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Centre of mass frame in nuclear reactions?

  1. Mar 7, 2014 #1
    I'm slightly confused as to how we discuss centre of mass frames in nuclear reactions.

    We have the reaction a(A,B)b. Before the reaction we can transfer to the centre of mass frame

    [itex] v_c = \frac{\sum m_i v_i}{\sum m_i} [/itex]

    and note that the total energy before the reaction is

    [itex] E_{CM} = E_{masses~before} + E_{kinetic~before~in~CM} [/itex]

    And the total energy after is

    [itex] E_{CM} = E_{masses~after} + E_{kinetic~after~in~CM} [/itex]

    The problem I'm having to understand is that these two before/after energies are often equated, but the centre of mass frame is not a constant since the masses are not constant in a nuclear reaction, so [itex] v_c [/itex] changes since the denominator [itex] \sum m_i [/itex] changes (numerator remains the same due to conservation of momentum).

    Am I wrong here? If I were to guess why we can do this, it's because the change in masses is so small that we can assume the centre of mass frame is constant before/after the reaction. But then this discussion breaks down when we're considering only light particles and the fractional mass change is larger, right?
  2. jcsd
  3. Mar 7, 2014 #2


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    Momentum is conserved. If you work in a frame where total momentum is zero before the interaction then total momentum will also be zero after the interaction.

    The formula:

    [itex] v_c = \frac{\sum m_i v_i}{\sum m_i} [/itex]

    is a non-relativistic approximation. It will not, in general, correctly tell you the relative velocity of the frame in which total momentum is zero.
  4. Mar 7, 2014 #3
    The relativistic formula for the speed of the center of mass is

    [itex] v_c = \frac{\sum m_i \gamma_i v_i}{\sum m_i \gamma_i}, [/itex]

    where [tex]\gamma_i = \frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}[/tex]
  5. Mar 7, 2014 #4


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    That seems not to account for the possibility of massless particles moving at light speed. May I assume that one would simply divide total momentum by total energy (using units in which c=1) to cover that case?
  6. Mar 7, 2014 #5
  7. Mar 8, 2014 #6
    So the frame in which the momentum is zero, is the same before and after a nuclear reaction, because the actual formula takes relativistic mass changes in to account?
  8. Mar 8, 2014 #7


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