Homework 1 - Problem 3

Problem

A mass $m$ is situated at the point $(x, y)$ in the plane; a mass $M$ is situated at the point $(X, Y)$. The two masses are joined by a linear spring with spring constant $k$ but are otherwise unconstrained. Assume the spring to have zero length when relaxed.

Write the equations of motion for this system as canonical Euler-Lagrange equations and as canonical Hamiltonian equations. Show explicitly that the quantity $p_x + p_y + p_X + p_Y$ Poisson commutes with the total energy $H$ and is thus conserved.

Solution

The first step is to determine the the Lagrangian equation, $L = T - V$, where $T$ is the total Kinetic Energy and $V$ is the total Potential Energy. For this system:

(1)
\begin{align} T = \frac {1}{2} m \left(\dot x^2 + \dot y^2 \right) + \frac {1}{2} M \left(\dot X^2 + \dot Y^2 \right) \end{align}
(2)
\begin{align} V = \frac {1}{2} k \left[\left(X-x\right)^2 + \left(Y-y\right)^2\right] \end{align}
(3)
\begin{align} L = \frac {1}{2} m \left(\dot x^2 + \dot y^2 \right) + \frac {1}{2} M \left(\dot X^2 + \dot Y^2 \right) - \frac {1}{2} k \left[\left(X-x\right)^2 + \left(Y-y\right)^2\right] \end{align}

Using the Lagrangian, the canonical Euler-Lagrange equations can be developed using:

(4)
\begin{align} \frac{\partial L}{\partial q^i} - \frac{d}{dt}\left(\frac {\partial L}{\partial \dot q^i}\right) = 0 \end{align}

where $q^i$ is used to refer to the generalized coordinates. Because there are four degrees of freedom for this system $\left( x, y, X, Y\right)$, there will be four Euler-Lagrange equations. For this problem, the four equations are:

(5)
\begin{align} \ddot x = \frac {k}{m}\left(X-x\right) \end{align}
(6)
\begin{align} \ddot y = \frac {k}{m}\left(Y-y\right) \end{align}
(7)
\begin{align} \ddot X = - \frac {k}{M}\left(X-x\right) \end{align}
(8)
\begin{align} \ddot Y = - \frac {k}{M}\left(Y-y\right) \end{align}

Now for the canonical Hamiltonian equations. First, define the momentum conjugate to $q^i$ as:

(9)
\begin{align} p_i = \frac {\partial L}{\partial \dot q^i} \end{align}

and the Hamiltonian, $H$, as:

(10)
\begin{align} H(q,p,t) = p_i q^i - L (q^i,\dot q^i,t) \end{align}

Applied to this problem, these equations are:

(11)
\begin{align} p_x = m \dot x \end{align}
(12)
\begin{align} p_y = m \dot y \end{align}
(13)
\begin{align} p_X = M \dot X \end{align}
(14)
\begin{align} p_Y = M \dot Y \end{align}
(15)
\begin{align} H = \frac {1}{2}\left(\frac{p_x ^2}{m}+\frac{p_y ^2}{m}+\frac{p_X ^2}{M}+\frac{p_Y ^2}{M}\right)+k\left[\left(X-x\right)^2+\left(Y-y\right)^2\right] \end{align}

Using these equations, the canonical Hamiltonian equations are defined as:

(16)
\begin{align} \dot p_i = - \frac {\partial H}{\partial q^i} \end{align}
(17)
\begin{align} \dot q^i = \frac {\partial H}{\partial p_i} \end{align}

Because there are four degrees of freedom, there will be eight Hamiltonian equations in canonical form. For this problem, those equations are:

(18)
\begin{align} \dot x = \frac {p_x}{m} \end{align}
(19)
\begin{align} \dot y = \frac {p_y}{m} \end{align}
(20)
\begin{align} \dot X = \frac {p_X}{M} \end{align}
(21)
\begin{align} \dot Y = \frac {p_Y}{M} \end{align}
(22)
\begin{align} \dot p_x = k\left(X-x\right) \end{align}
(23)
\begin{align} \dot p_y = k\left(Y-y\right) \end{align}
(24)
\begin{align} \dot p_X = -k\left(X-x\right) \end{align}
(25)
\begin{align} \dot p_Y = -k\left(Y-y\right) \end{align}

Hamilton's equations in canonical form are equivalent to the statement that $\dot f = \left\{f,H \right\}$ for all functions $f \left(q, p\right)$. If $\dot f = \left\{f,H\right\}=0$, then f is a conserved quantity. In order to check if the function $f = p_x+p_y+p_X+p_Y$ Poisson commutes with the total energy $H$, $\dot f$ was determined and was found to be:

(26)
\begin{align} \dot f = \dot p_x+\dot p_y+\dot p_X+\dot p_Y = k\left(X-x\right)+k\left(Y-y\right)-k\left(X-x\right)-k\left(Y-y\right)=0 \end{align}

Because $\dot f=0$, the quantity $f =p_x+p_y+p_X+p_Y$ is conserved.

Scott agrees, having corrected a few typographical errors — although the idea was to show that $\frac{d}{dt} (p_x+p_y+p_X+p_Y)$ equals zero by explicitly computing $\{p_x+p_y+p_X+p_Y, H\}$ to be zero, and not the other way around…

Discussion

Evidently I can't read/count… I guess I originally did this as problem 4, when it was supposed to be problem 3. I know I did the problem I claimed, I just mislabeled it. Thanks avakis2 for moving it for me :)

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