Calculating Photon Frequency From Relativistic Decay: Homework Solution

[Physgeek64]

Homework Statement

Hi- I've been doing some questions and came across the following

A high energy particle, traveling 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

Homework Equations

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 :)

 

[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.

 

[Physgeek64]

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.

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

 

[PeroK]

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

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

 

[Physgeek64]

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

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?

 

[PeroK]

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?

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

 

[Physgeek64]

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

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

 

[PeroK]

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

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

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

 

[Physgeek64]

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

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

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

 

[PeroK]

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

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?

 

[Physgeek64]

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?

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