Register to reply 
Christoffel symbols from definition or Lagrangian 
Share this thread: 
#1
Sep2807, 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 arclength [itex]ds^2 = E du^2 + 2F du dv + G dv^2[/itex]. Then the coefficients of of the corresponding EulerLagrange equations are essentially the Christoffel symbols. Are there any interesting examples where the Lagrangian method of computing Christoffel symbols breaks down? 


#2
Sep3007, 05:12 PM

P: 2,954

Pete 


#3
Sep3007, 07:39 PM

Sci Advisor
P: 2,340

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 44, 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 EulerLagrange 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{1r^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}{1r^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}{1r^2} \, \dot{r}^2  r \, (1r^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. 


#4
Oct1307, 05:01 AM

P: 75

Christoffel symbols from definition or Lagrangian
Hi Chris,
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. By the way, why did you take down Relwww? That was a really useful website. Thanks. 


#5
Oct1307, 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. 


#6
Oct1807, 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.



#7
Nov407, 10:23 PM

Sci Advisor
P: 2,340

Hi, noospace, I seem to have lost track of this thread, hence my tardiness.
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. 


Register to reply 
Related Discussions  
Christoffel Symbols  Advanced Physics Homework  12  
Christoffel Symbols  Advanced Physics Homework  5  
Christoffel symbols from definition or Lagrangian  Calculus & Beyond Homework  0  
From Christoffel symbols to connection 1forms  Differential Geometry  12  
Christoffel symbols examples  Differential Geometry  7 