# How to transform velocity 4-vectors to Zero Momentum Frame

I have a particle $$p$$ with speed $$u$$ in lab frame approaching a stationary particle $$q$$.

The $$p^{\mu}$$ and $$q^{\mu}$$ velocity 4-vectors are:

$$p_{LAB}^{\mu}=\gamma_u(c, u, 0, 0)$$ $$q_{LAB}^{\mu}=(c, 0, 0, 0)$$

To get to ZMF, I need a standard lorentz boost with speed $$v=u/2$$: $$p_{ZMF}^{\mu}= \begin{pmatrix} \gamma_{v} & -\gamma_v \beta_v & 0 & 0\\ -\gamma_{v} \beta_v & \gamma_v & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix}\begin{pmatrix} \gamma_{u}c\\ \gamma_{u}u\\ 0\\ 0\\ \end{pmatrix}=\gamma_u \gamma_{\frac{u}{2}} \begin{pmatrix} c-\frac{u^2}{2c}\\ \frac{u}{2}\\ 0\\ 0\\ \end{pmatrix}$$

and

$$q_{ZMF}^{\mu}= \begin{pmatrix} \gamma_{v} & -\gamma_v \beta_v & 0 & 0\\ -\gamma_{v} \beta_v & \gamma_v & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix}\begin{pmatrix} c\\ 0\\ 0\\ 0\\ \end{pmatrix}=\gamma_{\frac{u}{2}} \begin{pmatrix} c\\ -\frac{u}{2}\\ 0\\ 0\\ \end{pmatrix}$$

The magnitude of the first spatial component of $$p_{ZMF}^{\mu}$$ is a $$gamma_u$$ times more thanthe first spatial component of $$q_{ZMF}^{\mu}$$. I would expect that in the ZMF, they are opposite sign but otherwise equal. Is this expectation wrong, and if not, what am I doing wrong?

• Why do you think that a boost with speed $v=u/2$ would take you to the zero-momentum frame? Certainly, $q$ will now seem to be moving towards you with a speed $u/2$, but what about $p$? How does its velocity transform? Nov 24, 2020 at 14:10
• Oh yepp, I cannot just say that its going to be $u-\frac{u}{2}$, can't I, because I would need to use the velcoity addition formula between $u$ and $\frac{u}{2}$. Nov 24, 2020 at 14:13
• Correct! :) Giving up on "common-sense" physics is hard. ;) Nov 24, 2020 at 14:15
• You are really going to need to know something about the masses of p and q to get an answer.
– JEB
Nov 24, 2020 at 15:00

Your intuition is correct, in the zero momentum frame you'd assume both objects to be moving towards you at the same speed. (I'm assuming, of course, that the two objects are identical.) So the first question to ask yourself is "What should this speed be?" The assumption you made ($$v=u/2$$) is incorrect, as while $$q$$ will move towards you with a speed $$u/2$$, $$p$$ will not be moving with a speed $$u/2$$, because of the relativistic velocity addition law.

An easy way to figure it out is to actually equate the speeds of $$p$$ and $$q$$ in this new frame: if the frame is moving rightwards (say that's the direction of $$p$$'s motion) with a speed $$v$$, then it will see:

$$p\text{ moving with velocity }= \frac{u-v}{1-uv/c^2} \quad \text{and q moving with velocity}= -v.$$

Equating the magnitudes it's very easy to see that you get a quadratic equation in $$v$$ $$\frac{u}{c^2} v^2 - 2v + u = 0,$$ which you can solve for $$v$$. Interestingly, when the speeds are much smaller than the speed of light (i.e. in the "non-relativistic case) the first term vanishes since $$uv/c^2 \to 0$$, and you get back the classical result $$v = u/2$$!

You can solve the above quadratic equation to get two solutions for $$v$$, I invite you to find out why only one of them is physically acceptable. This is the boost velocity needed to get to the zero-momentum frame.

Here are alternative approaches to @Philip's answer (where the masses of your particles are assumed equal).

• Although Galilean-velocities add linearly, [real-world Minkowskian-]velocities don't...as you found. In fact, Euclidean-slopes also don't add linearly. (Linearity of Galilean-velocities is more the exception, rather than the rule.)

However, angles (arclengths of "unit circles", or twice-the-areas of "unit-circular"-sectors) are additive.

If you were given these two rays in ordinary euclidean geometry,
how would you find the ray in the "center" between these two rays?
...so that if you rotated to that center-ray, the other rays would be on opposite sides of it, oriented symmetrically. We would call that center-ray the "angle bisector".

In the diagram below, let's measure slopes with respect to the vertical.
If the slope of the angled-ray is $$u$$, then the slope of that center-ray ("angle bisector") would be $$m=\tan\left(\frac{\arctan(u)}{2}\right)$$ (This should agree with the result gotten using the relative-slope formula $$\tan\theta=\displaystyle\frac{m_2-m_1}{1+m_2m_1}$$ with @Philip's method.)
What would the analogous quantity be for special relativity? • The following approach also works when the masses are unequal (and is generalizable to multiple particles).
The "center-of-momentum" frame is the frame where the "spatial-component of the total 4-momentum" is zero, which means that this frame has a 4-velocity along the direction of the total-4-momentum.

So, one can determine this 4-velocity as the unit-vector along the total-4-momentum: $$\hat p_{COM} =\frac{\tilde p_{1}+ \tilde p_{2}}{ |\tilde p_{1}+ \tilde p_{2}| } =\frac{\tilde p_{1}+ \tilde p_{2}}{ M_{COM}c },$$ which can be written in terms of the invariant-mass of the system $$M_{COM}$$.