# Proof for $p=\gamma_Pmu$

As I'm reading about Relativistic Momentum, my book states the following:

$$p=m \frac{\Delta x}{\Delta t}=m\frac{\Delta x}{\sqrt{(1-u^2/c^2)}\Delta t}=\frac{mu}{\sqrt{1-u^2/c^2}}=\gamma_Pmu$$

"Whether this is the 'correct' expression for p depends on whether the total momentum P is conserved when the velocities of a system of particles are transformed with the Lorentz velocity transformation equations. The proof is rather long and tedious..."

I'm interested in seeing the proof that they are describing as "long and tedious."

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1
excellent question IMO :-) – David Zaslavsky♦ Jan 24 at 6:59
To be clear: Let $\Lambda$ be a Lorentz transformation and $P=\sum_i p_i$ the total momentum. Is the question whether $\Lambda(P) = \sum_i \Lambda(p_i)$? You would need to prove that Lorentz transformations act linearly on the momentum. This is trivial in the four-vector formalism where you extend the momentum 3-vector $p_i$ by adding a time component - the energy: $p_{i\mu} = (E_i, p_i)$. If you just focus on the spatial components it becomes hard because of the $\gamma$ factors and the energy gets involved. The easiest case to check would be when the momenta and the boost are all co-linear. – Michael Brown Jan 24 at 7:42
Your first $m$ looks like "relativistic" mass, your second $m$ rest mass, and therefore should be labelled differently. – Larry Harson Jan 24 at 19:15
Check my anwser. It might be the long way you want. – 71GA Jan 26 at 13:44

## 3 Answers

First you need to make sure that newton's momentum isn't conserved here so you test this in a case of two balls travelling towards eachother. What you will notice is that momentum is conserved in coordinate system $xy$ and is not conserved in coordinate system $x'y'$.

Here is how you prove that momentum is conserved in $xy$ ($p_1$ is momentum before collision and $p_2$ is momentum after collision):

$$\scriptsize \begin{split} p_1 &= \left[m_A v_{A1x} + m_{B} v_{B1x}\, \bigl| \, 0\right] = \left[m v- m v\, \bigl| \, 0\right] = m\left[v-v \, \bigl| \, 0 \right]=m\left[0 \, \bigl| \, 0 \right] \\ p_2 &= \left[m_A v_{A2y} + m_{B} v_{B2y} \, \bigl| \, 0 \right] = \left[m v - mv \, \bigl| \, 0 \right] = m\left[v-v \, \bigl| \, 0 \right]=m\left[0 \, \bigl| \, 0 \right] \\ \\p_1 &= p_2 \end{split}$$

I used quotes like $[~~|~~]$ just to sepparate $x$ and $y$ components of momentum like this $p_1=[p_{1x}|p_{1y}]$.

Now you proove that momentum in coordinate system $x'y'$ isn't conserved. Here you must pay a great deal on coordinate system travelling from left to right at all times - even after the collision (see the picture). This means that after collision both balls will allso have $x'$ components. Here is the proof of inequality:

Before collision: $$\scriptsize \begin{split} p_1' &= \left[ m_A v_{A1x}' + m_B v_{B1x}'\, \biggl| \, 0 \right] = \left[ m_A 0 + m_B \left( \frac{v_{B1x} - u}{1-v_{B1x}\frac{u}{c^2}} \right)\, \biggl| \, 0 \right]= m \left[\left( \frac{-v - v}{1+ v \frac{v}{c^2}} \right) \, \biggl| \, 0 \right] =~~~\\ &= m\left[ - 2v \left( \frac{1}{1+ \frac{v^2}{c^2}}\right) \, \biggl| \, 0 \right] \end{split}$$

After collision: $$\scriptsize \begin{split} p_2' &= \left[-2mv \, \biggl| \,m_A v_{A2y}' + m_B v_{B2y}'\right]=\\ &=\left[ -2mv \, \biggl| \, m_A \left( \frac{v_{A2y}}{\gamma \left(1 - v_{A2x} \frac{u}{c^2}\right)} \right) + m_B \left( \frac{v_{B2y}}{\gamma \left(1 - v_{B2x} \frac{u}{c^2}\right)} \right) \right]=\\ &= \left[ -2mv \, \biggl| \, m \left( \frac{v}{\gamma \left(1 - 0 \frac{v}{c^2}\right)} \right) - m \left( \frac{v}{\gamma \left(1 - 0 \frac{v}{c^2}\right)} \right)\right]= m\left[ -2v \, \biggl| \, \left( \frac{v}{\gamma} \right) - \left( \frac{v}{\gamma} \right)\right]=\\ &= m \left[ -2v \, \biggl| \, 0 \right] \end{split}$$

You will notice that equations differ for a factor of $1/(1+ v^2/c^2)$ and it means that classical momentum is not appropriate for relativity as it isnt conserved in all coordinate systems.

