Energy of scattered photon formula

[Reshma]

An electron of mass 'm' and speed 'v' collides with a gamma ray photon of initial energy hf₀, as measured from the laboratory frame. The photon is scattered in the electron's direction of travel. Verify that the energy of the scattered photon, as measured in the laboratory frame, is:
$$E = hf_0\left(1 + \frac{2hf_0}{mc^2}\sqrt{\frac{1 + v/c}{1 - v/c}}\right)^{-1}$$

Well this also seems to be a Compton effect problem. The relativistic Doppler equation for frequency is given by:
$$f = f_0\sqrt{\frac{1 - \beta}{1 + \beta}}$$
where $\beta = v/c$

Need guidance to apply this formula to obtain above result...

 

[Andrew Mason]

An electron of mass 'm' and speed 'v' collides with a gamma ray photon of initial energy hf₀, as measured from the laboratory frame. The photon is scattered in the electron's direction of travel. Verify that the energy of the scattered photon, as measured in the laboratory frame, is:
$$E = hf_0\left(1 + \frac{2hf_0}{mc^2}\sqrt{\frac{1 + v/c}{1 - v/c}}\right)^{-1}$$

Well this also seems to be a Compton effect problem. The relativistic Doppler equation for frequency is given by:
$$f = f_0\sqrt{\frac{1 - \beta}{1 + \beta}}$$
where $\beta = v/c$

Need guidance to apply this formula to obtain above result...

Use the Compton formuala to determine the energy of the photon after the collision in the frame of the electron before the collision. Then apply the Doppler equation to get the energy in the lab frame.

AM

 

[BlazzedTroll]

Thank you for the hints here on how to solve it but I have been working on this problem for several days now and I keep getting to similar places after using the compton scattering and doppler effect but I can't see how sqrt((1-v/c)/1+v/c)) is becoming sqrt((1+v/c)/(1-v/c))
can someone please explain this?

 

[Andrew Mason]

Thank you for the hints here on how to solve it but I have been working on this problem for several days now and I keep getting to similar places after using the compton scattering and doppler effect but I can't see how sqrt((1-v/c)/1+v/c)) is becoming sqrt((1+v/c)/(1-v/c))
can someone please explain this?

v is negative if the observer is moving away from the source and positive if the observer is moving toward the source.

AM

 

[paranormal]

To apply the relativistic Doppler equation to this problem, we can first consider the initial energy of the photon, hf0, as measured in the laboratory frame. This energy is related to the initial frequency of the photon, f0, through the equation E = hf0, where h is Planck's constant.

Next, we can consider the final energy of the scattered photon, E, also measured in the laboratory frame. This energy is related to the final frequency of the photon, f, through the same equation E = hf.

Now, using the relativistic Doppler equation, we can write:
f = f_0\sqrt{\frac{1 - \beta}{1 + \beta}}
where \beta = v/c. Since the photon is scattered in the electron's direction of travel, we can assume that the final velocity of the electron, v, is in the same direction as the initial velocity. This means that \beta = v/c = |v|/c.

Substituting this into the equation, we get:
f = f_0\sqrt{\frac{1 - |v|/c}{1 + |v|/c}}

We can then rearrange this equation to solve for f:
f = \frac{f_0}{\sqrt{\frac{1 + |v|/c}{1 - |v|/c}}}

Next, we can use the equation E = hf to relate the final energy of the photon, E, to its final frequency, f:
E = hf = hf_0\frac{1}{\sqrt{\frac{1 + |v|/c}{1 - |v|/c}}}

Finally, we can substitute this value for E into the original equation for the energy of the scattered photon, giving us:
E = hf_0\left(1 + \frac{2hf_0}{mc^2}\sqrt{\frac{1 + |v|/c}{1 - |v|/c}}\right)^{-1}

Since we assumed that the electron's velocity was in the same direction as the scattered photon, we can replace |v| with v in the equation. This gives us the final result:
E = hf_0\left(1 + \frac{2hf_0}{mc^2}\sqrt{\frac{1 + v/c}{1 - v/c}}\right)^{-1}

This verifies the energy of the