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Geodesic in a curved space

  1. Nov 1, 2007 #1
    Hi there:

    i have a question on geodesics in a Eculidean space equipped with a metric tensor \lambda(x)*I, where I is the identity matrix. Is any general statement that can be made towards the geodesic between two points in this modified space?

    My feel is that this space is quite special and should have some good properties but don't know how to address it.

    Thanks for any suggestion!!
     
    Last edited: Nov 1, 2007
  2. jcsd
  3. Nov 1, 2007 #2
    Aren't the geodesics just lines in that case? Incidentally, I don't think that space is curved.
     
  4. Nov 1, 2007 #3
    hi, why do you think geodesics are lines?

    According to Theodore Shifrin's book on line pp.86, for hyperbolic plane (u,v) equipped with metric tensor 1/v^2*I, the geodesics are not lines.

    I was just curious if any analytic relation between the geodesics and the lambda(x) function can be made (except for the PDE thing).

    I agree, it should not be called 'curved'.

    Thanks for your reply.
     
    Last edited: Nov 1, 2007
  5. Nov 1, 2007 #4
    I thought $\lambda$ was constant here
     
  6. Nov 3, 2007 #5

    Chris Hillman

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    Providing a buzzword

    of form [itex] g_{ab} = \lambda \, \eta_{ab}[/itex], where [itex]\lambda[/itex] is a scalar functon and where [itex]\eta_{ab}[/itex] is the metric tensor of Euclidean three-space, which in a Cartesian chart will appear as

    Such a metric tensor is said to define a Riemannian geometry which is conformally related to Euclidean geometry.

    It is, but I know more about the Lorentzian case than the Riemannian case so I'll limit myself to mentioning the buzzword "conformally flat".
     
    Last edited: Nov 3, 2007
  7. Nov 6, 2007 #6
    thanks for your reply.
     
  8. Nov 16, 2007 #7

    WWGD

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    Admin question, Please suggest: I have access to Sci-Workplace. If I wrote answers
    in SW and pasted them here, would the program read it effectively?.Again, sorry
    for not knowing to use Tex yet.


    I don't know what assumption we may be making here, but one way of seeing it
    (tho, I admit, I think I am missing an assumption here to have a full response)
    is that a geodesic in a manifold is a curve with acceleration 0, i.e, a curve whose
    covariant derivative is 0, and, using the Euclidean connection , lines, when
    parametrized by arc-length, have second derivative zero, i.e, acceleration
    zero (d^2/ds^2=d/ds(d/ds)=0 , means derivative of velocity with resp. to
    arc-length is zero, i.e, acc. is zero):

    Write your line in terms of arc-length (any curve, with few conditions
    can be parametrized by arc-length)

    L(s)=x_1=a_1+b_1s....x_n=a_n+b_1s , Sum(b_i)^2=1 (bcse. of arc-length par.)

    It follows that d^L(s)/ds^2==0


    The doubt I have is where we are making use of the Euclidean connection
    in here. (which I think is the only connection that is compatible with the
    metric dx^2+dy^2 and has zero torsion, i.e, the only Levy-Cine-Cita
    connection. I will try to prove this and get back )
     
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