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The Phenomenon

The lecture I am watching contained this figure:

enter image description here

It is supposed to show two particles colliding in $S$ and $S'$ frames.

I wanted to check the vectors on this slide using lecture notes, but I failed.


Context

Below I outline the context of which this problem above appears.

4-velocity basics

Relying on lecture notes:

Let's write the proper velocity, $u^{\mu}=\frac{dx^{\mu}}{d\tau}$ as $$u^{\mu}=\left(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau}\right)=\frac{dt}{d\tau}\left(c,\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)=\frac{dt}{d\tau}(c,\vec{u})$$

It is also said that proper time is the time measured by an ideal clock carried by the particle, and is related to the invariant path length by $$ds^2 = c^2 d\tau^2$$

The tangent vector to the wordline is the 4-velocity of the particle, and has components: $$u^{\mu}=\frac{dx^{\mu}}{d\tau}$$

Since proper time is an affine parameter, the length-squared of the 4-velocity, ie $\eta_{\mu\nu} u^{\mu} u^{\nu}$ is constant:

$$\eta_{\mu\nu} u^{\mu} u^{\nu} = \left(\frac{ds}{d\tau}\right)^2 = c^2$$

Where the first equality comes from the fact that the length of the tangent vector is the derivative of the proper path length $s$ along the curve with respect to the parametrisation, here $u$. Second equality comes from $ds^2 = c^2 d\tau^2$.

Using this equality we can find a relation between coordinate and proper time. We'll need: $\eta_{\mu\nu}=\text{DIAG}(1,-1,-1,-1)$, so:

$$\eta_{\mu\nu} u^{\mu} u^{\nu} = \left(\frac{dt}{d\tau}\right)^2(c^2-|\vec{u}|^2)$$

We end up with:

$$\frac{dt}{d\tau}=\left(1-\frac{|\vec{u}|^2}{c^2}\right)^{-\frac{1}{2}} = \gamma_u$$

Velocity transform laws

Between $S$ & $S'$, the 4-velocity transforms are described by this equation:

$$u'^{\mu}=\Lambda^{\mu\text{ }}_{\text{ }\nu}u^{\nu}$$

ie:

enter image description here

It is claimed that (and I have checked it and it is true) that from the first component of the equaition above we can have:

$$\frac{\gamma_u}{\gamma_{u'}}=\frac{1}{\gamma_v}\frac{1}{1-\frac{\vec{u}^1v}{c^2}}$$

Combine this above with the other components, get:

$$\vec{u}'^1=\frac{\vec{u}^1-v}{1-\frac{\vec{u}^1v}{c^2}}$$ $$\vec{u}'^2=\frac{\vec{u}^2}{\gamma_v(1-\frac{\vec{u}^1v}{c^2})}$$ $$\vec{u}'^3=\frac{\vec{u}^3}{\gamma_v(1-\frac{\vec{u}^1v}{c^2})}$$


My Solution

Before the collision, the left particle travels with $\gamma_u(u,0)$ in $S$, which is to my understanding, is the 2-velocity (ie x and y coordinates) part of the 4-velocity. Transform this to $S'$:

To transform the first component, ie $u$ to $S'$, I'll need the $\vec{u}^1$ formula above. $\vec{u}^1$, ie the x-directional velocity component of the particle in $S$ is $u$, $S'$ moves with respect to $S$ with speed $v$, standard configuration, using results above:

$$u' = \frac{u-v}{1-\frac{uv}{c^2}}$$

The problem

This is not equal to $$\gamma_u\gamma_v(u-v)$$ as we might have hoped based on the figure.


Question

What is going on here? Is my reasoning for the x-component of velocity in $S'$ to be $\frac{u-v}{1-\frac{uv}{c^2}}$ wrong, or is the figure wrong? or both the figure and my derivation is correct but I misunderstand what the figure is showing?

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The figure is correct. It shows the 4-velocity of the particles after collisions. Note that the spatial components of the 4-velocity are not the same as the spatial components of the velocity in $x$ and $y$ directions. x-directional 4-velocity is, using the matrix equation given in the question:

$$u^1=-\beta \gamma_v \gamma_u c + \gamma_v \gamma_u \vec{u}^1$$ $$ = \gamma_u \gamma_v (u-v)$$ as given on the figure.

What $u'$, ie $$u' = \frac{u-v}{1-\frac{uv}{c^2}}$$ is then? The x component of the velocity, not 4-velocity. Need to multiply up by $\gamma_{u'}$ to get the same spatial component of the 4-velocity. Ie we get:

$$u^1 = \gamma_{u'} \frac{u-v}{1-\frac{uv}{c^2}}$$ $$ =\frac{1}{\sqrt{1-\frac{u'^2}{c^2}}} \frac{u-v}{1-\frac{uv}{c^2}}$$ $$ =\frac{1}{\sqrt{1-\frac{\left(\frac{u-v}{1-\frac{uv}{c^2}}\right)^2}{c^2}}} \frac{u-v}{1-\frac{uv}{c^2}}$$

Which is numerically the same as the $\gamma_u \gamma_v (u-v)$ result. Possible to verify by algebra or just plotting both results.

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