Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [URL]http://www.cassiobury.net/images/momentum_diagram.jpg[/URL]

    I'm also using the velocity addition formulae which were derived in the previous chapter:
    \begin{align}
    & 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}\\
    \end{align}
    And the reverse:
    \begin{align}
    & 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}\\
    \end{align}
    So, from the unprimed frame it is clear that:
    \begin{align}
    & v_{xA} = 0\\
    \end{align}
    Also let:
    \begin{align}
    & v_{yA} = w\\
    \end{align}
    Therefore:
    \begin{align}
    & 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}
    \end{align}
    As the velocity of B is simply the opposite of A:
    \begin{align}
    & v'_{xB} = u\\
    & v'_{yB} = -w\sqrt{1-u^2/c^2}
    \end{align}
    Therefore, using the velocity addition formulae:
    \begin{align}
    & v_{xB} = \frac{2u}{1+u^2/c^2}\\
    & v_{yB} = \frac{-w(1-u^2/c^2)}{1+u^2/c^2}\\
    \end{align}
    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:
    \begin{align}
    & 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}\\
    \end{align}
    Now according ot Mr Feynman what I should be getting is:
    \begin{align}
    & \frac{m_w}{m_v}\sqrt{1-u^2/c^2} = 1
    \end{align}
    Can somebody please either point out a mistake in my reasoning/understanding, or tell me why it hasn't worked.

    Thanks,
    James
     
    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]

    therefore:

    [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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook