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A photon of frequency f, and another of frequency f' (take f' as given) collide to create an electron-positron pair. The frequency f is such that when the collision is head on, there is exactly enough energy to produce the electron-positron pair at rest in the center-of-mass frame. What is f)

So in the rest frame, $E=h(f+f'), p=h\frac{f-f'}{c}$. We know that in the center-of-mass frame, $E'=2m_ec^2$. So my idea was to find the velocity of the center of mass frame, and then $E=\gamma E'$, but this seems overly complicated, is there an easier way?

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might want to add the tag 'quantum mechanics' and/or 'modern physics'. – Dylan Sabulsky Dec 7 '12 at 4:52
Hm I think it's true that in the center of mass frame, $v=\frac{pc^2}{E}$, I think I can use that. – JLA Dec 7 '12 at 4:54
Going to be messy either way. You could do what you wanted or you could just use the conversation of energy to do this (use relativistic momentum). Minimum energy for pair production is 1.022 MeV. – Dylan Sabulsky Dec 7 '12 at 4:57
What do you mean by that? – JLA Dec 7 '12 at 5:01
One sec, trying to work out. – Dylan Sabulsky Dec 7 '12 at 5:14

Using the conservation of energy, we can write $$h(f+f')=2m_{e}c^{2} +T_{e+}+T_{e-}$$ where $T_{e+}$ is the kinetic energy in the positron and $T_{e-}$ is the kinetic energy of the electron. These values are zero for your case, if I'm reading it correctly. If the collision is head on, this makes this much simpler. We can identify $2m_{e}c^{2}$ as the minimum energy needed for Pair Production. The question seems to tell us that you then have $$h(f+f')=1.022 ~ \text MeV$$If the frequency is such that when hit head on there is enough energy for pair production then we have a case where the photon of frequency $f$ has the minimum energy needed for this process and hence has the maximum wavelength. $$\frac{hc}{\lambda_{max}}=2m_{e}c^{2}$$ So now we can say $$\lambda_{max}=\frac{h}{2m_{e}c}=1.21 \cdot 10^{-12}~\text m$$

Please let me know if I'm unclear. Hope this helps! Cheers

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In the center of mass frame, the frequency of each photon is $f''=m_e c^2/h$. Due to the relativistic Doppler effect, $f/f''=f''/f'$, if I am not mistaken.

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How does this help to determine what $f$ is? It is still indeterminate through what you described, right? – Dylan Sabulsky Dec 7 '12 at 6:18
@Dylan Sabulsky: $f'$ is given, $f''$ is known from the first formula in my answer, so $f=f''^2/f'$ from the second formula in my answer. – akhmeteli Dec 7 '12 at 7:54
fair enough! +1 – Dylan Sabulsky Dec 7 '12 at 15:34

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