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]

[itex]u_1,u_2[/itex] 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]

[itex]u_1,u_2[/itex] 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:

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

Also we know that [itex]E=\gamma mc^2[/itex], 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 [itex]\alpha/2[/itex] 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 ([itex]m,u_3,\beta[/itex]) and 3 equations [itex](E,p_x,p_y)[/itex], so the unknowns are determined. Just double check the definitions of everything before plugging into formulas, since you do have a lot of mistakes.

 

[paranormal]

I would first clarify the context of this problem. Is it related to nuclear fusion, particle physics, or some other field? This information would be important in determining the appropriate equations and assumptions to use in solving the problem.

Assuming this is related to nuclear fusion, I would approach the problem by first considering conservation of momentum and energy. The total momentum before the collision is equal to the total momentum after the collision, and the total energy before the collision is equal to the total energy after the collision. This can be expressed as:

m1u1 + m2u2 = (m1 + m2)v

and

m1c2 + m2c2 = (m1 + m2)c2 + KE

where v is the final velocity of the fused particle and KE is the kinetic energy of the fused particle.

Next, I would use the Law of Cosines to relate the angle α to the final velocity v:

v2 = u12 + u22 - 2u1u2cosα

Substituting this into the equations for conservation of momentum and energy, we can solve for the final mass m:

m = (m1m2u1u2cosα + m1m2c2) / (c2 - u1u2cosα)

This expression takes into consideration the angle α and the masses and velocities of the two particles before the collision. However, it is important to note that this is a simplified model and may not accurately reflect the complex processes involved in nuclear fusion. Further research and experimentation would be needed to fully understand the behavior of fused particles.