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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 (696 points)
edited Mar 18, 2019 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.