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Christoffel symbols from definition or Lagrangian

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noospace
#1
Sep28-07, 11:47 PM
P: 75
Hi,

Let [itex]\mathbf{x}(u,v)[/itex] be a local parametrization of a regular surface. Then the coefficients of [itex]\mathbf{x}_{uu},\mathbf{x}_{uv}[/itex] etc. in the basis of the tangent space are defined as the Christoffel symbols.

On the other hand, if we write the first fundamental form [itex]\langle,\rangle[/itex] in differential form we have an extremization problem of the arc-length

[itex]ds^2 = E du^2 + 2F du dv + G dv^2[/itex].

Then the coefficients of of the corresponding Euler-Lagrange equations are essentially the Christoffel symbols.

Are there any interesting examples where the Lagrangian method of computing Christoffel symbols breaks down?
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pmb_phy
#2
Sep30-07, 05:12 PM
P: 2,954
Quote Quote by noospace View Post
Then the coefficients of [itex]\mathbf{x}_{uu},\mathbf{x}_{uv}[/itex]
I don't understand your notation. What do those two symbols mean?
On the other hand, if we write the first fundamental form [itex]\langle,\rangle[/itex] in differential form we have an extremization problem of the arc-length

[itex]ds^2 = E du^2 + 2F du dv + G dv^2[/itex].

Then the coefficients of of the corresponding Euler-Lagrange equations are essentially the Christoffel symbols.
Where did you get that definition from?? The coefficients of the above expression are the components of the metric tensor, not the components of the Christoffel symbols

Pete
Chris Hillman
#3
Sep30-07, 07:39 PM
Sci Advisor
P: 2,340
Quote Quote by noospace View Post
Let [itex]\mathbf{x}(u,v)[/itex] be a local parametrization of a regular surface. Then the coefficients of [itex]\mathbf{x}_{uu},\mathbf{x}_{uv}[/itex] etc. in the basis of the tangent space are defined as the Christoffel symbols.

On the other hand, if we write the first fundamental form [itex]\langle,\rangle[/itex] in differential form we have an extremization problem of the arc-length

[itex]ds^2 = E du^2 + 2F du dv + G dv^2[/itex].

Then the coefficients of of the corresponding Euler-Lagrange equations are essentially the Christoffel symbols.

Are there any interesting examples where the Lagrangian method of computing Christoffel symbols breaks down?
Hi, noospace, I infer from your choice of notation that you are taking a course in surface theory (i.e. Gauss's theory of two dimensional surfaces in three dimensional Euclidean space), but unless you are using nonstandard definitions of the technical terms you mentioned, I think you must be a bit confused:

1. The second partials you mentioned in general do not agree with the Christoffel coefficients, even for a Monge patch. See Millman and Parker, Elements of Differential Geometry, Section 4-4, for a formula giving the second fundamental form (and thus the Christoffel coefficients) in terms of a Monge parameterization of a surface in [itex]E^3[/itex]. (Looks like you forgot about the normal vector to the surface and the inner product.)

2. The coefficients from the Euler-Lagrange equations (after normalizing to make them monic in the highest order derivatives [itex]\ddot{u}, \, \ddot{v}[/itex]) are precisely the Christoffel coefficients [itex]{\Gamma^a}_{bc}[/itex]. Perhaps part of the problem here is that you forgot to normalize after applying the Euler operator to the geodesic Lagrangian?

Suggestion: try both methods with some simple examples, such as the following local parameterization of the unit sphere:
[tex]
\vec{X}(r,\phi) = \left[
\begin{array}{c} \sqrt{1-r^2} \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array}
\right]
[/tex]
(This parameterizes the upper hemisphere.) You should get first fundamental form corresponding to the line element
[tex]
ds^2 = \frac{dr^2}{1-r^2} + r^2 \, d\phi^2, \; 0 < r < 1, \; -\pi < \phi < \pi
[/tex]
and second fundamental form corresponding the geodesic equations
[tex]
\ddot{r} + \frac{r}{1-r^2} \, \dot{r}^2 - r \, (1-r^2) \, \dot{\phi}^2 = 0
[/tex]
[tex]
\ddot{\phi} + \frac{2}{r} \, \dot{r} \, \dot{\phi} = 0
[/tex]
Note that the latter immediately yields a first integral which can be plugged into the former.

noospace
#4
Oct13-07, 05:01 AM
P: 75
Christoffel symbols from definition or Lagrangian

Hi Chris,

Quote Quote by Chris Hillman View Post
Hi, noospace, I infer from your choice of notation that you are taking a course in surface theory (i.e. Gauss's theory of two dimensional surfaces in three dimensional Euclidean space), but unless you are using nonstandard definitions of the technical terms you mentioned, I think you must be a bit confused:
I'm borrowing my notation from Do Carmo `Differential Geometry of Curves and Surfaces', it's a text on classical differential geometry.

