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## Homework Statement

## Homework Equations

Conservation of linear momentum

Conservation of energy

Energy-momentum relationship: ##E^2-(m_0c^2)^2=c^2p^2##

Lorentz transformation of energy: ##p=\gamma(v)(p'+vE'/c^2)##

##\mathbf{p}=\gamma(v)m_0v##, ##E=\gamma(v)m_0c^2##

## The Attempt at a Solution

My attempt to the first question:

In laboratory frame ##S##:

In a zero-momentum frame ##S'## moving with velocity ##v## relative to ##S##:

Components of ##p_3'## are:

$$ p_{3x}'=p_3'\cos\theta' \tag{1} $$

$$ p_{3y}'=p_3'\sin\theta' \tag{2} $$

By Lorentz transformations for energy and momentum,

$$ p_{3x}=\gamma(v)(p_{3x}'+vE_3'/c^2) \tag{3} $$

$$ p_{3y}=p_{3y}' \tag{4} $$

Since

$$ E_3'=\gamma(v_3')Mc^2 \tag{5} $$

$$ p_3'=\gamma(v_3')Mv_3' \tag{6} $$

Substitute (6) into (5),

$$ E_3'=p_3'c^2/v_3' \tag{7} $$

Substitute (1) and (7) into (3),

$$ p_{3x}=\gamma(v)p_3'(\cos\theta'+v/v_3') \tag{8} $$

Substitute (2) into (4),

$$ p_{3y}=p_3'\sin\theta' \tag{9} $$

So the angle of deflection of mass $M$ is given by

$$ \tan\theta=\frac{p_{3y}}{p_{3x}}=\frac{\sin\theta'}{\gamma(v)(\cos\theta'+v/v_3')} \tag{10} $$

If we can resolve ##\gamma(v)## and ##v_3'##, then we can take the derivative of ##\tan\theta## with respect to ##\theta'## and obtain its maximum value.

But I am facing two difficulties:

(1) How to obtain ##\gamma(v)##? My attempt:

$$ v_2'=-v \tag{11} $$

$$ v_1'=v_1-v \tag{12} $$

$$ \gamma(v_1)=M/m \tag{13} \quad \text{(given in the question)} $$

$$ \gamma(v_2')=\gamma(-v)=\gamma(v) \tag{14} $$

Since $S'$ is a zero-momentum frame,

$$ \gamma(v_1')Mv_1'=\gamma(v_2')mv_2' \tag{15} $$

Substitute (11) - (14) into (15),

$$ \gamma(v_1-v)M(v_1-v)=-\gamma(v)mv \tag{16} $$

Now we can solve for ##v## and then calculate ##\gamma(v)##, but the calculation is very complicated.

(2) How to obtain ##v_3'##? Apply energy-momentum relationship on mass ##M## in ##S'## after collision:

$$ (E_3')^2-(Mc^2)^2=(cp_3')^2 \tag{17} $$

Now we have equation (7) and (17) for the three variables ##E_3'##, ##p_3'## and ##v_3'##. We need one more relationship to solve for ##v_3'##.

Besides, it will take too much effort to solve the above equations. I guess there should be a simpler approach. Do you have any idea?

Note 1 : The author only briefly mentioned 4-vectors, so we are not supposed to use it to solve this problem.

Note 2: Another approach is as follows:

$$ \mathbf{p}_1=\mathbf{p}_3+\mathbf{p}_4 \tag{18} $$

$$ E_1+mc^2=E_3+E_4 \tag{19} $$

$$ E_1=\gamma(v_1)Mc^2 \tag{20} $$

$$ E_1^2-(Mc^2)^2=c^2p_1^2 \tag{21} $$

$$ E_3^2-(Mc^2)^2=c^2p_3^2 \tag{22} $$

$$ E_4^2-(mc^2)^2=c^2p_4^2 \tag{23} $$

By (18),

$$ (\mathbf{p}_1-\mathbf{p}_3)\cdot(\mathbf{p}_1-\mathbf{p}_3)=\mathbf{p}_4\cdot\mathbf{p}_4 \tag{24} $$

$$ p_1^2-2p_1p_3\cos\theta+p_3^2=p_4^2 \tag{25} $$

Substitute (13) and (19) - (23) into (25), take derivative of ##\cos\theta## with respect to ##p_3##, set it to zero, then we can find the value of ##p_3## that gives the maximum value of ##\theta##. But the computation is very complicated.

Note 3: A general expression for ##\theta## can be found on p.309 of Goldstein's Classical Mechanics. It seems that the answer in the question is incorrect.