Solving Curvature Tensor of Paraboloid Problem

• snoopies622
In summary: How is this circularity avoided?In summary, the conversation is about a person trying to compute the curvature tensor of a paraboloid using different coordinate systems. They are having trouble with the Christoffel symbols and are unsure of their calculation methods. Another person suggests using a different equation for the Christoffel symbols and offers an automated program for assistance. The conversation also delves into the definition of Christoffel symbols and their role in covariant differentiation.
snoopies622
Lately I've been trying to teach myself GR and it's been going fairly well, but yesterday for practice I decided to compute the curvature tensor of a paraboloid and it's not working. I've tried using three different coordinate systems, starting with what I thought would be the most obvious one,

x= r cos(theta)
y= r sin(theta)
z= r^2

but in every case the Christoffel symbols have failed. For example, using the above, the basis vectors are

e sub r = < cos(theta), sin<theta>, 2r >
e sub theta= <-r sin(theta), r cos(theta), 0 >

and the partial derivative of e sub r (for example) with respect to r = <0,0,2>. Yet there is no linear combination of these two basis vectors that will make <0,0,2>. The other two coordinate systems I've tried have had similar difficulties.

A paraboloid strikes me as a legitimate 2-dimensional manifold. What am I doing wrong?

Well, it turns out that changing z from r^2 to kr^2 allows for all the Christoffel symbols. The non-zero ones are

{r,theta theta}=-r
{theta, r theta}={theta, theta r}=1/r
{k, r r}=2k/r^2
{k, r k}={k, k r}= 2/r
{k, theta theta}=2k

where {r, theta theta} means the Christoffel symbol with r on top and two thetas on the bottom, etc.

Strangely, when I plug these into the curvature tensor equation, all the components (with r and theta indices only) are zero, as if this were a plane. Does that make sense? Maybe the Ricci scalar will be non-zero…

I really don't know how you went about solving this problem. The way I would go about solving it is to compute the metric tensor first, then use the metric tensor to compute the Christoffel symbols. Of course, I have an automated program that does the later task for me.

Given that
dx = cos(theta)*dr - r*sin(theta)*dtheta
dy = sin(tehta)*dr + r*cos(theta)*dtheta
dz = 2*r*dr

I get the line element

ds^2 = dx^2+dy^2+dz^2 = (4r^2+1)*dr^2 + r^2*dtheta^2

For the Christoffel symbols I get

$$\Gamma^r{}_{rr} = \frac{4r}{r^2+1}$$
$$\Gamma^\theta{}_{\theta r} = \Gamma^\theta{}_{r \theta} = \frac{1}{r}$$
$$\Gamma^r{}_{\theta \theta} = -\frac{r}{4 r^2 + 1}$$

I won't guarantee that I haven't made any errors, but those are the results I get for this problem.

Thanks, pervect. I also found the Christoffel symbols for

x= r cos(theta)
y= r sin(theta)
z= r^2

that way, but I am still left with a mystery. As I learned it (Relativity Demystified, David McMahon, chapter 4) the partial derivative of a basis vector is itself a vector, which can be expressed as a linear combination of coefficients (the Christoffel symbols) and basis vectors. For example (pretend that these d’s are actually partial derivative symbols),

d/dr (e sub r)= {r,rr}(e sub r) + {theta, rr}(e sub theta)

where the {}’s are Christoffel symbols and (e sub r) and (e sub theta) are the basis vectors < dx/dr, dy/dr, dz/dr > and < dx/d-theta, dy/d-theta, dz/d-theta >, respectively. I know that Christoffel symbols can be computed in different ways (including using the metric tensor) and I’ve read other GR material since McMahon, but I’ve taken the linear combination description above as the DEFINITION of a Christoffel symbol and up until now that hasn’t given me any problems. In this case, however, it doesn’t work. For, plugging into the equation above yields

d/dr < dx/dr, dy/dr, dz/dr > =
d/dr < cos(theta), sin(theta), 2r >=
< 0, 0, 2 > =
(4r/r^2+1)< cos(theta), sin(theta), 2r > + (0)< -r sin(theta), r cos(theta), 0>

which is obviously false. Have I been misled?

