rakuenso Posted April 22, 2009 Posted April 22, 2009 (edited) I have a surface Z, which is a function of the variables x1,x2,x3... etc. ie. Z(x1,x2,x3...) I have a point Z0 and a point Z1 which corresponds to some point on this surface. There is some line M that connects Z0 and Z1 on the Z surface, note that M does NOT have to be the shortest distance. However, M must be bound to the surface Z. How do I find the length of the line M on this surface? Ideally the expression would be somehow linked to the directional derivative, IE. I've been thinking of slicing the line M into tiny little components, and taking the directional directive at each point of M, the integrating with respect to.. something (maybe dx1, dx2, dx3... Any tips? Some recommend arc length, others recommended differential manifolds (Which I have NO IDEA) Edited April 22, 2009 by rakuenso
ajb Posted April 22, 2009 Posted April 22, 2009 My initial thought is to consider the induced metric and the corresponding geodesic equation. That is, I am thinking of this problem in terms on manifolds, which I feel you are trying to avoid. By length, you are assuming some metric. What metric? Where does this metric come from? 1
Bignose Posted April 22, 2009 Posted April 22, 2009 My initial thought is to consider the induced metric and the corresponding geodesic equation. This was exactly my thought, too. The reference I'd go to would be Synge & Schild Tensor Calculus. It is an "old-fashioned" tensor calculus text, but it develops right up to this exact problem, and if you know calculus and differential equations, you have the necessary prereqs for the book (I've found you need more prereqs for some of the more modern tensor calculus books). 1
rakuenso Posted April 22, 2009 Author Posted April 22, 2009 (edited) Thanks for the book advice, I haven't quite figured out what the metric space is yet, so I'm leaving it pretty generally. I'll probably be working on some connected Riemmann manifold Edited April 23, 2009 by rakuenso
ajb Posted April 23, 2009 Posted April 23, 2009 Thanks for the book advice, I haven't quite figured out what the metric space is yet, so I'm leaving it pretty generally. I'll probably be working on some connected Riemmann manifold If the space your surface is embedded in is [math]\mathbb{R}^{n}[/math] then it comes with the canonical Euclidean metric. You can then pull this back to the surface in question as you have the embeddings. If the space is a more general Riemannian manifold then recipe of pulling back the metric will still work.
rakuenso Posted April 23, 2009 Author Posted April 23, 2009 (edited) Ah, thanks. Unfortunately it doesnt look very euclidean at this point. I suppose physically x1,x2..xn represents a certain configuration of a system, and Z is the energy of that configuration. So the difference between x1,x2,...,xn and y1,y2....yn would be physically interpreted as the change in the configuration. While the difference in z0=z(x1,x2,...,xn) and z1=z(y1,y2...yn) would be the difference in energies of the configurations. The surface thus doesn't seem Euclidean Edited April 23, 2009 by rakuenso
ajb Posted April 23, 2009 Posted April 23, 2009 I don't fully follow, but are you thinking of surfaces of constant energy for a Hamiltonian system? I don't think that configuration spaces come with a metric generally.
rakuenso Posted April 23, 2009 Author Posted April 23, 2009 (edited) Well, I'm thinking of surfaces of only potential energy, and yes, I am precisely trying to find the metric the metric configuration space (or if doesn't come predefined with one, find something analagous) If I understand correctly (from what I've read about tensor calculus in a night) If we let the set [math]\left\{x^1,x^2,...,x^N\right\}[/math] as a point that represents a certain configuration, with [math]x^1,x^2,...,x^N[/math] variables themselves being the coordinates in configuration space a path (or curve) in configuration space can represented as: [math]x^r=f^r(t)[/math] where [math] r=(1,2,...N) [/math] if we add the dimensionality of energy, ie, we form the energy-configuration space, which is an N+1 dimension space, then a point in energy-configuration space becomes the set [math]\left\{x^1,x^2,...,x^N,x^z\right\} \; where \; x^z[/math] is the newly added dimension Q #1: Is an energy surface Z represented in this energy-configuration space, [math]u^i[/math] some parameter, written as: [math]x^s=f^s(u^1,u^2,...u^r)[/math] where [math] r=(1,2,...N) \; \; and \; \; s=(1,2,....N,z) [/math] ? eventually we'd be able to eliminate the parameters and write [math] x^z = g(x^1,x^2,...,x^N) [/math], forming a surface. What would these parameters [math]u^i[/math] be? Q #2: how would you represent the curve in the energy-configuration space bound to the surface Z? Would it be [math] x^s=f^s(t) [/math] where [math] s=(1,2,....N,z) [/math] still? this seems like the unbounded form, ie, there's nothing to say it has to be on the surface? How would I express it in a general form, but require it to be on the surface? I'm planning to add two more dimensions to this energy-configuration space later EDIT: Is it possible to find the "length" of a line connecting two points in space without defining the metric? Edited April 24, 2009 by rakuenso
ajb Posted April 24, 2009 Posted April 24, 2009 This sounds very much like (but not identical to) nonautonomous mechanics. Here the configuration space is [math]\{q,p\}[/math] is extended to include energy and time. We need both energy and time as we want to have a symplectic structure. I know people have tried slightly different things, I have an old book from the 60's that talks about various configuration and state spaces. I'll give you the title when I find, but I doubt you will be able to get hold of a copy. I suggest you have a look at the books by Abraham and Mardsen as well as Arnold on classical mechanics. On your final question, you need a metric to define length. But what you could do is define symplectic length by using a symplectic structure instead of a metric.
rakuenso Posted April 24, 2009 Author Posted April 24, 2009 (edited) nm just got your message, QTPH looks like a 6N+2 dimensional space Edited April 24, 2009 by rakuenso
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