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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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    One strategy is outlined in my answer to https://physics.stackexchange.com/questions/687451/lowest-kinetic-energy-of-particle-for-which-reaction-is-possible-invariant-mass – robphy Jan 14 '22 at 14:14
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    You have been taught about the relativistic invariant $E^2-p^2c^2$, which, as the name implies, is the same in every frame. So you never have to transf0rm anything. This quantity is the same in the "before" lab frame and the "after" CM frame. – Cosmas Zachos Jan 14 '22 at 15:22

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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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