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Inner product in curvilinear coordinates

  1. May 11, 2013 #1
    Hello,

    let's assume we have an admissible change of coordinates [itex]\phi:U\rightarrow \mathbb{R}^n[/itex]. I would like to know how the inner product on ℝn changes under this transformation. In other words, what is [itex]\left\langle \phi (u), \phi (v) \right\rangle[/itex] for some [itex]u,v \in U[/itex] ?

    I thought that if we consider the two corresponding vectors [itex]x(u)[/itex] and [itex]y(v)[/itex] where [itex]x(u) = \phi_1(u)e_1 + \ldots \phi_n(u)e_n[/itex] we obtain: [tex]\left\langle x,y \right\rangle = \left\langle \phi (u), \phi (v) \right\rangle = \left\langle \phi_1(u)\phi_1(v) + \ldots \phi_n(u)\phi_n(v) \right\rangle[/tex] but for some reason (that I would like to know) this does not seem to be the usual approach.
     
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  3. May 11, 2013 #2

    quasar987

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    What one usually considers (i.e. the notion that is relevant) is that of a field of inner products; i.e. an inner product on each tangent space of R^n, and then we ask what is the corresponding field of inner products on U if we identify U with R^n via [itex]\phi[/itex]. The tangent space TpU of U at a point p is just a copy of R^n "attached" to p (i.e. with the origin at p). So rigorously, that is TpU = {p} x R^n. And similarly for Tx(R^n). If the field of inner product on R^n you are interested in is the standard euclidiean inner product < , >, then the corresponding inner product on the tangent space TpU at the point p of U is [itex]\mathbb{R}^n \times \mathbb{R}^n\rightarrow \mathbb{R} : (V,V')\mapsto \left<(D\phi)_p(V),(D\phi)_p(V')\right>[/itex].

    The basic idea behind all this abstract nonsense is just this: [itex]\phi[/itex] puts the points of U in bijection with the points of R^n. In particular, for p a point of U and c, c' two paths in U passing through p, we ask what is the angle between the correspond curves in R^n (i.e. [itex]\phi\circ c[/itex] and [itex]\phi\circ c'[/itex])? Or somewhat more generally, what is the inner product of their tangent vectors at [itex]x=\phi(p)[/itex] knowing that the tangent vectors of c,c' at p are V, V' ? The answer is [itex]\left<(D\phi)_p(V),(D\phi)_p(V')\right>[/itex].
     
  4. May 12, 2013 #3
    Hi Quasar. Thanks a lot.
    I am still not very familiar with the concept of tangent space (though I know what it is) and I don't understand your notation with the big D.

    From your explanation however I think I got one important point: when we have a change of coordinates phi, for each point u we can obtain n curves (the coordinate curves) whose tangent vectors at phi(u) form a (local) basis for R^n. These are the column vectors of the Jacobian matrix J.

    The points at phi(u) and phi(v) can thus be represented as Ju and Jv, thus we have <phi(u), phi(v)>=<Ju,Jv>.

    Am I right?
     
  5. May 12, 2013 #4

    quasar987

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    D is for derivative. So my [itex]D\phi[/itex] is your J.

    This is correct.


    This is incorrect.
     
  6. May 12, 2013 #5
    Yes, I realized too late it was incorrect.
    I was basically trying to interpret (a bit too quickly) an old post I received here.

    I am still a bit puzzled about two things:

    1) The last formula you wrote is still a bit obscure to me. Even if I replace [itex]D\phi[/itex] with [itex]J[/itex] it is unclear how one could derive concretely an explicit formula for the inner product. I suspect the source of confusion lies in those V, V' as argument of a Jacobian (?!)

    2) In this particular problem, why should we complicate our lives by resorting to esoteric abstract concepts, when the problem is easily solvable with the formula I gave in the first post. It is not a critic to your elegant solution, but I am just missing what is the advantage of rewriting the formula for the inner product in terms of Jacobians of a transformation.

    Thanks.
     
    Last edited: May 12, 2013
  7. May 12, 2013 #6

    quasar987

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    You ask in the OP "how does the inner product changes under a change of coordinates"? Then you propose a formula and wonder why it isn't the "usual way to do it". I explained what the "usual way" to do it is. This boils down to what exactly one means by changes.

