# Why is the Lagrangian a function on the tangent bundle?

I understand that empirically the state of a dynamical system (at a given instant in time) is determined by specifying it's position and velocity, but I'm slightly unsure as to why the Lagrangian is defined on the tangent bundle? Is the point that the Lagrangian of a system is defined independently of any path $q(t)$ that the system takes through configuration space. Thus, at each instant in time we wish to have a function that encodes the dynamics of a system at that instant. This implies that such a function should be dependent on the point in configuration space, $q$ and also the possible velocities at that point (i.e. the tangent vectors at that point), $\dot {q}$, in other words it should be a function on the tangent bundle $$\mathcal{L} :TM\rightarrow \mathbb {R} \;\;, \;\;\mathcal{L}= \mathcal{L}(q, \dot{q})$$ By evaluating $\mathcal{L}$ on a particular curve through configuration space, $q(t)$, with corresponding velocity curve $\dot{q} (t) =\frac{dq} {dt}$ (i.e. $\mathcal{L}= \mathcal{L}(q(t), \dot{q} (t))$) we obtain a description of the dynamics of the system if it followed that particular path through configuration space?!

Sorry for the long-windedness of this post, just really want to get this concept sorted it in my head.

• That sounds right to me. – John Hughes Mar 28 '15 at 12:26
• Is this we can treat $q$ and $\dot {q}$ as independent variables, as they only become dependent once a particular curve has been chosen (via the relation $\dot{q} (t) \frac{dq} {dt}$)?! – Perpetual learner Mar 28 '15 at 12:33
• Are you saying the tangent vector consists of both $q$ and $\dot{q}$ information? – mvw Mar 28 '15 at 12:33
• Yes...e.g., when you know a pendulum is at some angle $\theta_0$, you don't know $\dot{\theta}$: $\theta_0$ might be the limit of travel (in which case $\dot{\theta} = 0$) or it might be mid-swing, for a larger-swing situation. But the total energy of the system (or better, the Lagrangian!) depends only on $\theta$ and $\dot{\theta}$. And since the tangent vector itself tells you $\dot{theta}$, while its basepoint (answering @mvw's question) tells you $\theta$, it makes sense to treat the Lagrangian as a function on $TM$. Roughly: it's a function on $TM$ because no second derivatives appear! – John Hughes Mar 28 '15 at 12:38
• I would have expected $q \in M$ and $\dot{q} \in TM$. Maybe it gets clearer if you can tell me what $M$ is. – mvw Mar 28 '15 at 12:40

Let's look at an example: a damped mass on a spring, constrained to move left-and-right only, in outer space so there's no gravity. I'm going to assign a coordinate $x$ that indicates the displacement of the mass from its rest position. So $x = 0$ is the rest position.

This makes the configuration space $M$ be just $\mathbb R$; a tangent vector $v$ to $\mathbb R$ can be written as a pair $(p; u)$, where $p$ is the "basepoint" of the vector, and $u$ is the "tangent vector" part -- the thing that indicates velocity. Thus for the curve $\gamma(t) = t^2 + 1$, we have, since $gamma'(t) = 2t$, that the tangent vector to $\gamma$ at $t = 3$ is the vector $(10; 6)$, where the $10$ indicates the position, and the $6$ is the "vector part". With this formulation (which works nicely because $M = \mathbb R$, we have that $TM \sim \mathbb R \times \mathbb R$, and $\pi: TM \to M : (p; u) \mapsto p$.

To any smooth path in $M$, there's an associated path in $TM$. Here's an example.

The path defined by $h(t) = t^2$ corresponds to the path $$H: \mathbb R \to TM : t \mapsto (t^2; 2t)$$ in $TM$. I wrote $H$ simply by using $h(t)$ as the first coordinate, and $h'(t)$ as the second.

What about the other direction? If we have a path in $TM$, is there a path in $M$ that corresponds to it in the way just described?

