# What are the degrees of freedom of a dumbbell?

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.

• Your edit has confused me. It seems like a different question. Note that $(x,y,z,\theta,\phi)$ are legitimate DOFs. But is there a sixth? That depends on your mechanics: classical or quantum. In classical mechanics, rotation around the axis is a sixth DOF. In quantum mechanics, rotation around the axis does not produce a new state, so there are five DOFs. How this relates to the rest of your question I don't quite see. Aug 7, 2021 at 15:38
• It might seem somewhat of a different question, but it is nearly the same as the second question in my last post. I didn't want to keep that because, well, this is the question I wanted to focus upon in this question (I was satisfied with the answers to my first question). Aug 7, 2021 at 15:58

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.

• I am certainly not comfortable, just starting out Goldstein, with the idea of replacing your set of coordinates every time you bring a new constraint equation. Like for rigid bodies one has $3N$ coordinates (of all $N$ particles) to start with, but gets reduced to just $6$ (COM co-ordinates and three angles) given $\frac{n(n-1)}{2}$ constraints. Its like you went into a shop wishing to buy $3N$ shirts and came out buying $6$ trousers instead. Aug 6, 2021 at 7:01
• You might be over-thinking this. Just use whatever coordinate system is easiest. If it means changing, then change. Aug 6, 2021 at 10:45
• @Anu3082 At the risk of making a terrible analogy, I think it's more like buying a prepackaged bundle of $3N$ widgets, with a 6-digit code that labels which bundle you are buying (eg: the "deluxe" bundle 123456 with cool widgets or the "basic" bundle 987654 with simple ones; but every bundle has $3N$ widgets). You only really get to choose which bundle you are buying, even though you end up with $3N$ widgets at the end of the day. Aug 6, 2021 at 13:17
• I mean, I guess it's more like encoding the product ID as a vector in $\mathbb{R}^6$, but let's not push a corny analogy to its breaking point :) Anyway I totally agree with @garyp that you are overthinking it. Aug 6, 2021 at 13:18
• And of course when I said a 6 digit code I should have said a 5 digit code!! Aug 6, 2021 at 13:28

"There is a new suggestion: $$(X_c,Y_c,Z_c,θ,ϕ)$$ where $$(X_c,Y_c,Z_c)$$ are the coordinates of the center-of-mass and θ and ϕ are the polar and the azimuthal angles. 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. Am I right?"

Not really. You can consider the set $$(X_c,Y_c,Z_c,θ,ϕ,r)$$, where $$r$$ is the distance between the two points. The constraint then simply becomes $$r = \sqrt{6}$$. This is a much cleaner approach because it addresses the ambiguity you posed in your first question.

If you want to use Cartesians as you suggested, then to address the ambiguity you need to apply a continuity or "no teleportation" condition. That is, if your choice is between $$x_2 = 3, -1$$, you pick the value which is closest to the value of $$x_2$$ at the preceding time step.

• I got your point regarding the continuity or the teleportation idea, that is good. But I don't understand your first point: why should i put $r$ as a parameter into the coordinates, I mean I can, but we don't have to. My first point was different though: One constraint equation $r=6$ didn't eliminate one redundant co-ordinate (out of the 6 present) but required me to invent a new set of (5) variables that would be my co-ordinates now. Its like you have brought six shirts, then you want to return one, but instead come back with 5 trousers, if that's ok. Aug 6, 2021 at 6:54
• The general approach to these problems is to first pick a set of coordinates for the unconstrained system. The unconstrained system is 6 DOF so you need to pick 6 coordinates. The choice $(X_c,Y_c,Z_c,θ,ϕ,r)$ is perfectly acceptable for the unconstrained system. Now, you want to impose the constraints. You want to "let the math know" that the two particles are actually tethered together. You express that constraint as $r = 6$. Now the degrees of freedom are $(X_c,Y_c,Z_c,θ,ϕ)$, which evolve in time and describe the motion of the system. Associated with the constrain $r = 6$ is a constraint...
– Evan
Aug 6, 2021 at 21:27
• ...force. This is the tension/compression force in the cable or rod connecting the two particles.
– Evan
Aug 6, 2021 at 21:29

You are looking for 5 DOF. This can be established two ways

• 6 DOF of a general rigid body, minus 1 rotational DOF due to symmetry
• 4 DOF of an infinite line in space, plus 1 for locating the object along the line.

Your second approach is the correct one as three position coordinates and two angle coordinates are needed to fully describe the pose of the object.

Mathematically you need a little linear algebra and you can discover the position, velocity and acceleration of any point attached to the object.

