# Relativistic decay

Tags:
1. Jan 2, 2016

### Physgeek64

1. The problem statement, all variables and given/known data
Hi- i've been doing some questions and came across the following

A high energy particle, travelling towards an observer with velocity v decays to produce a photon with
energy E in the rest frame of the particle.
Find the frequency at which this photon would be detected by the observer

2. Relevant equations

3. The attempt at a solution
so initially i tried:

conservation of energy in particles frame: mc^2=E
then transformed this into the lab frame
E'=gamma(E-vp)
E'=Esqrt((1-v/c)/1+v/c))

and then used E'=hc/lambda to get f=E/hcsqrt((1-v/c)/1+v/c))

but then here is my problem- surely the decay of a single particle into a single photon violates the conservation of momentum as in the particles frame it initially has momentum 0, but after the decay it has momentum E/c? I may be wrong, but I'm just slightly confused. Thanks :)

2. Jan 2, 2016

### PeroK

Are you sure the particle doesn't decay into two photons? By conservation of momentum a single particle cannot decay into a single particle with a different mass.

3. Jan 2, 2016

### Physgeek64

This is my problem.... but that is the exact question

4. Jan 2, 2016

### PeroK

It could decay into another unspecified particle and a photon. It doesn't say only a photon!

5. Jan 2, 2016

### Physgeek64

But surely the frequency of the photon would depend entirely on the mass of the other particle, so it would be odd not to mention it?

6. Jan 2, 2016

### PeroK

It gives you the energy of the photon. The mass of the original particle and the details of the decay are irrelevant.

7. Jan 2, 2016

### Physgeek64

ahh okay- so with the exception of my 'conservation of energy' line, my above answer should hold?

8. Jan 2, 2016

### PeroK

Are you sure that's right? What would Mr Doppler have to say?

Also, there's no need to introduce $\lambda$ as $E' = hf_{obs}$

9. Jan 2, 2016

### Physgeek64

I did initially think that, but the expression i've obtained is already that of the doppler shift? Or is that just a coincidence?

Oh yeah- silly me

10. Jan 2, 2016

### PeroK

It's no coincidence. The energy transformation for a photon leads to the relativistic Doppler formula. Is the energy/frequency of the observed photon (in your answer) greater than or less than the emitted photon? What should it be?

11. Jan 2, 2016

### Physgeek64

It should be greater, so the other way round. I've probably made an algebraic mistake somewhere. But thank you! thats really helpful :)