Classical Mechanics 2

Dynamics of Finite-Dimensional Euclidean Spaces and Particle Systems.

1 Dynamics in Finite-Dimensional Euclidean Space

We now extend our consideration to dynamics within finite-dimensional Euclidean space \(\mathbb{R}^n\). In this context, any quantity that is not explicitly a scalar will be denoted in boldface, such as the position vector \(\mathbf{x} = (x_1, x_2, \dots, x_n)\).

The position of a particle in \(\mathbb{R}^n\) is a vector-valued function of time \(\mathbf{x}(t)\). Following the standard conventions of vector calculus, the velocity is defined as the derivative of the position vector: \[\mathbf{v} = \mathrm{D}\mathbf{x} = \left( \frac{dx_1}{dt}, \dots, \frac{dx_n}{dt} \right).\] Newton’s second law in \(\mathbb{R}^n\) is expressed as the following system of second-order differential equations: \[m\mathrm{D}^2 \mathbf{x} = \mathbf{F}(\mathbf{x}, \mathrm{D}\mathbf{x}),\] where the force \(\mathbf{F}: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n\) depends on both the position and the velocity of the particle.

When the force depends solely on the position, the existence of a conserved energy is linked to the existence of a potential scalar field.

NoteProposition: Conservation of Energy in \(\mathbb{R}^n\)

Consider Newton’s law for a position-dependent force: \(m\mathrm{D}^2\mathbf{x}(t) = \mathbf{F}(\mathbf{x}(t))\). The energy function \[E(\mathbf{x}, \mathrm{D}\mathbf{x}) = \frac{1}{2} m |\mathrm{D}\mathbf{x}|^2 + V(\mathbf{x})\] is conserved if and only if there exists a potential function \(V\) such that \[-\nabla V = \mathbf{F}.\]

Caution

To prove the “if” part, assume \(\mathbf{F} = -\nabla V\). Differentiating \(E\) with respect to \(t\) along a trajectory \(\mathbf{x}(t)\) yields: \[\mathrm{D}_t E = \mathrm{D}_t \left( \frac{1}{2}m \langle \mathrm{D}\mathbf{x}, \mathrm{D}\mathbf{x} \rangle + V(\mathbf{x}) \right) = m \langle \mathrm{D}\mathbf{x}, \mathrm{D}^2\mathbf{x} \rangle + \langle \nabla V, \mathrm{D}\mathbf{x} \rangle.\] Substituting Newton’s law \(m\mathrm{D}^2\mathbf{x} = \mathbf{F}\) and the potential relation \(\nabla V = -\mathbf{F}\), we obtain: \[\mathrm{D}_t E = \langle \mathrm{D}\mathbf{x}, \mathbf{F} \rangle + \langle -\mathbf{F}, \mathrm{D}\mathbf{x} \rangle = 0.\] Conversely, if \(\mathrm{D}_t E = 0\) for all trajectories, then \(\langle \mathrm{D}\mathbf{x}, m\mathrm{D}^2\mathbf{x} + \nabla V \rangle = 0\). Since this must hold for any velocity \(\mathrm{D}\mathbf{x}\), it follows that \(m\mathrm{D}^2\mathbf{x} = -\nabla V\). Comparing this with Newton’s law \(m\mathrm{D}^2\mathbf{x} = \mathbf{F}\), we conclude \(\mathbf{F} = -\nabla V\).

In higher dimensions, we encounter topological and analytical complexities that do not exist in the one-dimensional case. Specifically, while every continuous function in \(\mathbb{R}^1\) is the derivative of some potential, a vector field in \(\mathbb{R}^n\) is not necessarily a gradient.

TipDefinition: Conservative Force

Let \(\mathbf{F}\) be a smooth vector field defined on an open region \(U \subseteq \mathbb{R}^n\). \(\mathbf{F}\) is said to be conservative if there exists a smooth scalar function \(V\) on \(U\) such that \(\mathbf{F} = -\nabla V\).

If the domain \(U\) is simply connected, the condition for conservativeness can be characterized by the vanishing of the field’s partial derivative commutators.

NoteProposition: Integrability Condition

Let \(U \subseteq \mathbb{R}^n\) be a simply connected region and \(\mathbf{F}\) be a smooth vector field on \(U\). Then \(\mathbf{F}\) is conservative if and only if the following symmetry condition holds for all \(j, k \in \{1, \dots, n\}\) at every point in \(U\): \[\mathrm{D}_{x_k} F_j = \mathrm{D}_{x_j} F_k. \]¹

¹ In the specific case of \(n=3\), this condition is equivalent to the vanishing of the curl: \(\text{curl } \mathbf{F} = \nabla \times \mathbf{F} = \mathbf{0}\).

Caution

The necessity (the “only if” part) follows from the equality of mixed partial derivatives (Clairaut’s Theorem). If \(\mathbf{F} = -\nabla V\), then \(F_j = -\partial V / \partial x_j\), and thus \(\partial F_j / \partial x_k = -\partial^2 V / \partial x_k \partial x_j = \partial F_k / \partial x_j\).

