- #1

sergiokapone

- 297

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\begin{align}

\ddot{r} - r\dot{\phi}^2 &= \frac{q}{m} r\dot{\phi}B\\

r \ddot{\phi} + 2\dot{r}\dot{\phi} &= -\frac{q}{m}\dot{r}B

\end{align}

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- Thread starter sergiokapone
- Start date

- #1

sergiokapone

- 297

- 13

\begin{align}

\ddot{r} - r\dot{\phi}^2 &= \frac{q}{m} r\dot{\phi}B\\

r \ddot{\phi} + 2\dot{r}\dot{\phi} &= -\frac{q}{m}\dot{r}B

\end{align}

- #2

- 24,067

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Summary:How to solve equations of motion for charged particle in a uniform magnetic field in a polar coordinates?

A solution of equations of motion for charged particle in a uniform magnetic field are well known (##r = const##, ## \dot{\phi} = const##). But if I tring to solve this equation using only mathematical background (without physical reasoning) I can't do this due to entaglements of variables. What trick should I know?

\begin{align}

\ddot{r} - r\dot{\phi}^2 &= \frac{q}{m} r\dot{\phi}B\\

r \ddot{\phi} + 2\dot{r}\dot{\phi} &= -\frac{q}{m}\dot{r}B

\end{align}

You could look for a solution with fixed ##r##. That might be a quick way.

Alternatively, you could find the general equation of a circle in polar coordinates and see what that looks like.

- #3

sergiokapone

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As I mentioned I know the silution. I want to solve equations and get the constant r.You could look for a solution with fixed rrr. That might be a quick way

- #4

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As I mentioned I know the silution. I want to solve equations and get the constant r.

Why do you think those equations yield a constant ##r##? Most circles in polar coordinates do not have constant ##r##.

- #5

sergiokapone

- 297

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Why do you think those equations yield a constant ##r##? Most circles in polar coordinates do not have constant ##r##.

Ah, ok, this depend on initial condition. I can always choose so.

- #6

sergiokapone

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- #7

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Ah, ok, this depend on initial condition. I can always choose so.

There is nothing in your equations so far that implies that the origin must be at the centre of your circle. What you get from those equations must be the general equation of uniform circular motion.

It might be interesting to calculate this equation and see how close it is to what you have already.

- #8

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A "guess" is different from a "hypothesis". If you have an equation such as:

##x^3 - 3x^2 - 2x + 4 = 0##

Then, you might try ##x = 1## and find it is a solution. That's very different from assuming ##x = 1## is a solution. Guessing a solution is perfectly legitimate, even in pure mathematics.

- #9

sergiokapone

- 297

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Yes. But it doesn't have much values, I think. For example I know how to get solution for the problem in Cartesian coordianates, using method of complex velocity (known from Landau and Lifshitz, V2).Guessing a solution is perfectly legitimate, even in pure mathematics.

- #10

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Yes. But it doesn't have much values, I think.

Don't underestimate the value of educated guesswork. Especially in physics. Didn't Dirac just guess the form of his famous equation?

In general, trajectories in polar coordinates are difficult. Even the equation of a straight line is complicated. Another good approach to this problem would be a change of coordinates:

##x = r \cos\phi, \ \ y = r\sin \phi##

Don't underestimate the value of a well-chosen coordinate transformation. Especially in physics.

- #11

sergiokapone

- 297

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Don't underestimate the value of a well-chosen coordinate transformation. Especially in physics.

Perhaps, I agree with you.

- #12

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- #13

sergiokapone

- 297

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Here my solution, but the text in Ukrainian, unfortunately

- #14

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- #15

sergiokapone

- 297

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