What are the degrees of freedom of a dumbbell

classical-mechanicsconstrained-dynamicscoordinate systemsdegrees of freedomnewtonian-mechanics

Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.

Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:

  • $(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.

  • Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle around this line as axis, $\theta'$ that a point not on this line makes with, say the horizontal.

My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ – is that sufficient to locate the rigid body?

If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.

Best Answer

But can we really know the 6th coordinate?

You are right -- given knowledge of five coordinates, there is a discrete choice (corresponding to a reflection) for the remaining coordinate. However, by convention we typically (but not always) use the word "degree of freedom" to mean a continuously varying quantity. This expresses the idea that you aren't free to choose the 6-th coordinate arbitrarily as an initial condition; you have only finite number of choices for it. Once you have chosen your initial conditions however, the subsequent evolution from Newton's laws is completely fixed and there are only 5 functions of time you need to solve for.

If this is the effect of the constraint relation, then it doesn't really eliminate one redundant coordinate, but replaces the entire set, with one less coordinate.

I agree with your description that what is happening is "replacing the entire set, with one less coordinate." However, I also agree with the "eliminate one redundant coordinate" perspective. We started with $6$ coordinates and one constraint, and ended up with $5$ coordinates and no constraints; regardless of exactly how this was done, it seems fair to describe the net result as "eliminating" one of the 6 coordinates using the constraint.

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