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Are Jacobi fields defined at intersection points?

  1. Jan 4, 2015 #1

    Matterwave

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    • Two similar threads merged as requested
    I have some questions with regards to conjugate points on a congruence of time-like geodesics (will be referring to Wald 9.3 throughout). First, we define ##\gamma## to be a time-like geodesic with tangent ##\xi^a## parametrized by ##\tau## and with ##p\in\gamma##. We consider the "congruence of all timelike geodesics passing through ##p##" and we want to show that ##q\in\gamma## is a conjugate point of ##p## if and only if the expansion, ##\theta##, defined as ##\theta=\nabla_a \xi^a##, approaches ##-\infty## at ##q##.

    Anyways, the Jacobi fields are defined as solutions ##\eta^a## to the equation ##\xi^a\nabla_a(\xi^b\nabla_b\eta^c)=-R_{abd}^{~~~~~~~c}\eta^b\xi^a\xi^d##. And at least one particular Jacobi field will vanish, by definition, at conjugate points. We define a set of orthogonal spatial vectors ##e^a_i## which are orthogonal to ##\xi^a## (with ##\xi^a## itself serving as our time-direction basis vector, as it is parametrized by the proper time, it is normalized to -1). Furthermore, for simplicity in notation we define ##B_{ab}=\nabla_b\xi_a##. This is just background information.

    Now on to my question. Specifically, in equation 9.3.5 Wald makes the following derivations:
    $$\frac{d\eta^\mu}{d\tau}=\xi^a\nabla_a\eta^\mu=\xi^a\nabla_a[(e_\mu)_b\eta^b]$$ $$\frac{d\eta^\mu}{d\tau}=(e_\mu)_b\xi^a\nabla_a\eta^b$$ $$\frac{d\eta^\mu}{d\tau}=(e_\mu)_b B^b_{~a}\eta^a$$ $$\frac{d\eta^\mu}{d\tau}=B^\mu_{~\alpha}\eta^\alpha$$
    My question is in going from the second line to the third line (counting the top two equalities as just 1 line).

    I believe the full version of that line requires the following deduction: ##\xi^a\nabla_a\eta^b=\eta^a\nabla_a \xi^b=\eta^a B^b_{~a}##. Of course, for this to be true, it must be that ##\mathcal{L}_\xi \eta^a=0##.

    So far in Wald, ##\mathcal{L}_\xi\eta^a=0## has been said to be true, but in slightly different contexts than here, which is where I am getting confused (Wald presented equation 9.3.5 without justifications for the steps, leaving me to work out the equalities myself). If I can recall correctly, the relation ##\mathcal{L}_\xi\eta^a=0## was said to be true (originally in chapter 3) because ##\xi^a## and ##\eta^a## were coordinate basis vectors ##\xi^a=(\partial t)^a## and ##\eta^a=(\partial s)^a##(in chapter 3) on the 2-dimensional submanifold ##\Sigma## which is foliated by the congruence ##\gamma_s(t)## (wherein each ##s## denotes a particular geodesic).

    Therein lies my problem. For if ##p## is a point through which many different geodesics pass, then the congruence is singular at ##p##. If the congruence is singular at ##p##, then I can't create this ##\Sigma## submanifold with the congruence ##\gamma_s(t)## since the map ##(t,s)\rightarrow \gamma_s(t)## will no longer be bijective at p. In addition, my coordinate ##s## is no good at ##p## since the coordinate basis vector ##\eta^a=(\partial s)^a## in fact vanishes at p by construction. With the inclusion of this singularity in my congruence, and thus my inability to construct the submanifold ##\Sigma##, how can I be sure that ##\mathcal{L}_\xi \eta^a=0## holds everywhere on my congruence? How do I justify going from line 2 to line 3 in equation 9.3.5?

    Am I over thinking this? Is it possible Wald did not have to use ##\mathcal{L}_\xi\eta^a=0## in that step? Or is this property still rigorously valid even with the presence of the singularity in the congruence at ##p##?

