# Relative Lorentz Factor

1. Jun 27, 2015

### CassiopeiaA

If two particles have velocities v1 and v2 in a rest frame, how do we prove that the relative lorentz factor is given by :
γ(r)=γ(1)γ(2)(1−v1.v2)

and why is this quantity lorentz invariant

2. Jun 27, 2015

### Orodruin

Staff Emeritus
Did you try simply checking that it is Lorentz invariant? It obviously reduces to the relative Lorentz factor for frames where one of the particles is at rest, so if it is Lorentz invariant it is the relative Lorentz factor.

Do you know how to use 4-vectors? (In particular 4-velocities.)

3. Jun 27, 2015

### harrylin

The original derivation is in §5 of https://www.fourmilab.ch/etexts/einstein/specrel/www/

4. Jun 27, 2015

### robphy

Here's an analogous identity.
$\cos(B-A)=\cos B\cos A(1+\tan B\tan A)$,
which doesn't depend on the axis used to measure angles.

Last edited: Jun 29, 2015
5. Jun 29, 2015

### vanhees71

Ok, let's have particles with three-velocities $\vec{v}_1$ and $\vec{v}_2$ wrt. to an inertial reference frame. The first step is to translate this to the appropriate covariant objects. In this case that's the four-velocities of the (on-shell) particles which are the four-velocities
$$u_j^{\mu}=\gamma_j (1,\vec{v}_j), \quad \gamma_j=\frac{1}{\sqrt{1-\vec{v}_j^2}}, \quad c=1, \quad j \in \{1,2 \}$$
They are related to the four-momenta of the particles by
$$p_j^{\mu}=m_j u_j^{\mu},$$
where $m_j$ are the invariant (or rest) masses of the particles, but this we don't need.

The relative velocity is now defined as the velocity of particle 2 in the rest frame of particle 1. Thanks to the covariant tensor formalism you don't need to perform the Lorentz transform here (although that's a nice exercise, you should do for yourself, but it's also a bit tedious to type here in the forum). The trick is to express everything covariantly. In the rest frame of particle 1 (in the following written with a tilde over the components of four-vectors) you have
$\tilde{u}_1^{\mu}=(1,0,0,0), \quad \tilde{u}_2^{\mu} = \tilde{\gamma}_2 (1,\vec{\tilde{v}_2}).$
From this it's clear that
$\tilde{\gamma}_2=u_2^0=\tilde{u}_{1,\mu} \tilde{u}_2^{\mu}.$
But now, this is a scalar expression (the Minkowski product of two four-vectors) and thus you have
$\tilde{\gamma}_2=u_1 \cdot u_2=\gamma_1 \gamma_2 (1-\vec{v}_1 \cdot \vec{v}_2),$
which is the formula you wanted to derive in posting #1.

One should keep in mind that the Lorentz transform is at the heart of deriving the Minkowski space as the space-time model, establishing the four-tensor formalism. For practical calculations you usually don't need Lorentz transforms, if everything is formulated in covariant form, and you can express the quantities you want to know in terms of covariant equations, which is the case for all physically interesting quantities. Sometimes one has to define appropriate quantities in a covariant way (e.g., temperature and other thermodynamic quantities and material constant, where usually you define them in the (local) rest frame of the matter and then express this definition in a manifestly covariant way or the invariant cross sections for processes in high-energy physics).

Last edited: Jun 29, 2015