Four momentum of fused particles

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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



Etot2 = 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.
 

Answers and Replies

  • #2
fzero
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[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.
 
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[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.
 
  • #4
fzero
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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.
 

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