What Happens in a Relativistic Elastic Collision with Equal Masses?

AI Thread Summary
In an elastic head-on collision between two equal masses, the final velocity of the target mass can be derived from the conservation of momentum and energy principles. The initial 4-momenta of both masses are calculated in the laboratory frame, and a Lorentz transformation is applied to find the center-of-momentum frame. The analysis shows that if the masses are equal, the projectile effectively comes to rest while the target mass acquires its velocity, leading to a straightforward transfer of momentum. The challenge lies in simplifying the equations to express the final velocities clearly when both masses are equal. This problem illustrates the complexities of relativistic collisions and the importance of understanding frame transformations.
runevxii
Messages
7
Reaction score
1
I've done the bulk of this problem (part A) but I still can't figure the last bit (part B):

"Consider an elastic head-on collision between a projectile with mass m_a, velocity (vector)v_a, energy E_a, and three-momentum (vector)p_a and a stationary target of mass m_b. (That the collision is head-on measn that the two particles emerge from the collision both moving along the line of the incident velocity (vector)v_a.)

a) What is the final velocity (vector)v_b of the target particle b?"

To solve this problem I first determined the initial 4-momenta of both a in b the laboratory frame:
P_ai = (P_t, P_x, P_y, P_z) = (E_a/c, p_a, 0, 0)
P_bi = (m_b*c, 0, 0, 0)
Where c is the speed of light.

Then I did a Lorentz tranformation into a frame where the total 3-momenta is 0 (keeping in mind that I don't know beta or gamma yet because I don't know the speed of this frame relative to the lab frame):
P'_ai = (y*E_a/c - y*B*p_a, y*p_a - B*y*E_a/c, 0, 0)
P'_bi = (y*m_b*c, -y*B*m_b*c, 0, 0)
Here, y is gamma and B is beta.

Then to find the relative speed of the frames, I add the spatial components of the 4-momenta (i.e. the 3-momenta) and I set them to 0:
P_total = 0 = y(-B(E_a/c + m_a*c) + p_a)
B = v_cm/c = p_a*c / (E_a + m_b*c)
Here, v_cm is the relative velocity of the center-of-momentum frame relative to the lab frame.

Now I try to find the final 4-momentum of mass m_b. Since we know the collision is elastic, |P'_bi| = |P'_bf|, and
P'_bf = (y*m_b*c, y*B*m_b*c, 0, 0)
The 3-momentum simply has the opposite sign.

Then I transformed P'_bf back into the lab frame by inverse Lorentz transfromation, which gave
P_bf = (y^2*m_b*c + y^2*B^2*m_b*c, 2*y^2*B*m_b*c, 0, 0) = (E_b/c, p_b, 0, 0)

Utilizing the relationship, (vector)v_b = p_b*c^2/E_b = 2*y^2*B*m_b*c^3/(y^2*m_b*c^2 + y^2*B^2*m_b*c^2) = 2*y^2*B*m_b*c^3/[y^2*m_b*c^2(1 + B^2)],
(vector)v_b final = 2*B*c/(1 + B^2), B = v_cm/c = p_a*c / (E_a + m_b*c) from before.

So that's part a, now comes what seems like the easy part:

"b) Describe what happens if the 2 masses are equal."

I suspect that m_a becomes rested and m_b picks up all it's speed, but I'm having a hard time putting it into numbers.

I tried seeing what happened if I took m_a = m_b into the equation for B, and got y*v/((y + 1)c) but when I try to plug that into (vector)v_b, I can't simplify it and gain no further insight.
 
Physics news on Phys.org
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top