Let's look at a particular path in $TM$, namely, $$s: \mathbb R \to TM : t \mapsto (t; 0),$$ or, less formally, $s(t) = (t; 0)$.

That describes the zero-vector at every point of $\mathbb R$. Now suppose that $f$ is a path in $M$. We know that the corresponding path in $TM$ looks like $$F(t) = (f(t), f'(t))$$ For this to be the path $s$, we'd need two things. For every $t$, we'd need $$f(t) = t \\ f'(t) = 0.$$ Those two things are inconsistent, so there's no such path $f$.

What's all this have to do with $q$ and $\dot{q}$? In what I've written above, a typical point of $TM$ is $(p; u)$, where both $p$ and $u$ are real numbers. I've used the name $\pi$ (for "projection to the basepoint") for the function defined by $\pi(p;u) = p$, and I haven't given a name to the function that sends $(p; u)$ to just $u$.

A physicist gives rather different names to these two functions. S/he calls the first one $q$, and the second (alas), $\dot{q}$. Normally we give our coordinate single-letter names like $x$ and $y$, but ... well, after a while, you get used to this. For now, just think of $\dot{q}$ as a new letter in some bizarre alphabet.

For the function $H$ above, we have $$q \circ H (t) = t^2\\ \dot{q} \circ H (t) = 2t.$$ For the function $s$, we have $$q \circ s (t) = t\\ \dot{q} \circ s (t) = 0.$$

It gets worse. For a function like $H$ or $s$, it's conventional to entirely suppress the function name, and just say that for $H$, we have

$$q(t) = t^2\\ \dot{q} (t) = 2t.$$ while for $s$ we have $$q(t) = t\\ \dot{q} (t) = 0.$$

In the first case, if you take the derivative of what's called $q(t)$, you find out that it's equal to $\dot{q}(t)$, making that suggestive notation a little more reasonable, maybe. For the second, that turns out not to be true.

To summarize so far: a path in the tangent space "corresponds" (in the sense demonstrated in the example of $h$ and $H$) to a path in the base space only if, in physicist notation, the function $t \mapsto \dot{q}(t)$ turns out to be the time-derivative of the function $t \mapsto q(t)$. THAT is what that cryptic thing in your text is trying to say.

What about the whole spring-and-mass thing? Well, so far I've said that certain paths in $TM$ are "nice" in the sense that they correspond to paths in $M$. Even so, not every nice path in $TM$ corresponds to something physically real. The claim of the Lagrangian formulation is that you can write down a function $L$ on $TM$ such that a path is "physically real" only if a certain expression involving $L$ and that path turns out to be zero. In the mass-spring system, those paths will turn out to be damped harmonic motion -- no surprise! But the main thing here is that the notions of the tangent space and "nice" paths are independent of the energy/Lagrangian of the situation, and are merely dependent on the constraints of the system. (Mine, for instance, made the motion one dimensional, so that the configuration space $M$ was just $\mathbb R$.)

• So is the point that we define a function on the tangent bundle that is $2n$-dimensional, and then the functions $q$ and $\dot{q}$ project this function onto the base point on the manifold, and the tangent space at that point, respectively? In your description, is the idea that the tangent vector $\dot{q}(t)$ may not be the the derivative of the curve $q$ at $q(t)$ (it is in the tangent space at that point, but $q$ is not in the equivalence class of curves that define it)? – Perpetual learner Mar 28 '15 at 16:12
• I find it really confusing that (in physics) $q$ and $\dot{q}$ are used to denote points on the tangent bundle, as well as functions, it makes it really difficult to discern what's going on :-/ – Perpetual learner Mar 28 '15 at 16:16
• ... Would you be able to recommend any books/notes that gives a good introduction on the subject and these concepts? – Perpetual learner Mar 28 '15 at 16:26
• I personally really like Spivak's Physics For Mathematicians, but it starts from a fairly advanced point, alas. – John Hughes Mar 28 '15 at 18:02
• To asnwer your first comment: your last sentence captures the idea exactly. – John Hughes Mar 28 '15 at 18:02