    EDIT: I should add that later, Wald basically asserts that ##\mathcal{L}_\xi \eta^a=0## is "true everywhere" without proof. On page 227 he basically says "...and, as usual, we have ##\mathcal{L}_\xi \eta^a=\eta^b\nabla_b \xi^a-\xi^b\nabla_b\eta^a=0## everywhere" - although here he has switched to using ##T^a## and ##X^a## instead of ##\xi^a## and ##\eta^a## I think because he removed the geodesic requirement on ##\gamma## (which he has switched to ##\lambda## here). I don't know how he can just say this is true now that there's a singularity at p though.
     
    Last edited: Jan 6, 2015
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  3. Jan 10, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 11, 2015 #3

    Matterwave

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    The only thing I can think of is at the point ##p##, we have ##\eta^a=0##, so perhaps by the linearity of the Lie Bracket one might be able to argue that ##[\xi^a,\eta^b]=0## there, and thus one is able to smoothly extend the requirement ##\mathcal{L}_\xi \eta^a=0## beyond ##p## (and also ##q##). I'm not sure if this argument is valid, and why Wald would not have brought up this point (unless it was trivial).
     
  5. Feb 4, 2015 #4
    Consider a family of geodesics ##\gamma_s(v)## with tangent vector
    $$V = \frac{d}{dv}$$
    and deviation vector
    $$S= \frac{d}{ds}.$$
    Within regions where the map ##(v,s) \mapsto \gamma_s(v)## is one-to-one, smooth and smoothly invertible, ##V## and ##S## are coordinate basis vectors and ##[V,S] = 0##. Since, for the levi-civita connection ##[V,S] = \nabla_V S - \nabla_S V##, we thus obtain in this region, from the definition of the Riemann tensor, that
    $$ (*) R(V,S)V := \nabla_V \nabla_S V - \nabla_S \nabla_V V - \nabla_{[V,S]} V = \nabla_V \nabla_S V = \nabla_V \nabla_V S,$$
    sometimes called the Jacobi equation (or equation of geodesic deviation).

    However, when a point where two geodesics intersect are included in the region under consideration, the map ##(v,s) \mapsto \gamma_s(v)## is no longer one-to-one and at the point of intersection ##S = 0## and does no longer serve as a coordinate chart. Furthermore, as the lie bracket measures the non-coordinaticity of the vectors,
    $$[V,S] \neq 0$$
    at ##p##. This means that the above derivation of the Jacobi equation does not hold at the intersection p.

    Questions: Jacobi fields are often defined as vectors ##S## which satisfies the Jacobi equation, and (at least in general relativity) one uses these fields to learn about points of intersection of geodesics. However, my argument shows that the jacobi equation is not derivable at such an intersection (with the standard method) as this derivation rests on the fact that ##[V,S] = 0## at p.

    1. Can the Jacobi equation be derived with another method that also holds at intersection points?
    2. Or are Jacobi fields, as solutions to the Jacobi equation not defined at intersections?
     
  6. Feb 4, 2015 #5

    Matterwave

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    I posted a very similar question in a thread a month back: https://www.physicsforums.com/threads/conjugate-points-of-a-convergence-wald-9-3.790614/ [Broken]

    But I have not received any responses to it yet. Perhaps we can merge our threads/questions?
     
    Last edited by a moderator: May 7, 2017
  7. Feb 5, 2015 #6

    lavinia

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    A Jacobi field is a solution to a differential equation on a geodesic. One does not need to have a variation through geodesics to define it. It is completely determined by its initial conditions and exists along entire geodesic..

    It is a theorem that every Jacobi field may be obtained through a variation through geodesics but its existence depends only on the differential equation and the initial conditions on the geodesic itself.
     
    Last edited: Feb 5, 2015
  8. Feb 5, 2015 #7
    Hi! So Jacobi fields cant be used to state something definite about points of intersection between geodesics?

    In Wald's book in GR he states on the first paragraph at page 224 that "Roughly speaking p and q are conjugate if an 'infinitesimally nearby' geodesic intersects ##\gamma## in p and q." However, he goes on to stress that there need not be an actual geodesic intersecting ##\gamma## in p and q.

