Re: Binary stars

[Leigh Palmer <palmer@SFU.CA>]

I think circular and cylindrical symmetry are different animals.
Cylindrical symmetry entails  also an ignorable *z-coordinate*. There
are quantum mechanical *solutions* to cylindrically symmetrical
physical systems which do have circular symmetry except only in the
sense that the phase of the wave function in the azimuthal direction
is arbitrary unless it is split by some externally applied term. m=0
corresponds to circular symmetry. Circular symmetry is a subgroup of
cylindrical symmetry. I is also called "azimuthal symmetry".

I think the terminology is important.

Leigh

> Yes, of course, although I would call it cylindrical symmetry, since we're
> dealing with a 3-dimensional object.  But I would not liken the dumbell
> shape to a circle.
>        Ludwik's circle is at a distance r from the origin and has
> a velocity vector that is confined to a plane.  The L=1 orbit hasa
> substantial part of its volume lying inside of a sphere with radius
> equal to its "classical radius".
>                        Regards,
>                                Jack
>
> On Sat, 18 Dec 1999, Leigh Palmer wrote:
>
> > I don't understand. I know the dumbbell shape of the L=1 orbital,
> > which I suppose you are calling a dipole here, but it still has
> > circular symmetry, hasn't it?
> >
> > Leigh
> >
> > > l=1 is a dipole.  But I mis-spoke, of course; l= 0 is spherical, not
> > > circular.  I stand by the rest.
> > >                                Regards,
> > >                                        Jack
> > >
> > > On Sat, 18 Dec 1999, Leigh Palmer wrote:
> > >
> > > > >        But angular momentum 1 is not a circular orbit it any
> > > > > sense of the word.  The nearest we get to a circular orbit is
> > > > > a very high n s-wave.
> > > >
> > > > Well, an L=1 "orbit" has circular symmetry. That is, the azimuthal
> > > > coordinate is ignorable. Do you mean that the orbit is not spherical?
> > > >
> > > > Leigh