Understanding the Conservation of Momentum in Special Relativity Collisions

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Homework Statement
The problem is from Problem book in relativity and gravitation by Alan P. Lightman et al.

2.2 When a photon scatters off a charged particle which is moving with a speed very nearly that of light, the photon is said to have undergone an inverse Compoton scattering. Consider an inverse Compton scattering in which a charged particle of rest mass ##m## and total mass-energy (as seen in the lab frame) ##E\gg m##, collides head-on with a photon of freqency ##\nu (h\nu \ll m)##. What is the maximum energy the particle can transfer to the photon?

2.12 Consider the elastic collision of a particle of mass ##m_1## with a stationary particle of mass ##m_2<m_1##. Let ##\theta_{max}## be the maximum scattering angle of ##m_1##. In nonrelativistic calculations, ##\sin \theta_{max} =m_2/m_1##. Prove that this result also holds relativistically.
Relevant Equations
The conservation of 4-momentum
Solutions are given in the book, but I could not understand some part of them.

For problem 2.2, denote the 4-momentum of the photon by ##\mathbf P_\gamma##, that of the particle by ##\mathbf P## and the values after scattering by primes.
Then by the conservation of momentum, we have ## \mathbf P_\gamma+\mathbf P=\mathbf P_\gamma^{'}+\mathbf P^{'}##. Also, ##\mathbf P^2=-m^2##. But then the solution gives a equation that $$\mathbf P_\gamma \cdot \mathbf P_\gamma^{'}=\mathbf P \cdot (\mathbf P_\gamma-\mathbf P_\gamma^{'}) .$$
I could not figure out what laws this equation stands for. Other than this equation, I could understand the solution.

For problem 2.12, the solution first consider the C.M. frame. It reads, since the collision is elastic, ##E_1^{CM}=E_{1'}^{CM}##, i.e., $$ \mathbf P_1 \cdot \mathbf u_{CM}= \mathbf P_{1'} \cdot \mathbf u_{CM}$$ which can be evaluated in the lab frame yielding $$-E_1 +\mathbf p_1 \cdot \mathbf v_{CM}=-E_{1'}+\mathbf p_{1'} \cdot \mathbf v_{CM} $$ where ##\mathbf v_{CM}=\mathbf p_1/(E_1+m_2)##. The last equation is my problem.
In classical mechanics, I know ##\mathbf v_{CM}=(\mathbf p_1+\mathbf p_2)/(m_1+m_2)##. But I have difficulties to generalize this result to SR.

PS. I am not sure what the convention to type 4-vectors, the spatial components of them, and ordinary vectors. So the variables may look scrambled. I apologize for that.

Thank you for your time.
 
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Haorong Wu said:
For problem 2.2, denote the 4-momentum of the photon by ##\mathbf P_\gamma##, that of the particle by ##\mathbf P## and the values after scattering by primes.
Then by the conservation of momentum, we have ## \mathbf P_\gamma+\mathbf P=\mathbf P_\gamma^{'}+\mathbf P^{'}##. Also, ##\mathbf P^2=-m^2##. But then the solution gives a equation that $$\mathbf P_\gamma \cdot \mathbf P_\gamma^{'}=\mathbf P \cdot (\mathbf P_\gamma-\mathbf P_\gamma^{'}) .$$
I could not figure out what laws this equation stands for. Other than this equation, I could understand the solution.
Try squaring ##(\mathbf P_\gamma - \mathbf P_\gamma^{'}) +\mathbf P = \mathbf P^{'}##.
 
Haorong Wu said:
For problem 2.12, the solution first consider the C.M. frame. It reads, since the collision is elastic, ##E_1^{CM}=E_{1'}^{CM}##, i.e., $$ \mathbf P_1 \cdot \mathbf u_{CM}= \mathbf P_{1'} \cdot \mathbf u_{CM}$$ which can be evaluated in the lab frame yielding $$-E_1 +\mathbf p_1 \cdot \mathbf v_{CM}=-E_{1'}+\mathbf p_{1'} \cdot \mathbf v_{CM} $$ where ##\mathbf v_{CM}=\mathbf p_1/(E_1+m_2)##. The last equation is my problem.
In classical mechanics, I know ##\mathbf v_{CM}=(\mathbf p_1+\mathbf p_2)/(m_1+m_2)##. But I have difficulties to generalize this result to SR.
For an object of mass ##m##, the momentum and energy are given by ##p=\gamma m v## and ##E = \gamma m##, and it's easy to see that ##v = p/E##. This same formula is being applied to the two-particle system.

Alternatively, you know that in the center-of-mass frame, ##p_1 = -p_2##. You know the momentum and energy of both masses in the lab frame. You can use the Lorentz transformation to solve for the velocity required so that the masses have momenta equal in magnitude but opposite in direction.
 
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