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Relativistic Momentum Derivation Help

  1. Aug 21, 2011 #1
    Hi everyone. So basically I am still struggling to find a description of the derivation of relativistic momentum (via relativistic mass) which explains itself properly (although that may be my fault for not understanding). So, I tried doing it with help from Feynman and can't work out what I'm doing wrong. It starts with a collision of two identical particles which, from what I've seen, seems to be the standard way to go:


    I'm also using the velocity addition formulae which were derived in the previous chapter:
    & v_x = \frac{v'_x+u}{1+uv'_x/c^2}\\
    & v_y = \frac{v'_y\sqrt{1-u^2/c^2}}{1+uv'_x/c^2}\\
    And the reverse:
    & v'_x = \frac{v_x-u}{1-uv_x/c^2}\\
    & v'_y = \frac{v_y\sqrt{1-u^2/c^2}}{1-uv_x/c^2}\\
    So, from the unprimed frame it is clear that:
    & v_{xA} = 0\\
    Also let:
    & v_{yA} = w\\
    & v'_{xA} = \frac{0-u}{1-0/c^2}\\
    & v'_{xA} = -u\\
    & v'_{yA} = \frac{w\sqrt{1-u^2/c^2}}{1-0/c^2}\\
    & v'_{yA} = w\sqrt{1-u^2/c^2}
    As the velocity of B is simply the opposite of A:
    & v'_{xB} = u\\
    & v'_{yB} = -w\sqrt{1-u^2/c^2}
    Therefore, using the velocity addition formulae:
    & v_{xB} = \frac{2u}{1+u^2/c^2}\\
    & v_{yB} = \frac{-w(1-u^2/c^2)}{1+u^2/c^2}\\
    If we say that the mass of each particle is some function of its velocity and denote the velocity of B as v (as that of A is clearly w) then, by conservation of momentum:
    & 2m_ww = -\frac{-2m_vw(1-u^2/c^2)}{1+u^2/c^2}\\
    & m_w = \frac{m_v(1-u^2/c^2)}{1+u^2/c^2}\\
    Now according ot Mr Feynman what I should be getting is:
    & \frac{m_w}{m_v}\sqrt{1-u^2/c^2} = 1
    Can somebody please either point out a mistake in my reasoning/understanding, or tell me why it hasn't worked.

    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Aug 21, 2011 #2
    I can't follow what you wrote. You're using the same symbol for velocities in two different frames.
  4. Aug 21, 2011 #3
    Sorry my mistake, the first two should be unprimed. I will edit my OP.
  5. Aug 21, 2011 #4
    If I remember that chapter correctly, he first analyzes the collision from the frame where the x-component of A's velocity is zero, then he analyzes it from the frame where the x-component of B's velocity is zero.

    I'll call the vertical component of A's velocity [itex]w[/itex] (as you did) in the frame where (calling the velocity of A in this frame [itex]v_A[/itex] and B's velocity [itex]v_B[/itex]) [itex]v_{xA}=0[/itex]. In order to consider the conservation of momentum, you need to know [itex]v_{yB}[/itex] in this frame. Since in a third frame (the first one drawn in your diagram) the collisions are symmetrical, you know that if you were to go to a frame where the x-component of B's velocity is zero then the vertical component of A's velocity in this frame is equivalent in magnitude to [itex]v_{yB}[/itex] in the other frame:

    [itex]v_{yB} = - v'_{yA}[/itex]

    where [itex]v'_A[/itex] is the velocity of A in the frame where [itex]v'_{xB}=0[/itex]

    Since you also know the law that transforms velocities in the y-direction:

    [itex]v'_A = \frac{w \sqrt{1-u^2/c^2}}{1+0/c^2} = w \sqrt{1-u^2/c^2}[/itex]

    where [itex]u[/itex] is the velocity between the frames.

    Now you can apply conservation of momentum:

    [itex]2m_w w = 2m_{v_B} w \sqrt{1-u^2/c^2}[/itex]


    [tex]m_{v_B} = \frac{m_w}{\sqrt{1-u^2/c^2}}[/tex]

    From there, I believe he lets [itex]w[/itex] go to zero. This gives [itex]m_w=m_0[/itex], and [itex]v_{yB} = 0[/itex]. Since the vertical component of [itex]v_B[/itex] is zero, the entire motion of B is horizontal. Since [itex]u[/itex] is defined as the velocity between this frame and the frame where the horizontal component of B's velocity is zero, [itex]v_{xB} = v_{B} = u[/itex]. This gives: [itex]m_{v_B}=m_u[/itex]. Finally, we arrive at:

    [tex]m_{u} = \frac{m_0}{\sqrt{1-u^2/c^2}}[/tex]
  6. Aug 21, 2011 #5
    Thank you so much. It seems so simple when it is explained in a way that just 'clicks'. Reading Feynman he usually has a knack of explaining it in such a way, but in this case it was your explanation that did it for me! Basically ignoring the intermediate frame (the first one in my diagram) is what helped me, because I kept losing track of what u was supposed to be! I am slightly confused about one thing which is: why did your post at first say it was by SuperString and then 20 minutes later elfmotat?
  7. Aug 21, 2011 #6
    No problem :). The reason it said that is because I made a new account (didn't like my old name), and for some reason the site randomly logs me in under that name sometimes. I didn't want the post showing up under that name, so I deleted it and re-posted it under this name.
  8. Aug 21, 2011 #7
    Haha, I thought it might be something like that! It was just confusing getting two nearly identical emails from PF with different usernames!
    Thanks again.
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