Showing that SpaceTime Intervals are invariant: EM II notes 2014_09_03

Summary:  Continuing notes on the tensor version of the Lorentz tranform.  It's time to start on the second set of examples.


The interval in four space is invariant under Lorentz transforms and is called the Lorentz scalar.


The Lorentz transform also applies to differential distances as,

$dx^{\prime\mu} = \Lambda^\mu_\nu x^\mu$


We were asked in class to work out $x^2+y^2+z^2-t^2 = x^{\prime 2}+y^{\prime 2}+z^{\prime 2}-t^{\prime 2}$

The transforms we'll use are:

$x = \gamma\left(x^\prime + vt^\prime\right)$

$t = \gamma\left(t^\prime + vx^\prime\right)$

Substituting these into the l.h.s. gives

$\gamma^2\left(x^\prime + vt^\prime\right)^2 - \gamma^2\left(t^\prime + vx^\prime\right)^2 = x^{\prime 2} - t^{\prime 2}$

$ = \gamma^2\left(x^{\prime 2} +2vtx + v^2t^{\prime 2}\right) - \gamma^2\left(t^{\prime 2} + 2vxt+v^2x^{\prime 2}\right)= x^{\prime 2} - t^{\prime 2}$

$ = \gamma^2\left(x^{\prime 2} + v^2t^{\prime 2}\right) - \gamma^2\left(t^{\prime 2} + v^2x^{\prime 2}\right) = x^{\prime 2} - t^{\prime 2}$

$\gamma^2 x^{\prime 2} - \gamma^2v^2x^{\prime 2} = \gamma^2 x^{\prime 2}\left(1 - v^2\right) = x^{\prime 2}$

Similarly

$\gamma^2 v^{\prime 2}t^{\prime 2} - \gamma^2 t^{\prime 2} = -\gamma^2 t^{\prime 2}\left(-v^2 + 1\right) = -t^{\prime 2}$

So

$x^{\prime 2} - t^{\prime 2} = x^{\prime 2} - t^{\prime 2}$

Done

Picture of the Day:
Here's a throwback Thursday pic from 1946.  These are the infamous Carter Boys.  My dad is the twin on the right.