Parametric equations for geodesics

  • Thread starter andybham
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  • #1
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Main Question or Discussion Point

What are you trying to do when you find parametric equations for a geodesic lines on a surface?

Take the metric ds^2 = dq^2 + (sinh(q)*dp)^2

Are you simply trying to get q as a function of s? and p as a function of s?

If so, why?

Thanks
 

Answers and Replies

  • #2
Chris Hillman
Science Advisor
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Clarifying meaning of geodesic equations

In the line element for the hyperbolic plane (this trig chart covers all but one point) which you wrote down,
[tex]
ds^2 = dr^2 + \sinh(r)^2 \, d\phi^2, \;
0 < r < \infty, \; -\pi < \phi < \pi
[/tex]
(I changed to a more suggestive notation), s is playing the role of arc length. The standard method of obtaining the geodesic equations is to read off what MTW call the "dynamical Lagrangian"
[tex] L = \dot{r}^2 + \sinh(r)^2 \, \dot{\phi}^2 [/tex]
Then apply the Euler-Lagrange equations:
[tex] 0 = \frac{\partial L}{\partial r} - \frac{d}{d s} \; \frac{\partial L}{\partial \dot{r}}, \; \;
0 = \frac{\partial L}{\partial \phi} - \frac{d}{d s} \; \frac{\partial L}{\partial \dot{\phi}} [/tex]
Simplify the result so that [itex]\ddot{r},\ddot{\phi}[/itex] are monic and collect terms in the first order derivatives. The result is the geodesic equations in their standard form and the coefficients are the Christoffel coefficients:
[tex]
\ddot{r} + \sinh(r) \, \cosh(r) \, \dot{\phi}^2 = 0, \;
\ddot{\phi} + 2 \, \coth(r) \, \dot{r} \, \dot{\phi} = 0
[/tex]
where "dot" means differentiation wrt the parameter s. Here, s is playing the role of an arc length parameter, so the solutions of these equations give the geodesics as arc length parameterized curves. The second equation has a first integral
[tex] \dot{\phi} = \frac{L}{\sinh(r)^2} [/tex]
which is obviously similar to the analogous first integral for a radial chart for the Euclidean plane (substitute [itex]\sinh(r) \mapsto r[/itex] and consider what happens for small r).

Exercise: consider a simple semi-Riemannian analogue:
[tex]
ds^2 = -dt^2 + \cosh(t)^2 \, d\phi^2, \;
0 < t < \infty, \; -\pi < \phi < \pi
[/tex]
(This is the upper half of [itex]H^{1,1}[/itex], the hyperboloid of one sheet in [itex]E^{1,2}[/itex], the signature 1,1 manifold of constant curvature, aka "two-dimensional tachyonic momentum space".) Why do the resulting equations give all geodesics, including null geodesics, as affine-parameterized curves? (This point recently confused someone in another thread.)
 
Last edited:
  • #3
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Thankyou for your reply.

Why do the resulting equations give all geodesics, including null geodesics, as affine-parameterized curves?
because we are optimising s hence finding the shortest path through space-time?

I have an example question (and the corresponding solution) for the metric I gave

ds^2 = dq^2 + (sinh(q)*dp)^2

that was used on a past exam paper.



I understand everything except the question asks 'find parametric equations for geodesic lines on this surface' and the solution produces

cosh(q) = cosh(q0)*cosh(s)

from the first integrals. SImilarly, it also gives

dp = (sinh(q0)*ds)/(cosh^2(q0)*cosh^2(s)-1) [I apologise, I don't know how to add tex code]

I understand the maths involved in getting these results, however, in order for me to be able to obtain similar expressions for different problems in the exam I need to understand what the question is asking when it says 'find parametric equations for geodesic lines on this surface'

Are you simply trying to get q as a function of s? and p as a function of s?

If so, why?
 
  • #4
pervect
Staff Emeritus
Science Advisor
Insights Author
9,775
992
The dynamic Lagrangian method automatically gives the geodesics in affine parameterization. It is the assumption that one has used an affine parameterization that allows the simple method of the dynamic Lagrangian to work.

Note that L is a quardratic form Q in the dynamic Lagrangian method

If you wish to find the geodesics in arbitrary coordinates directly, you must solve the Euler-Lagrange equations with L = sqrt(Q), rather than the Lagrangian given in the previous post by Chris Hillman which is specific to the dynamic Lagrangian method.

It is the simplification involved in getting rid of the square root (which is quite significant, if you actually go through the work of solving it) that makes the dynamic Lagrangian method attractive.

A very similar method is the Einbein method, which is usually used in other contexts than geodesics but also serves the function of getting rid of the square root in a Lagrangian.
See for instance:

https://www.physicsforums.com/showthread.php?t=127956

For a textbook discussion, if you happen to have MTW "Gravitation", pg 315-324 discuss both the direct method and the dynamic Lagrangian method.
 
Last edited:
  • #5
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Ah I see, the two equations provide the parametric representation of a typical geodesic.

Thankyou both for all your help!
 

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