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## Main Question or Discussion Point

This is a slightly physics oriented question, so apologies for that.

Basically, having started studying differential geometry it has started to become a little clearer to me why one can consider the Lagrangian as a function of position and velocity, but I don't feel I'm quite there yet.

My confusion arises with the tangent bundle. I understand that the state of a physical system at a given instant in time is completely specified by its position and velocity at that given instant, and as such, for the Lagrangian to characterise all possible dynamics of a physical system, it must therefore be a function on the tangent bundle (correct?!)

As I understand it, the tangent bundle [itex]\mathcal{TM}[/itex] is defined as the disjoint union of tangent spaces [itex]T_{p}\mathcal{M}[/itex] parametrised by the manifold [itex]\mathcal{M}[/itex] (of dimension [itex]n[/itex]), [tex]\mathcal{TM}=\lbrace (p,\mathbf{X}_{p})\;\;\vert\quad p\in \mathcal{M}, \;\;\mathbf{X}_{p} \in T_{p}\mathcal{M}\rbrace[/tex]

Would it then be correct to say that there is an [itex]n[/itex]-dimensional tangent space [itex]T_{p}\mathcal{M}[/itex] for each

I have been reading this set of notes http://www.physics.usu.edu/torre/6010_Fall_2010/Lectures/01.pdf and they have been enlightening me a bit on the subject, but I'm a little unsure that I'm understanding it correctly?!

I understand that points [itex]p[/itex] on the manifold [itex]\mathcal{M}[/itex] exist independently of any chosen curve, as do the tangent vectors [itex]\mathbf{X}_{p}[/itex] in the tangent space [itex]T_{p}\mathcal{M}[/itex] to those points, so would it be correct to say that each point in the tangent bundle represents a point on [itex]\mathcal{M}[/itex] through which a curve could pass through at a given velocity? [In the middle of page 6 the lecturer pauses to discuss a "common point of confusion" about the notation [itex]\dot{q}[/itex] (for a vector in the tangent space), stating that it is not the derivative of anything, but just a vector that exists at the point [itex]q[/itex]. Is this just the case because for each given point we can associate a set of vectors (which form a tangent space at that point); they are not derivatives of anything, but they do, in essence, contain information on what direction (and at what speed) curves can move through that point?!]

He then moves on to say that a given curve in the tangent bundle looks like $$q=q(t),\qquad\dot{q}=\dot{q}(t)$$ I assume by this that the curve is parametrised by some parameter [itex]t[/itex] and that a given value of [itex]t[/itex] corresponds to a given point [itex](q,\dot{q})[/itex] in the tangent bundle? Also, he says that such a curve may not actually correspond to any motion of the system and we need [itex]\dot{q}(t)[/itex] to actually represent the tangent to [itex]q(t)[/itex], that is, we need to choose [itex]\dot{q}(t) = \frac{dq(t)}{dt}[/itex]. By this does he mean that although a given value of [itex]t[/itex] will correspond to a point [itex]q=q(t)[/itex] and a tangent vector [itex]\dot{q}=\dot{q}(t)[/itex], the tangent vector associated with that value of [itex]t[/itex] may not be that of the curve passing through [itex]q[/itex], hence we require that, in actual fact, the given value of [itex]t[/itex] corresponds to a vector tangent to the point [itex]q[/itex] such that its value [itex]\dot{q}(t)[/itex] corresponds to the derivative of the curve evaluated at that point, i.e. [itex]\dot{q}(t)=\frac{dq(t)}{dt}[/itex]?

Having done all this, do we then say that the Lagrangian is a one-parameter family of functions (parameterised by [itex]t[/itex]) on the tangent bundle, i.e. $$\mathcal{L}=\mathcal{L}(q,\dot{q}, t)=\mathcal{L}(\lbrace q_{i}\rbrace, \lbrace\dot{q}_{i}\rbrace , t)$$ One may then choose a curve in configuration space [itex]q(t)=\lbrace q_{i}(t) \rbrace[/itex] such that when the Lagrangian is evaluated on that curve, for a given value of [itex]t[/itex] it returns a number [itex]\mathcal{L}(q(t),\dot{q}(t),t)[/itex] corresponding to the value of the Lagrangian evaluated at the point [itex]q=q(t)[/itex] on the curve, whose tangent vector is [itex]\dot{q}(t)=\frac{dq(t)}{dt}[/itex]?

Sorry for the long-windedness of this post, I'm just really keen to get these concepts firmly cemented (correctly) in my mind!

Basically, having started studying differential geometry it has started to become a little clearer to me why one can consider the Lagrangian as a function of position and velocity, but I don't feel I'm quite there yet.

My confusion arises with the tangent bundle. I understand that the state of a physical system at a given instant in time is completely specified by its position and velocity at that given instant, and as such, for the Lagrangian to characterise all possible dynamics of a physical system, it must therefore be a function on the tangent bundle (correct?!)

