rakuenso Posted May 3, 2009 Share Posted May 3, 2009 Hi, Suppose I have two points A, and B in some N-dimensional manifold. I would like to define distance (or metric) between A and B as follows: [math]d(A,B)=min | \int_\gamma f(s) ds | [/math] along some curve [math]\gamma[/math] In my head, it seems to be possible to evaluate a line integral without precisely defining distance. (ie. we just take tiny little points along this line, and evaluate f(s) at each tiny little point). However, all equations I've seen uses something like: [math]ds=\sqrt{{dx}^2+{dy^2}}[/math] which seems like a euclidean metric. So here are my ultimate questions, 1) is it possible to evaluate such a line integral without defining a metric, and 2) would make it sense to define such a line integral Link to comment Share on other sites More sharing options...

Killjoy Posted May 3, 2009 Share Posted May 3, 2009 Hi, Suppose I have two points A, and B in some N-dimensional manifold. I would like to define distance (or metric) between A and B as follows: [math]d(A,B)=min | \int_\gamma f(s) ds | [/math] along some curve [math]\gamma[/math] In my head, it seems to be possible to evaluate a line integral without precisely defining distance. (ie. we just take tiny little points along this line, and evaluate f(s) at each tiny little point). However, all equations I've seen uses something like: [math]ds=\sqrt{{dx}^2+{dy^2}}[/math] which seems like a euclidean metric. So here are my ultimate questions, 1) is it possible to evaluate such a line integral without defining a metric, and 2) would make it sense to define such a line integral I think you would need a metric to define ds in your first definition, which would then become circular. Link to comment Share on other sites More sharing options...

ajb Posted May 3, 2009 Share Posted May 3, 2009 You don't need a metric to define line integrals, you need a one-dimensional volume or a one-form. You need an invariant notion of integration. So, let us take [math]\omega|_{U} = \omega_{A}(x)dx^{A}[/math] with [math]U \subset M[/math] with local coordinates [math]\{x^{A}\}[/math] and [math]\gamma :[a,b] \rightarrow U[/math] where I will use the notation [math]\gamma^{*}x^{A} = x^{A}(s)[/math] where [math]s[/math] is the coordinate on the interval. Then [math]\int_{\gamma} \omega := \int_{a}^{b}\left( \omega_{A}(x(s))\frac{dx^{A}(s)}{ds} \right)ds. [/math] In practice you may need more than one coordinate patch. The integral is also independent of how you pick the coordinate patch(s). Note no metrics. You will need a metric if you want to have a notion of distance. In your original post I am not sure what f is or what norm you are using. Norms and metric are just about the same things in this context. Link to comment Share on other sites More sharing options...

rakuenso Posted May 4, 2009 Author Share Posted May 4, 2009 You don't need a metric to define line integrals, you need a one-dimensional volume or a one-form. You need an invariant notion of integration. So, let us take [math]\omega|_{U} = \omega_{A}(x)dx^{A}[/math] with [math]U \subset M[/math] with local coordinates [math]\{x^{A}\}[/math] and [math]\gamma :[a,b] \rightarrow U[/math] where I will use the notation [math]\gamma^{*}x^{A} = x^{A}(s)[/math] where [math]s[/math] is the coordinate on the interval. Then [math]\int_{\gamma} \omega := \int_{a}^{b}\left( \omega_{A}(x(s))\frac{dx^{A}(s)}{ds} \right)ds. [/math] In practice you may need more than one coordinate patch. The integral is also independent of how you pick the coordinate patch(s). Note no metrics. You will need a metric if you want to have a notion of distance. In your original post I am not sure what f is or what norm you are using. Norms and metric are just about the same things in this context. I wanted to define the metric function [math]d(A,B)[/math] using the aforementioned line integral but I was afraid of a circular argument like Killjoy said. For simplicity we can say f is the potential energy V, which is only dependent solely on some point in the configuration (Q) manifold. (Eventually I will redefine f on QP space as T-V) Link to comment Share on other sites More sharing options...

ajb Posted May 4, 2009 Share Posted May 4, 2009 The metric comes in when you define the norm. Link to comment Share on other sites More sharing options...

rakuenso Posted May 4, 2009 Author Share Posted May 4, 2009 Ｉ thought the space becomes a metric space whenever you arbitrarily define a function d that satisfies the axioms: 1. d(x, y) ≥ 0 (non-negativity) 2. d(x, y) = 0 if and only if x = y (identity of indiscernibles) 3. d(x, y) = d(y, x) (symmetry) 4. d(x, z) ≤ d(x, y) + d(y, z) (triangle inequality). My definition above seems to satisfy all 4. Link to comment Share on other sites More sharing options...

ajb Posted May 4, 2009 Share Posted May 4, 2009 Sorry, what I meant was that your definition requires a metric/norm viz your [math]||[/math]. What you can do on a Riemannian manifold is define the length of a curve viz [math]L(\gamma) = \int \sqrt{g_{AB} \frac{\partial x^{B}}{\partial s} \frac{\partial x^{A}}{\partial s}} ds[/math], (with the usual abuses of notation). Then, as you suggest you can equip the manifold itself (and not just give a norm on the tangent planes) with a metric space sructure. [math]d(a,b) = \inf L(\gamma)[/math] where we vary the paths connecting [math]a[/math] and [math]b[/math]. The important thing here is that I have use the Riemannian metric on [math]M[/math] (a map between the tangent and cotangent spaces at a point) to give a metric structure to the points on the manifold. I don't think the two structures should be seen as independent. This is my point/question about the norm you use. Link to comment Share on other sites More sharing options...

rakuenso Posted May 5, 2009 Author Share Posted May 5, 2009 (edited) Hmm by || i meant the absolute value, not the norm. Are they equivalent? Merged post follows: Consecutive posts mergedHmm by || i meant the absolute value, not the norm. Are they equivalent? Because in this case, my norm/length is defined as exactly: [math] ||x|| = \int_\gamma f(s) ds [/math] where x is a vector connecting two points.. edit: ajb, I think I'm starting to see what you mean =) Edited May 5, 2009 by rakuenso Consecutive posts merged. Link to comment Share on other sites More sharing options...

ajb Posted May 5, 2009 Share Posted May 5, 2009 Because in this case, my norm/length is defined as exactly: [math] ||x|| = \int_\gamma f(s) ds [/math] where x is a vector connecting two points.. I am not quite sure in your construction how you go from [math]X \in Vect(M)[/math] to [math]f \in C^{\infty}(M) (?)[/math], but yes I think that will involve a norm/metric. (In the way I presented we have [math]X^{A} = \frac{d}{ds} = \frac{\partial x^{A}(s)}{\partial s}[/math], again with all the abuse of notation. Then use the metric to give a function as [math]X\cdot X[/math]). The way one informally thinks of defining a length of a curve is to approximate it by lots of little arrows (vectors). You then add up the lengths of all these little vectors. But what is the length of the vectors? This is where the metric comes in. It allows you to define a length of a vector. With out the metric vectors have no length. This is then use to define the length of a curve. I don't know of anyway round this. Link to comment Share on other sites More sharing options...

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