Finding ds² on a Cone: How to Use Geodesic Equations for Parallel Transport

[LCSphysicist]

Homework Statement

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Relevant Equations

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I am having too much trouble to solve this exercise, see:

Using (R,phi,z)
ub is the path derivative
U is the path
V is the vector

$$V^{a};_{b}u^{b} = (\partial_{b}V^{a} + \Gamma^{a}_{\mu b} V^{\mu})u^{b}$$

$$U = (0,\theta,Z)$$

I am not sure what line element to use, i mean, a circle around a cone does reduce to a plane, but i think this isn't right...

 

[Office_Shredder]

What is $\alpha$?

 

[martinbn]

What is $\alpha$?

May be the angle at which the cone opens!

 

[LCSphysicist]

May be the angle at which the cone opens!

$2 \alpha$ is the angle at which the cone opens :)
Thank you

 

[stevendaryl]

To compute parallel transport around a cone, you can do it two different ways:

1. You can set up a coordinate system on the cone, and compute the connection coefficients for that coordinate system, and then use them for parallel transport.
2. (This is the really easy way) You can use the fact that a cone is just a section of the plane glued together in a particular way. Parallel transport on the plane is trivial using Cartesian coordinates. However, if you parallel-transport a vector over the "seam" where the two edges are glued together, there will be a discontinuous jump in the direction the vector is pointing (when viewed in Cartesian coordinates---the change is continuous in cone-based coordinates described in 1.)

 

[stevendaryl]

To compute parallel transport around a cone, you can do it two different ways:

1. You can set up a coordinate system on the cone, and compute the connection coefficients for that coordinate system, and then use them for parallel transport.
2. (This is the really easy way) You can use the fact that a cone is just a section of the plane glued together in a particular way. Parallel transport on the plane is trivial using Cartesian coordinates. However, if you parallel-transport a vector over the "seam" where the two edges are glued together, there will be a discontinuous jump in the direction the vector is pointing (when viewed in Cartesian coordinates---the change is continuous in cone-based coordinates described in 1.)

I apologize for the hand-drawn picture, because I don't have access to a drawing program. Suppose you have a disk on the plane with a sector missing with angle $Q$. There are two "cuts" shown: one vertical, and one that makes an angle of $Q$ relative to the vertical. Take a vector that is pointing straight up and transport it around the loop indicated by the dashed line. Initially, the vector points in the same direction as the first cut. After parallel-transporting it around the loop, it now makes an angle of $Q$ relative to the second cut.

Now, if we "glue" the two cuts together, then the cut disk becomes a cone, and the dashed path becomes a closed loop. So parallel transporting the vector around that loop causes a rotation through the angle $Q$.

(The angle $Q$ is not equal to your $2 \alpha$, but they are related.)

 

[LCSphysicist]

View attachment 278791

I apologize for the hand-drawn picture, because I don't have access to a drawing program. Suppose you have a disk on the plane with a sector missing with angle $Q$. There are two "cuts" shown: one vertical, and one that makes an angle of $Q$ relative to the vertical. Take a vector that is pointing straight up and transport it around the loop indicated by the dashed line. Initially, the vector points in the same direction as the first cut. After parallel-transporting it around the loop, it now makes an angle of $Q$ relative to the second cut.

Now, if we "glue" the two cuts together, then the cut disk becomes a cone, and the dashed path becomes a closed loop. So parallel transporting the vector around that loop causes a rotation through the angle $Q$.

(The angle $Q$ is not equal to your $2 \alpha$, but they are related.)

Hello, thank you for the reply. This way is obviously beautiful and simple, but i am interesting too in improving my abilities in using the geodesic equations, so i want to try the other way too.
Particularly, i am having too much trouble to know how to get an equation to find the ds² in this case, do you have any clue?thx

 

[stevendaryl]

Hello, thank you for the reply. This way is obviously beautiful and simple, but i am interesting too in improving my abilities in using the geodesic equations, so i want to try the other way too.
Particularly, i am having too much trouble to know how to get an equation to find the ds² in this case, do you have any clue?thx

Yes, you started down that path in your first post. Let's set up a coordinate system on the surface of the cone as follows:

• $r$ is the distance from the vertex of the cone to the point.
• $\phi$ is the "longitudinal" coordinate that goes from 0 to $2 \pi$ around the cone.

You can relate these coordinates to the cartesian coordinates $(x,y,z)$ as follows:

• $x = r sin(\alpha) cos(\phi)$
• $y = r sin(\alpha) sin(\phi)$
• $z = r cos(\alpha)$

This would be a cone with the vertex at the point $(x = 0, y=0, z=0)$ opening upward, with the axis of the cone parallel to the z-axis.

Now, consider two nearby points on the cone. The distance between those points would be given by:

$ds^2 = dx^2 + dy^2 + dz^2$
$= (dr sin(\alpha) cos(\phi) - r d \phi sin(\alpha) cos(\phi))^2$
$+ (dr sin(\alpha) sin(\phi) + r d \phi sin(\alpha) sin(\phi))^2$
$+ (dr cos(\alpha))^2$

(there's no $d \alpha$, since $\alpha$ is constant)

Compare this to $ds^2 = g_{rr} dr^2 + g_{\phi \phi} dr d\phi + g_{\phi r} d\phi dr + g_{\phi \phi} d\phi^2$

to figure out $g_{rr}$ and $g_{r \phi}$ (the other two are zero). Then you figure out the $\Gamma^\alpha_{\beta \gamma}$ from $g_{\mu \nu}$ and its derivatives. Then you use the parallel transport equation to find out how $V^\alpha$ changes as you move along a path.

It's a lot of work, but that's the plan.