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If a photon with sufficient energy interacts with a stationary proton, the reaction occurs $$\gamma+p\to p+\pi^0$$

What is the minimum energy of a photon $E_\textrm{min}$ for this reaction to occur.

I've heard that the way to find minimum energy is to use reference frame in which both proton and $\pi$ meson have zero velocity after the collision. I've tried this approach and I've even gotten the energy of the photon in this reference frame as $$E_\gamma'=\frac{m_\pi^2+2m_pm_\pi}{2(m_p+m_\pi)}c^2$$

However I'm having problems with transforming this energy back to the original reference frame. Is this the right approach to this problem?

Dio
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1 Answers1

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If you think geometrically (using the strategy of the answer I referenced in the comment: Lowest kinetic energy of particle for which reaction is possible (invariant mass)),
one need not "transform" to the COM-frame and back.

Draw an energy-momentum diagram.
Your problem in this threshold case is akin to a problem involving the doppler effect [Two timelike vectors with a common tail, whose tips are joined by a lightlike vector.].

When you say

reference frame in which both proton and meson have zero velocity after the collision.

that means that the 4-momenta of the final products are parallel (collinear). [That is, what is generally not-parallel is, in the threshold case, parallel. What would have been a quadrilateral with three timelike vectors (two of which have equal mangitude) and a lightlike vector, you instead have a triangle with two timelike vectors (one with the invariant-mass $m_p+m_{\pi}$) and a lightlike vector.]

The Doppler factor (Bondi k-factor) $k=\exp\theta$ can be expressed in terms of the proton mass and the invariant mass, $(m_p+m_{\pi})$. The $\theta$ is the rapidity [Minkowski-angle] so that velocity $v=\tanh\theta$ and time-dilation factor $\gamma=\cosh\theta$.

With $k$ known, the rapidity-angle $\theta$ can then be used to solve for an unknown leg of Minkowski right triangle using hyperbolic-trig functions of $\theta$.

It's like solving for an unknown component in a free-body diagram. Try it!

[Using $m_p=938$ and $m_\pi=135$, I get 144.715.]

Looking at your answer, it's close... hint: think Doppler effect.

robphy
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