Differential equation system, got stuck in a physics problem

In summary, the conversation discusses a complex problem involving calculating the trajectory of a charged particle in a custom magnetic field. The problem is represented by a system of equations and there is confusion about the correct form. The participants suggest using polar or spherical coordinates and share their progress in solving the equations. Eventually, a partial solution is presented for a specific case.
  • #1
ramses728
9
0
hi guys,
I'm solving a pretty complex problem: calculating a trajectory of a charged particle in a custom magnetic field. I arrive to the point where this very nice equation system blocks my way :P

http://ramses728.altervista.org/img/phys.jpg

hope you can help me somehow
 
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  • #2
It's not a valid link, there's no picture in it. I'm using Firefox as a webbrowser.
 
  • #3
Strange... But if you copy and paste it, it works 100%
 
  • #4
As I understand it, the equation is
[tex]m x'''(t) = \frac{q z''(t) y(t)}{r(t)}, m y'''(t) = \frac{q z''(t) x(t)}{r(t)}, m z'''(t) = \frac{ q\left( x'(t) x(t) + y'(t) y(t) \right) }{ r(t) }[/tex]
(the last [itex]y'''(t)[/itex] in the original should be a [itex]z'''(t)[/itex] I presume), where
[tex]r(t) = \sqrt{ x(t)^2 + y(t)^2 [/tex].

This looks very complicated. Are you sure it is correct? I would either try polar or spherical coordinates (especially if there is some kind of symmetry in the system), or otherwise to first solve [itex]x(r(t))[/itex] or at least the functions in a more convenient variable than [itex]t[/itex]. But actually, I have no idea :confused:
 
  • #5
Yeah it's complicated indeed... and you got that right with [tex]z'''(t)[/tex]

I actually know what it should look like when it's solved (graphically i mean) and there is some pretty serious symmetry in the system... I'll give a shot with the cilindrical coordinates, but there will be much work at it especially because I have to reconstruct the field function with polar coordinates (mabye esier... who knows...)

anyway if someone has any ideas how to solve this (also with maple or similar), I'd relly apreciate it

as soon as i get the cilindrical set of equations, I'll post them.

thanks, ramses

Edit: ah yeah forgot to say it's an equation system, so three equations together
 
  • #6
How did you get third derivatives? Kinematic equations always involve acceleration, the second derivative.
 
  • #7
... you're right... i made some confusion added a derivation grade for acceleration and velocity XD ... so, here are the correct ones:

[tex]
\left\{
\begin{array}[l]{l}
m x(t)'' = q z(t)' \frac{x(t)}{r(t)}\\
m y(t)'' = q z(t)' \frac{y(t)}{r(t)}\\
m z(t)'' = -\frac{q}{r(t)}(x(t)' x(t) + y(t)' y(t))\\
\end{array}
[/tex]

where

[tex]r(t)=\sqrt{x(t)^2 + y(t)^2}[/tex]

still working on those polar ones

thanks for your interest

(P.S. I finally realized that fourmulaes can be written with latex! XD)
 
  • #8
So, as

[tex]r^{\prime}(t) = \frac{1}{2} \frac{(2 x(t) x^{\prime}(t) + 2 y(t) y^{\prime}(t))}{\sqrt{x^2(t) + y^2(t)}} = \frac{x(t)x^{\prime}(t) + y(t) y^{\prime}(t)}{\sqrt{x^2(t) + y^2(t)}}[/tex]

You can write

[tex]m z^{\prime \prime}(t) = -q r^{\prime}(t) [/tex]

Hence, for z you get

[tex]z^{\prime}(t) = \kappa - \frac{q}{m} r(t) [/tex]

where [tex]\kappa[/tex] is a constant.

Now, for [tex]\kappa = 0[/tex], the solution is straightforward. You will then get

[tex]x^{\prime \prime} = -(\frac{q}{m})^2 x(t) [/tex]

and

[tex]y^{\prime \prime} = -(\frac{q}{m})^2 y(t) [/tex]

But in the more general case, [tex]\kappa[/tex] may not be taken as 0, so you'll get

[tex]x^{\prime \prime}(t) = \frac{q}{m} x(t) (\frac{\kappa}{r(t)} - \frac{q}{m})[/tex]

and

[tex]y^{\prime \prime}(t) = \frac{q}{m} y(t) (\frac{\kappa}{r(t)} - \frac{q}{m})[/tex]
 
Last edited:
  • #9
:bugeye: you got it! Well thanks very much!

just for curiosity are you a math student?
 
  • #10
I've only solved it for [tex]\kappa =0[/tex]. For [tex]\kappa \ne 0[/tex], it is still a nasty mess.

And, no, I'm not a student.
 
  • #11
It's still a great advance, because I can set the initial conditions for z velocity (so [tex]z(t)'[/tex]) to be 0 forcing the constant to be zero. So I have the problem solved for this specifical case. But as you say it still needs work for the more general case...

Hope you didn't take me thinking you were a student as offensive, I just thought that you were closely involved on math so supposed that. If any offense was taken I apologise. Thanks again for this partial solution
 
  • #12
No worries, mate. :cool:
 

1. What is a differential equation system?

A differential equation system is a set of equations that describe how a physical system changes over time. It involves derivatives, or rates of change, to represent the relationship between the variables in the system.

2. How do I solve a differential equation system?

There are various methods to solve a differential equation system, such as separation of variables, substitution, and integrating factors. It is important to identify the type of differential equation and choose the appropriate method to solve it.

3. What does it mean to be "stuck" in a physics problem involving differential equation systems?

Being "stuck" in a physics problem means that you are unable to find a solution using the methods you have tried. This can occur when the system is too complex or when the initial conditions are not known.

4. Can I use software to solve a differential equation system?

Yes, there are many software programs, such as Mathematica, MATLAB, and Maple, that can solve differential equation systems. However, it is still important to understand the underlying concepts and methods for solving these equations.

5. How can I apply differential equation systems in physics?

Differential equation systems are commonly used in physics to model and predict the behavior of physical systems, such as motion, heat transfer, and electrical circuits. They can also be used to analyze and design control systems in engineering applications.

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