I'm not very well versed in Abstract algebra and group theory, so this question might not make sense to begin with, but I got an idea when reading up on how to rigorously calculate relative velocities.

Suppose I'm traveling north with some velocity relative to the earth $ v_{me/earth}$, and an other car comes from north due south with velocity $ v_{him/Earth} < 0$ because of we orientated our coordinate system to increase towards north. My velocity relative to the other car is $$ v_{me/him} = v_{me/Earth}+v_{Earth/him} $$ where the last term is equal to $ (-1) \cdot v_{him/Earth} $

Now this form of velocity addition obeys a certain algebra similar to if we were to simply treat the indices as fractions and the addition as multiplication, meaning $$ v_{A/B}= v_{A/C} +v_{C/D}+v_{D/B} $$ because $$\frac{A}{B} = \frac{A}{C} \cdot \frac{C}{D} \cdot \frac{D}{B} $$

My question is: does this have anything to do with Group theory? Is there like a velocity group under addition?

  • 2
    $\begingroup$ I think you're looking for Galilean group. $\endgroup$
    – Ruslan
    Feb 20, 2020 at 13:07

1 Answer 1

  • As pointed out in the comments by Ruslan, there exists a group for transformations between coordinate frames differing in only relative velocity -- Galilean group. It represents transformations of the following form:

$$ \begin{align} x' & = x - vt \\ y' & = y \\ z' & = z \\ t' &= t \end{align} $$

  • As you can see for relativistic speeds, we need to consider $t' = \gamma(t - vx/c^2)$ and also $x' = \gamma(x-vt)$ with $\gamma = \left(1-\frac{v^2}{c^2}\right)^{-1/2}$ and hence is no more classical. The corresponding group is called Lorentz group. In $c \rightarrow \infty$, we recover Galilean group.
  • $\begingroup$ Is the strange or convenient way of computing indices the consequence of this? $\endgroup$
    – Johnny
    Feb 20, 2020 at 16:05
  • $\begingroup$ But yeah, you are right this is ultimately about Galilean transformation $\endgroup$
    – Johnny
    Feb 20, 2020 at 16:20

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