Now you write down an inequality just found out, do some algebra and you predict that you can change sign $\neq$ with $=$ if you multiply both sides of inequality equation by some function... i used notation $\gamma(v)$.

$$\scriptsize \begin{split} m\left[ -2v \left( \frac{1}{1+\frac{v^2}{c^2}} \right) \, \biggl| \, 0 \right] &\neq m \left[-2v \, \biggl| \, 0 \right]\\ -2 mv \left( \frac{1}{1+\frac{v^2}{c^2}} \right) &\neq -2 mv\\ \frac{-2 mv}{1+\frac{v^2}{c^2}} &\neq -2 mv\\ \gamma(v) \, \frac{-2 mv}{1+\frac{v^2}{c^2}} &= -2 mv \, \gamma(v)\\ \end{split}$$

We found out that the function $\gamma(v)$ is actually like this (i don't know exactly how scientists calculated it or derived it, but they found it):

$$\scriptsize \gamma(v) = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

CAUTION: $v$ is not a relative speed of coordinate systems mooving relative to eachother (which i denote usualy by $u$) but it is a full speed of an object in some coordinate system.

Now you only need a proof to proove that this function is the one. And here is the finale. Because in $\gamma(v)$ there is a full speed you first calculate full speed $v$ for an object before and after collision in coordinate system $x'y'$ and then use those to proove that if you multiply inequality equation above with $\gamma(v)$ you get equality. I choose only one object (that is ball with mass $m_A$) and check its momentum before and after the collision:

Full speed of $m_A$ in $x'y'$ before collision:

$$\scriptsize v = -2v \left( \frac{1}{1+\frac{v^2}{c^2}} \right)= \frac{-2v}{1+\frac{v^2}{c^2}}\\$$

Full speed of $m_A$ in $x'y'$ after collision:

$$\scriptsize \begin{split} v&=\sqrt{(-v)^2 + \left(\frac{v}{\gamma}\right)^2}=v \sqrt{1+ \frac{1}{\gamma^2}} =\\ &= v \sqrt{ 1 + \left(1 - \frac{u^2}{c^2}\right)} = v \sqrt{2 - \frac{v^2}{c^2}} \end{split}$$

And the proof that momentum of ball with mass $m_A$ is conserved now:

$$\begin{split} \gamma \! \left(\frac{-2v}{1+\frac{v^2}{c^2}}\right) \, \frac{-2mv}{1 + \frac{v^2}{c^2}} &= -2mv \, \gamma \! \left(v \sqrt{2-\frac{v^2}{c^2}}\right)\\ \frac{1}{\sqrt{1- \frac{\left\{-2v/\left(1+v^2\!/\!c^2\right)\right\}^2}{c^2}}} \, \frac{-2mv}{1 + \frac{v^2}{c^2}} &= -2mv \, \frac{1}{\sqrt{1-\frac{\left\{v\sqrt{2-v^2\!/\!c^2}\right\}^2}{c^2}}}\\ \frac{1}{\sqrt{1- \frac{\left\{-2v/\left(1+v^2\!/\!c^2\right)\right\}^2}{c^2}}} \, \frac{1}{1 + \frac{v^2}{c^2}} &= \frac{1}{\sqrt{1-\frac{\left\{v\sqrt{2-v^2\!/\!c^2}\right\}^2}{c^2}}}\\ \frac{1}{1- \frac{\left\{-2v/\left(1+v^2\!/\!c^2\right)\right\}^2}{c^2}} \, \frac{1}{\left(1 + \frac{v^2}{c^2}\right)^2} &= \frac{1}{1-\frac{\left\{v\sqrt{2-v^2\!/\!c^2}\right\}^2}{c^2}}\\ \frac{1}{1- \frac{4v^2}{c^2 \left(1+v^2\!/\!c^2\right)^2}} \, \frac{1}{\left(1 + \frac{v^2}{c^2}\right)^2} &= \frac{1}{1-\frac{v^2\left(2-v^2\!/\!c^2\right)}{c^2}}\\ \frac{1}{\frac{c^2 \left(1+v^2\!/\!c^2\right)^2-4v^2}{c^2 \left(1+v^2\!/\!c^2\right)^2}} \, \frac{1}{\left(1 + \frac{v^2}{c^2}\right)^2} &= \dots\\ \frac{{\bigl(1+\frac{v^2}{c^2}\bigl)}^2}{\frac{c^2 \left(1+v^2\!/\!c^2\right)^2-4v^2}{c^2}} \, \frac{1}{\left(1 + \frac{v^2}{c^2}\right)^2} &= \dots\\ \frac{1}{\frac{c^2 \left(1+v^2\!/\!c^2\right)^2-4v^2}{c^2}} &= \dots\\ \frac{1}{\frac{c^2 \left(1+2v^2\!/\!c^2+v^4\!/\!c^4\right)-4v^2}{c^2}} &= \dots\\ \frac{1}{\frac{c^2+2v^2+v^4\!/\!c^2-4v^2}{c^2}} &= \dots\\ \frac{1}{\frac{c^2-2v^2+v^4\!/\!c^2}{c^2}} &= \dots\\ \frac{1}{\frac{c^2}{c^2} - 2\frac{v^2}{c^2} + \frac{v^4}{c^4}} &= \dots\\ \frac{1}{1 - \frac{v^2}{c^2} \left( 2 - \frac{v^2}{c^2} \right)} &= \dots\\ \frac{1}{1 - \frac{v^2 \left( 2 - v^2\!/\!c^2 \right)}{c^2}} &= \frac{1}{1-\frac{v^2\left(2-v^2\!/\!c^2\right)}{c^2}}\\ \end{split}$$

