# Why is the momentum a covector?

Can someone tell me why the momentum is an element of the cotangent space?

More detailed: if we have some smooth manifold M and the cotangent space $T_{x}M^{*}$ I know that the momentum p is an element of $T_{x}M^{*}$, but I have no intuition why.

In my theoretical mechanics lecture my professor told us that the generalized momentum with components $p_{a}=\frac{\partial L}{\partial \dot{q}^{a}}$ from Hamilton mechanics is a covector, but we never spoke about the "regular" momentum $p=mv$. Is it a covector too?

Edit: I read that linked article before I wrote mine and I think mine is not a duplicate. The similarities are that we both want an intuitive explanation why the momentum is a covector. The only answer to his article is an explanation why the 1-form p acts linear on velocities. Indeed it is a good answer but it does not give any intuition why the momentum is a covector. Furthermore I asked if the "regular" momentum $p=mv$ is a covector too. This aspect is missing completely in the linked post.

• possible duplicate of Momentum a cotangent vector? Commented Jun 14, 2015 at 11:09
• If this thread gets closed due to duplication I would be thankful if someone could answer my question in the linked thread.
– user247741
Commented Jun 14, 2015 at 11:46
• Fair enough, I've retracted my close vote. Commented Jun 14, 2015 at 13:14
• Being a multiple of the velocity i.e. $p=mv$ it seems that the "regular" momentum is a vector. Wikipedi confirms this point of view: en.wikipedia.org/wiki/Momentum so which seems to be a covector is the generalized one. Commented Jun 14, 2015 at 13:42
• A short answer (with just multivariate calculus ideas, no manifold ideas): the (non-generalized) momentum is the gradient of the kinetic energy. The gradient of a scalar function is naturally a covector, because it acts on vectors to give back the scalar-valued directional derivative. Only through the inner product can we identify covectors with vectors.
– Ian
Commented Jun 14, 2015 at 16:13

Addendum: if you know Lagrangian mechanics, there the generalised momentum is defined to be $$\frac{\partial L(q,\dot{q}, t)} {\partial \dot{q}}$$ because this is the thing that is conserved if one of the coordinates is cyclic. This is clearly a linear function on the generalised velocities, so you can identify it with a covector.

For a free particle $L = \frac{1}{2} m \dot{q} ^2$ so that Lagrange's equation implies $$\frac{d}{dt}\frac{\partial L(q,\dot{q}, t)} {\partial \dot{q}} = 0 \implies m\dot{q} = const$$ Note that this is no longer a statement about $p=mv$ itself, but about a linear function of it. In other conditions there may be nothing interesting to say about $m\dot{q}$, but if the Lagrangian doesn't depend on $q$, $\partial L / \partial \dot{q}$ will still be conserved.

Short answer: for a coordinate system $(q^1,...q^n)$ on a manifold $M$ we let the generalised momenta $(p_1,...,p_n)$ be a basis for the contangent space which acts on $\lambda \in \pi^{-1}(M) \subset T^*M$ by $p_i(\lambda) = \lambda(\frac{\partial}{\partial q^i})$ where $\pi: T^*M \rightarrow M$ is the projection map . This gives the same results when $M$ is a vanilla vector space even though here momentum is not quite $p=mv$.

The underlying reason for this is that in Hamiltonian mechanics, the physics actually happens in the cotagent bundle, the 2n dimensional manifold parametrized by $(q^1 \circ \pi, ... , q^n \circ \pi, p_1, ..., p_n)$.

This formalism is motivated by the somewhat symmetric hole played by the $q^i$ and the $p_i$ in Hamilton's equation, so that we eventually forget about the base manifold and actually consider arbitrary 2n dimensional manifolds equipped with a anti-symmetrical non degenerate differential form (the symplectic form) which distinguishes the position from the momenta.

The topic is too big to explain in detail here, but looking at mechanics in this way gives you many deep results relatively easily. For instance, conservation on the symplectic form under motions implies conservation of volume of phase space. Also, due to the similarities between Hamilton's equations and the Cauchy Riemann equations, complex analysis methods can give some insight. This is the field of pseudoholomorphic curves.

For an introduction see the last chapters of Spivak's Physics for Mathematicians.

