Relativistic Momentum Derivation Help

[jimbobian]

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

 

[elfmotat]

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}

I can't follow what you wrote. You're using the same symbol for velocities in two different frames.

 

[jimbobian]

I can't follow what you wrote. You're using the same symbol for velocities in two different frames.

Sorry my mistake, the first two should be unprimed. I will edit my OP.

 

[elfmotat]

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 let's [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]

 

[jimbobian]

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 let's [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]

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?

 

[elfmotat]

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?

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.

 

[jimbobian]

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 as Superstring. I didn't want the post showing up under that name, so I deleted it and re-posted it under this name.

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.