Hope this helps
Hi, noospace, I seem to have lost track of this thread, hence my tardiness.
noospace said:
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
noospace said:
\vect{x} : U \subset \mathbb{R}^2 \to S 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 \mathbb{R}^3 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 E^3.
noospace said:
I didn't forgot the normal vector. The second derivatives of a local parametrization can be written in terms of the moving frame x_u,x_v, x_u\wedge x_v. Do Carmo defines the Christoffel symbols as the coefficients of the tangent vectors x_u,x_v. The normal vectors don't come into it.
Let's back up a bit. You didn't say so, but I assume that x(u,v) is a three-vector valued function, the parameterization of our two-dimensional surface. If so, x_u, \, x_v are the tangent vectors (the elements of the "coordinate basis" for the local coordinate chart we are about to construct, obtained by applying \partial_u, \; \partial_v to our parameterizing map x(u,v).). Taking the E^3 inner product of these vectors gives the metric induced on S from E^3. Hopefully this sounds familiar!
In your example, we can write the paraboloid (I made one small change in notation) as
<br />
x(r,\phi) = <br />
\left[ \begin{array}{c} \frac{1}{2} \, a \, r^2 \\ <br />
r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right]<br />
Then at each point x(r, \phi), the vectors
<br />
x_r = \left[ \begin{array}{c} a \, r \\ \cos(\phi) \\ \sin(\phi) \end{array} \right], \;<br />
x_\phi = \left[ \begin{array}{c} 0 \\ -r \, \sin(\phi) \\ r \cos(\phi) \end{array} \right]<br />
span the tangent plane to that point of S. Taking the E^3 inner product of these vectors gives the line element, written in a local coordinate chart using the parameters r,\phi as coordinates:
<br />
ds^2 = \left( 1 + a^2 \, r^2 \right) \, dr^2 + r^2 \, d\phi^2, \;<br />
0 < r < \infty, \; -\pi < \phi < \pi<br />
(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 \phi=\pm \pi, and one more local coordinate chart to cover r=0. 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 x_r \times x_\phi, 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
<br />
\vec{e}_1 = \frac{1}{\sqrt{1+a^2 \, r^2}} \, \partial_r, \;<br />
\vec{e}_2 = \frac{1}{r} \, \partial_\phi<br />
rather than the coordinate basis \partial_r, \; \partial_\phi. 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.
noospace said:
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!
noospace said:
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.
noospace said:
By the way, why did you take down Relwww?
For the reasons
stated here! Or IOW, everyone should just read a good book!
