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Four momentum of fused particles

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data

    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


    2. Relevant equations

    Etot2 = p2c2+m2c4


    3. 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.
     
  2. jcsd
  3. Jan 5, 2012 #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.
     
  4. Jan 9, 2012 #3
    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?

    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.
     
  5. Jan 9, 2012 #4

    fzero

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    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.


    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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