 #1
 16
 0
Homework Statement
Homework Equations
Conservation of linear momentum
Conservation of energy
Energymomentum 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 zeromomentum 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:
$$ v_2'=v \tag{11} $$
$$ v_1'=v_1v \tag{12} $$
$$ \gamma(v_1)=M/m \tag{13} \quad \text{(given in the question)} $$
$$ \gamma(v_2')=\gamma(v)=\gamma(v) \tag{14} $$
Since $S'$ is a zeromomentum frame,
$$ \gamma(v_1')Mv_1'=\gamma(v_2')mv_2' \tag{15} $$
Substitute (11)  (14) into (15),
$$ \gamma(v_1v)M(v_1v)=\gamma(v)mv \tag{16} $$
Now we can solve for ##v## and then calculate ##\gamma(v)##, but the calculation is very complicated.
(2) How to obtain ##v_3'##? Apply energymomentum 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 4vectors, 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),
$$ (\mathbf{p}_1\mathbf{p}_3)\cdot(\mathbf{p}_1\mathbf{p}_3)=\mathbf{p}_4\cdot\mathbf{p}_4 \tag{24} $$
$$ p_1^22p_1p_3\cos\theta+p_3^2=p_4^2 \tag{25} $$
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.
Attachments

54.8 KB Views: 1,105

23.6 KB Views: 1,741

28.1 KB Views: 899