Classical Mechanics 3
Particle Systems.
1 Conservation of Energy for Systems of Particles
Consider a system consisting of \(N\) particles, each moving in \(\mathbb{R}^n\). Let \(\mathbf{x}^j\) denote the position of the \(j\)-th particle. We define the configuration of the entire system as a single vector (or matrix) in \(\mathbb{R}^{nN}\), denoted by \(\boldsymbol{X} := (\mathbf{x}^1, \mathbf{x}^2, \dots, \mathbf{x}^N)\). Newton’s second law for this system is expressed as a collection of \(N\) vector equations: \[m_j \mathrm{D}^2 \mathbf{x}^j = \mathbf{F}^j(\boldsymbol{X}, \mathrm{D}\boldsymbol{X}), \quad j = 1, \dots, N,\] where \(m_j\) is the mass of the \(j\)-th particle and \(\mathbf{F}^j\) is the force acting upon it, which generally depends on the positions and velocities of all particles in the system.
In such a multi-particle system, the energy of an individual particle is typically not conserved due to the continuous exchange of energy through interactions. However, under suitable assumptions on the forces \(\mathbf{F}^j\), one can define a total energy function for the entire system that remains invariant over time.
Suppose the forces depend only on the positions \(\boldsymbol{X}\). We define the total energy \(E\) of the system as the sum of individual kinetic energies and a collective potential energy \(V\): \[E(\boldsymbol{X}, \boldsymbol{V}) = \sum_{j=1}^{N} \frac{1}{2} m_j |\mathbf{v}^j|^2 + V(\boldsymbol{X}).\]
To verify this, we differentiate the total energy \(E\) with respect to time \(t\) along the system’s path \(\boldsymbol{X}(t)\). Let \(\mathbf{v}^j = \mathrm{D}\mathbf{x}^j\). Using the multivariable chain rule, we have: \[\mathrm{D}_t E = \sum_{j=1}^N \left( m_j \langle \mathbf{v}^j, \mathrm{D}\mathbf{v}^j \rangle \right) + \sum_{j=1}^N \langle \nabla^j V, \mathbf{v}^j \rangle.\] Substituting the equations of motion \(m_j \mathrm{D}\mathbf{v}^j = \mathbf{F}^j\) and the potential relation \(\nabla^j V = -\mathbf{F}^j\), the expression becomes: \[\mathrm{D}_t E = \sum_{j=1}^N \langle \mathbf{v}^j, \mathbf{F}^j \rangle + \sum_{j=1}^N \langle -\mathbf{F}^j, \mathbf{v}^j \rangle = \sum_{j=1}^N \left( \langle \mathbf{v}^j, \mathbf{F}^j \rangle - \langle \mathbf{v}^j, \mathbf{F}^j \rangle \right) = 0.\] Thus, the total energy \(E\) is an integral of motion for the \(N\)-particle system.
Similar to the single-particle case, the existence of a global potential \(V\) for a given force field \(\boldsymbol{F} = (\mathbf{F}^1, \dots, \mathbf{F}^N)\) is determined by the topology of the configuration space and the symmetry of the force derivatives.
This condition is a direct application of the theory of closed and exact differential forms in \(\mathbb{R}^{nN}\). If such a \(V\) exists, the equality of mixed partial derivatives \(\frac{\partial^2 V}{\partial x_k^j \partial x_m^l} = \frac{\partial^2 V}{\partial x_m^l \partial x_k^j}\) necessitates the symmetry of the Jacobian of \(\boldsymbol{F}\). Conversely, on a simply connected domain \(U\), the vanishing of the exterior derivative (which is what the symmetry condition implies) ensures the existence of a global primitive \(V\) such that \(dV = -\sum_{j,k} F_k^j dx_k^j\). This \(V\) can be explicitly constructed via a line integral in the configuration space \(U\), and its independence of path is guaranteed by the given derivative symmetry and Stokes’ Theorem.
2 Conservation of Momentum
In the study of multi-particle systems, the concept of momentum provides a fundamental link between Newton’s laws and the translational symmetries of the system. For a system of \(N\) particles, we define the linear momentum of each constituent as follows.
Recalling that \(\mathrm{D}\mathbf{p}^j = m_j \mathrm{D}^2\mathbf{x}^j = \mathbf{F}^j\), Newton’s second law can be reformulated to state that the force acting on a particle is equal to the rate of change of its momentum.
Newton’s third law introduces a specific structure to these forces, suggesting that the force \(\mathbf{F}^j\) on the \(j\)-th particle arises from mutual interactions \(\mathbf{F}^{j,k}\) with other particles \(k\). If we assume these interactions are symmetric such that \(\mathbf{F}^{j,k} = -\mathbf{F}^{k,j}\), and that they are conservative, the total potential energy often takes the form: \[V(\boldsymbol{X}) = \sum_{j < k} V^{j,k}(\mathbf{x}^j - \mathbf{x}^k).\]
1 Except in trivial cases where no forces act on a specific particle, the individual momenta \(\mathbf{p}^j\) are generally not conserved due to the exchange of momentum via interactions.