The Christoffel symbols are the derivatives of basis vectors using the connection on the manifold
$$\nabla_{e_i}e_j = \Gamma^k_{ij}e_k. (*)$$
For a surface embedded in R3, which you are using, the connection $\nabla$ is *not* the same thing as the connection in R3. So you can't just differentiate them componentwise and apply (*).
You can obtain the connection working from the metric tensor, as pervect just did.
Alternatively, I think, you can define the connection on the surface by differentiating in R3 and then projecting back onto the tangent space of the surface.

$$e_r = (\cos\theta,\sin\theta,2r)$$
$$e_\theta=(-r\sin\theta,r\cos\theta,0)$$

A normal vector to the surface is $u=(2r\cos\theta,2r\sin\theta,-1)$, so

$$(d/dr)e_r=(0,0,2)=(4re_r-2u)/(4r^2+1)$$
$$(d/d\theta)e_r=(-\sin\theta,\cos\theta,0)=e_\theta/r$$
$$(d/dr)e_\theta=(-\sin\theta,\cos\theta,0)=e_\theta/r$$
$$(d/d\theta)e_\theta=-r(\cos\theta,\sin\theta,0)=-r(e_r+2ru)/(4r^2+1)$$

Projecting onto the tangent space maps u to 0, so

$$\nabla_{e_r}e_r=4re_r/(4r^2+1)$$
$$\nabla_{e_\theta}e_r=\nabla_{e_r}e_\theta=e_\theta/r$$
$$\nabla_{e_\theta}e_\theta=-re_r/(4r^2+1)$$

from which you can read off

$$\Gamma^r_{rr}=4r/(4r^2+1),\ \Gamma^\theta_{rr}=\Gamma^r_{r\theta}=\Gamma^r_{\theta r}=0,$$
$$\Gamma^\theta_{r\theta}=\Gamma^\theta_{\theta r}=1/r,\ \Gamma^r_{\theta\theta}=-r/(4r^2+1),\ \Gamma^\theta_{\theta\theta}=0.$$

Edit: although the definition is usually taken to be the (unique) torsion free connection with $\nabla g=0$, g being the metric tensor.

Last edited:
Thank you gel. I can't say that I completely understand at the moment, but you've given me a couple things to think about.

Say pervect,

I was just re-reading this old thread and became puzzled over this equation from gel's explanation:

$$\nabla_{e_i}e_j = \Gamma^k_{ij}e_k$$

It looks like covariant differentiation is being used to define Christoffel symbols, but Christoffel symbols are themselves needed to perform covariant differentiation in the first place, aren't they?

1. What is the curvature tensor of a paraboloid?

The curvature tensor of a paraboloid is a mathematical concept used to describe the curvature of a parabolic surface. It is a rank-4 tensor that contains information about the curvature in all directions on the surface.

2. How is the curvature tensor calculated for a paraboloid?

The curvature tensor is calculated using the partial derivatives of the first and second fundamental forms of the paraboloid. These forms represent the geometric properties of the surface, and the curvature tensor can be derived from them using various mathematical operations.

3. Why is the curvature tensor important in solving the paraboloid problem?

The curvature tensor provides a quantitative measure of the curvature of the paraboloid, which is crucial in solving the problem. It allows us to analyze the surface and determine its properties, such as the principal curvatures and directions of curvature.

4. What is the significance of solving the curvature tensor of a paraboloid problem?

Solving the curvature tensor of a paraboloid problem has various applications in mathematics, physics, and engineering. It can be used to understand the behavior of curved surfaces, design structures, and solve optimization problems.

5. Are there any real-world examples where the curvature tensor of a paraboloid is used?

Yes, the curvature tensor of a paraboloid is used in various real-world applications. For example, it is used in designing parabolic reflectors for satellite dishes, analyzing the shape of car windshields, and determining the optimal curvature for lenses in optical systems.

Replies
7
Views
2K
Replies
42
Views
2K
Replies
19
Views
3K
Replies
1
Views
1K
Replies
4
Views
826
Replies
10
Views
2K
Replies
5
Views
656
Replies
13
Views
2K
Replies
7
Views
2K
Replies
11
Views
2K