- #1
runevxii
- 7
- 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.
"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.