rakuenso

Has anyone experimentally obtained a picture like this: http://en.wikipedia.org/wiki/File:HAtomOrbitals.png If so, where? If not, how do we actually know it's correct?

eh? I've done calc II and we went over quite a bit of vector calculus. But all the stuff we did dealt with flows, and integrals of dot products (which is just a scalar), and stuff like stoke's theorem. here it seems like I'm integrating an actual vector as opposed to a dot product. So the problem is analagous to finding the integral curves for a vector field (which here is the gradient) using ODEs? nm I think I get what you mean now, if my vector field is [math]< \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} >[/math], say [math]V=\frac{1}{|x-y|} [/math] then I would form the autonomous system, as it is independent of time: [math] \frac{dx}{dt} = \frac{\partial V}{\partial x} = \frac{-1}{|x-y|(x-y)} [/math] [math] \frac{dy}{dt} = \frac{\partial V}{\partial y} = \frac{1}{|x-y|(x-y)} [/math] so I would have to look for an integral curve with components (x(t), y(t)) that obeys the above equations

Are [math]x^A[/math] my variables? like [math]x=x^1,y=x^2[/math] etc. And what erxactly is [math]X^A[/math]? How do you arbitrarily parametrize [math]x^A[/math] wrt to t? Or is this [math]x^A(t)[/math] exactly what I'm trying to solve? I also have difficulties seeing where the integration along a "vector" part comes in. For two dimensions, so far I've written the [math]n[/math] th point [math](x_n, y_n)[/math] as the following, over some interval tiny interval [math]t[/math] (in the case of the unit vector it is equal to the magnitude of the vector)": [math] x_n = x_0 + \frac{1}{t} [\frac{\partial f}{\partial x}(x_0,y_0) + \frac{\partial f}{\partial x}(x_1,y_1) + ... + \frac{\partial f}{\partial x}(x_n,y_n) ] [/math] [math] y_n = y_0 + \frac{1}{t} [\frac{\partial f}{\partial y}(x_0,y_0) + \frac{\partial f}{\partial y}(x_1,y_1) + ... + \frac{\partial f}{\partial y}(x_n,y_n) ] [/math] note that [math](x_1,y_1), (x_2,y_2) .. etc [/math] can be found using the above definition The path then would be comprised of the points [math] (x_0,y_0), (x_1,y_1) ... (x_n, y_n) [/math], which when taken over the [math]\displaystyle\lim_{n\to\infty}\displaystyle\lim_{t\to\infty}[/math] becomes continuous and never ending. Surely, given some function [math] f(x,y) [/math] this path can be traced out using numerical methods, but a method of finding an analytic solution seems much more difficult.

Suppose I have a function [math]f(x,y)[/math], along with a gradient at some arbitrary point [math](x_0, y_0)[/math] The surface is smooth and [math]C^1[/math], there thus also exists some gradient vector: [math]<\frac{df}{dx}(x_0, y_0) , \frac{df}{dy}(x_0, y_0)>[/math] I want to "follow" this vector to the next point [math](x_1, y_1)[/math], here there is another gradient vector [math]<\frac{df}{dx}(x_1, y_1) , \frac{df}{dy}(x_1, y_1)>[/math], I repeat this process for up to [math](x_n, y_n)[/math]. How do I write an expression for the curve? It seems like I would have to integrate along a vector somehow?

Bignose, under what conditions is [math]\frac{d}{dt}\frac{dx}{dx}=\frac{d}{dx}\frac{dx}{dt}[/math]? I tried to do some examples, say: [math] x = t^2 [/math] [math] \frac{dx}{dt} = 2t [/math] [math]\Rightarrow[/math] [math] \frac{d}{dx}\frac{dx}{dt} = \frac{1}{\sqrt{x}}[/math] since [math] \sqrt{x}=t [/math] However, [math] \frac{d}{dt}\frac{dx}{dx} = 0 [/math] So [math] \frac{d}{dt}\frac{dx}{dx}\neq \frac{d}{dx}\frac{dx}{dt} [/math]

Hi I have a function f: [math]f(x(t))=\frac{d(x(t))}{dt}+x(t)[/math] Now how would I differentiate f with respect to x

Hmm by || i meant the absolute value, not the norm. Are they equivalent? Merged post follows: Consecutive posts merged 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 =)

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.

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)

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

Is there a name for a family of parametric curves [math] x(t) , y(t) [/math]where... [math](\frac{dx}{dt})^2+(\frac{dy}{dt})^2=1[/math]? If possible, generalized to multiple dimensions... [math](\frac{dx_1}{dt})^2+(\frac{dx_1}{dt})^2+...+(\frac{dx_n}{dt})^2=1[/math]

nm just got your message, QTPH looks like a 6N+2 dimensional space

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?

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

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