It looks like you're using an Ad Blocker.
Please white-list or disable AboveTopSecret.com in your ad-blocking tool.
Thank you.
Some features of ATS will be disabled while you continue to use an ad-blocker.
originally posted by: delbertlarson
Equation (28) can likely be treated with numeric methods, but I am not expert on such, and would therefore appreciate any pointers to references that might help.
originally posted by: moebius
a reply to: delbertlarson
Well for what it is worth, setting the magnetic potential to zero will give you a second-order ode.
Mathematica even produces a solution, although not a pretty one.
Link:
www.wolframalpha.com/input/?i=DSolve%5By%27%27%5Br%5D+%2B+2+%2F+r+*+y%27%5Br%5D+%3D%3D+-((a+-+b+%2F+r)%5E2+-+c)+*+y%5Br%5D,+y%5Br%5D,+r%5D
originally posted by: Rosinitiate
I don't believe anything actually has a "hard core".
originally posted by: hubrisinxs
a reply to: delbertlarson
Here is a Link to a pdf on how to solve partial differential equations It is from Cornell university, I think and; hopefully, it might give you an idea on how to treat it though plotting it on something like Mathematica or Maple then finding numerical solutions.
Hope I was in anyway helpful, and good luck with you research.
originally posted by: AMPTAH
originally posted by: delbertlarson
Equation (28) can likely be treated with numeric methods, but I am not expert on such, and would therefore appreciate any pointers to references that might help.
Your equation is too complicated.
This means you can't get any useful "insights" into the nature of the underlying phenomena.
You can solve all these complicated equations using numerical methods, if you have "boundary conditions" for some region of space specified. You can even use "FEA" Finite Element Analysis, otherwise called "FEM" Finite Element Methods, to get a picture of the region of space governed by your complicated equation. Again, you have to specify boundary conditions and/or initial conditions.
But, at the end of the day, Physics is about trying to find the "simple" explanations for phenomena, because, otherwise, we get lost in the complexity of the mathematics, and nothing useful can be obtained from the effort.
originally posted by: moebius
a reply to: delbertlarson
Well for what it is worth, setting the magnetic potential to zero will give you a second-order ode.
Mathematica even produces a solution, although not a pretty one.
Link:
www.wolframalpha.com/input/?i=DSolve%5By%27%27%5Br%5D+%2B+2+%2F+r+*+y%27%5Br%5D+%3D%3D+-((a+-+b+%2F+r)%5E2+-+c)+*+y%5Br%5D,+y%5Br%5D,+r%5D
originally posted by: AMPTAH
originally posted by: moebius
a reply to: delbertlarson
Well for what it is worth, setting the magnetic potential to zero will give you a second-order ode.
Mathematica even produces a solution, although not a pretty one.
Link:
www.wolframalpha.com/input/?i=DSolve%5By%27%27%5Br%5D+%2B+2+%2F+r+*+y%27%5Br%5D+%3D%3D+-((a+-+b+%2F+r)%5E2+-+c)+*+y%5Br%5D,+y%5Br%5D,+r%5D
The standard procedure, if you have a closed solution for part of the problem, is to use "perturbation" theory to calculate the desired parameters for the full problem. So, put a small scaling parameter in front of the magnetic potential and treat it as a small perturbation to the system, to calculate what you want. Maybe you'll get somewhere.
originally posted by: micpsi
The mathematics is totally wrong. One cannot assume the relativistic wave function is a plane wave (equ. 6), work out what equation it satisfies and then claim that the equation is the general relativistic wave equation of a spinless particle! The general solution of a scalar wave equation with Lorentz invariance that describes a spinless particle moving in a potential field V is NOT a plane wave. That's the form of a FREE relativistic particle moving in the ABSENCE of an interaction potential V. It is mathematically wrong to work back from a particular solution in which V = 0 to an equation that is supposed to hold in the general case where V ≠ 0. The mathematical howlers in the analysis would not get past any physical journal editor. What has been misinderstood is that you can use equ. 1 with E^2 = p^2c^2 + (mc^2)^2 and equ. 6 only when the particle is free (V = 0). The relativistic wave equation in a potential V is well-known and can be found here:
en.wikipedia.org...