How does one write Newtons 2nd Law using the language of forms?

Question

Newton's second law says that $F=ma$.

Supposing that the force is conservative and can thus be expressed in terms of a potential $V$ we have that $F=-dV$.

We have that $V$, being a function, can also be considered as a 0-form; so $dV$, and therefore $F$ is a 1-form.

So we ought to consider that $ma$ needs to be expressed as a 1-form; the natural suggestion is $a dx$; but $a$ is a second derivative.

How do I express $a$ as a 1-form naturally?

Answer 1

To my mind, Newton's equation makes the most sense as an equation of vector fields. Let $(M,g)$ be a (Pseudo-)Riemannian manifold with Riemannian connection $\nabla$. Then the equations of motion for a position-dependent conservative force $F$ are given by $$m {}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}=F\circ \gamma=-(\nabla V) \circ \gamma,$$ where $\gamma: \mathbb{R} \supset I \to M$ is the wanted curve, ${}^{\gamma}\nabla_{\frac{d}{dt}}$ is the pullback of the Riemannian connection, and $\nabla V$ is the gradient vector field of $V$ defined through $dV(\cdot)=g(\nabla V, \cdot)$.

Now if you want to write this in terms of differential forms, you'd need to convert ${}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}$ into form language. Do you consider something like $$m (g\circ \gamma)({}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}, \cdot)=-(dV \circ \gamma)(\cdot)$$ as an equation for differential forms along $\gamma$ natural? I don't, and I cannot see a "natural" way of getting around this.

Answer 2

First of all, this is a good question. To formulate my answer I try to use Newton's philosophy with translation to the language to differential geometry.

How does one write Newtons 2nd Law using the language of forms?

There are some ways, depending of how you define momentum, but the natural expression of a force is using vectors. Fortunately, vector can be naturally identified with forms (see one construction bellow).

Newton's second law says that $F=ma$.

Newton's second law says that the time derivative of momentum is proportional to the force applied, this implies:

1. If the evolution of the system in the space of possibles momenta is parametrized by time then the image of the force is a tangent vector to this space.
2. At a given instant, we can say the magnitude and direction of the force without any evidence of its causes, therefore the law say nothing about the domain of the force.

Primitive notions

• The physical space:
  • The space is the Euclidean space, here treated as a 3-manifold $\mathcal M$. [0, Definitiones, Scholium, II]
  • The trajectory of the body is a curve in $\mathcal M$ parametrized by time, i.e., a function $\gamma:I\subset\mathbb R\rightarrow\mathcal M:t\mapsto\gamma(t)$, where $I$ is open, I also suppose that $\gamma$ is two times differentiable.
• The domain of the force:
  • Each possible state of the body is a point in an arbitrary (but well-defined) set $\mathcal P$ called phase space.
  • The evolution of the state of the body is a curve in $\mathcal P$ parametrized by time, i.e., a function $\Phi_z:I\rightarrow\mathcal P:t\mapsto \Phi_z(t)$, none hypothesis are made.

Without reference to coordinates

• Secondary notions:
  • The velocity of the body is the curve in $T\mathcal M$ (the tangent bundle) parametrized by time defined as the derivative of position, i.e., $\gamma':I\subset\mathbb R\rightarrow T\mathcal M:t\mapsto \frac{d}{dt}\gamma(t)$.
  • The mass is a bundle isomorphism $M:T\mathcal M\rightarrow B$ where the vector bundle $B$ is usually the tangent bundle $T\mathcal M$ itself or the cotangent bundle $T^*\mathcal M$ (I prefer use the cotangent bundle).
  • The momentum of the body is the curve in $B$ induced by the mass and the velocity, i.e., $\gamma^\flat:I\subset\mathbb R\rightarrow B:t\mapsto M\circ\gamma'(t)$.
• Newton's second law: the time derivative of momentum is proportional to the force and is in the same direction.[0, Lex II]
  • Since $\frac{d}{dt}\gamma^\flat(t)$ lies in $T B$ then the force is a map $F:\mathcal P\rightarrow T B$.
  • Newton's second law is written as: $\frac{d}{dt}\gamma^\flat(t) = F\circ\Phi_z(t)$.

