Derivation of the Geodesic equation using the variational approach in Carroll

silverwhale
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Hello Everybody,

Carroll introduces in page 106 of his book "Spacetime and Geometry" the variational method to derive the geodesic equation.

I have a couple of questions regarding his derivation.

First, he writes:" it makes things easier to specify the parameter to be the proper time τ instead of the general parameter λ". Why does he do that? How does it make things easier? And why CAN he do that? is this just a variable substitution? I don't get it.

Second, he makes a taylor expansion of the metric in page 107. He writes down:

g_{\mu \nu} \rightarrow g_{\mu \nu} +(\partial_\sigma g_{\mu \nu}) \delta x^\sigma.

Now, how can I make this taylor expansion? It seems to be a taylor expansion of a matrix but I never did that before. And what does the arrow mean? "Substitute by"?

Third and last, why is putting the term g_{\mu \nu} \frac{dx^\mu}{\tau} \frac{dx^\nu}{d \tau} in the Euler lagrange equations equivalent to making the substitution mentioned earlier?

Thanks for reading, and any help regarding one of these questions would be really appreciated!
 
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it makes things easier to specify the parameter to be the proper time τ instead of the general parameter λ".
you can do that for a particle which has mass.For those particles who have zero mass,an affine parameter must be chosen.
Also gμv are functions of x only.A taylor expansion can be done.Also for a displacement xμ+dxμ,you can write it in terms of those at xμ by using taylor expansion.( g before arrow is at xμ+dxμ).geodesic follows from the principal of maximum proper time.So you can use that for your principle.
 
silverwhale said:
Hello Everybody,

Carroll introduces in page 106 of his book "Spacetime and Geometry" the variational method to derive the geodesic equation.

I have a couple of questions regarding his derivation.

First, he writes:" it makes things easier to specify the parameter to be the proper time τ instead of the general parameter λ". Why does he do that? How does it make things easier? And why CAN he do that? is this just a variable substitution? I don't get it.

If you happen to have MTW's gravitation, you can find some discussion on pg 317.

The basic point though is that you can consider the general solutions r(lambda), t(labmda) to be "beads on a curve" that you can slide around by reparameterizing lambda - i.e lambda = f(lambda').

But the geodesic equations are much simpler if you reparameterize lambda so that it's equal to proper time, which implies the points are equally spaced.

MTW works out the general case in which the points are not equally spaced, you'll see how much messier it is and why Caroll decided to avoid it if you look it up or carry out the calcuationitsef.

Re point 3 - I suspect a typo here. Rather than go into detail, I'll just say that setting the beads (MTW's way of describing the reparameterization) to be proper time implies they're equally spaced along the curve, which is an additional contraint on lambda.
 
silverwhale said:
First, he writes:" it makes things easier to specify the parameter to be the proper time τ instead of the general parameter λ". Why does he do that? How does it make things easier? And why CAN he do that? is this just a variable substitution? I don't get it.

λ is just a general affine parameter. τ can also be used as an affine parameter, so the equation must hold with the substitution λ→τ.

silverwhale said:
Second, he makes a taylor expansion of the metric in page 107. He writes down:

g_{\mu \nu} \rightarrow g_{\mu \nu} +(\partial_\sigma g_{\mu \nu}) \delta x^\sigma.

Now, how can I make this taylor expansion? It seems to be a taylor expansion of a matrix but I never did that before.

It's easy to see via chain rule that it's just a small variation in the metric:

\delta(g_{\mu \nu})=\frac{\partial g_{\mu \nu}}{\partial x^\sigma}\delta x^\sigma

silverwhale said:
And what does the arrow mean? "Substitute by"?

Pretty much. Substitute the stuff after the arrow for the stuff before the arrow.

silverwhale said:
Third and last, why is putting the term g_{\mu \nu} \frac{dx^\mu}{\tau} \frac{dx^\nu}{d \tau} in the Euler lagrange equations equivalent to making the substitution mentioned earlier?

The variation of the action δI is equal to the action after the substitution minus the action before the substitution (because it's a tiny "change" in the action). Setting the simplified integrand of δI equal to zero is equivalent to plugging the integrand of I (the Lagrangian) into the Euler-Lagrange equations.
 
elfmotat said:
λ is just a general affine parameter. τ can also be used as an affine parameter, so the equation must hold with the substitution λ→τ.

I think the point of the OP's question is being missed. It's not "why choose proper time as the affine parameterization", it's more along the lines of "why are parameterizations generally restricted to affine parmaterizations of proper time, rather than something more general?" Which isn't what he asked (because he doesn't know yet that parameterizations are generally so restricted, so he's naturally thinking of a more general parameterization).

The short answer is that the familiar geodesic equations become much more complicated when you don't choose such an affine parameterization.

The result of any affine parameterization is, that if you specify regualar intervals of your affine parameter lambda, the points on the curve are equally spaced.

[add]So, if, and only if, you have an affine parameterization, does lambda =0, 1, 2,3, 4, ... result in a set of points r(lambda), t(lambda) that are equally spaced by proper time (which is of course the Lorentz interval between the points).

Making this true requires an auxillary condition.Non-affine parameterizations will give unequally spaced points with regular intervals of lambda.

Resticting arbitrary parameterizations to affine ones gives a secondary condition, which is undoubtedly the origin of part 3 of his question. This auxillary condition is what I called the "equal spacing" condition.

As Caroll says, the reason this is done is to make things easy. One has the freedom to impose another condition, one uses this freedom to choose a condition that makes the result more tractable.

I have to run, I hope this helps...
 
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Thanks Everybody for your answer, they were very very helpful. I'll go through them several times and if I still don't get it I'll ask!

Many thanks again everybody! :)
 
Derive geodesic equation using variational principle.
 
Prove bianchi identity.
 
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