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Geodesics in a 2D embedded space

  1. Feb 5, 2008 #1
    For a course on tensor analysis, we were asked to perform some calculations regarding a 2D space embedded in 'regular' Euclidean 3D space. I've gotten some results, but I've hit a snag...

    1. The problem statement, all variables and given/known data

    The following functions relate the coordinates in the embedding (3D, Euclidean) space to the coordinates in the embedded (2D) space (they describe the 'shape' of the embedded space):

    [tex]X^1 = r \sin{\phi}[/tex] (1)
    [tex]X^2 = r\cos{\phi}[/tex] (2)
    [tex]X^3 = \log{r}[/tex] (3)

    where the [itex]X^n[/itex] coordinates represent the coordinates in the embedding space, and [itex]r[/itex] and [itex]\phi[/itex] represent the radial and angular coordinates on the (apparently funnel-shaped) embedded space.

    From this information, the following relations can be derived after some calculation (L is a constant):

    [tex]\frac{d\phi}{ds} = \frac{L}{r^2}[/tex] (4)

    and

    [tex](1+r^2)\left(\frac{dr}{ds}\right)^2 = r^2 - L^2[/tex]. (5)

    I was asked to derive these results, which I did (intermediate results given below) - starting from calculating the metric in the embedded space, then calculating Christoffel symbols of 1st and 2nd kinds, and using them calculating the shape of the geodesic equations and the kinematic condition.

    Now, the question that's giving me some trouble is:

    - Show that, for a given L > 0, there exists a geodesic along which r(s) is a constant, and calculate this constant.

    2. Relevant equations

    Induced metric from embedding:
    [tex]g_{\mu\nu}(x)=G_{AB}(X(x))\frac{\partial F^A(x)}{\partial x^{\mu}}\frac{\partial F^B(x)}{\partial x^{\nu}}[/tex]

    Relation between contra- and covariant metric components:
    [tex]g^{\mu\nu}g_{\nu\alpha}=\delta^{\mu}_{\alpha}[/tex]

    Christoffel symbol of the first kind:

    [tex]\Gamma_{\mu\nu\alpha}=\frac{1}{2}(g_{\mu\nu,\alpha} + g_{\mu\alpha,\nu}-g_{\nu\alpha,\mu})[/tex]

    Christoffel symbol of the second kind:

    [tex]\Gamma^{\mu}_{\nu\alpha}=g^{\mu\beta}\Gamma_{\beta\nu\alpha}[/tex]

    Geodesic equation (u is position differentiated w.r.t. arc length s, i.e. 'proper motion'):

    [tex]\dot{u}^{\mu} + \Gamma^{\mu}_{\alpha\beta}u^{\alpha}u^{\beta}=0[/tex]

    Kinematic condition:

    [tex]u_{\mu}u^{\mu}=1[/tex]

    3. The attempt at a solution
    Intermediate results, using the standard formulae:

    The components of the (covariant) metric for the embedded space:

    [tex]g_{11}=1 + \frac{1}{r^2}[/tex]
    [tex]g_{22}=r^2[/tex]
    [tex]g_{12} = g_{21} = 0[/tex]

    Contravariant metric components:

    [tex]g^{11} = \frac{r^2}{r^2 + 1}[/tex]
    [tex]g^{22}=\frac{1}{r^2}[/tex]
    [tex]g^{12}=g^{21}=0[/tex]

    Christoffel symbols of the first kind:

    [tex]\Gamma_{111}=\frac{-1}{r^3}[/tex]
    [tex]\Gamma_{122}=-r[/tex]
    [tex]\Gamma_{212} = \Gamma_{221} = r[/tex]
    (other Christoffel symbols of the 1st kind are 0)

    Christoffel symbols of the second kind:

    [tex]\Gamma^{1}_{11}=\frac{-1}{r^3 + r}[/tex]
    [tex]\Gamma^{1}_{22}=\frac{-r^3}{r^2+1}[/tex]
    [tex]\Gamma^{2}_{12} = \Gamma^{2}_{21}=\frac{1}{r}[/tex]
    (other Christoffel symbols of the 2nd kind are 0)

