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How do I use the geodesic equation for locations on earth

  1. Dec 19, 2015 #1
    So I've gone through the process of deriving the geodesic equation, I thought I understood it. I hoped that once the equation was obtained I'd be able to do simple replacements and find the shortest path between two locations on earth. I'm really stuck right now though so does anyone know how I'd go about doing this ?
     
  2. jcsd
  3. Dec 19, 2015 #2
    It would be helpful if you could supply more details about the part(s) where you got stuck and any relevant equations.

    If you had treated the earth as a two-sphere, you should be able to solve the geodesic equations to find that the shortest path between two points lies on the arc of a great circle.
     
  4. Dec 19, 2015 #3

    https://www.physicsforums.com/file:///C:\Users\acer\AppData\Local\Temp\msohtmlclip1\01\clip_image002.gif [Broken]

    This is where i'm stuck, I only want to solve this problem for the assumption that earth is a perfect sphere, to get this i followed the steps from http://ocw.mit.edu/courses/physics/...paces-spacetime-metric-and-geodesic-equation/
    to be frank i don't know where to go from here to solve the problem, the two sphere thing is new to me , will it require me to go into Christoffel symbols?
     
    Last edited by a moderator: May 7, 2017
  5. Dec 20, 2015 #4
    I'm afraid the image isn't showing up correctly. No, it is not necessarily to delve into Christoffel symbols, although I think the geodesic equations are much cleaner in that notation.
    You should get two differential equations, one for [itex]\theta[/itex] and one for [itex]\phi[/itex]. The solution of these coupled differential equations will give you the geodesic between two points.
     
  6. Dec 20, 2015 #5

    lavinia

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    How are you representing the metric on the sphere?
     
  7. Dec 21, 2015 #6
    Tensors
     
  8. Dec 21, 2015 #7
    That's not the most illuminating answer; its a metric tensor after all. I'm pretty sure what lavinia's referring to is the explicit form of the metric components that you have.
    Could you try to fix the image or typeset the relevant equations? If not, we will find it quite tricky to guide you through.
     
  9. Dec 21, 2015 #8

    lavinia

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    Right. That is what I meant.
     
  10. Dec 21, 2015 #9
  11. Dec 21, 2015 #10
    Since you managed to arrive there, I will assume that you are familiar with the Einstein summation convention then? To proceed from that equation, you need to insert the explicit expressions for your metric tensor [itex]g_{ij}[/itex] and your coordinates [itex]x^{i}[/itex] into the expression. Since the two-sphere is a two-dimensional manifold, it is parameterised by the two angles [itex](\theta,\phi)[/itex].
    Could you please try doing that and see where that gets you (and try to show us your work - so we can advise accordingly)?

    (P.S. if any staff happens to see this, I think this would be better placed in the homework section)
     
  12. Dec 22, 2015 #11
    Yes. I'll do that and get back to you shortly thank you:)
     
  13. Dec 30, 2015 #12
    Hey,
    So I'm working on expaning the summation and this, "k," is confusing me. It looks like something that should be summed over but besides it being a representation of spsce curvature I have no idea what it represents (is it a constant?)
     
  14. Dec 30, 2015 #13
    I presume the [itex]k[/itex] you're referring to is the one in the Robertson-Walker metric? It's just a constant. In fact, only the sign of k matters, because the scale factor can always be redefined such that [itex]|k| = 1[/itex] (except when [itex]k = 0[/itex], of course).

    On a side note - have you been successful in your original problem here of deriving the geodesics on a two-sphere?
     
  15. Dec 30, 2015 #14
    Well I'm still working on that as k also showed up in the geodesic equation(the form i sent the link of) I found out that it is guassian curvature, and I think and it's one over radius squared for a sphere. It seems different to the k in the robertson walker metric but there are a lot of similarities(both k's can tell us if a universe is open or closed.)
    You can see it in that link I sent earlier and it's complicating the summation.
     
  16. Dec 30, 2015 #15
    I think you must have misunderstood something somewhere - the notion of curvature certainly exists for all metrics, but there shouldn't be any 'k' that appears explicitly for the two-sphere. Also, the 'k' in the FLRW metric is a constant, and shouldn't complicate the summation in any way.
    Perhaps you want to illustrate your concerns more clearly so we can figure out what the issue is?
     
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