Quote Quote by Chris Hillman View Post
1. The second partials you mentioned in general do not agree with the Christoffel coefficients, even for a Monge patch. See Millman and Parker, Elements of Differential Geometry, Section 4-4, for a formula giving the second fundamental form (and thus the Christoffel coefficients) in terms of a Monge parameterization of a surface in [itex]E^3[/itex]. (Looks like you forgot about the normal vector to the surface and the inner product.)
[itex]\vect{x} : U \subset \mathbb{R}^2 \to S[/itex] is a local parametrization of a regular surface, ie a smooth nonsingular map which is a homeomorphism onto a nbhd of some point. A regular surface is just a subset of [itex]\mathbb{R}^3[/itex] for which each point in S has a local parametrization.

I didn't forgot the normal vector. The second derivatives of a local parametrization can be written in terms of the moving frame [itex]x_u,x_v, x_u\wedge x_v[/itex]. Do Carmo defines the Christoffel symbols as the coefficients of the tangent vectors [itex]x_u,x_v[/itex]. The normal vectors don't come into it.


Quote Quote by Chris Hillman View Post
2. The coefficients from the Euler-Lagrange equations (after normalizing to make them monic in the highest order derivatives [itex]\ddot{u}, \, \ddot{v}[/itex]) are precisely the Christoffel coefficients [itex]{\Gamma^a}_{bc}[/itex]. Perhaps part of the problem here is that you forgot to normalize after applying the Euler operator to the geodesic Lagrangian?
Not according to my definitions. If you treat the first fundamental form as a Lagrangian than you can retrieve the Christoffel symbols from the Lagrange equation. I guessed there must be a theorem that this holds in general, you seem to indicate otherwise. My question is whether or not there are any couterexamples.

Quote Quote by Chris Hillman View Post
Suggestion: try both methods with some simple examples, such as the following local parameterization of the unit sphere:
[tex]
\vec{X}(r,\phi) = \left[
\begin{array}{c} \sqrt{1-r^2} \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array}
\right]
[/tex]
(This parameterizes the upper hemisphere.) You should get first fundamental form corresponding to the line element
[tex]
ds^2 = \frac{dr^2}{1-r^2} + r^2 \, d\phi^2, \; 0 < r < 1, \; -\pi < \phi < \pi
[/tex]
and second fundamental form corresponding the geodesic equations
[tex]
\ddot{r} + \frac{r}{1-r^2} \, \dot{r}^2 - r \, (1-r^2) \, \dot{\phi}^2 = 0
[/tex]
[tex]
\ddot{\phi} + \frac{2}{r} \, \dot{r} \, \dot{\phi} = 0
[/tex]
Note that the latter immediately yields a first integral which can be plugged into the former.
How can this be so? Surely they argree for a simple sphere?

By the way, why did you take down Relwww? That was a really useful website.

Thanks.
noospace
#5
Oct13-07, 05:17 AM
P: 75
I haven't had time to check your counterexample. Here is an example where the claim holds, for a paraboloid [itex]z=ar^2[/itex],

[itex]x(r,\phi) = (r\cos\phi,r\sin\phi,ar^2)[/itex].

Then the Christoffel symbols according to Do Carmo's definition are

[itex]\Gamma^r_{rr} = 0,\Gamma^r_{\phi\phi}=0,\Gamma^\phi_{r\phi} = 0,\Gamma^\phi_{rr} = -\frac{r}{1+4a^2r^2},\Gamma^{r}_{r\phi} = \frac{1}{r},\Gamma^{\phi}_{\phi\phi} = \frac{4a^2r}{1+4a^2r^2}[/itex]

which are the same as what you get if you use the Lagrangian for the paraboloid in cylindrical coordinates.
noospace
#6
Oct18-07, 07:06 PM
P: 75
I'm still interested to find any counterexamples where the tangential components of the second derivatives of a local parametrization fail to be the Christoffel symbols as defined by Chris Hillman.
Chris Hillman
#7
Nov4-07, 10:23 PM
Sci Advisor
P: 2,340
Hi, noospace, I seem to have lost track of this thread, hence my tardiness.

Quote Quote by noospace View Post
I'm borrowing my notation from Do Carmo `Differential Geometry of Curves and Surfaces', it's a text on classical differential geometry.
Right, unfortunately I don't have it front of me! If you have more questions you may want to provide a bit more detail, e.g. when you wrote

Quote Quote by noospace View Post
[itex]\vect{x} : U \subset \mathbb{R}^2 \to S[/itex] is a local parametrization of a regular surface, ie a smooth nonsingular map which is a homeomorphism onto a nbhd of some point. A regular surface is just a subset of [itex]\mathbb{R}^3[/itex] for which each point in S has a local parametrization.
I am not sure without checking your textbook what S is supposed to be; from such clues as "parameterization" and first letter of "surface" I guess that indeed you are learning surface theory, and that S is a surface immersed in [itex]E^3[/itex].

