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I am doing Lagrangian mechanics and working with Noether's theorem. Please, could you explain the difference between the configuration space and the tangent space?

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  • $\begingroup$ For a single point particle, the configuration space is the tangent space. For any other systems (such as multiple particles or fields), it will be something else. $\endgroup$ – Slereah Mar 2 '18 at 10:47
  • $\begingroup$ Thank you @Slereah . could you please elaborate a bit on what you mean by 'something else'? $\endgroup$ – DeepLearner Mar 2 '18 at 10:51
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    $\begingroup$ For a field described by the fiber bundle $E$, the configuration space will be the jet bundle $J^2E$ $\endgroup$ – Slereah Mar 2 '18 at 11:42
  • $\begingroup$ More precisely, for a free Newtonian particle in 3D, the configuration space is $\mathbb R^3 $. Its tangent space at any point P is isomorphic to $\mathbb R^3 $. $\endgroup$ – DanielC Mar 2 '18 at 12:12
  • $\begingroup$ @Slereah Can you elaborate on this a bit? Why isn’t the configuration space just the space of sections of $E$? $\endgroup$ – BRT Mar 2 '18 at 12:40
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Let's build this by example. Consider a particle in three dimensional space. We want to talk about the set of all possible configurations for this system. But what is a configuration? For a single particle it is just its position, encoding how the system is seem in space. If the particle is free of constraints this is obviously $\mathbb{R}^3$. If the particle is constrained to be on the surface of a sphere, then it is $S^2\subset \mathbb{R}^3$.

Now think about $K$ particles free of constraints. A certain configuration of this system is built by bringing together all positions of all $K$ particles. This obviously contains the informations of relative positions as well and how the system "looks like". Mathematically the set of all configurations is now

$$\mathbb{R}^{3K}=\underbrace{\mathbb{R}^3\times\cdots\times \mathbb{R}^3}_{\text{$K$ times}}.$$

Obviously now imposing constraints as in the case above will restrict the possible configurations to a subset of $\mathbb{R}^{3K}$.

Now we would like to describe the configurations by using coordinates. These coordinates need not be the coordinates labeling points in space, but must be coordinates which are adapted to the problem in question.

And when analyzing problem sometimes there are certain quantities which obviously encode in the simpler possible manner possible, whereas other coordinates seems awkward.

For the problem of $K$ particles free of constraints, at first sight each of them could be anywhere and hence we can actually use the cartesian coordinates of each particle to give coordinates to the configuration of the system. In that case the configurations of the system are given exactly by

$$(x^1,y^1,z^1,\dots,x^K,y^K,z^K)$$

where $(x^i,y^i,z^i)$ are the coordinates in space of the $i$-th particle. You might say: but you already said that when you said that the set of possible configurations is $\mathbb{R}^{3K}$, but bear with me, the fact that $\mathbb{R}^{3K}$ is already the set of all collections of $K$ three tuples of cartesian coordinates is just one special situation, and in general it is nice to make a clear distinction between the space and the coordinates laid on it.

Now for the problem of the particle constrained to move on the sphere, it is obviously better to take a set of coordinates adapted to the sphere, like angular coordinates $(\theta,\phi)$. One thing we obviously expect is that these coordinates are assigned smoothly to configurations.

On the other hand, now forgetting coordinates a bit, we want to talk meaningfuly about curves of configurations. These are functions $\gamma(t)$ which for every value of the parameter $t\in \mathbb{R}$ gives you a configuration. Why do we want this? The obvious reason is that we want to talk about the time evolution of a system, and this is an assignment of a configuration for every instant of time, i.e., a mapping $t\mapsto \gamma(t)$. Obviously we want some notion of continuity and smoothness for this kind of curve of configurations.

The fact that we need a set of configurations, on which we can put coordinates smoothly and on which we can talk about smooth curves suggests we need the structure of a smooth manifold. Thus

The configuration space of a classical system is a smooth manifold $Q$ whose points represent the possible configurations of the system. We usually take $Q$ to be such that the constraints that may exist are already imposed on its definition. For the particle restricted to the sphere $Q=S^2$.

Now given a smooth manifold $Q$, for every $a\in Q$ we can talk about a vector space $T_a Q$ called tangent space to $Q$ at $a$ which contains *all possible directions on $Q$ that can be followed when at point $a$. So if the system is on the configuration $a\in Q$ an element $v\in T_a Q$ represents a possible direction on which the system could evolve. This is called a infinitesimal virtual displacement. We can also think of this $v$ as a possible velocity with which the configuration could be changing with.