For the sufficiency, if \(U\) is simply connected, we can define a potential function \(V(\mathbf{x})\) by the line integral \(V(\mathbf{x}) = -\int_{\gamma} \mathbf{F} \cdot d\mathbf{r}\), where \(\gamma\) is any path from a fixed point \(\mathbf{x}_0\) to \(\mathbf{x}\). By Stokes’ Theorem, the condition \(\mathrm{D}_{x_k} F_j = \mathrm{D}_{x_j} F_k\) ensures that the integral is path-independent, making \(V\) well-defined and ensuring \(\nabla V = -\mathbf{F}\).

It is crucial to note that the topological requirement of simple connectivity is indispensable. If \(U\) contains “holes,” a field may satisfy the local symmetry condition but fail to be globally conservative.

ImportantExample: The Vortex Field

Consider the vector field \(\mathbf{F}\) defined on the non-simply connected region \(U = \mathbb{R}^2 \setminus \{(0,0)\}\) by: \[\mathbf{F}(x, y) = \left( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right).\] One can verify that \(\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y} = \frac{y^2 - x^2}{(x^2 + y^2)^2}\) throughout \(U\). However, the line integral of \(\mathbf{F}\) along a unit circle \(C\) centered at the origin is: \[\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left( \frac{-\sin \theta}{1}, \frac{\cos \theta}{1} \right) \cdot (-\sin \theta, \cos \theta) d\theta = \int_0^{2\pi} 1 d\theta = 2\pi \neq 0.\] Since the closed-loop integral is non-zero, \(\mathbf{F}\) cannot be represented as the gradient of a single-valued potential \(V\) on \(U\), despite satisfying the local integrability condition.

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2 Velocity-Dependent Conservative Forces

In higher dimensions, we encounter a phenomenon that has no counterpart in the one-dimensional case: a velocity-dependent force can still result in a conservative system, provided the force satisfies a specific orthogonality condition.

NoteProposition: Generalized Conservation of Energy

Consider a particle in \(\mathbb{R}^n\) moving under the influence of a force \(\mathbf{F}(\mathbf{x}, \mathbf{v})\) of the form: \[\mathbf{F}(\mathbf{x}, \mathbf{v}) = -\nabla V(\mathbf{x}) + \mathbf{F}_2(\mathbf{x}, \mathbf{v}),\] where \(V\) is a smooth potential function and the additional force component \(\mathbf{F}_2\) satisfies the orthogonality condition: \[\langle \mathbf{v}, \mathbf{F}_2(\mathbf{x}, \mathbf{v}) \rangle = 0\] for all \((\mathbf{x}, \mathbf{v}) \in \mathbb{R}^n \times \mathbb{R}^n\). Then the total energy function \(E(\mathbf{x}, \mathbf{v}) = \frac{1}{2}m|\mathbf{v}|^2 + V(\mathbf{x})\) is constant along every trajectory \(\mathbf{x}(t)\).

Caution

To prove this, we compute the time derivative of the energy \(E\) along an arbitrary orbit \(\mathbf{x}(t)\) with velocity \(\mathbf{v}(t) = \mathrm{D}\mathbf{x}(t)\). Applying the chain rule, we obtain: \[\mathrm{D}_t E = m \langle \mathbf{v}, \mathrm{D}\mathbf{v} \rangle + \langle \nabla V, \mathbf{v} \rangle.\] Substituting Newton’s law \(m\mathrm{D}\mathbf{v} = \mathbf{F}(\mathbf{x}, \mathbf{v}) = -\nabla V(\mathbf{x}) + \mathbf{F}_2(\mathbf{x}, \mathbf{v})\), the expression becomes: \[\mathrm{D}_t E = \langle \mathbf{v}, -\nabla V + \mathbf{F}_2 \rangle + \langle \nabla V, \mathbf{v} \rangle.\] Distributing the inner product yields: \[\mathrm{D}_t E = -\langle \mathbf{v}, \nabla V \rangle + \langle \mathbf{v}, \mathbf{F}_2 \rangle + \langle \nabla V, \mathbf{v} \rangle.\] The first and last terms cancel due to the symmetry of the inner product. By the given hypothesis \(\langle \mathbf{v}, \mathbf{F}_2 \rangle = 0\), the remaining term vanishes: \[\mathrm{D}_t E = 0.\] Thus, the energy \(E\) is an integral of motion for the system.

The most prominent physical example of such a force is the magnetic component of the Lorentz force acting on a charged particle.

TipExample: Motion in a Magnetic Field

Consider a particle with charge \(q\) moving in \(\mathbb{R}^3\) through a static magnetic field \(\mathbf{B}(\mathbf{x})\). The force exerted by the magnetic field is given by the cross product: \[\mathbf{F}_2(\mathbf{x}, \mathbf{v}) = q(\mathbf{v} \times \mathbf{B}(\mathbf{x})).\] Using the fundamental property of the triple scalar product, we observe that the power delivered by this force is: \[\mathbf{v} \cdot \mathbf{F}_2 = q \mathbf{v} \cdot (\mathbf{v} \times \mathbf{B}).\] Since the vector \(\mathbf{v} \times \mathbf{B}\) is, by definition, orthogonal to \(\mathbf{v}\), it follows that \(\mathbf{v} \cdot (\mathbf{v} \times \mathbf{B}) = 0\) for any velocity \(\mathbf{v}\). Consequently, a purely magnetic field does no work on a charged particle and does not change its kinetic energy, acting only to deflect its direction of motion.

This result implies that the speed of a particle remains constant when moving in a purely magnetic field, even if the field \(\mathbf{B}(\mathbf{x})\) is spatially non-homogeneous.