What Happens in a Relativistic Elastic Collision with Equal Masses?

In summary, the problem involves an elastic head-on collision between a projectile and a stationary target, where the final velocity of the target is being calculated. The initial 4-momenta of both particles are determined and a Lorentz transformation is done to find the relative speed of the frames. The final 4-momentum of the target is then calculated and used to find the final velocity. For part B, when the masses are equal, the projectile comes to rest and the target gains all of its speed, but further calculations are needed to determine the exact numbers.
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runevxii
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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.
 
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FAQ: What Happens in a Relativistic Elastic Collision with Equal Masses?

What is a relativistic elastic collision?

A relativistic elastic collision is a type of collision between two particles or objects in which both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system before and after the collision remain the same.

How does a relativistic elastic collision differ from a non-relativistic elastic collision?

In a non-relativistic elastic collision, the kinetic energy of the particles is conserved, but the total momentum may change. In a relativistic elastic collision, both kinetic energy and momentum are conserved, taking into account the effects of special relativity.

What is the equation for calculating the kinetic energy in a relativistic elastic collision?

The equation for calculating the kinetic energy in a relativistic elastic collision is:

KE = [(m1m2)c^2/(m1+m2)c^2] * [(c^2-u1^2)^1/2 + (c^2-u2^2)^1/2]

Where m1 and m2 are the masses of the two particles, c is the speed of light, and u1 and u2 are the initial velocities of the particles.

How does the speed of light affect the outcome of a relativistic elastic collision?

The speed of light plays a significant role in relativistic elastic collisions. As the speed of the particles approaches the speed of light, the kinetic energy and momentum equations become more complex and the outcomes of the collision may differ significantly from those in non-relativistic collisions.

Can a relativistic elastic collision ever result in a change in mass or energy?

No, in a relativistic elastic collision, the total mass and energy of the system are always conserved. This is a fundamental principle of special relativity and is supported by experimental evidence.

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