Given the 5 DOF variables $$(X_c,Y_c,Z_c,\varphi,\theta)$$ and their derivatives here is what you do

• Position Kinematics

$$\begin{array}{l} \text{Position vector for center of mass}\\ \boldsymbol{P}_{c}=\begin{pmatrix}X_{c}\\ Y_{c}\\ Z_{c} \end{pmatrix}\\ \hline \text{Rotation matrix}\\ \mathbf{R}=\mathrm{RZ}(\varphi)\mathrm{RX}(\pi-\theta) \\ =\begin{bmatrix}\cos\varphi & \sin\varphi\cos\theta & \sin\varphi\sin\theta\\ \sin\varphi & -\cos\varphi\cos\theta & -\cos\varphi\sin\theta\\ 0 & \sin\theta & -\cos\theta \end{bmatrix}\\ \hline \text{Position of point }\boldsymbol{p}=(x,y,z)\text{ local to body.}\\ \boldsymbol{P}=\boldsymbol{P}_{c}+\mathbf{R}\,\boldsymbol{p} \end{array}$$

• Velocity Kinematics

$$\begin{array}{l} \text{Velocity vector for center of mass}\\ \boldsymbol{V}_{c}=\begin{pmatrix}\dot{X}_{c}\\ \dot{Y}_{c}\\ \dot{Z}_{c} \end{pmatrix}\\ \hline \text{Rotational velocity}\\ \boldsymbol{\Omega}=\boldsymbol{\hat{k}}\dot{\varphi}+{\rm RZ}(\varphi)\boldsymbol{\hat{i}}(-\dot{\theta})=\begin{pmatrix}-\dot{\theta}\cos\varphi\\ -\dot{\theta}\sin\varphi\\ \dot{\varphi} \end{pmatrix}\\ \hline \text{Velocity of point }\boldsymbol{p}\\ \boldsymbol{V}=\boldsymbol{V}_{c}+\boldsymbol{\Omega}\times\left(\boldsymbol{P}-\boldsymbol{P}_{c}\right) \end{array}$$

• Acceleation Kinematics

$$\begin{array}{l} \text{Acceleration vector for center of mass}\\ \boldsymbol{A}_{c}=\begin{pmatrix}\ddot{X}_{c}\\ \ddot{Y}_{c}\\ \ddot{Z}_{c} \end{pmatrix}\\ \hline \text{Rotational Acceleration}\\ \boldsymbol{\alpha}=\boldsymbol{\hat{k}}\ddot{\varphi}+{\rm RZ}(\varphi)\boldsymbol{\hat{i}}(-\ddot{\theta})+\boldsymbol{\Omega}\times\boldsymbol{\hat{k}}\dot{\varphi}\\ +\left(\boldsymbol{\Omega}+\boldsymbol{\hat{k}}\dot{\varphi}\right)\times{\rm RZ}(\varphi)\boldsymbol{\hat{i}}(-\dot{\theta})\\ \hline \text{Acceleration of point }\\ \boldsymbol{A}=\boldsymbol{A}_{c}+\boldsymbol{\alpha}\times\left(\boldsymbol{P}-\boldsymbol{P}_{c}\right)+\boldsymbol{\Omega}\times\left(\boldsymbol{V}-\boldsymbol{V}_{c}\right) \end{array}$$

Nobody said 3D dynamics was easy. But if you are methodical with every step and pay attention to details you can figure it out.

• I can see that $(X_c, Y_c, Z_c, \theta, \phi)$ are the $6$ Dof's of the dumbbell. Thanks for showing that all necessary physical quantities can be drawn from these $6$ variables. Aug 7, 2021 at 8:52
• @Anu3082 I see only five DOFs in your list of variables. Aug 7, 2021 at 15:24
• Because this comment was about the dumbbell, not about a rigid body in general, as I have mentioned in the edit above Aug 7, 2021 at 15:55
• @Anu3082 - you miscounted that is 5. Three distances and two angles = 5. A dumbbell is a rigid body, with 1 rotational symmetry axis. Aug 7, 2021 at 20:08

For simplification I will take the 2D space .

The constraint equation is:

$$\left( x_{{2}}-x_{{1}} \right) ^{2}+ \left( y_{{2}}-y_{{1}} \right) ^ {2}-{L}^{2} =0\tag 1$$

thus you obtain three generalized coordinates

solving equation (1) for $$x_2$$

$$x_2=x_1\pm\sqrt{L^2-(y_2-y_1)^2}$$

you get two solution for $$x_2$$ but this doesn't affected the equations of motion . to solve the equations of motion the initial conditions must fulfilled the constraint equation. you get this two initial conditions configurations one with $$+x_2~$$ and one with $$-x_2$$ , but notion is wrong with this two solution. • Say you found out three of the four co-ordinates $(x_1, x_2, y_1)$ by solving the differential equations whatever it may have been. How do you find out the fourth co-ordinate $y_2$? Won't there be two answers for it? Sure $y_2$ doesn't enter the DE's but can you be satisfied with two answers for $y_2$ once you have found out the rest? Aug 7, 2021 at 9:12
• Lets take my case $~x_1(t)~,y_1(t)~y_2(t)$ are the generalized coordinates you obtain the EOM's and you have two configuration for the initial conditions the red one and the blue one the solution for blue IC is $x_2(t)=x_1(t)+\sqrt{L^2-(y_2(t)-y_1(t))^2}~$ and for the red IC is $x_2(t)=x_1(t)-\sqrt{L^2-(y_2(t)-y_1(t))^2}$, by giving the IC you decide which case you want to simulate? But I don't think is something wrong with this ?
– Eli
Aug 7, 2021 at 11:00