rwsoft Posted September 23, 2008 Share Posted September 23, 2008 Hello all I'm new to these forums and hoping for a little help. I'm a software engineer and CNC machinist with a passion for electronics and physics. Currently I studying Maxwell's original treatise on Electromagnetism. I've been working on creating computer models based on his original work, but I'm having problems getting my head around the partial differentiations. Since I've been teaching myself the calculus used in his formulas, first I guess I would like to verify I'm on the right track. First I understand the Hamilton operator as follows: [math]H = -\nabla V[/math] Given the vector/quaternion position P, using the previous formula would solve as: [math]H = -\left(\frac{V}{P_{x}}, \frac{V}{P_{y}}, \frac{V}{P_{z}}\right)[/math] [math]H = -\left(i \frac{V}{P_x} + j \frac{V}{P_y} + k \frac{V}{P_z}\right)[/math] Now for an electric field: [math]E = -\nabla \phi -\frac{\partial A}{\partial t}[/math] Now I'm having serious problems understanding how the solve the [math]\frac{\partial A}{\partial t}[/math] part. Every example/text refers to partial differentiation as a formula and how to derive the resulting formula, but in the above formula I don't understand what the initial formula is to derive the resulting formula from... Is there a good tutorial/example that might explain how this works, I'm sure I just need another way of looking at this to get my head wrapped around it Thanx Link to comment Share on other sites More sharing options...
Bignose Posted September 23, 2008 Share Posted September 23, 2008 I don't understand what you mean by "solve the [math]\frac{\partial{A}}{\partial{t}}[/math] part". That part represents the change of A with respect to time, just like the gradient (which you say you understand) represents the change of something with respect to space. t is just another independent variable like x,y, and z. Link to comment Share on other sites More sharing options...
NeonBlack Posted September 24, 2008 Share Posted September 24, 2008 (edited) I also do not understand your confusion. As bignose said, t is (pretty much) no different than x y or z. Maybe it would help if you noticed that the "del" operator acting on a scalar is typically written as: [math]\nabla V = [ \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} ][/math] Edited September 24, 2008 by NeonBlack punctuation Link to comment Share on other sites More sharing options...
rwsoft Posted September 25, 2008 Author Share Posted September 25, 2008 Sorry for taking so long to respond, but I just wanted to rethink this and I think I found my source of confusion. My mathematics book explains; give the function [math]w = f(x, z) = x^2 + xz^2[/math] The partial derivation of w with respect to x is obtained when z is held constant is: [math]\frac{\partial w}{\partial x} = 2x + z^2[/math] or with respect to z: [math]\frac{\partial w}{\partial z} = 2xz[/math] So for my model I was attempting with the [math]\frac{\partial A}{\partial t}[/math] as per the text: [math]A = A(x, y, z, t?) = ?[/math] Therefore: [math]\frac{\partial A}{\partial t} = ?[/math] But as Bignose said "That part represents the change of A with respect to time", I believe I lost sight that it's just the change in A I need. Therefore, for my computer models, [math]\frac{\partial A}{\partial t} = A_{current frame} - A_{previous frame}[/math], which would be the change of A for the duration of the frame. I guess in very simple terms, a partial differentiation is a very explicit definition of, in this case, of [math]\delta A[/math]. I been a very long day and I'm very tired so I hope this is making sense... Link to comment Share on other sites More sharing options...
NeonBlack Posted September 25, 2008 Share Posted September 25, 2008 (edited) I think I understand. In principle, you can use any arbitrary A. Though for realistic models, you want to make sure that [math]\nabla \bullet A=0[/math] If you want to study a particular system, the magnetic vector potential is defined as [math]B=\nabla \times A[/math], or maybe a more useful form, for line currents is [math]A=\frac{\mu_0 I}{4 \pi} \int \frac{dl}{r}[/math] This probably wasn't as simple as you had hoped. If you want to, you can start off by considering DC currents (I know), where [math]\frac{\partial A}{\partial t}=0[/math] Edited September 25, 2008 by NeonBlack Link to comment Share on other sites More sharing options...
