# Paraboloid problem

1. Feb 16, 2008

### 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?

2. Feb 19, 2008

### snoopies622

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…

3. Feb 19, 2008

### pervect

Staff Emeritus
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.

4. Feb 19, 2008

### snoopies622

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?

5. Feb 19, 2008

### gel

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: Feb 19, 2008
6. Feb 20, 2008

### snoopies622

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.

7. Feb 24, 2008

### snoopies622

Say pervect,

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