To prove this, we differentiate the total momentum \(\mathbf{P}\) with respect to time \(t\): \[\mathrm{D}_t \mathbf{P} = \mathrm{D}_t \sum_{j=1}^N \mathbf{p}^j = \sum_{j=1}^N \mathbf{F}^j = \sum_{j=1}^N \sum_{k \neq j} \mathbf{F}^{j,k}.\] We can rewrite this double sum as a sum over all pairs \((j, k)\) with \(j < k\): \[\mathrm{D}_t \mathbf{P} = \sum_{j < k} \left( \mathbf{F}^{j,k} + \mathbf{F}^{k,j} \right).\] Applying the condition \(\mathbf{F}^{j,k} = -\mathbf{F}^{k,j}\), each term in the parentheses vanishes: \[\mathrm{D}_t \mathbf{P} = \sum_{j < k} (\mathbf{0}) = \mathbf{0}.\] Thus, \(\mathbf{P}\) remains constant throughout the motion.
This principle can be generalized beyond simple pairwise forces. In a gradient system, the conservation of momentum is deeply connected to the geometry of the potential \(V\).
Assume the potential is translationally invariant. For a fixed configuration \(\boldsymbol{X}\), consider the function \(g(s) = V(\mathbf{x}^1 + s\mathbf{a}, \dots, \mathbf{x}^N + s\mathbf{a})\). By hypothesis, \(g(s)\) is constant, so its derivative at \(s=0\) must vanish: \[\left. \frac{d}{ds} \right|_{s=0} V(\mathbf{x}^1 + s\mathbf{a}, \dots, \mathbf{x}^N + s\mathbf{a}) = \sum_{j=1}^N \langle \nabla^j V, \mathbf{a} \rangle = \left\langle \sum_{j=1}^N \nabla^j V, \mathbf{a} \right\rangle = 0.\] Since this holds for any vector \(\mathbf{a} \in \mathbb{R}^n\), we must have \(\sum_{j=1}^N \nabla^j V = \mathbf{0}\). Recalling the definition of force in a potential field, \(\mathbf{F}^j = -\nabla^j V\), we find: \[\mathrm{D}_t \mathbf{P} = \sum_{j=1}^N \mathbf{F}^j = -\sum_{j=1}^N \nabla^j V = \mathbf{0}.\] Conversely, if \(\mathrm{D}_t \mathbf{P} = \mathbf{0}\), then \(\sum \nabla^j V = \mathbf{0}\), which implies that the directional derivative of \(V\) along any translation vector \(\boldsymbol{A}\) is zero, thereby establishing translational invariance.
As we have seen, the conservation of momentum is a direct consequence of the system’s translational symmetry—the fact that the physics of the interaction depends only on the relative distances between particles, and not on the absolute position of the system in Euclidean space.
3 Center of Mass and the Two-Body Problem
The center of mass provides a simplified perspective for analyzing the collective motion of a multi-particle system.
By differentiating the center of mass with respect to time, we obtain the velocity of the center of mass: \[\mathrm{D}\mathbf{c} = \frac{1}{M} \sum_{j=1}^{N} m_j \mathrm{D}\mathbf{x}^j = \frac{\mathbf{P}}{M},\] where \(\mathbf{P}\) is the total momentum of the system. This leads to a fundamental observation regarding the inertia of the system as a whole.
The concept of the center of mass is particularly powerful when applied to the two-body problem. For a two-particle system, if the potential energy \(V(\mathbf{x}^1, \mathbf{x}^2)\) is invariant under simultaneous translation, it can be expressed solely as a function of the relative displacement: \[V(\mathbf{x}^1, \mathbf{x}^2) = \tilde{V}(\mathbf{x}^1 - \mathbf{x}^2),\] where \(\tilde{V}(\mathbf{a}) = V(\mathbf{a}, \mathbf{0})\). By introducing the relative position vector \(\mathbf{y} := \mathbf{x}^1 - \mathbf{x}^2\), we can decouple the motion. The individual positions \(\mathbf{x}^1\) and \(\mathbf{x}^2\) can be recovered from \(\mathbf{c}\) and \(\mathbf{y}\) as: \[\mathbf{x}^1 = \mathbf{c} + \frac{m_2}{M}\mathbf{y}, \quad \mathbf{x}^2 = \mathbf{c} - \frac{m_1}{M}\mathbf{y}.\]
To prove this, we consider the difference between the accelerations of the two particles. From Newton’s second law, we have: \[\mathrm{D}^2\mathbf{x}^1 = -\frac{1}{m_1}\nabla^1 V = -\frac{1}{m_1}\nabla \tilde{V}(\mathbf{y}),\] \[\mathrm{D}^2\mathbf{x}^2 = -\frac{1}{m_2}\nabla^2 V = \frac{1}{m_2}\nabla \tilde{V}(\mathbf{y}),\] where we have used the chain rule \(\nabla^1 \tilde{V}(\mathbf{x}^1 - \mathbf{x}^2) = \nabla \tilde{V}\) and \(\nabla^2 \tilde{V}(\mathbf{x}^1 - \mathbf{x}^2) = -\nabla \tilde{V}\). Subtracting the two equations yields: \[\mathrm{D}^2\mathbf{y} = \mathrm{D}^2\mathbf{x}^1 - \mathrm{D}^2\mathbf{x}^2 = -\left( \frac{1}{m_1} + \frac{1}{m_2} \right) \nabla \tilde{V}(\mathbf{y}).\] Multiplying both sides by \(\mu\) gives the desired result \(\mu \mathrm{D}^2\mathbf{y} = -\nabla \tilde{V}(\mathbf{y})\).
Consequently, when the total momentum of a two-body system is conserved, the evolution of the relative position is mathematically equivalent to a single-particle system with mass \(\mu\). The motion of the center of mass becomes trivial, allowing us to focus entirely on the internal dynamics of the pair.