# Varying an Action

1. Aug 16, 2013

### dm4b

Hi,

I'm reading Zee's new GR book right now and ran across an action I am having trouble "varying". It's the first term in Eq (9), page 244. Looks like this:

$S=-m\int d\tau \sqrt{-\eta _{\mu \nu }\frac{dx^{\mu }}{d\tau }\frac{dx^{\nu }}{d\tau }}$

I'm familiar with the trick that gets rid of the square root as outlined by guys like Carrol, as well as Zee. But, I want to tackle this thing head on the hard way w/o the trick ;-)

But, I'm getting stuck reproducing what Zee has in Eq (10)

$\delta \left ( -m\int d\tau \sqrt{-\eta _{\mu \nu }\frac{dx^{\mu }}{d\tau }\frac{dx^{\nu }}{d\tau }} \: \right )=m\int d\tau \; \eta _{\mu \nu }\frac{dx^{\mu }}{d\tau }\frac{d\delta x^{\nu }}{d\tau }$

I get the same thing, but with a 1/L included, because the root won't go away.

Anybody know the trick? Or does Zee have a typo? Can't be me, right? ;-)

2. Aug 16, 2013

### dm4b

I think I figured out what is going on here, but it only raised another question.

On page 128 he varies the exact same action, but with the Minkowski metric replaced with the full-blown GR metric dependent upon X.

Sure enough, he has the 1/L factor here and the exact same result I get when I start to vary this action too.

But he then goes on to say we should exploit the freedom in choosing the parametrization, by choosing length parametrization, thereby making L=1.

I don't get why that makes L = 1, in either problem. Although, I guess it explains where it went to in the equations I cited in the OP.

Anybody know why L=1 here?

3. Aug 16, 2013

### ProfDawgstein

I don't know, but maybe this helps...

http://home.comcast.net/~peter.m.brown/gr/geodesic_equation.htm [Broken]

Last edited by a moderator: May 6, 2017
4. Aug 16, 2013

### dm4b

I actually ended up figuring this out, the answer was in the beginning of Chapt 2. Page 125.

5. Aug 16, 2013

### samalkhaiat

Write
$$d \lambda = \left( d x_{ \mu } d x^{ \mu } \right)^{ \frac{ 1 }{ 2 } }$$
then
$$\delta ( d \lambda ) = \frac{ d x_{ \mu } }{ d \lambda } \ \delta ( d x^{ \mu } ) = \frac{ d x_{ \mu } }{ d \lambda } \ \frac{ d }{ d \lambda } \left( \delta x^{ \mu } \right) \ d \lambda$$
Now, (because of reparameterization invariance of the action) you can set $\lambda = \tau$ in the integrand on the RHS.

Sam