    Namely, we are interested (and by that I mean, in questions pertaining to differential geometry) in the situation where we have a point p of U and c, c' two paths in U passing through p, and we ask what is the inner product of the tangent vectors to the curves ϕ∘c and ϕ∘c' at x=ϕ(p) knowing that the tangent vectors of c,c' at p are V, V'? The answer to this question is [itex]\left<(D\phi)_p(V),(D\phi)_p(V')\right>[/itex] and we say that the inner product

    [tex](V,V') :=\left<(D\phi)_p(V),(D\phi)_p(V')\right>[/tex]

    is how the inner product < , > changes under the coordinate change ϕ:U-->R^n
     
  8. May 12, 2013 #7

    Office_Shredder

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    mnb, you seem to have a conceptual block that is very common for this type of thing. Rn is several things:
    A vector space
    A manifold which has Rn as a tangent space at every point
    Its own dual space


    And often these ideas are used with only the surrounding context describing what is happening. If you have an inner product on Rn, you could be referring to that as an inner product on a vector space, or an inner product on a manifold. When you ask how an inner product changes under a change of coordinates, it's not clear which type of inner product is being referred to. If the change of coordinates is not a linear change of coordinates however, it is assumed that you want to think of Rn as a manifold, not as a vector space. (because vector spaces transforming under nonlinear functions have basically no mathematics available to them)

    So you can't use the points in Rn that you are plugging into [itex]\phi[/itex] and getting out of it in your inner product, because you are treating them as elements of a manifold, not as elements of a vector space.
     
  9. May 13, 2013 #8
    Hi Office_Shredder,

    you are right: I probably have a conceptual block (or more) that is making it difficult to understand these things. The more we go on, the more confused I feel. I am sorry to be boring, but please allow me to start again from scratch.

    Let's go step-by-step.
    Let's assume that we have now a linear function [itex]\phi:R^n\rightarrow R^n[/itex]
    I will make a sequence of independent and logically disconnected claims, and hopefully we'll see where the conceptual block is.

    1) [itex]\phi[/itex] is linear from ℝn to ℝn, so we can write [itex]\phi(u)[/itex] as [itex]F\mathbf{u}[/itex] where F is the matrix associated with [itex]\phi[/itex]

    2) [itex]\phi[/itex] describes a change of coordinates.

    3) [itex]\phi[/itex] can be also interpreted as a (regular) coordinate patch for the manifold [itex]R^n[/itex].

    4) it is not allowed to write things like [itex]<F\mathbf{u},F\mathbf{v}>[/itex] because F is a function that maps point from a coordinate-patch into points of an abstract manifold, but we have not given that manifold any metric yet.

    5) If we give a metric to the manifold ℝn we obtain a Riemannian manifold.

    6) A metric is a (smooth) function that defines a different inner product for each location of the manifold.

    7) A common (but not necessarily the unique) way to assign a metric is to use the function [itex]\phi[/itex]. For instance, a metric can be obtained in our case by [itex]F^T F[/itex].

    8) The metric tensor FTF is obtained by assigning at each location the dot product between all the pairs of column vectors in the Jacobian matrix (they are tangent vectors belonging to the tangent space at a particular location).

    9) F is linear, thus the metric [itex]F^T F[/itex] does not depend on the position in the manifold: it is constant. In fact, the Jacobian matrix in this case is F itself: J=F.

    10) POSSIBLE CONCEPTUAL BLOCK. Now our inner product, although constant, is defined only locally. In principle it should not be allowed to take "global" inner products like [itex]<F\mathbf{u},F\mathbf{v}>[/itex] because Fu and Fv are vectors pointing to different positions of the manifold...on the other hand everybody knows that this is indeed possible :confused: ... why?


    ...Let's stop here for the moment. If I receive help at this stage I'll try to proceed to the more general case of [itex]\phi[/itex] nonlinear.
     
  10. May 13, 2013 #9

    WannabeNewton

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    Parallel transport of vectors from tangent space to tangent space in ##\mathbb{R}^{n}##, with respect to a global cartesian coordinate system, is trivial in the sense that a vector has the same components at every tangent space in the cartesian coordinate basis. Therefore, the tangent space it is attached to has no real significance given the above coordinate representation. In such a case, we can just extend the inner product to all of ##\mathbb{R}^{n}##.
     
  11. May 14, 2013 #10
    Hi WannabeNewton,

    thanks for participating to this thread!
    After reading the answer you gave and after re-reading the first post of quasar, it seems to me that if we have a non-linear coordinate patch [itex]\phi:U\rightarrow \mathbb{R}^n[/itex] that maps coordinates from an open set U into points of the manifold [itex]\mathbb{R}^n[/itex], and if we assign a field of inner products on each tangent space of ℝn, in general it does not make any sense to write anything like [itex]<\phi(u), \; \phi(v)>[/itex] for any u,v[itex]\in[/itex]U. Am I right?

    In such a scenario, the only thing that makes sense, as quasar pointed out, is to calculate the angle (or inner product) of the tangent vectors of two curves passing through the same point.

