In Classical Mechanics there's this notion of configuration manifold. Although I've heard about that a lot and although I often use that concept, I'm not sure I really understand them well because I've found no book talking about that, except Spivak's Physics for Mathematicians.

So my understanding is the following: the configuration manifold of a system in Classical Mechanics is basically one smooth manifold $M$ whose points are possible states of the system. In that case, for one particle in three dimensions, each state can be considered the point in space the particle is and so $M=\mathbb{R}^3$ is the configuration manifold.

Now, reading about that on Spivak's book it seems he only talks about configuration manifolds when talking about constraints. So what is the relationship between configuration manifolds and constraints? The configuration manifold must already include the constraints in some way?

I thought before reading this that the configuration manifold would be simply a manifold we choose whose points label states and that a constraint would be to restrict the allowed states to a submanifold of the first one.

What really is the precise definition of configuration manifold and how it relates to constraints?

  • $\begingroup$ did you ever find a good answer to this question? if $M$ is the "unconstrained'' configuration manifold and $Q$ is the "physical'' configuration manifold, then I feel like the definition of $Q$ should be something like $Q= \{ x\in M \,|\, \varphi_i(x)=0 \}$ where $\varphi_i:M\to \mathbb{R}$ are some set of smooth holonomic constraint functions. But I want more than this. The definitions of immersions, submersions, and embeddings probably come into play... $\endgroup$
    – J Peterson
    Commented Jan 30, 2023 at 17:39
  • $\begingroup$ I guess the answer by @QMechanic is the most accurate, this is about terminology and so is really context-dependent. When reading other people's works the best policy is to check how they are defining things and when writing your own the proposal in his answer seems like a very reasonable one. $\endgroup$
    – Gold
    Commented Jan 30, 2023 at 18:05

3 Answers 3


I think that your description that the points of the configuration manifold are possible states of the system is as close to a precise definition as one will find. So for $n$ particles in three dimensions, the configuration manifold is just $(\mathbb{R}^3)^n$.

As for how this relates to constraints, consider the simplest example: two particles attached with a rigid rod with length $L$, in three dimensions. Let the particles have positions $x_i$ and $y_i$, $i=1,2,3$. Then the constraint of a rigid rod is that $$L = \sum_i(x_i-y_i)^2 \tag{1}.$$ The configuration manifold of this system is that subset of $(\mathbb{R}^3)^2$ that verifies (1), that is, a level set of the distance function.

It is a general principle that the level sets of a smooth function of the coordinates are also smooth manifolds. Hence we can say that imposing a constraint is picking out a submanifold of the configuration manifold for an unconstrained system.


OP is essentially asking about terminology. As usual, be prepared that different authors call different notions differently.

Well, here is a suggestion: Call the configuration space before (after) the constraints are implemented for the extended (physical) configuration space, respectively.

More generally, if an author is talking about a configuration space, it must be deduced from context whether he is talking about the extended or the physical configuration space.


Usually people call configuration space, $\mathcal{M}$, to the space of all posible coordinates needed to determine your system (although,I would include the velocities too).

In non-relativistic theories

The coordinates are three... and we need to provide three coordinates per (point) particle, i.e., for $n$ particles one needs $3n$-coordinates describing a $\mathcal{M}=\mathbb{R}^{3n}$ space.

If you have two particles contrained to be separated by a distance $d$, this means that you will need one less coordinate to determine your system,${}^\dagger$ that is you need five intead of six, say three Euclidean coordinates to determine the position of the first particle, and two angles to determine the position of the second (wrt the first one), then $\mathcal{M} = \mathbb{R}^3 \times S^2$.

${}^\dagger$ The constraint kills only one degree of freedom because it is a scalar constraint.


Your Answer

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