I Carroll GR: Geodesic Eq from Var Principles

[hawkdron496]

TL;DR Summary

Trying to understand a trick Carroll uses to derive the geodesic equation by extremizing proper time

On pages 106-107 of Spacetime & Geometry, Carroll derives the geodesic equation by extremizing the proper time functional. He writes:

What I am unclear on is the step in 3.47. I understand that the four velocity is normalized to -1 for timelike paths, but if the value of f is fixed, how can we vary it at all?

 

[jbergman]

I've dug into this question before and my conclusion is that what Carroll is doing here doesn't make sense. Not that it doesn't give you the right answer, but just that he hasn't justified his approach.

There are actually quite a few threads on this on physics.stackexchange.com. Here is one example, https://physics.stackexchange.com/q...ation-of-the-geodesic-equations/486049#486049

There is also some discussion in some Riemannian Geometry books as to when you can do the above. If I have more time, I will try and dig up a reference tomorrow.

 

[strangerep]

What I am unclear on is the step in 3.47. I understand that the four velocity is normalized to -1 for timelike paths, but if the value of f is fixed, how can we vary it at all?

I don't much care for that sleight of hand either. It's not necessary. Crunching through with either EL equation is not excessively difficult (and one may even learn a little bit about the properties of homogeneous functions). Every serious student of GR should do it.

In the 1st case (with $\sqrt{-f}\,$), one finds that the 4 EL eqns are not all independent and one cannot write those EL eqns in the form $\,\ddot x^\mu = \dots\,$.

In the 2nd case (with $f$), one finds EL eqns which, when contracted with $\dot x^\mu$, give an auxiliary eqn of the form $df/ds = 0$, i.e., $f = const$, and one can choose the constant depending on whether we're working with timelike, lightlike or spacelike paths.

 

[Orodruin]

I disagree that there is sleight of hand here. You are well within your right to reparametrise the world line such that $f$ becomes a constant of motion and when you do the variation of the expression eith the square root reduces to the same EL equation as that without.

Summary: Trying to understand a trick Carroll uses to derive the geodesic equation by extremizing proper time

What I am unclear on is the step in 3.47. I understand that the four velocity is normalized to -1 for timelike paths, but if the value of f is fixed, how can we vary it at all?

It is not about not varying it. It is about choosing a reparametrization of the curve such that it is constant when solving the equations of motion. It is easy to show that the proper time is invariant under such reparametrizations.

one cannot write those EL eqns in the form x¨μ=….

Yes you can. The difference to the geodesic equations is that you will obtain an additional term proportional to $\dot x^\mu$ and involving the derivative of $\ln (-f)$.

 

[strangerep]

Yes you can. The difference to the geodesic equations is that you will obtain an additional term proportional to $\dot x^\mu$ and involving the derivative of $\ln (-f)$.

The logic I relied on is as follows. Let me first change the notation slightly. Denote $F = \sqrt{-f}\,,$ and $u^\mu := \dot x^\mu \equiv d x^\mu/ds$. Then the EL eqn in the 1st case is $$0 ~=~ E_\mu[F] ~:=~
\frac{\partial^2 F}{\partial u^\mu \partial u^\nu}\, \dot u^\nu
+ \frac{\partial^2 F}{\partial x^\nu \partial u^\mu}\, u^\nu
- \frac{\partial F}{\partial x^\mu} ~~~~ (1)$$
The 1-homogeneity of $F$ wrt $u^\mu$ (indirectly wrt $s$) implies that $$u^\mu E_\mu[F] ~=~ 0 ~~~~ (2)$$ identically. Hence the 4 eqns in (1) are not independent. Moreover, by homogeneity, $$u^\mu \; \frac{\partial^2 F}{\partial u^\mu \partial u^\nu} ~=~ 0 ~, ~~~~ (3)$$ which means that
$$rank\left( \frac{\partial^2 F}{\partial u^\mu \partial u^\nu}\right) ~<~ 4 ~,$$ so the matrix of 2nd order derivatives of $F$ occurring in (1) is non-invertible in general.

 

[Orodruin]

The logic I relied on is as follows. Let me first change the notation slightly. Denote $F = \sqrt{-f}\,,$ and $u^\mu := \dot x^\mu \equiv d x^\mu/ds$. Then the EL eqn in the 1st case is $$0 ~=~ E_\mu[F] ~:=~
\frac{\partial^2 F}{\partial u^\mu \partial u^\nu}\, \dot u^\nu
+ \frac{\partial^2 F}{\partial x^\nu \partial u^\mu}\, u^\nu
- \frac{\partial F}{\partial x^\mu} ~~~~ (1)$$
The 1-homogeneity of $F$ wrt $u^\mu$ (indirectly wrt $s$) implies that $$u^\mu E_\mu[F] ~=~ 0 ~~~~ (2)$$ identically. Hence the 4 eqns in (1) are not independent. Moreover, by homogeneity, $$u^\mu \; \frac{\partial^2 F}{\partial u^\mu \partial u^\nu} ~=~ 0 ~, ~~~~ (3)$$ which means that
$$rank\left( \frac{\partial^2 F}{\partial u^\mu \partial u^\nu}\right) ~<~ 4 ~,$$ so the matrix of 2nd order derivatives of $F$ occurring in (1) is non-invertible in general.

