6
$\begingroup$

As per my knowledge, degrees of freedom of any physical system are the number of independent quantities(coordinates) which need to be specified in order to specify the state of a system uniquely. However, my question is, why is there a unique number of degrees of freedom for any system?

Or, to put it differently, if a set of $n$ independent coordinates describes the state of the system completely, why is it true that any other set of $n$ independent coordinates also describes the state of the system completely?

And why can there not be a set of $m$ independent coordinated which also describe the state of the system completely, where $m>n$ ?

$\endgroup$

2 Answers 2

5
$\begingroup$

I think that the term "degree of freedom" had been initially derived from the observation that the system may be completely described by $n$ independent coordinates - they said: "well, if this $n$ is really a characteristic of the system - let's give it a name!". It is my personal belief that the name comes from a classical kinematics: if a body can't move "freely" in some direction - this reduces $n$ by one.

Therefore, your question may be reduced to the following one: "why the number of independent coordinates which describe the system completely does not depend on coordinates themselves?". Am I right here?

Mathematical answer:

We describe systems by means of linear algebra. There is a concept of a basis in linear algebra: a (minimal) set of independent vectors is a basis, while the projection of any vector from a state-space of a system on this set preserves the norm of the state-space vectors (there are many alternative definitions).

What the above definition says is that if you find a set of (independent) vectors, and can represent any state of the system in terms of these vectors without loosing any information - this set of vectors is a basis.

Why do we require the set of vectors to be minimal? Well, you can add another independent vector to this set, but if the system has no component in this additional coordinate - this vector is redundant. Real life example: you can describe the position of a body using three vector coordinates (x, y, z). You can expand this set of three vectors with a vector of velocity of the wind (which is clearly independent with any of the positional vectors of the body), but we need not worry what the wind is. We are measuring the position of a body.

So, we have a basis, what now? Well, there is a theorem in linear algebra which proves that if you have a basis for the state-space, and you find another basis, then the dimensions of these bases are the same. This means that you always need exactly $n$ independent coordinates to describe the system completely (without loss of information).

Intuitive answer:

This one is very tricky, because all the discussion about "coordinates" and "independence" is bound to the framework of linear algebra. Really, I find it very hard to reason the above statements intuitively.

Maybe this: assume that there is finite number of properties of interest in the system (as it is usually the case). You are given $n$ boundary conditions which are not related - none of them can be derived from the rest. In this case, each boundary condition allows you to express a single property of your system. If you find that these $n$ boundary conditions describe all the properties of interest - this means that you have $n$ properties of interest.

Now, if you're given with $m > n$ unrelated boundary conditions, you can express $m$ properties of your system. However, we have already shown that you have just $n$ properties of interest. Conclusion: the $m$ unrelated boundary conditions describe $m - n$ parameters which you are not interested in. Throw them away because they obscure your view.

$\endgroup$
1
$\begingroup$

For a slightly different take on the same ideas presented by Vasiliy Zukanov...

Start with a large enough set of coordinates to describe the state of the system. These might be the $3N$ components of the positions of $N$ point particles, or they might be angles between lines, or they might be something else entirely. In general you start with $\ell$ such coordinates, and the resulting configuration space is an $\ell$-dimensional manifold.

If you apply $k$ independent holonomic constraints to the system, you restrict the space of valid configurations to a submanifold of dimension $n = \ell - k$. This is formally shown, for instance, in Arnold's Mathematical Methods of Classical Mechanics. The dimension of a manifold, like that of a vector space, is is independent of the way you parametrize it: each point resides in a neighborhood that is diffeomorphic to $\mathbb{R}^n$, and each point's tangent space is isomorphic to $\mathbb{R}^n$. If the submanifold also had dimension $m \neq n$, then you could construct a (differentiable) map with (differentiable) inverse placing $\mathbb{R}^n$ and $\mathbb{R}^m$ in bijection.

You can think of $n$ as the number of "directions" you could move - whatever you call them, or however you rearrange them, there cannot be more or fewer.

For a more concrete example, consider the motion of a single point. Initially maybe you start with Cartesian coordinates, which are obviously just parametrizing the manifold $\mathbb{R}^3$. Now someone tells you the point is constrained to the surface of Earth, perhaps with the constraint $$ x^2 + y^2 + z^2 = R_\oplus^2. $$ You are now restricted to a $2$-dimensional manifold. No matter where you are on Earth, you can locally parametrize your neighborhood (restricted to the surface) with a set of two coordinates, whatever they may be.

Of course that example makes the result seem trivial. The leap comes in understanding that the same holds even for less intuitive coordinates. You might have two angles involved, for instance, in parametrizing a double Foucault pendulum, but then someone tells you the middle joint is constrained to move in a plane. You just went from a 4D configuration space ($S^2 \times S^2$, which is only locally but not globally like $\mathbb{R}^4$) to a 3D one ($S^2 \times S^1$, which locally could be recast using, say, two heights and a horizontal position so that it looks like a region of $\mathbb{R}^3$ - the point is that your choice of coordinates can't change the dimension).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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