Quote Quote by noospace View Post
I didn't forgot the normal vector. The second derivatives of a local parametrization can be written in terms of the moving frame [itex]x_u,x_v, x_u\wedge x_v[/itex]. Do Carmo defines the Christoffel symbols as the coefficients of the tangent vectors [itex]x_u,x_v[/itex]. The normal vectors don't come into it.
Let's back up a bit. You didn't say so, but I assume that [itex]x(u,v)[/itex] is a three-vector valued function, the parameterization of our two-dimensional surface. If so, [itex]x_u, \, x_v[/itex] are the tangent vectors (the elements of the "coordinate basis" for the local coordinate chart we are about to construct, obtained by applying [itex] \partial_u, \; \partial_v[/itex] to our parameterizing map [itex]x(u,v)[/itex].). Taking the [itex]E^3[/itex] inner product of these vectors gives the metric induced on S from [itex]E^3[/itex]. Hopefully this sounds familiar!

In your example, we can write the paraboloid (I made one small change in notation) as
[tex]
x(r,\phi) =
\left[ \begin{array}{c} \frac{1}{2} \, a \, r^2 \\
r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right]
[/tex]
Then at each point [itex]x(r, \phi)[/itex], the vectors
[tex]
x_r = \left[ \begin{array}{c} a \, r \\ \cos(\phi) \\ \sin(\phi) \end{array} \right], \;
x_\phi = \left[ \begin{array}{c} 0 \\ -r \, \sin(\phi) \\ r \cos(\phi) \end{array} \right]
[/tex]
span the tangent plane to that point of S. Taking the [itex]E^3[/itex] inner product of these vectors gives the line element, written in a local coordinate chart using the parameters [itex]r,\phi[/itex] as coordinates:
[tex]
ds^2 = \left( 1 + a^2 \, r^2 \right) \, dr^2 + r^2 \, d\phi^2, \;
0 < r < \infty, \; -\pi < \phi < \pi
[/tex]
(In general, we need to impose restrictions on the ranges of the coordinates to avoid singularities, multiple values, and other possible problems. In the case of a surface of revolution we will in general need another local coordinate chart made like this one, in order to cover the "date line" singularity at [itex]\phi=\pm \pi[/itex], and one more local coordinate chart to cover [itex]r=0[/itex]. Or we can change coordinates to find a global coordinate chart, if this is possible, which in general need not be the case.)

Now, what can you say about the cross product [itex]x_r \times x_\phi[/itex], geometrically speaking?

Can you follow the procedure I outlined (probably also covered in Do Carmo's book) to compute the geodesic equations and compare it with the procedure you learned (mislearned?) from somewhere?

For computations it is usually best to adopt the frame field
[tex]
\vec{e}_1 = \frac{1}{\sqrt{1+a^2 \, r^2}} \, \partial_r, \;
\vec{e}_2 = \frac{1}{r} \, \partial_\phi
[/tex]
rather than the coordinate basis [itex] \partial_r, \; \partial_\phi[/itex]. Applying the former to our parameterizing map gives unit tangent vectors spanning the tangent plane at a given point on S.

Scalar multiplying to make a vector into a unit length vector is often called "normalization"; so is dividing through some equation to make the highest order term monic. Perhaps my usage of these two slightly different meanings of "normalization" without comment was confusing; if so, I apologize.

Quote Quote by noospace View Post
If you treat the first fundamental form as a Lagrangian than you can retrieve the Christoffel symbols from the Lagrange equation.
Right, this is the "method of the geodesic Lagrangian", which is usually the most computationally convenient way to obtain the geodesic equations from a given line element. However, the Euler-Lagrange operator applied to the geodesic Lagrangian will in general need to be normalized, by making the unique second order terms monic, before you can identify the coefficients of the quadratic terms in first order derivatives with the Christoffel coefficients!

Quote Quote by noospace View Post
I guessed there must be a theorem that this holds in general, you seem to indicate otherwise. My question is whether or not there are any couterexamples...How can this be so? Surely they argree for a simple sphere?
Sorry, I can't guess what is troubling you. It seems you might have misunderstood the minor point about normalization; the more important point is that you can always compute the geodesic equations from a Lagrangian; indeed, this is often the most efficient route.

Quote Quote by noospace View Post
By the way, why did you take down Relwww?
For the reasons stated here! Or IOW, everyone should just read a good book!


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