# Invariance of the speed of light

1. Aug 1, 2013

### littleHilbert

Hello!

Consider the law of addition of velocities for a particle moving in the x-y plane:

$u_x=\frac{u'_x+v}{1+u'_xv/c^2},\, u_y=\frac{u'_y}{\gamma(1+u'_xv/c^2)}$

In the book by Szekeres on mathematical physics on p.238 it is said that if u'=c, then it follows from the above formulae that u=c, i.e. invariance of the speed of light under the given Lorentz boost, which is of course exactly what we wish, since a Lorentz boost must preserve the null cone.

The weird thing is that when I start with $c^2=(u')^2=(u'_x)^2+(u'_y)^2$ and try to apply the above formulae to simply get $(u'_x)^2+(u'_y)^2=(u_x)^2+(u_y)^2$, i.e. working backwards towards the invariance of the null cone, I get quickly lost in the actual computation, because it seems to be leading nowhere…nothing cancels out. It doesn't matter which boost one takes…it doesn't seem to work (at the moment).

It can't be that hard. I don't know what I'm doing wrong, and can't imagine that I missed some concept. Did anybody already see how the computation goes? Is there anything one should pay attention to? Thanks

2. Aug 1, 2013

### George Jones

Staff Emeritus
On the same page, Szekeres outlines a method that shows this.

3. Aug 1, 2013

### littleHilbert

If you mean using the angles in $u_x=u \cos \theta,\, u_y=u \sin \theta, \, u'_x=u' \cos \theta',\, u'_y=u' \sin \theta'$ …does it really help? I tried to plug them in, too…but same thing…the computation gets lengthier and seems to be getting nowhere. There is actually no more on that page, except for the relation between the angles.

4. Aug 1, 2013

### George Jones

Staff Emeritus
Yes, use $u'_x = c \cos \theta'$ and $u'_y = c \sin \theta'$ in

$\frac{u_x}{c} = \frac{u'_x+v}{c+u'_xv/c},\, \frac{u_y}{c}=\frac{u'_{y}/\gamma}{(c+u'_xv/c)}$,

and calculate $\left( u_{x}/c \right)^2 + \left( u_{y}/c \right)^2$.

In $u_y$, keep the $1/\gamma$ in the numerator, so that $1/\gamma^2 = 1 - v^2 / c^2$. Expand the squares, and write $\sin^2 \theta' = 1 - \cos^2 \theta'$.

5. Aug 1, 2013

### littleHilbert

Oh yes, that sounds much more promising! I'll try this now…thanks in advance! :-)

6. Aug 1, 2013

### littleHilbert

OK, that was trivial...I knew it must be easy…if one arranges things the right way. I too was using the pythagorean theorem in order to get rid of some terms. But at the same time I was computing the difference $u^2_x+u^2_y-c^2$ to get 0…and yeah…somehow fell asleep.
Many thanks, George.