    When [itex]\phi:U\equiv\mathbb{R}^n\rightarrow \mathbb{R}^n[/itex] is linear we have the peculiar (very) special case in which calculating the angle between the tangents of two curves at their point p of intersection is exactly the same as calculating the angle of the two tangent vectors "transported" to the origin, because the field of inner products is constant across ℝn.

    If this is right, then (almost) everything should be clear.
     
    Last edited: May 14, 2013
  12. May 14, 2013 #11

    quasar987

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    I would say that it makes sense in that [itex]<\phi(u), \; \phi(v)>[/itex] is a well-defined quantity. It's just not a very relevant quantity (again, for question pertaining to differential geometry). Notice however, that when you switch to manifolds, i.e. in the situation where [itex]\phi[/itex] is a map M-->N btw two manifolds (or more generally btw two open subsets of M and N), and there is a riemannian metric < , > on N, then the expression [itex]<\phi(u), \; \phi(v)>[/itex] no longer makes any sense since [itex]\phi(u) [/itex] and [itex]\phi(v)[/itex] are points of N while < , > eats elements of TpN. (Correctly interpreted, [itex]\left<(D\phi)_p(V),(D\phi)_p(V')\right>[/itex] still makes sense though.)

    Yes, this is right and very well put!
     
  13. May 14, 2013 #12
    When I said that it does not make sense I was implicitly assuming that we have already assigned a position-dependent metric to ℝn. In such a context it seems that we agree that the expression does not make sense.
    Surely, if we decide to assign the standard Euclidean metric to ℝn, then that quantity is still well-defined.

    If what I am saying is correct, then you have removed another conceptual block of mine. Namely, until yesterday I was assuming that the function [itex]\phi[/itex] automatically "creates/imposes" a metric on ℝn. I hope this was just a great misconception because from what I understood, we are pretty much free to assign any metric that we like to ℝn.




    And this almost where I wanted to head with my original question, but I was unable to formulate it properly. I wanted to know what happens in the following scenario:


    *) ℝn is a manifold equipped with the standard Euclidean inner product.

    **) I consider the quantity <x,y> for some x,y[itex]\in[/itex]ℝn.

    ***) I introduce a coordinate patch [itex]\phi:U \rightarrow \mathbb{R}^n[/itex] (for example [itex]\phi[/itex] is the mapping from polar to Cartesian coordinates).

    ****) I assign the new position-dependent metric JTJ to ℝn, where J is the Jacobian associated with [itex]\phi[/itex].

    *****) QUESTION: Is it possible to express <x(u),y(v)> in terms of the new metric tensor?


    If [itex]\phi[/itex] is linear the answer is yes: <x(u),y(v)> = [itex]\mathbf{x}(\mathbf{u})^T (J^T J) \mathbf{y}(\mathbf{v})[/itex] because J does not depend on the position.
    But what if [itex]\phi[/itex] is non-linear (e.g. polar to cartesian mapping)?
     
  14. May 14, 2013 #13

    quasar987

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    You realize this is a horrendously unnatural thing to consider? If I understand you properly, you want to consider the field of metric on R^n whose value at a point p of R^n is [itex](V,V')\mapsto V^TJ_{\phi^{-1}(p)}^TJ_{\phi^{-1}(p)}V'[/itex].

    Again I feel compelled to point out that this expression is well-defined only because R^n is a vector space and U is an open subset of R^n and hence all the tangent spaces are naturally identified with one another. At the level of manifolds, there is no analog of your "construction" because [itex]J_{\phi^{-1}(p)}[/itex] is a map [itex]T_{\phi^{-1}(p)M}\rightarrow T_pN[/itex] and [itex]V,V'\in T_pN[/itex]

    By x(u) do you mean Ju?
     
  15. May 14, 2013 #14
    Yes ... unfortunately it arises in a problem I am facing, so I can't really avoid it. It is not something I am trying out of curiosity.



    I think we totally agree on this!




    Uhm...I am not completely sure about this.
    However I just remember that some time ago I found from a book of tensor analysis a generalized formula for the inner product involving the components of the metric tensor. The formula can be found here and I am wondering if we can use it in this particular case of [itex]\phi : U\rightarrow \mathbb{R}^n[/itex] ...
     
  16. May 14, 2013 #15

    quasar987

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    A metric tensor and an inner product are two names for the same thing. The formula on wiki is just the familiar [itex]<U,V> = |U||V|\cos\theta [/itex].
     
  17. May 15, 2013 #16
    Yes.
    Thanks for your help. I think the original question has been answered satisfactorily.
    I would have an additional question related to the problem, but for the sake of clarity I think we should end this thread to avoid potential confusion to other readers.
     
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