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From what I remember, one of the first steps in finding the equations of motion for an orbiting body is to argue that the body's motion has to be restricted to a plane, because the central force has no component perpendicular to the plane defined by the initial position and velocity vectors. (Or equivalently, because the direction of the total angular momentum has to be constant, since there is no torque.) This means that you can choose coordinates such that $\dot\phi = 0$, which makes the third term of this general Lagrangian vanish:

$$L = \frac{1}{2}m\dot r^2 + \frac{1}{2}mr^2\dot\phi^2 + \frac{1}{2}mr^2\dot\theta^2\sin^2\phi - U(r)$$

For example, this is done in this derivation around equation (11). (Note: in this post I use the Zwillinger convention for spherical coordinates listed on MathWorld.)

Now, I could be wrong, but I think there should be a total of three conserved quantities in this problem. If I use the argument in the previous paragraph to impose a coordinate system in which $\dot\phi = 0$, then I can easily identify the energy as the Noether invariant corresponding to time translation invariance, and the angular momentum as the Noether invariant corresponding to rotational invariance around the $z$ axis ($\theta\to\theta + \epsilon$). But I can't think of a third one, and I suspect that the restriction $\dot\phi = 0$ (by the reasoning of the first paragraph) eliminates that third conserved quantity.

So what if I don't use the physical argument from the first paragraph, so that I'm not limited to $\dot\phi = 0$? From the general Lagrangian I've written above, if my intuition is right, I should be able to obtain three conserved quantities from Noether's theorem alone. Of course, the energy is one, and that's still easy to compute; another one should be angular momentum, although that no longer corresponds to $\theta\to\theta + \epsilon$ but to some more complicated transformation. So I guess my question has two parts,

  1. Prerequisite: What is the symmetry transformation that generates angular momentum when $\dot\phi \neq 0$?
  2. Main question: Is there another spatial symmetry that generates another conserved quantity? If so, what is it?

P.S. I do know there's basically no practical value to this question, since you can always choose coordinates such that $\dot\phi = 0$, but I'm just curious.

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Wait, isn't that precisely the angular momentum? You just said angular momentum has to be constant before that, so I assumed you already figured it out. I do know there is a Laplace-Runge-Lenz vector, but if I remember correctly, that one is associated to the ellipticity of the trajectory. –  Raskolnikov Nov 30 '10 at 22:02
@Raskolnikov: well, it might be the angular momentum. I guess what is actually bothering me is the feeling that there should be another conserved quantity that we miss out on by setting $\dot\phi=0$. I think I can edit the question to clarify it. (Thanks for the comment) –  David Z Nov 30 '10 at 23:24
@David: well, choosing coordinates where $\dot{\phi} = 0$ is really the same thing as setting $\theta = \pi/2$ (you can see this by obtaining the same Lagrangian in both cases). So this is just restriction to the plane of motion. –  Marek Nov 30 '10 at 23:53
@Marek: I think we're using different conventions for the coordinates. Here I meant $\phi$ to refer to the polar angle and $\theta$ to refer to the azimuthal angle (the one that varies with time as the body orbits). –  David Z Dec 1 '10 at 0:09
@David: no, this is not about terminology. I am actually saying that by enforcing $\dot{\phi} = 0$ you are effectively swapping $\theta$ and $\phi$ coordinates. Because what does it mean to enforce that condition? Well, you said it yourself "This means that you can choose coordinates such that $\dot{\phi}˙=0$". Yes you can! But those are precisely the coordinates where $\phi$ has the meaning of an azimuthal angle. And you can see that $\theta$ becomes the polar angle from the restricted Lagrangian. –  Marek Dec 1 '10 at 0:30
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2 Answers

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I think you make some quite confusing statements, so let me be a little bit too explicit. First, for any system whose laws don't depend explicitly on time one obtains conserved energy as an integral of motion.

Central force systems are invariant under the action of $SO(3)$. This is so because both kinetic and potential energy are scalars. Noether's theorem then tells us that the generators of this action must be conserved. Now, $SO(3)$ is three-dimensional and it can be shown that the corresponding generators are e.g. $L_x$, $L_y$, $L_z$ components of angular momentum vector $\mathbf L$. These generators correspond to infinitesimal rotations around $x$, $y$ and $z$ axis respectively. You can check that your Lagrangian is indeed invariant with respect to these transformations (this calculation is a bit tedious though).

So we have in total four conserved quantities $E$ and $\mathbf L$. Now, important fact to note is that setting $\dot{\phi} = 0$ amounts to redefining coordinates and taking $\phi$ as an azimuthal parameter. So instead let's just work with the condition $\theta = \pi/2$. This indeed eliminates two quantities (not just one as you thought) because this restricts our configuration variety to a plane and we remain only with action of $SO(2)$ group (the surviving generator being $L_z$). What is being implicitly used is the fact that $\mathbf L$ is orthogonal to this plane and so the fact that $L_y$ and $L_z$ is conserved is automatically satisfied because they are both zero.

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That may be what I'm looking for - I'd forgotten about the other components of angular momentum. I'll have to think about this. –  David Z Dec 1 '10 at 2:08
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The plane of motion at any given instant is the space spanned by the orbiting body's position vector and velocity vector. To get a normal to this plane, take the cross product of those two vectors. The normal to the plane equally can be used to define what plane we're talking about because a given vector is only normal to one plane.

The cross product of position and velocity is just a constant times the angular momentum, so the conservation of angular momentum implies that the plane of motion doesn't change (and is stronger).

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While this argument is essentially correct, it would take some time to make it mathematically precise. It would go probably something like: consider the surface swept by position vector (considered as a straight line) as the particle evolves in time. This surface can then be arbitrary well triangulated by triangles consisting of position and velocity vectors and so each triangle (and therefore also whole surface) would have the same normal vector. Now because the surface is smooth it follows that it lies in the plane. –  Marek Nov 30 '10 at 23:04
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