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Special relativity: maximum angle of deflection in an elastic collision

3 votes

A problem from A.P. French's Special Relativity:

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My attempt to the first question:

In laboratory frame $S$:
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In a zero-momentum frame $S'$ moving with velocity $v$ relative to $S$:
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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.

asked Jun 29, 2018 in Physics Problems by redoopi (130 points)
You have not made any use yet of the approximation $M \gg m$. Perhaps you can put $mc^2 \approx 0$ in eqns 19 & 23.
We cannot do this in eqn 23 because $E_4$ is also very small. Besides, the computation is still too complicated, which is weird for an introductory textbook. I have found a general expression for ##\theta## on p.309 of Goldstein's Classical Mechanics. It seems that the answer in the question is incorrect.
Eqn 12 and 16 are incorrect as I should have used relativistic velocity addition. I am having a discussion with some guys in https://www.physicsforums.com/threads/max-angle-of-deflection-in-relativistic-elastic-collision.950954/

1 Answer

1 vote

The Lorentz factor (EQ0):

$$\gamma =\frac{1}{\sqrt{1-(\frac{v}{c})^2}}$$

Apply conservation of energy (EQ1 & EQ2 respectively):

$$E_{1} + mc^{2} = E_{3} + E_{4}$$

$$\frac{M}{\sqrt{1-(\frac{v_{1}}{c})^2}}c^2 + mc^{2} = E_{3} + E_{4}$$

Using the energy-momentum relation (EQ3 & EQ4 respectively):

$$E_{3}^2 = P_{3}^2c^2 +M^2c^{4}$$

$$E_{4}^2 = P_{4}^2c^2 + m^2c^{4}$$

Apply conservation of momentum in x direction (EQ5):

$$ \frac{M}{\sqrt{1-(\frac{v_{1}}{c})^2}}v_{1} = \frac{M}{\sqrt{1-(\frac{v_{3}}{c})^2}}v_{3}\cos\theta + \frac{m}{\sqrt{1-(\frac{v_{4}}{c})^2}}v_{4}\cos\phi$$

$$ \frac{M}{\sqrt{1-(\frac{v_{1}}{c})^2}}v_{1} = P_{3}\cos\theta + P_{4}\cos\phi$$

Apply conservation of momentum in y direction (EQ6):

$$\frac{M}{\sqrt{1-(\frac{v_{3}}{c})^2}}v_{3}\sin\theta = \frac{m}{\sqrt{1-(\frac{v_{4}}{c})^2}}v_{4}\sin\phi$$

You obtain EQ7 and EQ8:

$$P_{4}=P_{3}\frac{\sin\theta}{\sin\phi} = \frac{\gamma^2mv_{1}\sin\theta}{\sin(\theta+\phi)}$$

$$P_{3} = \frac{\gamma^2mv_{1}\sin\phi}{\sin(\theta+\phi)}$$

Let's subtract EQ3-EQ4:

$$E_{3}^2 - E_{4}^2 = (P_{3}^2-P_{4}^2)c^2 - (M^2-m^2)c^4$$

$$(E_{3}+E_{4})(E_{3}-E_{4}) = (P_{3}+P_{4})(P_{3}-P_{4})c^2 + (M^2-m^2)c^4$$

$$(\frac{M}{\sqrt{1-(\frac{v_{1}}{c})^2}}c^2+ mc^2)(E_{3}-E_{4}) = (P_{3}+P_{4})(P_{3}-P_{4})c^2 + (M^2-m^2)c^4$$

Using EQ0 and EQ7 you obtain EQ9 :

$$E_{3} - E_{4} = \frac{1}{m(\gamma^2+1)}[P\_{3}^2(1-\frac{\sin^2\theta}{\sin^2\phi}) + m^2c^2(\gamma^2-1)]$$

Knowing that $\gamma = \frac{M}{m}$ EQ2 equals to EQ10:

$$E_{3} + E_{4} = mc^2(1+\gamma^2)$$

Adding EQ9 + EQ10 you obtain EQ11

$$E_{3} = \frac{mc^2\gamma^2}{2(\gamma^2+1)}[3+\gamma^2+\frac{v\_{1}^2\gamma^2(\sin^2\phi-\sin^2\theta)}{c^2\sin^2(\theta+\phi)}]$$

Plugging in $\phi = \frac{\pi}{2}$ and you can obtain scattering angle $\theta = \frac{m}{\sqrt{3}M}$

NOTE: I am not an expert in special relativity. I discussed this problem before posting the solution. Please let me know if there is something you disagree with.

answered Jul 7, 2018 by Jorge Daniel (578 points)
edited 6 days ago by sammy gerbil
I don't understand how you obtain $\theta$ from EQ11. Besides, I finally did the calculations and it seems that $\theta_{max}$ should be $\sin^{-1}(m/M)$. Please refer to https://www.physicsforums.com/threads/max-angle-of-deflection-in-relativistic-elastic-collision.950954/
I will have a look at it, I may have got wrong in the calculation.