Note that:

$$ \gamma(1+\frac v c) = \frac{1+\frac v c} {\sqrt{1-(\frac v c)^2}}$$ $$ = \frac{1+\frac v c}{\sqrt{(1+\frac v c)(1-\frac v c)}} = \sqrt{\frac{1+ \frac v c}{1-\frac v c}}$$

which is the standard Doppler shift formula.


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