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Taylor expansion of the square of the distance function

  1. Sep 28, 2015 #1
    Does it make a sense to define the Taylor expansion of the square of the distance function? If so, how can one compute its coefficients? I simply thought that the square of the distance function is a scalar function, so I think that one can write
    $$
    d^2(x,x_0)=d^2(x'+(x-x'),x_0)=d^2(x',x_0) + (x-x')^\mu\partial_\mu d^2(x',x_0)\\+1/2(x-x')^\mu(x-x')^\nu\nabla_\mu\partial_\nu d^2(x',x_0)+\dots
    $$
    Is this anywhere near correct?
     
  2. jcsd
  3. Sep 28, 2015 #2
    I think the Taylor expansion of any polynomial is just the original polynomial.
     
  4. Sep 28, 2015 #3
    I couldn't get your point. Could you please explain it more?! Where am I doing anything wrong?! I think that I miss some points about the distance and displacement..
     
  5. Sep 28, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    What is the definition of your distance function? The Euclidean distance is just a polynomial so the taylor expansion is trivial.
     
  6. Sep 28, 2015 #5
    I'd like to consider the taylor expansion of the square of the distance function (geodesic distance) ##d^2:M\times M\to R ## on a smooth Riemannian manifold ##(M,g)##. BTW, AFAIK, one can only define the Euclidean distance on a Euclidean space, right?!
     
  7. Oct 5, 2015 #6
    Before we can have a Taylor expansion on a manifold, it's necessary to specify a system of local coordinates in which to express this expansion. For instance, say we want to express the squared distance D(p,q) between two points on the unit sphere S2 in ℝ3. And suppose we are considering the coordinate patch that has a positive z=coordinate, and is defined by projection (X,Y,Z) → (X,Y).

    Then for any pair of points (x,y) and (u,v) in local coordinates, we can calculate their squared distance on the sphere:

    D((x,y),(u,v)) = (arccos(xu + yv + √((1-x2-y2) (1-u2-v2)))2

    Like any smooth function of finitely many variables, this has a Taylor series. But calculating it explicitly looks complicated.
     
  8. Oct 8, 2015 #7
    You are right! For this, I used normal coordinates, then wrote
    $$
    d^2(x,x_0)=d^2(x'+(x-x'),x_0)=d^2(x',x_0) + (x-x')^\mu\partial_\mu d^2(x',x_0)\\+1/2(x-x')^\mu(x-x')^\nu\nabla_\mu\partial_\nu d^2(x',x_0)+\dots
    $$
    I see your points.. but the thing is that I'm not sure the squared distance is a smooth function..
     
  9. Oct 8, 2015 #8
    Good idea to use a smart choice of coordinates. On a manifold M, I am sure that the squared distance function

    dsqq: M →

    via dsqq(p) := dist(p,q)2 from a given point q is (unlike the unsquared distance) a smooth function in some neighborhood of q.

    Which is just the condition we need for dsqq to have a Taylor series at q.

    I will try to come up with a proof later.

    But, the squared distance function is never differentiable at every point p of any compact manifold. (E.g., consider a circle of circumference = 1 and graph its squared distance function. When p is at the maximum distance = 1/2 from q, the function dsqq is clearly not differentiable there.)

    (It is a difficult theorem that, if there exists some point q such that a manifold's squared distance function dsqq is differentiable at all points of the manifold, then the manifold is diffeomorphic to Euclidean space n.)
     
    Last edited: Oct 8, 2015
  10. Oct 9, 2015 #9
    As promised, here is a proof that the squared distance function dsqq (notation as above) on a manifold M is smooth in some neighborhood of the point q:

    The exponential map in differential geometry:

    expq: TqM → M​

    taking any vector v in some neighborhood U of the origin in the tangent space TqM at the point q, to the manifold M, via

    expq(v) := γv(1),

    where γv denotes the unique distance-minimizing constant-speed geodesic curve with γv(0) = q, whose velocity γ'v(0) at time equals v.* (For an arbitrarily large neighborhood, there is usually no unique distance minimizing curve.)

    It is a standard fact in differential geometry that this exponential map is a diffeomorphism from the open neighborhood of the origin U of TqM, onto its image γv(U), which we call W. W will be an open set of M. For simpler notation, let's call this diffeomorphism h:

    h: U → W​

    Every diffeomorphism has an inverse diffeomorphism and h is no exception:

    hinv: W → U.​

    Also, the map

    lensq: U →
    via

    lensq(v) := ||v||2
    is easily shown to be smooth. Finally, the composition

    lensq hinv: W →

    gives the distance of any point in the neighborhood W (of q in M) to the point q, and since it is the composition of two smooth maps, must be smooth itself.
    __________________
    * If all geodesics can be extended indefinitely in M, then M is called a complete Riemannian manifold. In this case, the exponential map is then defined for all real multiples of v via

    expq(tv) := γv(t)​

    for any t in , where γv is the same uniquely defined geodesic curve as above. In this case, the exponential map expq will be onto the manifold M, but in most cases will not be a diffeomorphism TqM → M.
     
  11. Oct 11, 2015 #10

    Ssnow

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    Gold Member

    Assuming that ## M## is a Riemannian manifold, fixing a local coordinate system (a normal coordinate system, a preferred coordinate system...) and using the exponential map it is possible to find the expansion of the distance that is a polynomial with coefficients that probably depends by Riemannian invariants of the manifold ## M## as the Riemann scalar curvature ## R ## the Ricci tensor ## R_{ij} ## or the Riemann tensor ##R_{ijkl}##...
     
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