Alternative definitions of geodesic

[andrewkirk]

In the terminology of my text (Gelfand and Fomin 'Calculus of Variations'), for the saddle point that I linked the term 'variation' has no meaning, because the saddle point depicted is a function (from $\mathbb{R}^2$ to $\mathbb{R}$) and 'variation' is a property of a functional, at a function. Based on their definition of 'variation' I would expect - based on the above argument - that the variation of the length functional at the spacelike geodesic is non-vanishing. They define 'vanishing' to mean identically zero. At least I think that's what they mean, but there's some uncertainty because they say '$\delta J_y$ is vanishing at $y$' means that $\delta J_y[h]=0$ 'for all admissible h', and they don't explain what they mean by 'admissible' anywhere that I can find (h is the incremental function added to the function at which the variation is zero.

What text are you using? I have to say that I'm not in love with Gelfand and Fomin, as they keep saying things like 'the functional J[h]' whereas J is the functional and J[h] is a real number that is obtained when one applies the functional to function h. It caused me no end of confusion.

 

[PAllen]

In the terminology of my text (Gelfand and Fomin 'Calculus of Variations'), for the saddle point that I linked the term 'variation' has no meaning, because the saddle point depicted is a function (from $\mathbb{R}^2$ to $\mathbb{R}$) and 'variation' is a property of a functional, at a function. Based on their definition of 'variation' I would expect - based on the above argument - that the variation of the length functional at the spacelike geodesic is non-vanishing. They define 'vanishing' to mean identically zero. At least I think that's what they mean, but there's some uncertainty because they say '$\delta J_y$ is vanishing at $y$' means that $\delta J_y[h]=0$ 'for all admissible h', and they don't explain what they mean by 'admissible' anywhere that I can find (h is the incremental function added to the function at which the variation is zero.

What text are you using? I have to say that I'm not in love with Gelfand and Fomin, as they keep saying things like 'the functional J[h]' whereas J is the functional and J[h] is a real number that is obtained when one applies the functional to function h. It caused me no end of confusion.

The book I am using is a monograph by Gilbert Ames Bliss. What you describe is similar to part of his presentation, except that he does define admissibility of h. The h functions need only be continuous, meet boundary conditions, and be c2 smooth except for a finite number of points. If, without assuming anything specific about h, the derivative of the integral with respect to the parameter applied to it is zero, it is said the variation is zero. This is equivalent to satisfying the Euler Lagrange equations. Spacelike geodesics clearly satisfy these equations, so they have stationary (zero) variation. However, since they are not locally extremal, they are some type of saddle point.

 

[bcrowell]

Assume for simplicity that, in a given coordinate system the two spacetime points differ only in a single spatial coordinate, say x. Then there is a basis for the vector space of paths between the two points, that is made up of basis functions each of which is confined to one of the x-t, x-y or x-z planes.

Here you seem to be assuming that spacetime can be covered by a single chart and that it has the structure of a vector space. Neither of these is true in GR.

 

[stevendaryl]

Here you seem to be assuming that spacetime can be covered by a single chart and that it has the structure of a vector space. Neither of these is true in GR.

Well let's suppose that we have a definition of geodesic, "locally geodesic", that applies in a small enough simply-connected, open subset of the manifold. Then wouldn't that extend to arbitrary geodesics by saying that the path is locally geodesic in every sufficiently small simply-connected open subset of the manifold that the geodesic passes through?

 

[andrewkirk]

Here you seem to be assuming that spacetime can be covered by a single chart and that it has the structure of a vector space. Neither of these is true in GR.

Re the coverage by a single chart: only locally. The intended context of the comment is two spacelike separated points inside the same geodesic ball. That fits in with the overall picture via the finite set of points $h_k$ you posed in the OP that break up the full geodesic. If the points are close enough together, each pair can be in a geodesic ball.

