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Old Mar 5th 2013, 03:50 PM   #1
Pmb
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Degrees of freedom

Howdy,

I'm refreshing my quantum mechanics by working through Liboff's text. I'd like someone to double check my work. Here's the homework for chapter one and my answers. I think they're all correct but wish a second opinion, just to make sure. This stuff can get tricky.

1.1 For each of the following systems, specify the number of degrees of freedom and a set of good coordinates.

(a) A bead constrained to move on a closed circular loop.
(b) A bean constrained to move on helix of constant pitch and constant radius.
(c) A particle on a right circular cylinder.
(d) A pair of scissors on a plane.
(e) A rigid rod in space.
(f) A rigid cross in space.
(g) A linear spring in space.
(h) Any rigid body with one point fixed.
(i) A Hydrogen atom
(j) A lithium atom
(k) A compound pendulum (two pendulums attached end to end)

Answer:

(a) Distance along loop from arbitrary fixed point on loop. 1 degree of freedom.
(b) Distance along helix from an arbitrary fixed point on helix. 1 degree of freedom.
(c) Cylindrical coordinates. 2 degrees of freedom.
(d) 3 numbers to locate center of scissors. One for angle scissors make with chosen axis. One for angle scissors is open. 5 degrees of freedom.
(e) 3 numbers to locate center of rod in space. Two numbers to orient rod in space, typically q and f.5 degrees of freedom.
(f) 3 numbers to locate center of rod in space. Two numbers to orient the rod in space. Two numbers to rotate about both axes in space. 6 degrees of freedom.
(g) Three numbers to locate center of spring in space, two numbers to orient spring in space and one number for amount spring is stretched. 5 degrees of freedom.
(h) 3 numbers to locate body in space. 2 numbers to orient body and 2 numbers about each axis of rotation. 7 degrees of freedom
(i) 3 numbers to locate proton in space. 3 numbers to locate the electron in space. 6 degrees of freedom.
(j) 3 numbers to locate the nucleus in space. 3 numbers for each electrons in space. 12 degrees of freedom.
(k) 2 degrees of freedom for first pendulum. 2 degrees of freedom for second pendulum.
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Old Mar 5th 2013, 05:27 PM   #2
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a, b, c, k are okay.

d) The scissors are on a plane, so only 2 degrees to get CoM.

e) - h) Do you need to assume that, for example, the rod is composed from a cylinder? My assumption would be that the cylinders are made of a string. Otherwise it looks good.

i) and j) These are a bit tricky. I assume that you haven't gone over the QM version of the atom? You can't localize the center of the atom in QM. The electons, then, can be found for quantum numbers n, l, m, and m_s for each electron. If you are doing the Bohr model of the atom then after you find the CoM you need to locate the three electrons on a plane. So you would have 2 degrees to incline the plane, then 1 degree for each electron to locate it on the plane in it's orbit.

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Last edited by topsquark; Mar 5th 2013 at 05:36 PM.
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Old Mar 5th 2013, 06:18 PM   #3
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Originally Posted by topsquark View Post
a, b, c, k are okay.
Thanks.

Originally Posted by topsquark View Post
d) The scissors are on a plane, so only 2 degrees to get CoM.
There are 2 degrees to locate the scissors in the plane. Another degree to rotate the scissors in the plane and one more to describe how wide the scissors are open.

Originally Posted by topsquark View Post
e) - h) Do you need to assume that, for example, the rod is composed from a cylinder?
I assumed that it is.

Originally Posted by topsquark View Post
My assumption would be that the cylinders are made of a string. Otherwise it looks good.
Thanks.

Originally Posted by topsquark View Post
i) and j) These are a bit tricky. I assume that you haven't gone over the QM version of the atom?
Yes. I have. As I said, this is a refresher. I took QM as an undergraduate and a full QM series in graduate school. But that was over 15 years ago and I'm rusty. Hence the refresher.


Originally Posted by topsquark View Post
You can't localize the center of the atom in QM.
Good point. I toolk the center of the atom is taken as the center of mass. I believe that's a valid approach. Remeber, when we measure these things they can have an exact position. They can fall into a position eigenstate.

Originally Posted by topsquark View Post
The electons, then, can be found for quantum numbers n, l, m, and m_s for each electron. If you are doing the Bohr model of the atom then after you find the CoM you need to locate the three electrons on a plane. So you would have 2 degrees to incline the plane, then 1 degree for each electron to locate it on the plane in it's orbit.
Three electrons? In a hydrogen atom??

Last edited by Pmb; Mar 5th 2013 at 06:20 PM.
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Old Aug 19th 2018, 09:22 PM   #4
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The usual constraint on pendulum problems is that they lie in a plane. (the strings are taut or rigid, massless, etc.)

There are six degrees of freedom on two particles (the masses) in space. But from this we must deduct the constraints. So the convention they are all in a plane takes the z coordinate away. That leaves four.

If they are in a plane, the top mass must be on an arc of a circle with the center where the string is tied to the ceiling and radius the length of the top string. Being a particle constrained to be on (part of) a circle, it has only one degree of freedom. This could be expressed as a distance from some point on the arc, but for pendulum problems, it is convenient to make it the angle the string forms with the verticle.

Once the top mass is set, the argument shows the bottom mass in on an arc of circle with center at the top mass and radius the length of its string. That too can be expressed as an angle with the verticle.

So that is only two degrees of freedom.
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Old Aug 20th 2018, 05:11 AM   #5
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Just to be awkward...

For the Atoms,
Might it be reasonable to suggest that the electron(s) only have a single degree of freedom?
This would be their energy level within the atom.
The 3 dimensional "position" of the electron is undefined
(or defined only as a probability).
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