# Relativistic doppler effect with photons

Suppose $$2$$ photons move along the $$x$$-axis in opposite directions and collide head on. They fuse to form a single particle with rest mass $$M$$, which then decays to produce $$2$$ particles moving in opposite directions along the $$y$$-axis. The frequency of the electromagnetic wave associated with the photon is given as $$f_0=E/h$$.

Say an observer stands at $$x=\infty$$ moving with velocity half that of light in the positive $$x$$-direction. Using the relativistic Doppler effect, how can I find the frequency the observer measures of the electromagnetic wave associated with the photon moving towards the observer? Say in terms of $$M$$, $$h$$ and $$c$$.

I know that the velocity of the initial photon relative to the observer would be $$v=c$$ (could have used relativistic velocity addition, but it is obvious since the speed of light is the same in all frames). But if I sub this into the formula for the Doppler effect, it tells me the observed frequency is $$f=f_0\sqrt{\frac{c+v}{c-v}}$$, I get $$f=\infty$$ which is clearly not true.

• It would be a lot easier to figure out where you went wrong if you kinda show the math. I think you went wrong by substituting v=c but v is the velocity of the observer with respect to the source (so c/2) and not the velocity of light. May 31, 2020 at 1:19
• Are the two particles going in the y-axis photons? It would be helpful to know more about your question, like do you want to know the frequency of the initial photons (and which), and whether the initial photons have the same frequency before they collide, etc.
– PNS
May 31, 2020 at 7:38
• Also, another detail: is the observer coming near or going away from the photons? Sorry, I did not quite get the phrasing in your question.
– PNS
May 31, 2020 at 7:40
• @PradyothShandilya I have added some detail. But my notes say that $v$ in the formula is the speed of the source relative to the observer, so I thought I can't say that the photon speed is $c/2$ here since the speed of light is the same in all frames? May 31, 2020 at 12:00
• @PNS The phrasing of the question is confusing me a bit too, and doesn't really clarify whether the last $2$ particles are photons or not, sorry about that. But to answer your other question, of the initial $2$ photons one is moving away (negative $x$ direction) from the observer, and one is moving toward or 'catching up' (positive $x$ direction) with the observer. The photon I am concerned with here is the one moving in the same direction, 'catching up' with the observer. May 31, 2020 at 12:04

The relative velocity in the Doppler effect equation is that between the source and the observer and not the wave and the observer. In your case, assuming the source is stationary, $$v=c/2$$ as the observer is moving at a speed of $$c/2$$. Therefore, the denominator in the right-hand-side of your equation would not be 0 and so, the frequency observed is not infinite.

Ok, I am going to make a few assumptions:

• The two initial photons which collide to form the particle have the same frequency.

• Since the particle has not been specified, let's go ahead and say that there is no loss of energy in the process.

• The particle formed stays at rest.

Good. Now, the energy of the particle is $$E = Mc^2$$. So $$\frac{1}{2}E$$ is the energy of one photon. So, we get that the frequency of each of the photons is $$f_0 = \frac{E}{2h} = \frac{Mc^2}{2h}$$

where $$h$$ is the Planck's constant. Now, here we leverage some symmetry. The equation above (in the question) works when both the observer and the light is moving. But, in this case, the observer can say that they are at rest. I think you can use the relativistic addition formula and get the $$v$$, then proceed to te plug-ins.

If you have massive particles ($$m$$), you can transform their 4-velocities:

$$u^{\mu} = (\gamma c, \gamma\vec v)$$

between frames, and that means you can transform their 4-momenta:

$$p^{\mu} = mu^{\mu} =(\gamma mc, \gamma m\vec v) =(E/c, \vec p)$$

and that means you can transform their wave-vectors:

$$k^{\mu} = \frac 1 {\hbar} p^{\mu}= (\omega/c, \vec k) = 2\pi(\nu/c, \hat k/\lambda)$$

which is a relativistic Doppler shift from the velocity addition formula.

But for light, there is a problem:

$$\nu = c/\lambda$$

so that

$$E = pc$$

meaning

$$v = c$$

and

$$\gamma \rightarrow \infty$$

which breaks the Lorentz transformation.

While the velocity addition formula still works with one of the velocities equal to $$c$$, the connection to frequency or wavelength is lost.

So for your problem, you want to transform a wave vector:

$$k^{\mu} = 2\pi(f, f, 0, 0)$$

along the $$x$$-axis by $$\beta=c/2$$:

$$k'^{\mu} = 2\pi(\gamma f-\gamma \beta b, -\gamma \beta b f +\gamma f, 0, 0)$$

So that:

$$f' = f\gamma (1-\beta) = \frac{1-\beta}{\sqrt{(1-\beta)(1+\beta}}=f\sqrt{\frac{1-\beta}{1+\beta}}=f/\sqrt 3$$

which is the standard Doppler shift formula.