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Vector Calculus

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1. Homework Statement

At time t = 0, the vectors [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] are given by [itex] \textbf{E} = \textbf{E}_0 [/itex] and [itex] \textbf{B} = \textbf{B}_0 [/itex], where te unit vectors, [itex] \textbf{E}_0 [/itex] and [itex] \textbf{B}_0 [/itex] are fixed and orthogonal. The equations of motion are

[tex] \frac{\mathrm{d}\textbf{E}}{\mathrm{d}t} = \textbf{E}_0 + \textbf{B}\times\textbf{E}_0[/tex]

[tex] \frac{\mathrm{d}\textbf{B}}{\mathrm{d}t} = \textbf{B}_0 + \textbf{E}\times\textbf{B}_0[/tex]

Find [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] at a general time t, showing that after a long time the directions of [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] have almost interchanged.

3. The Attempt at a Solution

Now this looks like a "simple" coupled differential equations, instead of using the formal method to solve te system, I differentiate with respect to time and get

[tex] \ddot{\textbf{E}} = \dot{\textbf{B}}\times\textbf{E} = \textbf{B}_0(\textbf{E}_0\cdot\textbf{E})+\textbf{B}_0\times\textbf{E}_0[/tex]

And similar equation for [itex] \ddot{\textbf{B}} [/itex] (replace E's with B's). But I don't know what to use this result for, this is again a coupled differential equations. Is there a better way to solve this problem? I want to find E and B as functions of time.
Any hint would be appreciated.
 
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Answers and Replies

HallsofIvy
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1. Homework Statement

At time t = 0, the vectors [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] are given by [itex] \textbf{E} = \textbf{E}_0 [/itex] and [itex] \textbf{B} = \textbf{B}_0 [/itex], where te unit vectors, [itex] \textbf{E}_0 [/itex] and [itex] \textbf{B}_0 [/itex] are fixed and orthogonal. The equations of motion are

[tex] \frac{\mathrm{d}\textbf{E}}{\mathrm{d}t} = \textbf{E}_0 + \textbf{B}\times\textbf{E}_0[/tex]

[tex] \frac{\mathrm{d}\textbf{B}}{\mathrm{d}t} = \textbf{B}_0 + \textbf{E}\times\textbf{B}_0[/tex]

Find [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] at a general time t, showing that after a long time the directions of [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] have almost interchanged.

3. The Attempt at a Solution

Now this looks like a "simple" coupled differential equations, instead of using the formal method to solve te system, I differentiate with respect to time and get

[tex] \ddot{\textbf{E}} = \dot{\textbf{B}}\times\textbf{E} = \textbf{B}_0(\textbf{E}_0\cdot\textbf{E})+\textbf{B}_0\times\textbf{E}_0[/tex]

And similar equation for [itex] \ddot{\textbf{B}} [/itex] (replace E's with B's). But I don't know what to use this result for, this is again a coupled differential equations. Is there a better way to solve this problem? I want to find E and B as functions of time.
Any hint would be appreciated.
Well, you have "uncoupled" B and E- that's a start. The equations for the various components of E are still coupled. Go ahead and solve them by "uncoupling" again. How complicated that is will depend upon the constant vectors E0 and B0. If they had one or more components 0, that would simplify things a lot.
 
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Well, you have "uncoupled" B and E- that's a start. The equations for the various components of E are still coupled. Go ahead and solve them by "uncoupling" again. How complicated that is will depend upon the constant vectors E0 and B0. If they had one or more components 0, that would simplify things a lot.
So I assume there is no simple "shortcut" to solve this? Because there is no more information then what I wrote in #0, about the start vectors, so I can't assume some components are 0. This seems more like a diff. equation problem than a vector calculus. Ofcourse this problem is relatively easy to solve when only considering it as a system of coupled differential equations. I thought that there would be some "tricks" to solve this another way.
 

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