# How do i calculate change of momentum when I send a photon in the direction of travel

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?

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

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.