    Can one think about Jacobi fields as stemming from the wish to be able to derive the Jacobi equation for a congruence on (singular) intersection points? Since this is not possible, one does the next best thing and focuses on a central curve and defines Jacobi fields to be the solution to this equation on that curve?

    P.S: would you happen to know the precise definition of a "deviation vector"? (See https://www.physicsforums.com/threads/what-defines-a-deviation-vector.796093/)
     
  9. Feb 5, 2015 #8

    lavinia

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    Yes they can. Two points are conjugate along a geodesic if there is a non-zero Jacobi field along it that is zero at both points. As I understand it, in Riemannain geometry at least, all Jacobi fields can be obtained through a variation through geodesics. So in order to have conjugate points,there must be a variation through geodesics whose end points are fixed. For example rotation of a sphere keeping the north and south poles fixed is a variation through geodesics (great circles intersecting the two poles). The Jacobi fields along these geodesics are all zero at the two poles. For a higher dimensional sphere one can rotate in independent directions to get n linearly independent Jacobi fields.


    This seems strange to me. For Riemannian manifolds every Jacobi field can be obtained through a variation through geodesics. This must be something to do with the Lorentz metric. Maybe null geodesics ?


    I think that Jacobi fields arise naturally from variational problems. I am not sure exactly what you are asking.
    In Riemannian geometry one looks at the path space of a Riemannian manifold and analyzes variations of a path keeping end points fixed. These variations are taken over piecewise smooth paths - not necessarily geodesics.

    The one I have studied is the variation of the "energy integral" ( the integral of the square of the length of the tangent vector to a curve.)

    The fundamental theorem is that a critical point of the energy integral is a geodesic - and visa versa. Since the energy integral is defined on paths one must make the path space into something like a manifold. A path in the path space is a variation. Jacobi fields determine the null space of the Hessian of the energy integral. This takes a bit of explaining.


    [/QUOTE]

    I don't know General Relativity but it seems like a deviation vector is the tangent to a variation through geodesics. Think of the variation as locally a two parameter mapping into the manifold where the first parameter is the time parameter of the geodesics. The derivative along the second parameter keeping the time fixed seems to be a deviation vector. Sorry I don't know this stuff yet.

    BTW:

    - Geodesics can intersect without determining a Jacobi field. For instance the equator and a polar great circle intersect in two points but there is no Jacobi field.

    Jacobi fields come from variations through geodesics.

    - Locally a geodesic minimizes length (in Riemannian geometry). Beyond a conjugate point it no longer minimizes length.

    So I would guess - and this is totally a guess - that the same idea holds in GR, a time like geodesic fails to maximize wrist watch (proper time?) time beyond a conjugate point. This makes intuitive sense to me since maximizing wrist watch time is a variational statement about the integral of the metric along time like paths.
     
    Last edited: Feb 5, 2015
  10. Feb 5, 2015 #9
    I do not see how the fact that Jacobi fields stems from variations through geodesics saves us from problems at the endpoints (where geodesics being varied intersect). For example in "Introduction to curvature" by John Lee (Chapter 10), he derives the Jacobi equation by considering
    $$\nabla_S \nabla_T T - \nabla_T\nabla_S T,$$
    where the variation is defined by a map ##\Gamma: (- \epsilon,\epsilon) \times [a,b] \to M##; where ##(s,t) \in (- \epsilon,\epsilon) \times [a,b]## (t being the parameter along the geodesic); and ##T = \tfrac{\partial}{\partial t} ##, ##S = \tfrac{\partial}{\partial s} ##. The proof relies on the "symmetry lemma" which essentially uses that
    $$\frac{\partial^2 x^\mu}{\partial s \partial t} = \frac{\partial^2 x^\mu}{\partial t \partial s}.$$

    But I do not think this holds on the points ##\Gamma(s,a) = p## and ##\Gamma(s,b)##? Or does it? I would think that the right side was zero, while I can not see why the left hand side should be.
     
    Last edited: Feb 5, 2015
  11. Feb 6, 2015 #10

    lavinia

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    I am having a little trouble understanding your question and have erased two replies.

    But perhaps this is what you were talking about. Please let me know if I am off base.

