What Happens in a Relativistic Elastic Collision with Equal Masses?

Click For Summary
SUMMARY

The discussion focuses on the analysis of a relativistic elastic collision between two particles of equal mass. The initial 4-momenta for both particles are established in the laboratory frame, followed by a Lorentz transformation to a center-of-momentum frame. The final velocity of the target particle is derived using the relationship between momentum and energy, leading to the conclusion that when the masses are equal, the projectile comes to rest while the target acquires its speed. The participant struggles with simplifying the equations to derive numerical insights for equal masses.

PREREQUISITES
  • Understanding of relativistic mechanics and elastic collisions
  • Familiarity with 4-momentum and Lorentz transformations
  • Knowledge of gamma (γ) and beta (β) factors in relativity
  • Basic grasp of energy-momentum relationships in physics
NEXT STEPS
  • Study the derivation of 4-momentum in relativistic collisions
  • Learn about Lorentz transformations in detail
  • Investigate the implications of equal mass collisions in relativistic physics
  • Explore numerical methods for solving relativistic collision problems
USEFUL FOR

Physics students, researchers in relativistic mechanics, and anyone interested in understanding the dynamics of elastic collisions in high-energy physics.

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.
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org

Similar threads

Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K