Relativistic billiard balls.

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Assuming equal mass billiard balls and elastic collisions, classical physics shows that after any collision the motion of the colliding balls will be orthogonal. How does that situation change under SR ? More generally for an elastic collision between objects m and M with m<M, is maximum angle of deflection ( given classically by sin(x)=m/M ) affected ?
 
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  • #2
PeroK
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Assuming equal mass billiard balls and elastic collisions, classical physics shows that after any collision the motion of the colliding balls will be orthogonal. How does that situation change under SR ? More generally for an elastic collision between objects m and M with m<M, is maximum angle of deflection ( given classically by sin(x)=m/M ) affected ?
I can give you a quick answer, as I'm just going offline for a while.

You can follow the same method by considering the COM frame. In that frame, the two particles have the same speed before and after the collision and move in opposite directions.

However, if one particle is at rest in your frame and the other is moving at a speed ##u##, then the COM frame is not moving at speed ##u/2##. Can you work out what speed it is moving at?

So, when you transform the resulting velocities back to your frame you get slightly different expressions and the angle is not a right angle.

Do you know enough SR to work this out yourself?
 
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I can give you a quick answer, as I'm just going offline for a while.
So, when you transform the resulting velocities back to your frame you get slightly different expressions and the angle is not a right angle.

Do you know enough SR to work this out yourself?
Thanks for your response - unfortunately I don't. The query arises out of following discussion which I think has reached an incorrect conclusion - in light of what you are saying.

https://www.physicsforums.com/threa...vistic-elastic-collision.950954/#post-6056515

If I read you correctly we DO get angular 'distortion' when SR equations are applied to collision problems. And therefore maximum angle of deflection (elastic collision) is NOT the same as it would be when calculated using classical physics. Although as is usual with SR the classical equations will work fine for v<<c.
 
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Thanks for your response - unfortunately I don't. The query arises out of following discussion which I think has reached an incorrect conclusion - in light of what you are saying.

https://www.physicsforums.com/threa...vistic-elastic-collision.950954/#post-6056515

If I read you correctly we DO get angular 'distortion' when SR equations are applied to collision problems. And therefore maximum angle of deflection (elastic collision) is NOT the same as it would be when calculated using classical physics. Although as is usual with SR the classical equations will work fine for v<<c.
I haven't had time to digest that thread. It's mostly talking about the unequal mass collision. Let me check my working and get back to you. I may have time today.
 
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Can we apply Eqn 4.80 to solve the problem on maximum angle of deflection for a relativistic collision as per original problem posed in that thread ? Would be interested to see if we can find where that √3 comes in.

I think I'm correct in saying that if mass M (>m) is incident on stationary mass m with collision angle = θ(max) the post collision trajectories are also orthogonal per classical theory.
 
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vela
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Can we apply Eqn 4.80 to solve the problem on maximum angle of deflection for a relativistic collision as per original problem posed in that thread ? Would be interested to see if we can find where that √3 comes in.
Equation 4.80 was derived using the assumption ##m=M##. The problem in the thread you're looking at considers a collision between unequal masses.

From the expression for ##\theta_1## right before 4.80, you can write
$$\tan \theta_1 = \frac{\sin\theta_0}{\gamma(\cos\theta_0+1)} = \frac{2\sin\frac{\theta_0}{2}\cos\frac{\theta_0}{2}}{\gamma(2\cos^2 \frac{\theta_0}{2})} = \frac 1\gamma \tan\frac{\theta_0}{2}.$$ For any value of ##\gamma##, you should be able to convince yourself that ##\theta_{1\text{max}}=\pi/2##.
 
  • #9
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From the expression for ##\theta_1## right before 4.80, you can write
$$\tan \theta_1 = \frac{\sin\theta_0}{\gamma(\cos\theta_0+1)} = \frac{2\sin\frac{\theta_0}{2}\cos\frac{\theta_0}{2}}{\gamma(2\cos^2 \frac{\theta_0}{2})} = \frac 1\gamma \tan\frac{\theta_0}{2}.$$ For any value of ##\gamma##, you should be able to convince yourself that ##\theta_{1\text{max}}=\pi/2##.
I don't think there's any problem with this conclusion - it's showing maximum recoil angle = 90. Which is true for any collision between any mass m<=M. Doesn't it just mean there's no forward scattering of object m ?
 
  • #10
vela
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I don't think there's any problem with this conclusion - it's showing maximum recoil angle = 90. Which is true for any collision between any mass m<=M. Doesn't it just mean there's no forward scattering of object m ?
I thought you were trying to figure out where the ##\sqrt 3## comes from when ##M \gg m##.
 
  • #11
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Yes I was but I think I'm getting rapidly 'out of depth' here. Can't convince myself at all unless θο = π when the expression has a zero denominator. But then the other angle will be zero and yet we are saying they don't sum to π/2. If I sound confused , I am !

In the original problem, don't forget the stipulation that γ = M/m.

I am aware that the formulas in the Oxford reference deal with an elastic collision between equal masses but I was hoping to see what would change (under SR) in that instance and then try to adapt it for the problem in the other post. Under SR it seems that angles have some dependence on γ - hence on velocity. Whereas in classical physics both the π/2 (sum of angles) and maximum deflection sinθ = m/M results are independent of velocity.
 

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