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.


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:

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


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:


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.


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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.