We can then bring together all these spaces in a coherent way. We define

$$TQ=\bigcup_{a\in Q}\{a\}\times T_a Q$$

to be the space of all pairs $(a,v)$ where $v\in T_a Q$. Thus we have

The tangent bundle $TQ$ of the configuration space $Q$ is the space of all possible directions at all possible configurations. In other words it is the possible of all possible velocities at all possible configurations. If $(q,U)$ is a coordinate system on the configuration space which assigns coordinates $q^1,\dots,q^n$ to configurations, then there is one natural coordinate system on $TM$ which we denote $(q,\dot{q},TU)$ which is just the result of bringing together $q^1,\dots, q^n$ the coordinates of the configurations and $\dot{q}^1,\dots, \dot{q}^n$ the components of the velocity. In the $S^2$ case we have coordinates $(\theta,\phi,\dot{\theta},\dot{\phi})$ on $TS^2$.

In that setting the Lagrangian of a system is defined as $L : TQ\to \mathbb{R}$ or $L : \mathbb{R}\times TQ \to \mathbb{R}$ when one needs a time-dependent Lagrangian. In other words: the Lagrangian needs information not just about the configuration of the system, but also about the velocity.

And indeed, as shown by Michael Spivak in the book "Physics for Mathematicians" working with infinitesimal virtual displacements one can derive the Euler-Lagrange equations, which shows how $L = T - V$ appears naturally.

So the answer to your question is: the configuration space is a manifold encoding all configurations of the system, the tangent space at each configuration is a vector space containing all possible directions in which said configuration can change, i.e., all velocities and finally the tangent bundle is the space of all configurations together with their available velocities.

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The wording of your question suggests a misconception or two.

For a given physical system, it makes sense to refer to the configuration space $Q$. However, it does not make sense to refer to the tangent space. Rather, we speak of the tangent space to a manifold (in this case, probably $Q$) and at a point $p\in Q$. This is reflected in the notation $T_pQ$, which reads "the tangent space to $Q$ at the point $p$."

If you'd like to visualize it, you can imagine that $T_pQ$ is the set of vectors which live in the tangent plane to $Q$ at the point $p$:

(taken from here)

From here, it's useful to consider the tangent bundle $TQ$, which is simply the disjoint union of all of the $T_pQ$'s. Roughly speaking, an element $v\in TQ$ is a vector which "knows where it lives" - that is, there is a natural projection $\pi : TQ \rightarrow Q$ which maps a vector in $TQ$ to the point of $Q$ where its tangent space is attached.

The Lagrangian, then, can be thought of as a function $$ L : Q \times TQ \times \mathbb{R} \rightarrow \mathbb{R}$$ $$ (x,v,t) \mapsto L(x,v,t)$$

which eats a point $x\in Q$, a vector $v \in TQ$, and a real number $t$ (which denotes the time) and spits out a real number.


Now that that's cleared up, it seems like you're asking about the distinction between $Q$ and some tangent space $T_pQ$.

The most immediate difference which comes to my mind is that $T_pQ$ is a vector space while $Q$ typically is not. If $Q=\mathbb{R}^n$, then it's possible to define "position vectors" and add them together component-wise. On the other hand, maybe $Q=S^2$ (this arises for the spherical pendulum) - in this case, what does it mean to add two points on a sphere?

It is true that $Q$ and $T_pQ$ are homeomorphic in a neighborhood of $p$ - this is due to the definition of a manifold. However, the two spaces generically have different structure. $T_pQ$ is a vector space which is isomorphic to $\mathbb{R}^d$; $Q$ is generically not a vector space and may have a wildly different global topology.

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    $\begingroup$ $TQ$ is already the space of pairs $(q,v)$, no need for$Q\times TQ$ $\endgroup$ – Slereah Mar 2 '18 at 16:27
  • $\begingroup$ It's not quite just the set of such pairs, and I thought the distinction would add needless confusion to the answer, but point taken. $\endgroup$ – J. Murray Mar 2 '18 at 16:50
  • $\begingroup$ Thanks @J.Murray . That actually does clear my misconception! I appreciate it! :) $\endgroup$ – DeepLearner Mar 3 '18 at 4:46
  • $\begingroup$ @J.Murray Not a set of pairs in the sense that it lacks a canonical projection to the second factor, but as you said yourself, the elements of $TQ$ already know where they are from. However locality is very important in the Lagrangian, and no Lagrangian will ever assign a number to a point-velocity pair in which the velocity is in a different point, so in your notation, $L$ woult have to be a $L:Q\times_QTQ\times\mathbb R\rightarrow\mathbb R$ function, which is of course, redundant, since the information carried in the 1st factor is already in the 2nd. $\endgroup$ – Bence Racskó Mar 4 '18 at 14:02

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