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Conservations laws relativity

  1. Aug 5, 2006 #1
    A particle of rest mass Ma, decays into two massles particles of B abd C of energy Eb and Ec respecitvely. The momenta of particles B and C are separated by an angle [tex]\theta[/tex]. Calculate the combinded momentum and combined energy of B and C and hence show that particle A has a mass given by,
    Ma = 1/[tex]c^2[/tex] . [sqrt(2EbEc(1- cos [tex]\theta[/tex])]

    Pa = 0 = Pb + Pc = [tex]\gamma[/tex]MbVb + [tex]\gamma[/tex]McVc

    E = Ea = Eb + Ec
    = [tex]\gamma[/tex]Ma[tex]c^2[/tex]

    I know that Eb and Ec can be easily found using the [tex]E^2[/tex] formula, but am not sure how to take the equations and find Ma.
     
    Last edited by a moderator: Aug 5, 2006
  2. jcsd
  3. Aug 5, 2006 #2
    The idea here is just what you started. But use 4-momentums I think.

    Using 4-momentums (I will use capital P for a 4-momentum and a lower case p for 3 momentums so there is no ambiguity)

    Please note I am using units in which c=1
    [tex]
    (P_A)^2 = (P_B + P_C)^2
    [/tex]
    From the LHS of the equation:
    [tex]
    (P_A)^2 = m_A^2
    [/tex]
    From the RHS of the equation:
    [tex]
    (P_B + P_C)^2 = m_B^2 + m_C^2 + 2 P_B \cdot P_C
    [/tex]


    [tex]
    p_B = (E_B,\vect{p}_b)
    [/tex]
    [tex]
    p_C = (E_C,\vect{p}_c)
    [/tex]

    So:
    [tex]
    (P_B + P_C)^2 = m_B^2 + m_C^2 + 2 (E_B E_C - p_B p_C cos(\theta))
    [/tex]

    But:
    [tex]
    m_B = m_C = 0
    [/tex]

    Which implies that:
    [tex]
    p_B = \sqrt{E_B^2 + m_B^2} = E_B
    [/tex]
    [tex]
    p_C = \sqrt{E_C^2 + m_C^2} = E_C
    [/tex]

    So therefore:
    [tex]
    m_A^2 = (P_B + P_C)^2 = 2 E_B E_C(1-cos(\theta))
    [/tex]

    Now just go back and put your c's in!
     
    Last edited: Aug 5, 2006
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