Equation of motion in polar coordinates for charged particle

[sergiokapone]

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}

 

[PeroK]

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.

 

[sergiokapone]

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

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

 

[PeroK]

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 $r$.

 

[sergiokapone]

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.

 

[sergiokapone]

But what if I do not know is a circle? I need to solve equations in a right way, without any hypothesis.

 

[PeroK]

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.

 

[PeroK]

But what if I do not know is a circle? I need to solve equations in a right way, without any hypothesis.

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.

 

[sergiokapone]

Guessing a solution is perfectly legitimate, even in pure mathematics.

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).

 

[PeroK]

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.

 

[sergiokapone]

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

Perhaps, I agree with you.

 

[vanhees71]

Well, cylinder coordinates are simply not the best choice of generalized coordinates for this problem. Try again with Cartesian coordinates. If you want to use the Hamilton principle also the choice of gauge can help a lot. E.g., you can choose a gauge, where $\vec{A}$ depends only on one coordinate so that two become cyclic!

 

[sergiokapone]

I had already solve similar, but slightly harder problem in Cartesian coordinates. The charged particle in uniform magnetic field with friction.

Here my solution, but the text in Ukrainian, unfortunately

 

[vanhees71]

Looks good, though I can only read the formulas. You have everything solved since this problem is just a special case of your more general problem. Just set $k=0$.

 

[sergiokapone]

Yes, I checked $k=0$ case. My post originated from this problem, I thought I can obtain the solution in polar coordinates. But when I was simplified problem (removed the friction) I realized I can't solve even this simple case. Yes, I agree with you, the Cartesian coordinates is a best choice for this problem.