rwsoft Posted September 26, 2008 Author Share Posted September 26, 2008 Currently I'm mostly interested in the magnetic fields themselves, but will need all aspects of it. More or less it's known that the magnetic field, electric field and the "force field" all are perpendicular to each other. So initially the computer model is the help visualize this. But eventually there's three scenarios I need to test further: First is the easy one, AC/DC current. Here I just need to get myself back up to speed with the math again... The second scenario is the one that got me started on this. With electric dipoles (like batteries), some simple testing shows that the electric field and magnetic field are one and the same, not perpendicular, but I've been wrong before The third is the electron beam. Here, I'm sure that I will have to do my own testing, but initial research shows that the beam itself creates no magnetic field. My research maybe wrong in area or there is an explanation I've yet to discover... If these three ideas hold true, then I'm assuming that all magnetic fields are generated only by dipoles and this is really what I'm after and wish to model. This is the foundation for a little more far fetched theory, but I sure this is not the forum and probably belongs in the forum of "Wannabe Physicists and Tinfoil Hats" Thanks for all the help Link to comment Share on other sites More sharing options...
NeonBlack Posted September 26, 2008 Share Posted September 26, 2008 Then you may be interested in this previous discussion on how beams and moving charges create electric and magnetic fields. http://www.scienceforums.net/forum/showthread.php?t=34610 Link to comment Share on other sites More sharing options...
Bignose Posted September 26, 2008 Share Posted September 26, 2008 Therefore, for my computer models, [math]\frac{\partial A}{\partial t} = A_{current frame} - A_{previous frame}[/math], which would be the change of A for the duration of the frame. This opens up a whole new can of worms, because numerical integration of these equations isn't always straightforward. Consider the equation [math]\frac{\partial{A}}{\partial{t}} = R[/math] where R represents all the functions and terms on the right hand side. First, you approximate [math]\frac{\partial{A}}{\partial{t}}[/math] by [math]\frac{\Delta A}{\Delta t}[/math] Now, your choice of [math]\Delta t[/math] can be critical, and can be determined by how you evaluate R. For example, let's write [math]\Delta A = A^{n+1} - A^{n}[/math] where the superscript n stands for the time level. So, n would be the current time, and n+1 would be one time step into the future. The question becomes do you evaluate R at time n, or at time n+1? Each has it's advantages. Evaluating R at time n is easier -- you have all the values of the variables at time n, so it is straightforward. But, you will be limited to taking really small [math]{\Delta t}[/math]'s because if you take too large of steps, the solution won't be very accurate and may even become unstable (oscillating). So, the other choice is to evaluate R at time n+1. But, then because you don't have all the variables at time n+1, you will have to invert a matrix, which is very computationally expensive and time expensive. But, you can take larger [math]{\Delta t}[/math]'s. This method is also usually much more stable. The best method can also depend on whether the partial differential equation is parabolic, hyperbolic, or elliptic. To complicate things further, partial differential equations can be mixed; that is parabolic in one variable and elliptic in another (or any other mixture). Obviously, there are many more issues (and the discussion on the above can get much more in depth). But, I don't want to go too deep into it here. I'd suggest you get a good book on numerical simulation. A good one is Lapidus and Pinder's Numerical Solution of Partial Differential Equations in Science and Engineering, though there are many more out there that are good because this is an important and difficult subject. It has almost become it's own separate field. There is the research into the physics that are described by the equations and then there is the research into determining the best way of using a computer to generate solutions to those equations. The big point is that care has to be taken in choosing your method of solving these equations numerically, because you want to be absolutely sure that what the computer outputs is based on the physics of the problem, and not are artificial errors based on the method of solution. Link to comment Share on other sites More sharing options...
rwsoft Posted September 27, 2008 Author Share Posted September 27, 2008 The devil is always in the details. I'll definitely try to get my hands on that book. Also thanks for the thread. I remember reading about this in the paper "Physics Without Einstein" by Dr. Harold Aspden. He gets in depth about the electric and magnetic fields of electrons. I had forgot about this paper... Once again guys, thanks for all the help, it's much appreciated. Link to comment Share on other sites More sharing options...
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