The vector space to which I referred is an infinite-dimensional vector space of functions, not a 4D one of spacetime directions. Each element of the vector space denotes a path (not necessarily geodesic) between the two points. We choose a local coordinate system in which the $x$ two points have coordinates $(a^t,a^x,a^y,a^z)$ and $(a^t,b^x,a^y,a^z)$. Then the vector space mentioned is the set of suitably nice (ie continuous etc) functions $f$ from the real interval $[a^x,b^x]$ to $\mathbb{R}^3$ such that $f(x')$ is the $t, y, z$ coordinates of the point on the path that has $x$ coordinate $x'$. The functional being considered is $f\mapsto L(\pi^{-1}\circ g_f([a^x,b^x]))$ where

• $g_f$ is the function $x\mapsto (x,f(x))$
• $\pi$ is the coordinate map for the geodesic ball
• $L$ is the length function for paths in $M$

I think I have almost convinced myself that the variation of the length functional really is zero at the spacelike geodesic, but I need to think it through more before I can have any confidence that I understand why.

 

[stevendaryl]

I think I have almost convinced myself that the variation of the length functional really is zero at the spacelike geodesic, but I need to think it through more before I can have any confidence that I understand why.

Are you defining geodesic to mean a parametrized path satisfying the geodesic equation (or the definition in terms of parallel transport, which is mathematically equivalent for a connection defined in terms of the metric)? In that case, doesn't the usual Lagrange-Euler equations imply that (spacelike or timelike) geodesics have zero variation?

If we consider a parametrized curve $x^\mu(s)$, then the invariant length of the curve is given by $\int ds \sqrt{g_{\mu \nu} U^\mu U^\nu}$, where $U^\mu = \frac{dx^\mu}{dx^\nu}$.If you treat this like an action integral, with lagrangian $L = \sqrt{g_{\mu \nu} U^\mu U^\nu}$, then the Lagrange-Euler equations for a path with zero variation leads to

$\dfrac{d}{ds} \dfrac{\partial L}{\partial U^\mu} - \dfrac{\partial L}{\partial x^\mu} = 0$

After reparametrizing to use an affine parameter, this becomes the geodesic equation.

I made the restriction to spacelike or timelike because the derivation of Euler-Lagrange equations breaks down if the path is a null path. That's because the quantity

$\dfrac{\partial L}{\partial U^\mu} = \dfrac{g_{\mu \nu} U^\nu}{L}$

is undefined whenever $L = 0$, which is always the case for null paths.

 

[bcrowell]

Well let's suppose that we have a definition of geodesic, "locally geodesic", that applies in a small enough simply-connected, open subset of the manifold. Then wouldn't that extend to arbitrary geodesics by saying that the path is locally geodesic in every sufficiently small simply-connected open subset of the manifold that the geodesic passes through?

Yes. As written, that sounds like a much weaker condition that the definition I proposed in #1, but I assume they're actually equivalent.

 

[Allin]

Geodesics is land surveying at the scale of the whole planet.

 

[DrGreg]

Geodesics is land surveying at the scale of the whole planet.

I think you mean "geodesy" or "geodetics". That's not what we are talking about here.

 

[andrewkirk]

the Lagrange-Euler equations for a path with zero variation leads to

$\dfrac{d}{ds} \dfrac{\partial L}{\partial U^\mu} - \dfrac{\partial L}{\partial x^\mu} = 0$

After reparametrizing to use an affine parameter, this becomes the geodesic equation.

That sounds like a promising approach that ought to work. I tried doing this but, because of the complexity of L, the algebra soon became very ugly and I ran out of paper ( or patience - one or the other).

Maybe I took a wrong turn somewhere. Have you worked it through?

 

[stevendaryl]

That sounds like a promising approach that ought to work. I tried doing this but, because of the complexity of L, the algebra soon became very ugly and I ran out of paper ( or patience - one or the other).

Maybe I took a wrong turn somewhere. Have you worked it through?

It's easy to get into a blind alley, but I don't think this is that bad.

