Christoffel symbols and Geodesic equations.

In summary, the conversation discusses a 2-dimensional manifold with a given line element and explores its properties. It is determined that the line element is well defined for all values of y and z, except for z=0. The non-vanishing Christoffel symbols and the geodesic equations are obtained. The conversation also discusses solving the geodesic equations and suggests an improved coordinate system. It is mentioned that there is no general method for solving differential equations, but there are various resources available online. Finally, the conversation addresses a clarification regarding a change of variables.
  • #1



Homework Statement

(a) Consider a 2-dimensional manifold M with the following line element


For which values of z is this line element well defined.

(b) Find the non-vanishing Christoffel symbols

(c) Obtain the geodesic equations parameterised by l.

(d) Solve the geodesic equations and suggest an improved coordinate system. What is the metric in the new coordinates? What lines describ the geodesic geometrically?

(e) What can you say about the Riemann curvature tensor, the Ricci tensor and the Ricci scalar of this manifold.

Homework Equations

The Attempt at a Solution

(a) The line element is well defined for all values of y and z other then z=0

(b) gzz,z= -2/z3

The only non vanishing christoffel symbol is,

Czzz= -1/z

(c) The geodesic equations are given by,



Stuck here.
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  • #2
To solve the differential equation in (d), try the change of variables z=e^v. Think about applying this change of variables to the original coordinate system.
  • #3
Hey Dick thanks for the hint. Haven't managed to get anywhere with it yet, but will give it a go again tomorrow.

If I may ask you another question however, there a general method for solving the differential equations/geodesic equations? If so it would be really useful if you guide me to an online resource for the same.
  • #4
For some forms of differential equations there are methods specific to that form. But there is no one method. There's tons of stuff online. Just google 'solving differential equations' and pick your favorite.
  • #5
Hi dick, could you possibly have meant e^l rather than e^v?
  • #6
trv said:
Hi dick, could you possibly have meant e^l rather than e^v?

No, I meant substitute z(l)=e^(v(l)). What does the differential equation for v(l) look like?

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