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1effect
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I have a system of differential equatiuons (see http://www.savefile.com/files/1458343 ). How can I solve it? Is it even solvable? Thank you
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Crosson said:Did you notice that you can attach .doc files to your posts at physicsforums?
Even better, you could learn LaTeX which is the main way we communicate formulas on these forums.
I made some progress, see http://www.savefile.com/files/1458817 . It seems that a symbolic (not analytic) solution can be found. Analytic means something else.In this case, you even could have stated the problem in words:
"Is it possible to solve the relativistic equations of motion for a charged particle in a constant electric and magnetic field?"
In this case no one has found analytical solutions to the full nonlinear system. Since we know the exact solution for the case [itex]\gamma = 1 [/itex] the best hope for analytical results in your case is to use perturbation theory together with the first or second term of the power series expansion for gamma. Maybe someone has done this before.
nicksauce said:Most people on these boards are going to download a file in .doc format like that. It's just a virus waiting to happen.
nicksauce said:Most people on these boards are going to download a file in .doc format like that. It's just a virus waiting to happen.
1effect said:I know how to use LaTex, I just preferred to use an editable file that people could write into.
I made some progress, see http://www.savefile.com/files/1458817 .
It seems that a symbolic (not analytic) solution can be found. Analytic means something else.
Crosson said:[tex]
\frac{\frac{-\frac{8 c^2 \text{vx}'(t) \text{vx}''(t) m^2}{B^2 q^2}-4 \text{vx}(t) \left(c^2-\text{vx}(t)^2\right) \text{vx}'(t)}{2
\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}-2 \text{vx}(t) \text{vx}'(t)}{2
\sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}}
[/tex]
So what we have now is the second-order equation involving only one unknown function [itex]v_x[/itex] (where I have no substituted your expression for [itex]\gamma[/itex], for the sake of brevity):
[tex]
\text{}\leftB \text{vx}(t) q-\frac{\text{E} q}{\gamma ^2}+\frac{m \left(\frac{-\frac{8 c^2 \text{vx}'(t)
\text{vx}''(t) m^2}{B^2 q^2}-4 \text{vx}(t) \left(c^2-\text{vx}(t)^2\right) \text{vx}'(t)}{2
\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2 q^2}}}-2 \text{vx}(t) \text{vx}'(t)\right)
\gamma }{2 \sqrt{c^2-\text{vx}(t)^2+\sqrt{\left(c^2-\text{vx}(t)^2\right)^2-\frac{4 c^2 m^2 \text{vx}'(t)^2}{B^2
q^2}}}}=0\right
[/tex]
Or did you want to be even more pedantic and say "a solution that can be represented in terms of arithmetic operations, root extractions, trigonometric/exponential functions, or other cataloged special functions."
In any case, I stand by my original statement that this equation cannot be solved "in the way that you are looking for." Did you consider my advice to apply perturbation theory? I am sure the results would be interesting.
1effect said:I am not familiar with the term, do you mean using Taylor expansions for [tex]v_x(t)[/tex] and [tex]v_y(t)[/tex]? Can you take me thru the first steps?
Thank you, there is a small error in your derivative but I get the gist.
This is what symbolic means. Analytic means that the function is continuous with continuous derivatives.
Crosson said:This is a good introduction to the method:
http://www.sm.luth.se/~johanb/applmath/chap2en/index.html
I might try to work it out another time.
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