The underlying reason of course being that the functional is independent under reparametrisations of the world line such that there is a flat direction. Imposing an affine parametrisation with a normalised tangent fixes this.

 

[jbergman]

I disagree that there is sleight of hand here. You are well within your right to reparametrise the world line such that $f$ becomes a constant of motion and when you do the variation of the expression eith the square root reduces to the same EL equation as that without.

The square root action is reparametrization invariant. The energy functional is not.

See, https://physics.stackexchange.com/q...e-squared-lagrangian-equivalent/149204#149204

 

[Orodruin]

The square root action is reparametrization invariant. The energy functional is not.

See, https://physics.stackexchange.com/q...e-squared-lagrangian-equivalent/149204#149204

This is besides the point. The world lines are the same.

 

[vanhees71]

The square root action is reparametrization invariant. The energy functional is not.

See, https://physics.stackexchange.com/q...e-squared-lagrangian-equivalent/149204#149204

The square-root action is indeed reparametrization invariant by construction, and this is the reason, why it describes only 3 independent degrees of freedom and not 4 as it might look at the first glance. It's a case of "gauge symmetry", and this can be used to get to the "square form", i.e., you can partially fix the gauge by imposing a constraint.

From a geometrical point of view it's suggestive to use the constraint that the parameter is an affine parameter, i.e., to impose the constraint $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\text{const}$. This you can do using a Lagrange parameter, but it's easier to directly observe that the Lagrangian
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}$$
automatically does the job. First it obeys the "affinity constraint" automatically. Since $L$ is not explicitly $\lambda$ determined (where $\lambda$ is the arbitrary parameter of the curves), you have
$$p_{\mu} \dot{x}^{\mu} -L=L=\text{const},$$
and it's easy to see that the Euler-Lagrange equations of motion coincide with the Euler-Lagrange equations of motion for the square-root form in the case where you choose an affine parameter, fulfilling the constraint condition. That's why the square form and the square-root form lead to the same equations of motion, i.e., the geodesics in spacetime you are looking for.

The physical cases are of course time-like geodesics for the description of massive point particles moving through a given gravitational field or the light-like geodesics for the calculation of "light rays" (in the literature usually called "photons", which I find a bit misleading, but that's maybe sementics) in a given gravitational field.

 

[jbergman]

This is besides the point. The world lines are the same.

Well, the real point is that Carroll doesn't justify why one can do this which is bad pedagogy. No one is disputing the second point but a valid argument should be given like in @vanhees71 notes on special relativity.

 

[pervect]

Summary: Trying to understand a trick Carroll uses to derive the geodesic equation by extremizing proper time

On pages 106-107 of Spacetime & Geometry, Carroll derives the geodesic equation by extremizing the proper time functional. He writes:
View attachment 313663
What I am unclear on is the step in 3.47. I understand that the four velocity is normalized to -1 for timelike paths, but if the value of f is fixed, how can we vary it at all?

The value of f is fixed at -1 on the geodesic path, which is the solution of the variational equations. It's not precisely constant and equal to -1 for paths near the geodesic path, but it's close to -1. One can imagine expanding the difference in f along a nearby path by some multivariable taylor series. What's important is that 3.48 is correct to the first order in $\delta f$. It doesn't need to be correct to higher order. A first-order change in f will contribute only second order terms to the results of 3.48. It's not a biggie though - if you don't like this result, with enough effort you can proceed without using this trick.

 

[Demystifier]

It's really about getting rid of the square root in the action. Extremizing the functional $\int d\lambda \sqrt{f(\lambda)}$ is equivalent to extremizing the functional $\int d\lambda F(f(\lambda))$, where $F(f)$ is an arbitrary strictly increasing function. Choosing $F(f)=f$ gives the desired result, without fixing anything. However, the new functional is no longer invariant under reparametrisations $\lambda\to\lambda'=h(\lambda)$. Hence we think of it as a functional with fixed $\lambda=\tau$, but this should not be viewed as a fixing of $f(\tau)$.

 

[jbergman]

It's really about getting rid of the square root in the action. Extremizing the functional $\int d\lambda \sqrt{f(\lambda)}$ is equivalent to extremizing the functional $\int d\lambda F(f(\lambda))$, where $F(f)$ is an arbitrary strictly increasing function.

Can you provide a reference or state why the above is true?

 

[Demystifier]

Can you provide a reference or state why the above is true?

Consider the functional
$$A=\int d\lambda \, F(f(\lambda))$$
where $F(f)$ is an arbitrary function obeying
$$\frac{\partial F(f)}{\partial f}>0. \;\;\;\;\; (1)$$
Since
$$\delta \int d\lambda \, F(f(\lambda))= \int d\lambda \,\delta F(f(\lambda))$$
the functional $A$ is extremized when
$$\delta F(f(\lambda))=0.$$
But
$$\delta F(f(\lambda))=\frac{\partial F(f)}{\partial f}\delta f(\lambda)$$
so (1) implies that the functional $A$ is extremized when
$$\delta f(\lambda)=0.$$
Thus, the extremization of $A$ does not depend on the choice of $F$, provided that it obeys (1).