My question : What does it mean when we arrive at negative values for distance variables like $\rho$ in cylindrical coordinates? (after some discussion here,I revised the question, at the end of the post)

My specific example:

For the motion of an ion in a quadrupole Paul trap, we arrive at the same equation (mathieu equation) (below) for both $\rho$ and $z$ components (the potential has only $\rho$ and $z$ dependance ): $$\frac{d^2u}{d\tau^2} + [a_u-2q_u\cos (2 \tau) ]u=0 \qquad\qquad $$

here $u$ represents a coordinate variable and may be $r$ , $\rho$ in cylindrical coordinates or $x , y ,z$ in Cartesian coordinates.

The problems is that , this equation has an oscillatory behavior and it's solution has negative values too. What is the meaning of negative $\rho$ in cylindrical coordinates?

Here is a sample solution for $a_z=0$ and $q_z=0.4$

A sample solution for $a_z=0$ and $q_z=0.4$

EDIT 1: after some answers posted , I found out that the problem has something to do with our derivation of equations.Here is how correct equations are.

The potential is only a function of $r$ (or $\rho$) and $z$ in cylindrical coordinates,so we don't have any force in $\hat \theta$ direction: $$\mathbf F=-\nabla V(r,z)=-(\hat r{\partial V \over \partial r} +\hat \theta \frac{1}{r} {\partial V \over \partial \theta} +\hat z {\partial V \over \partial z})$$

so the equations of motion will be (I assumed the mass to be 1) $$\ddot r - r\dot{\theta}^2=-{\partial V \over \partial r} ***$$ $$r \ddot{\theta} +2 \dot r \dot \theta=0 \to \dot{(r^2\dot \theta)} = 0$$ $$\ddot z=-{\partial V \over \partial z}$$ so the equations for $z$ and $r$ will not be the same.

Now , the problem is : how can we arrive at those equations (mathieu equation, above)(in the literature on Paul traps they use this equations) from these equations of motion ; eliminate $\theta$ dependance and instead of that let $r$ get negative values and interpret it as a positive $r$ with $\theta_0+\pi$?

  • $\begingroup$ Presumably $\{-|\rho|,\theta\}$ should be identified with $\{|\rho|,\theta+\pi\}$ $\endgroup$ Commented May 5, 2013 at 16:15
  • $\begingroup$ @zhermes I think so, but I don't know why. $\endgroup$
    – Zorich
    Commented May 5, 2013 at 16:18
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    $\begingroup$ @Monopole, if you are ok with $x,y,z$ coodinates, then look at the transition formulae -- these formulae do not know that you think that $\rho>0$, and they are used to write the equation in cylindircal coordinates. In this sense it is the same as identifying $\theta$ with $\theta+2\pi$ $\endgroup$ Commented May 5, 2013 at 18:45
  • $\begingroup$ @Peter in combination with what zhermes said, that could be an answer. $\endgroup$
    – David Z
    Commented May 5, 2013 at 19:42
  • $\begingroup$ cc @zhermes on above comment $\endgroup$
    – David Z
    Commented May 5, 2013 at 19:42

2 Answers 2


Consider an equation that you have written in Cartesian coordinates $x,y,z$ which gives you these coordinates as a function of time or something like that. Then you want to write it in, say, cylidrical coordinates. Then you write something like: $$ x=r\cos\phi\\ y=r\sin\phi\\ z=z $$ and plug it into your equation. Now just forget the word coordinates. It is just a mathematical transformation. Then you arrive at some equation for $r,\phi,z$, and you solve it. Then you can use the above formulae to reconstruct the original $x,y,z$ variables. This means that when $\phi$ is, say, $3.2\pi$, it means the same as $1.2\pi$ (check the eqns) and when $r=-|r|$ it means the same as $r=|r|$, while $\phi$ is shifted by $\pi$. Or any other explanation you prefer. The point is that you just have to plug it into the transformation given above.

The confusion stems from the fact that you are used to think about $r,\phi,z$ as coordinates, which is different from an arbitrary change of variables. Coordinates have to be in 1 to 1 correspondence with the space points. This is why you say $r\ge 0,\phi\in[0,2\pi)$ and some subtleties with $r=0$. When you are trying to solve an equation, this is completely unnecessary.

In fact, when you try to think about these new variables as of coordinates, you can note that $r$ should never pass to negative values -- instead, $\phi$ should jump (while you are, at the moment, thinking that it is constant). So, $\phi$ is not constant, and in this interpretation your equations are wrong (because $\phi$ jumps occasionally). In order to stick to the 'coordinates' interpretation, you have to modify the equations, they will be different, and the new equation will have non-negative $r$ and nonconstant $\phi$ as the solutions. This is a way too complicated, so it is easier to think about $r,\phi,z$ as of a convenient mathematical substituiton and not bother about any restrictions on the range. You only have to remember the formulae that link you to the original problem.