Problem: Find a differential 1-form $F^\flat$ defined in $B$ such that $F^\flat|_{\gamma^\flat(t)} = \mu\circ F\circ\Phi_z(t)$ for some bundle isomorphism $\mu:TB\rightarrow T^*B$.

To solve the problem, pay attention to the principle of inertia:

• Newton's first law: In absence of applied forces, the trajectory that solves $\frac{d}{dt}\gamma^\flat(t) = F\circ\Phi_z(t)$ is the trivial solution or the straight line. [0, Lex I]
  • This law govern a force of inertia $F = F_0$ such that the solutions are geodesics.[0, Definitio III]
  • If $B = T^*\mathcal M$ then this force of inertia is called Hamiltonian geodesic field because one can shown that $F_0 \lrcorner \omega = d\mathcal H_0$, where the Hamiltonian function $\mathcal H_0:T^*\mathcal M\rightarrow\mathbb R:\vec p\mapsto\frac{1}{2}\vec p(M^{-1}(\vec p))$ is the kinetic energy, $\lrcorner$ denote the interior product and $\omega$ is the canonical symplectic 2-form on $T^*\mathcal M$ defined as $\omega = -d\theta$, where the 1-form $\theta$ is defined such that for all $\vec p\in T^*\mathcal M$ we have $\theta\vert_{\vec p} = \vec p \circ d\pi\vert_{\vec p}$, $\pi:T^*\mathcal M\rightarrow \mathcal M$ is the bundle projection and $d\pi:TT^*\mathcal M\rightarrow T\mathcal M$ is the induced tangent map.

Finally, one write Newton's second law using the language of forms:
Theorem. Let $\mathcal P \cong B = T^*\mathcal M$, for any force $F$ there are a 1-form $F^\flat$ on $B$ such that $$ F\lrcorner\omega = F^\flat + d\mathcal H_0, \quad \text{or, equivalently,} \quad (F-F_0)\lrcorner\omega = F^\flat. $$

Proof: Informally, this is a consequence of the fact that $\omega$ is non-degenerate and induces a bundle isomorphism. Q.E.D.

Local expression in coordinates

To better understand the above construction I propose an attention to a local coordinate chart defined in $B$. Let $B = T^*\mathcal M$.

• Local coordinate chart:
  • Let $U\subset\mathcal M$ be a open set and in the frame or reference of some observer $(x^1, x^2, x^3):U\subset\mathcal M\rightarrow\mathbb R^3$ be a (possibly curvilinear) coordinate chart.
  • Let $(q, p) = (q^1, q^2, q^3, p_1, p_2, p_3)$ the coordinate chart on $\pi^{-1}(U)$ defined as follows: $q^i = x^i \circ \pi$ and $p_i:\pi^{-1}(U)\subset T^*\mathcal M\rightarrow\mathbb R:\sigma\mapsto \sigma\left(\frac{\partial}{\partial q^i}\right)$.
• Notation:
  • Denote the force as $F = \sum_{i = 1}^3 \dot q^i \frac{\partial}{\partial q^i} + \dot p_i \frac{\partial}{\partial p_i}$.
  • Denote its differential 1-form as $F^\flat = \sum_{i = 1}^3 v^i(q, p) dp_i - f_i(q, p) dq^i$.
  • The canonical symplectic 2-form is written as $\omega = \sum_{i = 1}^3 dq^i\wedge dp_i$.
  • The kinetic energy is written as $\mathcal H_0 = \frac{1}{2}\sum_{i = 1}^3\sum_{j = 1}^3 M^{ij} p_i p_j$, where $M^{ij} = dx^i(M^{-1}(dx^j))$.
• The equation:
  • $F\lrcorner\omega = F^\flat + d\mathcal H_0$ implies $$ \sum_{i = 1}^3 \dot q^i dp_i - \dot p_i dq^i = \sum_{i = 1}^3 \left(\frac{\partial}{\partial p_i}\mathcal H_0 + v^i(q, p)\right) dp_i - \left(f_i(q, p) - \frac{\partial}{\partial q^i}\mathcal H_0\right) dq^i, $$
  • In dynamical equilibrium $v_i(q, p) = f^i(q, p) = 0$ holds.
  • For a Hamiltonian force, there exists a function $\mathcal H_1$ such that $v_i(q, p) = \frac{\partial}{\partial q^i}\mathcal H_1$ and $f^i(q, p) = -\frac{\partial}{\partial p_i}\mathcal H_1$, consequently $\mathcal H_0 + \mathcal H_1$ is a constant of motion.
  • For a conservative force $v_i(q, p) = 0$ and there exists a function $V(q)$ such that $f^i(q, p) = -\frac{\partial}{\partial q^i}V$.