• You said that each $p_i \in T^*M$, isn'it ? I am asking you because usually elements of $T^*M$ acts on vectors i.e. elements of $TM$. But as far as I read yours $p_i$ act on $\lambda \in T^*M$. Could you please make clear this point. Commented Jun 14, 2015 at 15:46
• @Holonomia In the LHS $p_i$ is the projection function into i-th component, and on the RHS a point $\lambda \in T^*M$ is acting on a tangent vector. Spivak sticks to this rather confusing notation throughout his books. Commented Jun 14, 2015 at 15:58
• OK, I think I got it: the generalized momentum is the differential of the Lagrangian w.r.t. the coordinates $\dot{q}$ hence is a covector being a differential. Commented Jun 14, 2015 at 16:28
• Thank you for your answer. The notation with $p_{i}$ and $\lambda$ is indeed a bit confusing, but still a really helpful post.
– user247741
Commented Jun 14, 2015 at 16:58
• It depends on the Lagrangian. For the free particle the momentum acts by multiplying by the mass. For a central force and generalised position (angle) $\theta$ it acts as $p(\dot{\theta}) = mr^2\dot{\theta}$, which is angular momentum. Your last equation is almost right. To get the action on the vector you must evaluate the momentum at that vector, so that $p(X^i \partial / \partial x^i |_q) = \frac{\partial L(q, \dot{q}, t)}{\partial \dot{q}} |_{\dot{q}=(X^1,...,X^n)}$. Commented Jun 14, 2015 at 21:18

I think the point is $$p=mv$$ being very very special case is a misleading idea about momentum. Even a point system of particle in $$R^3$$ for some same reasons is misleading: space have non-important (for our discussion) flat structure and we add up momentums of particles as vector and get a vector! In the general case (real case!) on configuration manifold at a specific point $$(q,\dot q)$$, momentum is not a vector parallel to $$\dot q$$ at all. Rather it is a question: if I'd like to disturb evolution of system from $$\dot q$$ in a specific direction how much change in Lagrangian will be seen? (in this sense intuitively general momentum have common feature by special case $$mv$$ in some inertia property). Then, for every specific direction on manifold (a vector) you get a real number and so momentum, being that question, is a 1-form.

• nice answer. helped me grasp the idea Commented Jun 23, 2020 at 10:14
• I had never thought of covectors as questions before, but that's a great intuition! Commented Jun 9, 2023 at 0:35

The Lagrangian $$L(q, v)$$ is a function on the tangent bundle $$TM$$. The Hamiltonian function is defined as the Legendre transform of the Lagrangian: \begin{align*} H(q, p) = \sum_i p_i v_i - L \end{align*} The Hamiltonian is a scalar. Thus, the right hand side has to be a scalar. Now $$v_i$$ transforms like components of a vector, and thus $$p_i$$ has to to transform like components of a covector. Thus $$p \in T^*_qM$$, and $$(q, p) \in T^*M$$.

Another way to think about this is that the momentum transforms as a covector, not as a vector. Given generalized coordinates $$q^1,\dots,q^n$$, then under a coordinate transformation $$\mathbf{q}\mapsto \mathbf{q}' = \mathbf{q}'(\mathbf{q})$$, the generalized velocities transform as a vector (i.e. elements of the tangent space $$TQ$$): $$\begin{equation*} \dot{q}'^\mu = \frac{\partial q^\mu}{\partial q'^\nu}\dot{q}'^\nu, \end{equation*}$$ and so the operator $$\frac{\partial}{\partial\dot{q}^\mu}$$ transforms with the inverse, i.e. as a covector: $$\begin{equation*} \frac{\partial}{\partial\dot{q}'^\mu} = \frac{\partial q'^\nu}{\partial q^\mu}\frac{\partial}{\partial\dot{q}^\nu}. \end{equation*}$$ With this we see that the momenta transform as a covector too: $$\begin{equation*} p'_\mu = \frac{\partial L}{\partial\dot{q}'^\mu} = \frac{\partial q'^\nu}{\partial q^\mu}\frac{\partial L}{\partial\dot{q}^\nu}= \frac{\partial q'^\nu}{\partial q^\mu}p_\nu. \end{equation*}$$

This means that the momenta cannot simply be rescalings of the velocities, since their transformation properties are fundamentally different. But even more, they transform like covectors, i.e. elements of the cotangent space $$T^* Q$$.

• I think the OP is more interested in WHY the momentum transform as a covector, and not how does a momentum transform. Commented Jul 28, 2021 at 18:13