Wohoho now we have equality all of a sudden. Well thats what we wanted right? This is the proof that equation $p=\gamma(v)mv$ is the one to replace $p=mv$.

This is really step by step using easy algebra. I used ˝$\dots$˝ where i didn't change anything compared to a line above.

CAUTION: If you read this you own me a beer hehe. This and more of stuff like this will be in my new book. It is $1/4$ done at the moment.

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The answer to this question depends on what you mean by "derive." If you were to define relativistic momentum by the expression $\gamma m v$, then you could show, in real, physical experiments, that the relativistic momentum of an isolated system of particles is conserved. Moreover, one can show mathematically that this conservation law is Lorentz-covariant in the sense that it holds in all inertial frames, and that it reduces to the Newtonian expression at low speeds.

The covariance property is what the quote is referring to as far as I can tell, and I think this is what Michael Brown is referring to in his comment as well. The idea is that in writing an expression for the relativistic momentum, we want one that leads to conservation of this quantity in all inertial frames, and one that reduces to the Newtonian expression $mv$ in the limit of small velocities. If we were to find an expression for relativistic momentum that weren't to satisfy these criteria, then we would be inclined to call it "incorrect."

Having said this, here is how you would proceed with the proof of Lorentz-covariance of relativistic momentum in the case of a single space dimension and two particles colliding (the generalization to higher dimensions and more particles is more tedious but doesn't really add to understanding).

First, one shows that relativistic momentum transforms as follows under a change of inertial frame from frame $S$ to frame $S'$:

$p' = \gamma(p - v E/c^2)$

where $p$ and $p'$ are the momenta in the respective frames, $v$ is the relative velocity of the frames, and $E$ is the energy ($\gamma m c^2$). Then one notes that if we have conservation of energy and momentum in frame $S$

$p_{1i} + p_{2i} = p_{1f} + p_{2f}, \qquad E_{1i} + E_{2i} = E_{1f} + E_{2f}$

then in $S'$ one has

$p'_{1f} + p'_{2f} = \gamma(p_{1f} + p_{2f} - v(E_{1f}+E_{2f})/c^2) =\gamma(p_{1i} + p_{2i} - v(E_{1i}+E_{2i})/c^2) = p'_{1i} + p'_{2i}$

so that momentum is conserved in $S'$, and similarly for energy.

Please let me know if there are typos. Hope this helps!

Cheers!

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$$p=m \frac{\Delta x}{\Delta t}=m_o\frac{\Delta x}{\sqrt{(1-u^2/c^2)}\Delta t}=\frac{m_ou}{\sqrt{1-u^2/c^2}}=\gamma_Pm_ou$$

m is relativistic mass m$_o$ is rest mass m= m$_o\gamma$ where the $\gamma$ is lorentz factor

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This doesn't answer the question. Saying that relativistic momentum can be written as $mv$ where one has defined $m = m_0 \gamma$ is the same as simply saying that relativistic momentum equals $\gamma m_0 v$. Also, the concept of relativistic mass is antiquated. The mass of the object IS it's rest mass; there's no need to introduce a notion of relativistic mass in relativity. – joshphysics Jan 24 at 16:51
@joshphysics "no need to introduce a notion of relativistic mass in relativity." then can you tell me what is difference between m and m$\gamma$ introduce a notion of relativistic mass in relativity is necessary. – king Jan 24 at 19:07
I recommend avoiding relativistic mass as long as possible. – 71GA Jan 24 at 19:48

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@king In the modern treatment, $m$ denotes what you're calling the rest mass of the particle, and $\gamma m = \frac{1}{\sqrt{1-v^2/c^2}}m$. When I say that there's no need to introduce the notion of relativistic mass, I'm not saying that you're somehow forbidden from calling the quantity $\gamma m$ the "relativistic mass" of a particle, but this often leads people to conceptual confusion (in my experience), and one can precisely state every result in relativity, and perform any calculation he/she would like, without introducing such a term. So why not insist on terminological simplicity? – joshphysics Jan 24 at 22:17
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