    Along a geodesic, a Jacobi field is completely determined by its initial conditions. Since it satisfies a second order linear differential equation it requires two initial conditions.

    One can state them at time zero or also at two separate times provided that these times do not occur at conjugate points along the geodesic.

    One then can create a variation though geodesics whose derivatives at the two points equal the two values of the Jacobi field at these two times. This gives you the Jacobi field between the two times but also determines it for all time. So a variation through geodesics along a segment of the geodesic will determine the Jacobi field along the entire geodesic.

    Since the geodesic is defined on a closed interval one may extend this variation - using compactness of the interval - to the entire geodesic.

    Suppose for simplicity that the only intersection points of the geodesics of this variation are at the end points and assume that these endpoints are conjugate points along the geodesic. In this case all of the geodesics in the variation intersect at the two conjugate points. In the open interval one has a vector field that satisfies the Jacobi differential equation. But this field is continuous along the entire interval so its values at the end points are determined by the values in the open interval.
     
    Last edited: Feb 6, 2015
  12. Feb 6, 2015 #11

    Matterwave

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    I think the crux of Center of Bass's problem is the same as my problem. How do we show that ##[V,S]=0## or in my notation ##[\xi,\eta]=0## at conjugate points? Is the answer in post #3 sufficient?
     
  13. Feb 6, 2015 #12

    lavinia

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    My point was that you don't need to know anything about the endpoints except that the Jacobi field converges continuously to zero at both points. Between the two points you know that the field satisfies the Jacobi differential equation. So you know that you have a Jacobi field along the whole geodesic.
     
  14. Feb 6, 2015 #13

    Matterwave

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    So that would be roughly the answer I tried in post #3 wouldn't it?
     
  15. Feb 6, 2015 #14
    Many authors derive the Jacobi equation from
    $$(*)\nabla_T \nabla_S T - \nabla_S \nabla_T T = \nabla_T \nabla_S T = R(T,S)T.$$
    We can do that at every point excluding the endpoints by the fact that ##[S,T] = 0##, indeed then
    $$\nabla_T \nabla_S T = \nabla_T \nabla_T S$$
    at these points.
    Now, considering the endpoints, can we conclude that
    $$\nabla_T \nabla_S T =0$$
    since the Jacobi field ##S=0## here (certainly the RHS of (*) is zero), and thus that the Jacobi equation is satisfied here also?
     
  16. Feb 6, 2015 #15

    lavinia

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    The Jacobi equation is satisfied at the end points because it is already determined by the variation away from the end points. Any piece of the Jacobi fields that is non-zero completely determines the rest of it. This is just a property of linear differential equations. Since the field resulting from the variation is continuous it must be a Jacobi field at the end points as well.




    '
     
  17. Feb 6, 2015 #16

    lavinia

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    At the end points the time derivative is multivalued which is what you have been pointing out. So while T exists along each geodesic it does not have a single value at the endpoints. So what Lie bracket are you taking? I think there is a way around this but you must generalize the idea of vector field.

    Start with a smooth 2 parameter mapping,##f##, from a domain,##U##, in ##R^2## into the manifold, ##M##.
    ##f## need not be one to one i.e. coordinate curves might intersect. It only needs to be smooth.

    One can define a "vector field along ##f##"as an element of ##TM_{(x,y)}## at each point ##(x,y)## in ##U##.

    Two such vector fields are ##df(∂/∂x)## and ##df(∂/∂y)## . These vector fields exist on ##U##. Call these ##∂f/∂x## and ##∂f/∂x##


    One can take covariant derivatives along the curves (x,b) and (a,y) to get new vector fields along ##f##.

    It is not hard to prove that

    ##∇_{∂f/∂x}∂f/∂y## = ##∇_{∂f/∂y}∂f/∂x ##

    if the connection is symmetric, which it is.

    So as Matterwave said, if one could extend the variation a little beyond the conjugate point, this theorem would apply. I guess you can do this just by extending the geodesics in the variation a little beyond the conjugate point.

    Tell me what you think. I am not sure that this helps.
     
    Last edited: Feb 6, 2015
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