Relativistic momentum

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  • #1
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I have been trying to derive why relativistic momentum is defined as ##p=\gamma mv##.

I set up a collision between 2 same balls (##m_1 = m_2 = m##). Before the collision these two balls travel one towards another in ##x## direction with velocities ##{v_1}_x = (-{v_2}_x) = v##. After the collision these two balls travel away from each other with velocity ##{v_1}_y = (-{v_2}_y) = v##. Coordinate system travells from left to right with velocity ##u=v## at all times (after and before collision).

Please see the pictures below where picture (a) shows situation before collision and picture (b) after collision.

http://shrani.si/f/2A/m3/4kDXDQo1/momentum.png [Broken]

Below is a proof that Newtonian momentum ##mv## is not preserved in coordinate system ##x'y'##. I used ##[\, | \,]## to split ##x## and ##y## components. ##p_z'## is momentum before collision where ##p_k'## is momentum after collision.

[itex]
\scriptsize
\begin{split}
p_z' &= \left[ m_1 {v_1}_x' + m_2 {v_2}_x'\, \biggl| \, 0 \right] = \left[ m_1 0 + m_2 \left( \frac{{v_2}_x - u}{1-{v_2}_x\frac{u}{c^2}} \right)\, \biggl| \, 0 \right]= \left[ m \left( \frac{-v - v}{1+ v \frac{v}{c^2}} \right) \, \biggl| \, 0 \right] \\
p_z' &= \left[ - 2mv \left( \frac{1}{1+ \frac{v^2}{c^2}}\right) \, \biggl| \, 0 \right]
\end{split}
[/itex]

[itex]
\scriptsize
\begin{split}
p_k' &= \left[-2mv \, \biggl| \,m_1 {v_1}_y' + m_2 {v_2}_y'\right]=\left[ -2mv \, \biggl| \, m_1 \left( \frac{{v_1}_y}{\gamma \left(1 - {v_1}_y \frac{u}{c^2}\right)} \right) + m_2 \left( \frac{{v_2}_y}{\gamma \left(1 - {v_2}_y \frac{u}{c^2}\right)} \right) \right]\\
p_k' &= \left[ -2mv \, \biggl| \, m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right) - m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right)\right]\\
p_k' &= \left[ -2mv \, \biggl| \, 0 \right]
\end{split}
[/itex]

It is clear that ##x## components differ by factor ##1/\left(1+\frac{v^2}{c^2}\right)##.

QUESTION: I want to know why do we multiply Newtonian momentum ##p=mv## by factor ##\gamma = 1/ \sqrt{1 - \frac{v^2}{c^2}}## and where is the connection between ##\gamma## and factor ##1/\left(1+\frac{v^2}{c^2}\right)## which i got?

FURTHER EXPLAINATION:
In the proof above I used velocity transformations derived below (derivation is taken from here):

[itex]
v_x' = \frac{dx'}{dt'}=\frac{\gamma (d x - u d t)}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \frac{d x - u d t}{d t - d x \frac{u}{c^2}} = \frac{\frac{d x}{d t} - u \frac{d t}{d t}}{\frac{d t}{d t} - \frac{d x}{d t} \frac{u}{c^2}} \Longrightarrow \boxed{v_x' = \frac{v_x - u}{1- v_x \frac{u}{c^2}}}
[/itex]


[itex]
v_y' = \frac{dy'}{dt'}=\frac{d y}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \frac{\frac{dy}{dt}}{\gamma \left(\frac{dt}{dt} - \frac{dx}{dt} \frac{u}{c^2} \right)} \Longrightarrow \boxed{v_y' = \frac{v_y}{\gamma \left(1 - v_x \frac{u}{c^2} \right)}}
[/itex]
 
Last edited by a moderator:

Answers and Replies

  • #2
Nugatory
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Look carefully at the x component of the momentum in the first expression in the first line of your derivation of [itex]p_k^{'}[/itex]
 
  • #3
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Look carefully at the x component of the momentum in the first expression in the first line of your derivation of [itex]p_k^{'}[/itex]

You mean this expression?

[itex]
p_k' = \left[ -2mv\, \biggl| \, m_1 {v_1}_y + m_2 {v_2}_y \right]
[/itex]

Coordinate system and observer in its origin move with ##u=v## from left to right at all times. This is why after collision observer in ##xy## would say that balls don't move in ##x## direction, while observer in ##x'y'## would state, that they both move in opposite direction. This is where i got my ##x## component ##m_1 {v_1}_x' + m_2 {v_2}_x' = m (-v) + m(-v) = -2mv##.
 
Last edited:

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