    The two geodesic equations (dots denote differentiation with respect to s, the arc length):

    [tex]\ddot{r} - \frac{1}{r^3+r}\dot{r}^2 - \frac{r^3}{r^2+1}\dot{\phi}^2 = 0[/tex]
    [tex]\ddot{\phi} + \frac{2}{r}\dot{r}\dot{\phi}=0[/tex]

    The kinematic condition:

    [tex](1+\frac{1}{r^2})\dot{r}^2 + r^2\dot{\phi}^2=1[/tex]

    From the second geodesic equation, equation (4) can be obtained by multiplying with r^2 and treating the result as a total derivative:

    [tex]r^2\ddot{\phi} + 2r\dot{r}\dot{\phi}=\frac{d}{ds}(r^2\dot{\phi})=\frac{d}{ds}(L)=0[/tex]

    Equation (5) follows from the kinematic condition (introducing L as described above and shuffling some terms around).

    Now, to get back to the original question: for r(s) to be a constant means that all derivatives of r with respect to s must vanish. Looking at the first geodesic equation, the only way in which that can happen is when the derivative of [itex]\phi[/itex] with respect to s is also zero. And that means the only possible solution is a stationary point (i.e. r and [itex]\phi[/itex] both are constants). But that is in direct conflict with the kinematic condition, which would in that case state 0 + 0 = 1. So it seems there is no solution to the posed question!

    Am I making a mistake somewhere?
     
    Last edited: Feb 5, 2008
  2. jcsd
  3. Feb 5, 2008 #2

    kdv

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    A quick question: how do you get from dL/ds=0 to equation 4?
     
  4. Feb 5, 2008 #3
    [tex]L=r^2\dot{\phi}[/tex],

    so:

    [tex]\dot{\phi}=\frac{d\phi}{ds}=\frac{L}{r^2}[/tex].
     
  5. Feb 5, 2008 #4

    kdv

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    Sorry for the dumb question. I was thinking that a dot denoted a derivative with respect to a proper time. :blushing:
     
  6. Feb 5, 2008 #5

    kdv

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    I checked all the steps starting from your Christoffel symbols and I don't see any mistake. There is something puzzling indeed. Hope someone else can spot the problem.
     
  7. Feb 5, 2008 #6

    NateTG

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    Homework Helper

    I'm unfamiliar with the notations of differential geometry, but it seems like equation 5 should lead to a situation where:
    [tex]L=r \rightarrow \left(\frac{dr}{ds}\right)=0[/tex]

    Thinking geometrically, the surface is a funnel with an axis that goes through the origin, so it seems like rings around the funnel should have constant radius.
     
  8. Feb 6, 2008 #7
    Thank you for your replies so far. It seems to me that there's an error in the problem posed - I also really can't see where I could have made a mistake. I had also expected, from the shape of the embedded space, that geodesics with constant radius had to be circular trajectories around the 'throat' of the funnel - but those solutions are not allowed by the first geodesic equation. And since by my knowledge all geodesics have to obey both geodesic equations (and the kinematic condition), there is no solution. Nasty, because that makes it a trick question. I hadn't expected a trick like that in this assignment. :)
     
  9. Feb 6, 2008 #8

    NateTG

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    Yeah, the normals to the surface look like:
    [tex]<\sin \phi, \cos \phi, - r>[/tex]
    which always has a component in the [itex]z[/itex] dimension, but the derivatives of curves that fix [itex]r[/itex] never do.
     
  10. Feb 8, 2008 #9

    kdv

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    NateTG, I am trying to understand your point. You are saying that this shows that the lines of constant r are not geodesics? Is that what you show with the above argument? If that's the case, I am trying to understand the reason, intuitively. Why would lines of constant r not be geodesics? I think that it becomes obvious, after some thought, but I may be in the left field. If one "walks" from a point to another point at the same r, it is advantageous (in terms of distance) to "move down" the funnel a bit, where it is more narrow. In other words, it is advantageous to chaneg the value of r a bit in order to minimize the distance travelled. I am not sure if this is a correct reasoning.
     
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