As I understand it, the tangent bundle [itex]\mathcal{TM}[/itex] is defined as the disjoint union of tangent spaces [itex]T_{p}\mathcal{M}[/itex] parametrised by the manifold [itex]\mathcal{M}[/itex] (of dimension [itex]n[/itex]), [tex]\mathcal{TM}=\lbrace (p,\mathbf{X}_{p})\;\;\vert\quad p\in \mathcal{M}, \;\;\mathbf{X}_{p} \in T_{p}\mathcal{M}\rbrace[/tex]

Would it then be correct to say that there is an [itex]n[/itex]-dimensional tangent space [itex]T_{p}\mathcal{M}[/itex] for each

*fixed*point [itex]p\in\mathcal{M}[/itex], and as such, for a fixed point, there are an infinite number of tangent vectors that one could choose from? On the face of it I see how, as the manifold formed by the tangent bundle is [itex]2n[/itex]-dimensional, one can specify a point on the this manifold uniquely by specifying a point [itex]p\in\mathcal{M}[/itex] and a vector in the tangent space to that point, [itex]\mathbf{X}_{p}\in T_{p}\mathcal{M}[/itex], however, I'm finding it difficult to see how they can be treated as independent variables, as surely one is required to specify a point [itex]p\in\mathcal{M}[/itex] before one can determine the tangent space to that point (and thus, the possible tangent vectors that one can choose from)?I have been reading this set of notes http://www.physics.usu.edu/torre/6010_Fall_2010/Lectures/01.pdf and they have been enlightening me a bit on the subject, but I'm a little unsure that I'm understanding it correctly?!

I understand that points [itex]p[/itex] on the manifold [itex]\mathcal{M}[/itex] exist independently of any chosen curve, as do the tangent vectors [itex]\mathbf{X}_{p}[/itex] in the tangent space [itex]T_{p}\mathcal{M}[/itex] to those points, so would it be correct to say that each point in the tangent bundle represents a point on [itex]\mathcal{M}[/itex] through which a curve could pass through at a given velocity? [In the middle of page 6 the lecturer pauses to discuss a "common point of confusion" about the notation [itex]\dot{q}[/itex] (for a vector in the tangent space), stating that it is not the derivative of anything, but just a vector that exists at the point [itex]q[/itex]. Is this just the case because for each given point we can associate a set of vectors (which form a tangent space at that point); they are not derivatives of anything, but they do, in essence, contain information on what direction (and at what speed) curves can move through that point?!]

He then moves on to say that a given curve in the tangent bundle looks like $$q=q(t),\qquad\dot{q}=\dot{q}(t)$$ I assume by this that the curve is parametrised by some parameter [itex]t[/itex] and that a given value of [itex]t[/itex] corresponds to a given point [itex](q,\dot{q})[/itex] in the tangent bundle? Also, he says that such a curve may not actually correspond to any motion of the system and we need [itex]\dot{q}(t)[/itex] to actually represent the tangent to [itex]q(t)[/itex], that is, we need to choose [itex]\dot{q}(t) = \frac{dq(t)}{dt}[/itex]. By this does he mean that although a given value of [itex]t[/itex] will correspond to a point [itex]q=q(t)[/itex] and a tangent vector [itex]\dot{q}=\dot{q}(t)[/itex], the tangent vector associated with that value of [itex]t[/itex] may not be that of the curve passing through [itex]q[/itex], hence we require that, in actual fact, the given value of [itex]t[/itex] corresponds to a vector tangent to the point [itex]q[/itex] such that its value [itex]\dot{q}(t)[/itex] corresponds to the derivative of the curve evaluated at that point, i.e. [itex]\dot{q}(t)=\frac{dq(t)}{dt}[/itex]?

Having done all this, do we then say that the Lagrangian is a one-parameter family of functions (parameterised by [itex]t[/itex]) on the tangent bundle, i.e. $$\mathcal{L}=\mathcal{L}(q,\dot{q}, t)=\mathcal{L}(\lbrace q_{i}\rbrace, \lbrace\dot{q}_{i}\rbrace , t)$$ One may then choose a curve in configuration space [itex]q(t)=\lbrace q_{i}(t) \rbrace[/itex] such that when the Lagrangian is evaluated on that curve, for a given value of [itex]t[/itex] it returns a number [itex]\mathcal{L}(q(t),\dot{q}(t),t)[/itex] corresponding to the value of the Lagrangian evaluated at the point [itex]q=q(t)[/itex] on the curve, whose tangent vector is [itex]\dot{q}(t)=\frac{dq(t)}{dt}[/itex]?

Sorry for the long-windedness of this post, I'm just really keen to get these concepts firmly cemented (correctly) in my mind!

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