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Apparently one identifies the configuration space in physics often with a manifold $M$. The tangent bundle $TM$ is then the space of all possible positions and velocities.

Furthermore, many sources seem to claim that $T^*M$ can be regarded as the phase space, where $(q,p) \in T^*M$ satisfies by definition that $p \in T_q^*M.$

Again by definition this means that $p:=\partial_2L$ takes velocities as arguments and is linear(!) in them. Unfortunately, I don't see from the definition of the momentum by the Lagrangian why this should be a linear functional. So something is confusing me here.

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  • $\begingroup$ 'Apparently' and 'seem to claim' are pretty denigrating phrases. V.I. Arnold's famous book would be a good reference to check. An online source is math.ucr.edu/home/baez/classical, in particular math.ucr.edu/home/baez/classical/cm05week06.pdf $\endgroup$
    – David Roberts ♦
    Commented Apr 17, 2015 at 0:55
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    $\begingroup$ @DavidRoberts: I think you're reading something into the question. It looks to me like the OP is just being cautious ... $\endgroup$
    – Nik Weaver
    Commented Apr 17, 2015 at 0:59
  • $\begingroup$ OK, let me phrase my response like this: yes, the configuration space in physics is a symplectic manifold $(M,\omega)$ en.wikipedia.org/wiki/Symplectic_manifold, and "in the Hamiltonian formulation of classical mechanics...this manifold's cotangent bundle describes the phase space of the system." (ibid). References there should help also (apologies for grumpy-seeming tone). $\endgroup$
    – David Roberts ♦
    Commented Apr 17, 2015 at 1:02
  • $\begingroup$ @DavidRoberts yes, that was my impression from reading the references: A phase space is a cotangent bundle, but then I thought that elements in the cotangent space need to be linear as it is a dual space and so I got confused. The reason is: I don't see why a momentum has to be linear in the velocities just by the definition via the Lagrangian or is this some additional physical input here? $\endgroup$
    – Physicist 2.0
    Commented Apr 17, 2015 at 1:09
  • $\begingroup$ Did you mean to write $p\in T^*_qM$ instead of $p\in T^*_pM$? $\endgroup$
    – Josh Burby
    Commented Apr 17, 2015 at 5:12

2 Answers 2

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The Lagrangian is a function on the tangent bundle $L:TM\rightarrow\mathbb{R}$. Given a point $q\in M$ and a Lagrangian, we can define a function $L_q:T_qM\rightarrow \mathbb{R}$ using the simple formula $L_q(v_q)=L(v_q)$, where $v_q\in T_qM$ is a tangent vector at $q\in M$. Notice that $L_q$ is a mapping between the vector spaces $T_qM$ and $\mathbb{R}$. We may therefore consider the Frechet derivative $DL_q:T_qM\rightarrow L(T_qM,\mathbb{R})$, where $L(T_qM,\mathbb{R})$ is the space of linear maps between $T_qM$ and $\mathbb{R}$. Notice that $DL_q$ is essentially the partial derivative of $L$ in the velocity direction.

Here is how the cotangent bundle comes in. The set $L(T_qM,\mathbb{R})$ is precisely the cotangent space at $q$, i.e. $L(T_qM,\mathbb{R})=T^*_qM$. Therefore the Frechet derivative of $L_q$ is a map of the form $DL_q:T_qM\rightarrow T^*_qM$. This observation can be used to construct a mapping of $TM$ into $T^*M$ given by $v_q\mapsto (DL_q)(v_q)\equiv p_q\in T^*_qM$, which is known as the Legendre transform.

Does this help?

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    $\begingroup$ Are you sure that's called the Legendre transform? The Legendre transform relates the Lagragian $L$ to the Hamiltonian $H$. (See, e.g., en.wikipedia.org/wiki/… ) $\endgroup$
    – José Figueroa-O'Farrill
    Commented Apr 17, 2015 at 10:29
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    $\begingroup$ @JoséFigueroa-O'Farrill I guess it depends on who you ask. In Abraham and Marsden on p. 219 they say "The transformation $\mathbf{F}L:TQ\rightarrow T^*Q$ thus maps the Lagrange equations into the Hamilton equations. In the literature $\mathbf{F}L$ itself is sometimes called the Legendre transformation (e.g. Sternberg [1964]), while classically the name is usually reserved for the map that takes...[$L$ to $H$]." The Sternberg reference is this I think: amazon.com/Lectures-Differential-Geometry-Chelsea-Publishing/dp/…. $\endgroup$
    – Josh Burby
    Commented Apr 17, 2015 at 14:27
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    $\begingroup$ I should also say that $\mathbf{F}L(v_q)=(DL_q)(v_q)$. $\endgroup$
    – Josh Burby
    Commented Apr 17, 2015 at 14:34
  • $\begingroup$ @JoshBurby, very interesting, I am starting to learn about this stuff, would you have any good reading recommendations? $\endgroup$
    – roi_saumon
    Commented Apr 9, 2019 at 12:27
  • $\begingroup$ @JoshBurby doesn't the Frechet derivative need the concept of norm? $\endgroup$
    – roi_saumon
    Commented Jun 1, 2019 at 18:26
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$p$ is the differential of L with respect to the second variable ($\dot q$), so it represents a linear functional on the tangent space at $q, \dot q$), given by

$$ (u,v)\to \frac{d}{dt} L(q,\dot q+tv)|_{t=0}.$$

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  • $\begingroup$ but the statement was that it is a linear functional on the tangent space at $q$? $\endgroup$
    – Physicist 2.0
    Commented Apr 17, 2015 at 1:16
  • $\begingroup$ More accurately it should be a linear functional on $T_{(q,\dot q)} TM$, which can be canonically identified as $T_qM$. $\endgroup$
    – Fan Zheng
    Commented Apr 17, 2015 at 1:35
  • $\begingroup$ What is $u$? Also, I thought the momentum of a particle at $q\in M$ is a linear functional on the tangent space $T_qM$, as the question suggests, and not a linear functional on $T_{(q,\dot{q})}TM$. Note that the dimension of $T_{(q,\dot{q})}TM$ is twice the dimension of $M$, whereas the classical formula $p_i=\partial L/\partial q^i$ suggests that the momentum should have only $\text{dim}(M)$ components. $\endgroup$
    – Josh Burby
    Commented Apr 17, 2015 at 5:07
  • $\begingroup$ Yes, your expression is clearer. I should have discarded the $u$ and restricted $L$ to $T_qM$. It's not used anyway. $\endgroup$
    – Fan Zheng
    Commented Apr 17, 2015 at 5:27

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