  • $\begingroup$ The process of derivation of equations doesn't have anything with transformation of variables. I will edit the question to include some points about derivation of the equations. $\endgroup$
    – Zorich
    Commented May 5, 2013 at 21:33
  • $\begingroup$ @Monopole. You state 'and may be r , ρ in cylindrical coordinates or x,y,z in Cartesian coordinates.'. So I suppose that I can write your equation in Cartesian. Then I apply the transformation of variables to cylindrical. My point is that it is different -- either you think that values of the coordinate should be restricted or not. In the first case you will get an equation different from yours. $\endgroup$ Commented May 5, 2013 at 21:53
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    $\begingroup$ @Monopole, no, it does not change. Of course, you should remember that your $V$ is defined for positive $r$, so you should in this case write $V(|r|)$. Mathematically, initially (in Cartesian) you had $V(\sqrt{x^2+y^2+z^2})$ which turns into $V(|r|)$, but not $V(r)$. $\endgroup$ Commented May 5, 2013 at 22:33
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    $\begingroup$ @Monopole: I mean that if I start from Cartesian coordinates, then everything is automatically ok: suppose I had $V(x,y,z)=k\sqrt{x^2+y^2}$, then after substituition of $x,y,z$ I will get $V(r)=k\sqrt{r^2}=k|r|$. If you start from cylindrical formulation then you should be aware that you can allow any values of $r,\theta$ only if you ensure that you define your $V$ outside the traditional range so that $V(r,\theta)=V(r,\theta+2\pi)=V(-r,\theta+\pi)$. Since you usually have your potential defined in the standard range $r>0,0\leq\theta<2\pi$, you should extend the definition so that it is ok. $\endgroup$ Commented May 5, 2013 at 22:47
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    $\begingroup$ @Monopole: so if you have initially $V(r)=kr$ and you now want to allow $r$ to be negative, you have to additionally define (you didnt allow $r<0$ before, so there was no definition of $V$ for this case) $V(r)=V(|r|)=k|r|$. This is in case when $V$ does not depend on $\theta$. If $V$ depends on theta, say, $V(r,\theta)=kr\cos\theta$, then you have to say that $V(-|r|,\theta)=V(|r|,\theta+\pi)=-k|r|\cos\theta$. $\endgroup$ Commented May 5, 2013 at 22:56

You have to be careful when you derive your equations of motion in cylindrical versus cartesian coordinates. For example, if your Hamiltonian (I'm ignoring the $z$ direction because it is the same in cylindrical to cartesian) is

$ H = \frac{p_x^2}{2} + \frac{p_y^2}{2} + V(x, y)$

then in cylindrical coordinates you get

$ H = \frac{p_r^2}{2} + \frac{p_\theta^2}{2 r^2} + V(r, \theta)$

That $\sim 1/r^2$ term represents a repulsive term which prevents $r$ from getting negative, physically. If your equations of motion are derived physically, getting a negative radius should be impossible.

  • $\begingroup$ You are right. The problem is with derivation of equations. But I don't know how to explain what we do to simplify the situation.I will edit the question to include your suggestions and explain more about this comment. $\endgroup$
    – Zorich
    Commented May 5, 2013 at 21:35
  • $\begingroup$ The term you are talking about indeed saves you, but only if the angular momentum $p_\theta$ is not zero, which is the physical reason why you cant reach $r=0$. If $p_\theta=0$ then your system can pass through zero. Your $H$ will give you negative $r$ (consider the harmonic oscillator $V=r^2/2$). $\endgroup$ Commented May 5, 2013 at 21:57
  • $\begingroup$ This is just because there is NO reason to think that $r>0$ besides the fact that you are taught that it is so. When you do the transformation to cylidrical coordinates, what you really do is just a substituition of new variables, that are not limited in their values. If you say that these variables are coordinates then there are subtleties with transformation of derivatives (at points where some coordinate jumps), and you will get different, more involved equations (with some delta terms etc), that will always give $r>0$, but that is not what you are doing. $\endgroup$ Commented May 5, 2013 at 22:01
  • $\begingroup$ Fair enough, but for a Paul trap this is not the case. Also, as you noted above, inverting the equations is singular at the origin, so you're stuck in a rather sticky situation when you get there anyway. Of course, if I start out at the origin with zero angular momentum, what exactly is $p_\theta^2/r^2$ anyway? $\endgroup$
    – webb
    Commented May 5, 2013 at 23:29

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