Question

Consider the Hamiltonian equation $$ \dot{q}=\frac{\partial H}{\partial p}, \quad \dot{p}=-\frac{\partial H}{\partial q} . $$ Let \(H(p, q)=T(p)+W(q)\), where \(T\) is the kinetic energy and \(W\) is the potential energy. If \(H\) is analytic in \(p\) and \(q\) and \(H(0,0)=0\), prove the Lagrange theorem: a position of stable equilibrium is a point at which the potential energy is a relative minimum.

Short answer

Answer: In Hamiltonian mechanics, a position of stable equilibrium is achieved when the first derivatives of the potential energy with respect to position vanish, and its second derivative with respect to position is positive. This indicates that the potential energy has a relative minimum at the equilibrium position, which proves the Lagrange theorem.

Step-by-step solution

Write down the Hamiltonian and its derivatives

The Hamiltonian is given as \(H(p, q) = T(p) + W(q)\). We need to find the partial derivatives of the Hamiltonian with respect to \(p\) and \(q\): $$ \frac{\partial H}{\partial p} = \frac{\partial T}{\partial p}, \quad \frac{\partial H}{\partial q} = \frac{\partial W}{\partial q} $$

Equate derivatives to zero to find the equilibrium

For equilibrium, the time derivatives of the positions and momenta should be zero. Using the Hamiltonian equations, we obtain: $$ \frac{\partial T}{\partial p} = 0, \quad -\frac{\partial W}{\partial q} = 0 $$

Second derivatives of the potential energy

To determine the stability of the equilibrium, we need to look at the second derivative of the potential energy. Calculate the second derivative of the potential energy with respect to \(q\): $$ \frac{\partial^2 W}{\partial q^2} $$

Establish the stability condition

For a position of stable equilibrium, the second derivative of the potential energy must be greater than zero: $$ \frac{\partial^2 W}{\partial q^2} > 0 $$ This condition ensures that the potential energy is minimized in the equilibrium position.

Proof of Lagrange theorem

We have shown that in Hamiltonian mechanics, a position of stable equilibrium corresponds to a point where the first derivative of the potential energy vanishes, and the second derivative is positive. This implies that the potential energy has a relative minimum at this point, which proves the Lagrange theorem.

Key concepts

Hamiltonian Equations

Understanding the Hamiltonian equations is key in the study of classical mechanics. They are a set of differential equations governing the evolution of a physical system. In the simplest terms, the Hamiltonian equations describe how a system changes over time, and they reflect the relationship between a system's coordinates and momentum. The exercise presented focuses on these equations with respect to the kinetic and potential energies of a system. In mathematical language, the Hamiltonian equations are expressed as
\[ \frac{d}{dt}q = \frac{abla H}{abla p}, \]
and
\[ \frac{d}{dt}p = -\frac{abla H}{abla q}, \]
where \( q \) and \( p \) represent the generalized coordinates and momenta respectively. The 'nabla' symbol indicates a derivative with respect to a variable, and \( H \) represents the Hamiltonian function, which is a sum of the system's kinetic energy (\( T \)) and potential energy (\( W \)). By evaluating these equations at equilibrium points, where the time derivatives of positions and momenta are zero, one can draw insights about the system's behavior and stability.

Stable Equilibrium

When dealing with physical systems, the concept of stable equilibrium is vital. A system is said to be in stable equilibrium when it tends to return to its equilibrium position after a small disturbance. From the standpoint of Hamiltonian mechanics, stability can be determined mathematically.
In the solutions provided, stable equilibrium is associated with the conditions derived from the Hamiltonian equations when the derivatives of kinetic and potential energy with respect to their respective variables equal zero. This means that there are no net forces acting on the system, which is indicative of an equilibrium state.
To assess if this equilibrium is stable, we examine the second derivative of the potential energy. If this second derivative with respect to position is positive, it implies that a system, when slightly moved from this point, experiences a restoring force pushing it back to the equilibrium. This characteristic is crucial because it tells us that the equilibrium point is not only a point of balance but also a point from which the system can recover, an essential hallmark of stability.

Potential Energy Minimum

The potential energy of a system is a measure of its stored energy, which has the capability to do work purely because of the system's position or arrangement. In the context of Hamiltonian mechanics and the stable equilibrium of a system, the concept of potential energy minimum plays a central role. The equilibrium of a system is considered stable if, after a small shift from this position, the system tends to return, which can only happen if the equilibrium position coincides with a minimum of potential energy.
In the exercise we are examining, the requirement that the second derivative of the potential energy with respect to position (\( q \)) is greater than zero is paramount. Mathematically, it is expressed as
\[ \frac{abla^2 W}{abla q^2} > 0. \]
When this inequality holds, it indicates that the potential energy achieves a relative minimum at the equilibrium position. This aligns with the intuitive idea that in a valley (a minimum), any movement up the sides will result in the system rolling back down, thus returning to the starting point. Therefore, when the potential energy is at a minimum, the system's equilibrium is stable, confirming Lagrange's theorem as described in the exercise. Moreover, this is why when learning about equilibrium in Hamiltonian mechanics, identifying the potential energy minimum becomes such a fundamental step in solving problems.