With $L = \sqrt{g_{\mu \nu} U^\mu U^\nu}$,

• $\dfrac{\partial}{\partial U^\mu} L = \dfrac{1}{L} g_{\mu \nu} U^\nu$
• $\dfrac{\partial}{\partial x^\mu} L = \dfrac{1}{2L} \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu} U^{\mu'} U^\nu$

So the Euler-Lagrange equations give:

$\dfrac{1}{L} \dfrac{d}{ds} (g_{\mu \nu} U^\nu) + g_{\mu \nu} U^\nu \dfrac{d}{ds} \dfrac{1}{L} = \dfrac{1}{2L} \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu} U^{\mu'} U^\nu$

Now, the great simplification comes from assuming $\dfrac{d}{ds} \dfrac{1}{L} = 0$, so $L =$ a constant along the path. With this assumption, the equation simplifies to (multiplying both sides by $L$)

$\dfrac{d}{ds} (g_{\mu \nu} U^\nu) = \dfrac{1}{2} \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu} U^{\mu'} U^\nu$

We expand the left-hand side to get:

$(\dfrac{d}{ds} g_{\mu \nu}) U^\nu + g_{\mu \nu} \dfrac{d}{ds} U^\nu = \dfrac{1}{2} \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu} U^{\mu'} U^\nu$

Then you use: $\dfrac{d}{ds} g_{\mu \nu} = (\dfrac{\partial}{\partial x^{\mu'}} g_{\mu \nu}) \dfrac{d}{ds} x^{\mu'} = (\dfrac{\partial}{\partial x^{\mu'}} g_{\mu \nu}) U^{\mu'}$ (the chain rule for derivatives)

So we now have:
$(\dfrac{\partial}{\partial x^{\mu'}} g_{\mu \nu}) U^{\mu'} U^\nu + g_{\mu \nu} \dfrac{d}{ds} U^\nu = \dfrac{1}{2} \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu} U^{\mu'} U^\nu$

Rearranging gives:

$g_{\mu \nu} \dfrac{d}{ds} U^\nu = - ( (\dfrac{\partial}{\partial x^{\mu'}} g_{\mu \nu}) U^{\mu'} U^\nu - \dfrac{1}{2} \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu} U^{\mu'} U^\nu)$

Finally, get rid of the $g_{\mu \nu}$ from the left-hand side by multiplying by the inverse matrix, $g^{\mu \nu'}$ and summing over $\mu$. this gives:

$\dfrac{d U^{\nu'}}{ds} = - \frac{1}{2} g^{\mu \nu'} (2 \dfrac{\partial g_{\mu \nu}}{\partial x^{\mu'}} - \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu}) U^{\mu'} U^\nu$

So if we define $Q^{\nu'}_{\nu \mu'} = \frac{1}{2} g^{\mu \nu'} (2 \dfrac{\partial g_{\mu \nu}}{\partial x^{\mu'}} - \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu})$, then this becomes:
$\dfrac{d U^{\nu'}}{ds} = -Q^{\nu'}_{\nu \mu'} U^{\mu'} U^\nu$

Almost there! The usual connection coefficient is $\Gamma^{\nu'}_{\nu \mu'} = \frac{1}{2} g^{\mu \nu'} (\dfrac{\partial g_{\mu \nu}}{\partial x^{\mu'}} + \dfrac{\partial g_{\mu \mu'}}{\partial x^{\nu}} - \dfrac{\partial g_{\mu' \nu}}{\partial x^\mu})$, So we can write:

$Q^{\nu'}_{\nu \mu'} = \Gamma^{\nu'}_{\nu \mu'} + \frac{1}{2} g^{\mu \nu'} (\dfrac{\partial g_{\mu \nu}}{\partial x^{\mu'}} - \dfrac{\partial g_{\mu \mu'}}{\partial x^{\nu}})$

Notice that the last term on the right is antisymmetric under the exchange $\nu \Rightarrow \mu'$. On the other hand, $U^{\mu'} U^\nu$ is symmetric under that exchange. The product of an antisymmetric tensor with a symmetric tensor is zero. So:

$Q^{\nu'}_{\nu \mu'} U^{\mu'} U^\nu = \Gamma^{\nu'}_{\nu \mu'} U^{\mu'} U^\nu$

So the Euler-Lagrange equations boil down to:

$\dfrac{d U^{\nu'}}{ds} = -\Gamma^{\nu'}_{\nu \mu'} U^{\mu'} U^\nu$

which is the geodesic equation (whew!). Okay, I guess it was pretty bad...