Four momentum of fused particles

[LASmith]

Homework Statement

two particles of masses m₁ and m₂ move at speeds u₁, u₂ respectively collide and fuse. If α is the angle between the two directions of motion before the collision, find an expression for the new mass m, in terms of m₁ m₂ u₁ u₂ and α

Homework Equations

E_tot² = p²c²+m²c⁴

The Attempt at a Solution

(m₁² u₁² c² + m₁ ² c⁴)^1/2 + (m₂² u₂² c² + m₂² c⁴)^1/2 = mc²

However this does not take into consideration the angle α, so I am unsure as to where this comes in.

 

[fzero]

u_1,u_2 are speeds in the laboratory frame and, since there is an angle between the initial particles, there is net momentum of the system. So the final state particle cannot be at rest in the laboratory frame. You must also consider momentum conservation to solve the problem.

 

[LASmith]

u_1,u_2 are speeds in the laboratory frame and, since there is an angle between the initial particles, there is net momentum of the system.

From this can I re-write the RHS of the equation to be = (m²u₃²c² + m²c⁴)^1/2

Squaring both sides I obtain

m₁²(u₁²+c²) + m₂²(u₂²+c²) + 2(m₁²m₂²(u₁²u₂²+u₁²c²+u₂²c²+c⁴))^1/2 = m²(u₃²+c²)

Is this correct?

You must also consider momentum conservation to solve the problem.

So does momentum hold the same in special relativity as it does in general mechanics?
Therefore I would obtain

m₁u₁sin(α/2) - m₂u₂sin(α/2) = mu₃sinβ

&

m₁u₁cos(α/2) + m₂u₂cos(α/2) = mu₃cosβ

However this gives too many unknowns, I have attempted to find either u₃ or β in terms of u₁, u₂ or α but have failed to find a connection, as the mass m is not just simply m₁ + m₂

Any helps would be much appreciated.

 

[fzero]

From this can I re-write the RHS of the equation to be = (m²u₃²c² + m²c⁴)^1/2

Squaring both sides I obtain

m₁²(u₁²+c²) + m₂²(u₂²+c²) + 2(m₁²m₂²(u₁²u₂²+u₁²c²+u₂²c²+c⁴))^1/2 = m²(u₃²+c²)

Is this correct?

You're missing Lorentz factors:

\vec{p} = \gamma m \vec{u}, ~~~\gamma= \frac{1}{\sqrt{1-(u/c)^2}}.

Also we know that E=\gamma mc^2, so we can simplify much of this.

So does momentum hold the same in special relativity as it does in general mechanics?
Therefore I would obtain

m₁u₁sin(α/2) - m₂u₂sin(α/2) = mu₃sinβ

&

m₁u₁cos(α/2) + m₂u₂cos(α/2) = mu₃cosβ

However this gives too many unknowns, I have attempted to find either u₃ or β in terms of u₁, u₂ or α but have failed to find a connection, as the mass m is not just simply m₁ + m₂

Any helps would be much appreciated.

Having \alpha/2 in the formulas looks wrong. You can set the path of one of the incoming particles to be the x-axis, then the angle only appears in the momentum of the other two particles.

You have 3 unknowns (m,u_3,\beta) and 3 equations (E,p_x,p_y), so the unknowns are determined. Just double check the definitions of everything before plugging into formulas, since you do have a lot of mistakes.