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What are degrees of freedom

by aaaa202
Tags: degrees, freedom
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aaaa202
#1
Aug23-12, 04:01 PM
P: 1,005
I'm starting on lagrangian mechanics and is a little puzzled by the use of generalized coordinates. Shortly, what is a degree of freedom?
And what I find harder to understand, why is it that a holonomic constraint allows you to remove a degree of freedom? Consider for instance two particles between which the distance is fixed. This gives 5 degrees of freedom, at least so I heard. Because that is kind of weird to me. As far as I see it the particles can still move anywhere on the x,y and z axis can't they? I can see that in terms of rotations you can only make 2 different ones, and then you can translate the two particles in 3 different directions. But when is it that rotations comes into the picture, because for a collection of N particles you would just have 3N dof, which correspond to movement in three different directions in a euclidean coordinate system.

Talking about the problem with 2 particles with constant distance between them, is it then possible directly, mathematically from the constrain r=c to show that only 5 dof are needed? And can anyone do it?
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Aimless
#2
Aug24-12, 03:04 AM
P: 128
An easy way to see that this particular picture has five degrees of freedom is to view things from the perspective of the midpoint between the two particles (the center of mass if we assume the particles are equal mass).

In an unconstrained system, you can write the six total degrees of freedom as the position [itex](x,y,z)[/itex] of the center point, and the length and orientation [itex](l , \theta , \phi )[/itex] of the line between the two particles. This description is isomorphic to simply describing the system by the position of the two particles [itex] (x_1,y_1,z_1)[/itex] and [itex](x_2,y_2,z_2)[/itex], and it is obvious that they both have the same number of degrees of freedom (6).

However, now let's add back in the constraint. If we define the constraint based on the position of the two particles, the constraint equation is [itex] \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2} = L [/itex], where [itex]L[/itex] is the constant separation between the two particles. To see how this constraint reduces the number of degrees of freedom of the system, you can solve for one of the coordinates in terms of the other five; this constrains the system to a given five-dimensional hypersurface within the six-dimensional space defined by [itex] (x_1,y_1,z_1,x_2,y_2,z_2)[/itex]. However, exactly what that surface is can be a little hard to see.

To make it clearer, let's switch back to the other coordinate system, [itex](x,y,z,l , \theta , \phi )[/itex]. Here, the constraint is much easier to visualize: [itex]l=L[/itex], so our hypersurface is given by [itex](x,y,z,L , \theta , \phi )[/itex]. There are five independent coordinates which describe this hypersurface, [itex](x,y,z, \theta , \phi )[/itex], so therefore there are five degrees of freedom.

At the end of the day, that's what we mean by degrees of freedom. If you define the phase space of your system (that is, the space of allowed parameters) including your constraints, then the degrees of freedom of the system are the number of independent basis vectors which span the constraint surface within the phase space - which is a fancy way of saying "the number of independent ways in which the system can evolve."
voko
#3
Aug24-12, 06:17 AM
Thanks
P: 5,869
Quote Quote by Aimless View Post
If you define the phase space of your system (that is, the space of allowed parameters)
This is not the phase space. This is the configuration space. The phase space includes (generalized) momenta in addition to coordinates.

Aimless
#4
Aug24-12, 06:21 AM
P: 128
What are degrees of freedom

Quote Quote by voko View Post
This is not the phase space. This is the configuration space. The phase space includes (generalized) momenta in addition to coordinates.
Right. I should be more careful with terminology. Oops.

That said, the statement still applies.


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