# Classical Mechanics and Differential Geometry

Recall Newton's second law which describes the motion of a particle in the presence of a conservative force field: $m \frac{d^2 x}{dt^2} + \nabla V(x) = 0$ where $$m$$ is mass the mass of the particle, $$x \in \mathbb{R}^3$$ is the position of the particle and $$V$$ is the potential energy function. Newton's second law specifies a set of curves in $$\mathbb{R}^3$$. In this post, we are going to describe the same set of curves in a coordinate independent way.

### The Euler-Lagrange equations

Let $$M$$ be a manifold and $$L : TM \to \mathbb{R}$$ a function which we will call the Lagrangian. Define $$\Omega(a,b;M)$$ to be the set of paths from $$a$$ to $$b$$ in $$M$$. We define the action functional $$S : \Omega(a,b;M) \to \mathbb{R}$$ by the equation $S(x) = \int L\left( x,\frac{dx}{dt} \right) dt.$ It is natural to ask which curves are local minima for the action functional. This is a local condition on the curve: If $$x$$ is a local minima, then any small section of $$x$$ is also a local minima for its corresponding action functional. Therefore, we can compute in local coordinates what it means for $$x$$ to be a local minima of the action functional. Let $$x(s,t)$$ be a curve in $$\Omega(a,b;M)$$ which equals $$x$$ when $$s=0$$. It follows from the chain rule that $\frac{d}{ds} S(x) = \int \frac{\partial L}{\partial x^i} \frac{\partial x^i}{\partial s} dt + \int \frac{\partial L}{\partial v^i} \frac{\partial^2 x^i}{\partial t \partial s} dt$ where $$x^i$$ are coordinates on $$M$$ and $$v^i$$ are the corresponding coordinates on the fiber of $$TM$$. Using integration by parts and the fact that $$x(t_0,s) =a, x(t_1,s)=b$$, the second integral equals $\left[ \frac{\partial L}{\partial v^i} \frac{\partial x^i}{\partial s} \right]^{t_1}_{t_0} - \int \frac{d}{dt} \frac{\partial L}{\partial v^i} \frac{\partial x^i}{\partial s} dt = - \int \frac{d}{dt} \frac{\partial L}{\partial v^i} \frac{\partial x^i}{\partial s} dt.$ This implies that $\frac{d}{ds} S(x) = \int \left(\frac{\partial L}{\partial x^i} - \frac{d}{dt} \frac{\partial L}{\partial v^i} \right) \frac{\partial x^i}{\partial s} dt.$ Since our choice of path $$x(s,t)$$ in $$\Omega(a,b;M)$$ was arbitrary, evaluating at $$s=0$$ gives $\left\langle \frac{\partial L}{\partial x^i} - \frac{d}{dt} \frac{\partial L}{\partial v^i},y \right\rangle_{L^2}=0 \; \text{for all loops y}$ which implies that $\frac{\partial L}{\partial x^i} = \frac{d}{dt} \frac{\partial L}{\partial v^i} \; \text{for each i}.$ These are the Euler-Lagrange equations. If we use the Newtonian Lagrangian $L = \frac{1}{2} m (v^1 v^1 + v^2 v^2 + v^3 v^3) - V(x^1,x^2,x^3)$ then the Euler-Lagrange equations simplify to Newton's second law.

### Classical Mechanics on a Riemannian Manifold

Let $$(M,g)$$ be a Riemannian manifold and $$V : M \to \mathbb{R}$$ a function. We can generalize the Newtonian Lagrangian to this case: $L = \frac{1}{2} m g(v,v) - V(x)$ We call $$V(x)$$ the potential energy and $$\frac{1}{2} m g(v,v)$$ the kinetic energy. Let $$x^i$$ be coordinates on $$M$$ and $$v^i$$ the corresponding coordinates on the fibers of $$TM$$. If $$g = g_{ij} dx^i \otimes dx^j$$ then $$g(v,v) = g_{ij} v^i v^j$$ which implies $\frac{\partial L}{\partial v^i} = m g_{ij} v^j.$ Using the Riemannian metric g, we can identify $$TM$$ and $$T^*M$$ via the isomorphism $$v \mapsto g(-,v)$$. In coordinates, it sends the vector $$v^j$$ to $$p_i = g_{ij} v^j$$. Therefore we can write the Euler Lagrange equations as $\frac{d p_i}{dt} = - \frac{1}{m} \frac{\partial V}{\partial x^i}.$ From the equation $$v^i = g^{ij} p_j$$ we have that $\frac{d x^i}{dt} = g^{ij} p_j$ where $$g^{ij}$$ is the induced metric on $$T^*M$$. Therefore, on the cotangent bundle, the Euler-Lagrange equations specify the flow curves of the vector field $\left( g^{ij}p_j, - \frac{1}{m} \frac{\partial V}{\partial x^i} \right).$

### The Symplectic Structure on the Cotangent Bundle

The vector field we saw at the end of the last section is very special, but to see why, we need to understand the canonical symplectic structure on the cotangent bundle of the manifold $$M$$. Let $$\pi : T^*M \to M$$ be the projection. We can define a 1-form $$\alpha$$ on $$T^*M$$ by the equation $\alpha_{\eta}(V) = \eta(\pi' X).$ Let $$x^i$$ be coordinates on $$M$$ and $$p_i$$ the induced coordinates on the fibers of $$T^*M$$. Then we have $\alpha(\partial / \partial p_i) = 0 \quad \alpha(\partial / \partial x^i) = p_i.$ This implies that $$\alpha = p_i dx^i$$. We define $w = d \alpha = d p_i \wedge d x^i = d p_i \otimes d x^i - d x^i \otimes d p_i.$ This is called the canonical symplectic form on $$T^*M$$. The induced map $$w : TM \to T^* M$$ is $\frac{\partial}{\partial x^i} \mapsto d p_i \quad \frac{\partial}{\partial p_i} \mapsto - dx^i$ If $$H : T^*M \to \mathbb{R}$$ is a function, then $X_H = w^{-1} dH = \frac{\partial H}{\partial p_i} \frac{\partial}{\partial x^i} - \frac{\partial H}{\partial x^i} \frac{\partial}{\partial p_i}$ is called the symplectic gradient of $$H$$.

### The Hamiltonian

Now lets return to classical mechanics on a Riemannian manifold. We have have seen that solutions for the Euler-Lagrange equation are flow curves of the vector field $\left( g^{ij}p_j, - \frac{1}{m} \frac{\partial V}{\partial x^i} \right)$ This vector field is exactly the symplectic gradient of $H = \frac{1}{2} g^{ij} p_i p_j + \frac{1}{m} V.$