Notes

1. As you probably as noted, the hypothesis that $\mathcal M$ is Euclidean is completely irrelevant.
2. If the manifold $\mathcal M$ have a inner product, i.e., a metric tensor $\eta$, then $M$ is naturally defined such that $[M(X)](Y) = m~\eta(X, Y)$, where $m$ is a real positive constant, the mass of the body.
3. In reference [1] a similar approach are made using $B = T\mathcal M$.

References

[0] - Newton, Isaac, Philosophiæ Naturalis Principia Mathematica, London, 1687; Cambridge, 1713; London, 1726.
[1] - Newton's second law in field theory: https://doi.org/10.1016/j.difgeo.2021.101814
[2] - Mathematical Methods of Classical Physics: https://arxiv.org/pdf/1612.03100.pdf
[3] - Christoph (https://physics.stackexchange.com/users/6389/christoph), Representing forces as one-forms, URL (version: 2013-02-25): https://physics.stackexchange.com/q/54948

Answer 3

I think there is a need for a gentler introduction. If all you want is to insist that the potential version of N2L be written in the language of forms, then the crucial step is to realise that N2L is not naturally a vector equation. It is naturally a 1-form equation, because the canonical momentum (that is conserved when the coördinate is missing from the Lagrangian/Hamiltonian) is naturally a covariant quantity, not a contravariant quantity like vectors are. In concrete terms, $\mathrm d x^\mu F_\mu = \mathrm d x^\mu \frac{\mathrm d \ }{\mathrm d t} p_\mu$, and the usual $\frac{\mathrm d\ }{\mathrm d t} p_\mu = m g_{\mu\nu} a^\nu$ is a physically unnatural simplification for high schoolers. Note that it is physically the case that covariant momentum components $p_\mu$ are the ones that are conserved, but cannot be visualised as vectors pointing in any sensible direction. The contravariant momentum components $p^\nu = g^{\nu\mu}p_\mu$ make a vector that can point in some direction, but is not conserved. The easiest way to see this is true is if you just consider angular momentum in 2D.

So, if you really want to insist, then you can express N2L as $\mathrm d x^\mu \frac{\mathrm d\ }{\mathrm d t} p_\mu = - \mathrm d V$, and maybe there is a way to express this in terms of the symplectic one-form $\omega$.

One pleasing part of this is that forces have dropped out of this form. Yes, you may insist to bring it back, but given that we already know that forces are so ill-defined as to be the first on the chopping block when the quantum revolution came, it is fun to bid it goodbye here. Even as early as Hertz, the classical physics opposition to the expression of classical mechanics in terms of forces.

However, this is still not the smart thing to scrutinise. The reason is that the mathematically sensible thing to try and generalise would be the Lagrangian and Hamiltonian mechanics. This is where the much more elaborate answers that the other people have given come in. I might disagree with some of the characterisation of the stuff, but the main parts should be ok.

Edit: I forgot to point out that while in high school classical mechanics we are interested in particle trajectories (and in Lagrangian mechanics that can be in configuration space), it is actually somewhat more natural to think of things as an entire field. In Hamiltonian flow, it is the entire phase space that is flowing that is natural to express. The above arguments often invoke paths, and that might be why they are so ugly.

Answer 4

You can write

$$\text dV = - m v \text d v$$

where $v$ is the velocity.