Using $3N-f=d$ we can evaluate the degree of freedom or independent coordinates of a system. But how can we know which coordinates are actually independent? (Here $n$ = number of particles, $f$ = number of constraint equations and $d$ = degree of freedom or number of independent coordinates.)

If we take the case of double Atwood machine, we get 2 dof. So which two coordinates should be said to be independent? $x$ and $y$?

Update: If I take the case of, "A particle falling under gravity", the dof will be 1. So there should be only one independent coordinate with which we can describe the situation. If I take the fall of the particle in $y$ direction, then that one independent coordinate will be $y$?

  • 1
    $\begingroup$ you can always choose your coordinates. take (x,y,z) and (r, theta, phi). The only requirement is independence. $\endgroup$ – anna v Oct 17 '16 at 6:34
  • $\begingroup$ I want to know if I got two degrees of freedom which two coordinates are independent? Whichever coordinate system we use, there will be only two dof or 2 independent coordinates, so what are they in the case of double Atwood machine or can you give me any other example? $\endgroup$ – shivani Oct 17 '16 at 6:37
  • $\begingroup$ sorry, I do not know the atwood machine, maybe someone who knows what it is could answer. Generally one checks for independence by taking dot products and seeing that the angle is 90degrees $\endgroup$ – anna v Oct 17 '16 at 6:42
  • $\begingroup$ "Using $3n−f=d$ we can evaluate the degree of freedom or independent coordinates of a system." Please define all symbols used in your questions on this site. Otherwise, we don't know what you're asking. $\endgroup$ – DanielSank Oct 17 '16 at 7:01
  • $\begingroup$ As far as I can understand (OP should have a look), $N$ is the number of bodies (assuming three-dimensional space), $f$ is the number of constraints and $d$ is the number of independent degrees of freedom. For a two-dimensional system we would have $2N-f=d$ and for $N=2$ bodies, if $d$ is to be two as well, we need $f=2$ constraints. $\endgroup$ – Photon Oct 17 '16 at 7:09

If you have a system of $N$ particles with $f$ equations of constraints, then the effective degrees of freedom reduces from $3N$ to $3N-f$. This means you don't have to worry about all the $3N$ coordinates, but just focus on the $3N-f$ coordinates to study the dynamics of the system. This idea is based on a simple logic as I will try to explain it in terms of a circular motion.

Suppose a free particle exhibits a circular motion in the $xy$-plane. The solution for the equation of motion will be in the form

$$z(t)=0$$ $$x(t)^2+y(t)^2=r^2$$

where $x(t)$, $y(t)$ and $z(t)$ are the $x$, $y$ and $z$ coordinates at time $t$ and $r$ is the radius of the circular path and is assumed to be a constant. At first sight, a free particle has $3$ degrees of freedom. But the above equation of constraint($x(t)^2+y(t)^2=r^2$) and the restriction of the particle's motion confined to a plane ($z=0$) reduces the degree of freedom. Hence there is in effect $3-2=1$ degree of freedom of the system.

This means you only need any of the coordinates- $x(t)$ or $y(t)$- to describe the mechanics of the system. The choice is yours. If you choose $x(t)$ as the independent coordinate, then since $r$ is a constant, once you fix a value of $x(t)$, the value of $y(t)$ get fixed automatically because of the constraint equation. In other words, $y(t)$ depends on $x(t)$ by the above equation.

Now, you may ask that if you change the coordinate system from Cartesian to some other, say Spherical polar coordinates, then will the dof changes? No, it will not. The choice of a coordinate system will not affect the dynamics of the system. The above circular motion in plane polar coordinates can be written as:

Put $x=rcos\theta$ and $y=rsin\theta$ in the previous equation and we get

$$\phi(t)=0$$ $$(rcos\theta)^2+(rsin\theta)^2=r^2$$

Here, we have $r=\text{constant}$ and hence the only variable that changes with time is $\theta(t)$. So there is only one degree of freedom. You choose $\theta(t)$ as your independent coordinate. As you can see, the degree of freedom is still one.

In the case of free fall of a particle, the solution is given by:


where $y(t)$ is the position of the object in the $t^{th}$ second. Here, the degree of freedom is one. You only need $y$ to spot the particle at any time $t$.

Update: Degree of freedom of Double Atwood Machine

enter image description here

In the case of a simple Atwood machine, there is only one degree of freedom. Now, you replace one of the masses by another Atwood machine to form a double Atwood machine (or sometimes called compound Atwood machine). Here the system has two degrees of freedom:
1. one is the freedom of mass $1$ (and the attached movable pulley) to move up and down about the fixed pulley, and
2. one is the freedom of mass $2$ (and the attached mass $3$) to move up and down about the movable pulley

How to do this in terms of constraints?

To describe the configuration of the system, we need 3 coordinates each for masses $m_1$, $m_2$ and $m_3$ and another three for the movable pulley. i.e., a total of 12 coordinates. But thanks to the constraints present. There are $10$ constraints present here:

  • $8$ of which limit the motion of all the coordinates in a single direction ($x$ I'am taking here);

  • the remaining $2$ are given by
    $(x_p+x_1)=l$ and $(x_2-x_p)+(x_3-x_p)=l'$

where $x_p$ and $x_i$ are the vertical positions of the pulley and the masses $m_i$ respectively.

Hence the effective degree of freedom of the system reduces to $2$.

In a simple way, one degree of freedom is that of the mass $m_1$ and the other is that due to the mass $m_2$. As you can see from the figure, their motions are independent to each other. Hence $x$ and $x'$ are the independent coordinates here.

  • $\begingroup$ It is the most satisfying answer I ever got on stackexchange. Thanks a lot. I just wanted to clear one more thing. If you agree that the degree of freedom of a double Atwood machine is 2 (Here is my another question physics.stackexchange.com/q/281504) and as I evaluated using Newtonian mechanics, all the constraint equations include x=0 z=0, so only y changes. Now which 2 coordinates I should chose here. It seems that only changing of the y coordinate can describe the system then why 2 degrees of freedom? $\endgroup$ – shivani Oct 17 '16 at 13:06
  • $\begingroup$ shivani, please see the update to the answer. $\endgroup$ – UKH Oct 17 '16 at 14:14
  • $\begingroup$ When I first attended the class on constraint equations, I thought what will be the degrees of freedom of a sine wave and an electromagnetic wave. And this morning I came across a sentence in a book, "Electromagnetic field has an infinite degrees of freedom, roughly speaking two for each point of space-time." Its really awesome, but is it really the answer of question I asked or its telling something else? Thanks again. $\endgroup$ – shivani Oct 21 '16 at 15:06
  • 1
    $\begingroup$ I think the two degrees of freedom refers to the two state of polarization of the field. $\endgroup$ – UKH Oct 22 '16 at 2:24

Technically, when you choose your $n$ generalized coordinates $q^1,\ldots,q^n,$ among the $3N$ position coordinates ${\bf r}_1,\ldots,{\bf r}_N,$ of $N$ point particles, with $n\leq 3N$, you should make sure that the $3N\times n$ rectangular matrix

$$ \frac{\partial {\bf r}_i}{\partial q^j}, \qquad i\in\{1,\ldots N\},\qquad j\in \{1,\ldots n\}, $$

has maximal rank, i.e. has rank $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.