Homework 1 Problem 2

#### Problem

Solution by: Andy Bosiljevac

Express the equation of motion

(1)
\begin{align} 2x \ddot{x} + \dot{x}^2 + gx = 0 \end{align}

for the “falling chain” described in class in canonical Hamiltonian form.

#### Solution

Let: µ = mass per unit length
l = length of the chain
x = displacement of the chain from the x-axis

P.E. for the left half:

(2)
\begin{align} V_L = - mgh = - \mu \left( {l - x} \right)g\left( {\frac{{l - x}}{2}} \right) \\ = - \frac{1}{2}\mu g\left( {x^2 - 2lx + l^2 } \right) \\ \end{align}

P.E. for the right side:

(3)
\begin{align} V_R = - mgh = - \mu xg\left( {l - 2x + \frac{x}{2}} \right) \\ = - \mu gx\left( {l - \frac{3}{2}x} \right) \\ \end{align}

K.E. for the left side is equal to zero.
K.E. for the right side:

(4)
\begin{align} T_R = \frac{1}{2}mv^2 = \frac{1}{2}\mu x\left( { - 2\dot x} \right)^2 = 2\mu x\dot x^2 \end{align}

Therefore, the Lagrangian L is

(5)
\begin{align} L = T - V = T - V_R - V_L \\ = 2\mu x\dot x^2 - \mu g\left( {x^2 - \frac{{l^2 }}{2}} \right) \\ \end{align}

Now, to differentiate all the terms:

(6)
\begin{align} \frac{{\partial L}}{{\partial \dot x}} = 4\mu x\dot x \\ \\ \end{align}
(7)
\begin{align} \frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial \dot x}}} \right) = 4\mu x\ddot x + 4\mu \dot x^2 \\ \\ \end{align}
(8)
\begin{align} \frac{{\partial L}}{{\partial x}} = 2\mu \dot x^2 - 2\mu gx \\ \\ \end{align}

Combining terms and dividing by all constants, we get

(9)
\begin{align} \frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial \dot x}}} \right) - \frac{{\partial L}}{{\partial x}} = 0 = 4\mu x\ddot x + 4\mu \dot x^2 - 2\mu \dot x^2 + 2\mu gx \\ \\ \end{align}
(10)
\begin{align} 2x\ddot x + \dot x^2 + gx = 0 \\ \end{align}

Now, to move on to the Hamiltonian mechanics portion, start with

(11)
\begin{align} \frac{d}{{dt}}(\frac{\partial L}{{\partial \dot{x} }} )= \frac{\partial L}{{\partial x}} \end{align}

Since

(12)
\begin{align} p = \frac{\partial L}{{\partial \dot{x} }} \end{align}

, then

(13)
\begin{align} p = 4\mu x \dot{x} \end{align}

Now, apply the simplified version of the definition of the Hamiltonian:

(14)
\begin{align} H(x,p,t) = p\dot{x} - L(x,\dot{x},t) = (4\mu x\dot{x})\dot{x} - 2\mu x\dot{x^2} + \mu g(x^2-\frac{l^2}{{2}}) \end{align}

Making the substitution based on equation 13

(15)
\begin{align} H(x,p,t) = 4\mu x (\frac{p}{{4\mu x}})^2 - 2\mu x (\frac{p}{{4\mu x}})^2 +\mu g(x^2 - \frac{l^2} {{2}} ) \end{align}

FINALLY, the Hamiltonian comes out to be:

(16)
\begin{align} H(x,p,t) = \frac{p^2}{{8\mu x}} + \mu g(x^2 - \frac{l^2}{{2}}) \end{align}

Lastly, to verify what we have found,

(17)
\begin{align} \frac{dx}{{dt}} = \frac{\partial H}{{\partial p}} = \frac{p}{{4\mu x}} = \frac{4\mu x \dot{x}}{{4\mu x}} = \dot{x} \end{align}

VERIFIED!!

and

(18)
\begin{align} \frac{dp}{{dt}} = -\frac{\partial H}{{\partial x}} \end{align}
(19)
\begin{align} \frac{dp}{{dt}} = 4\mu x \ddot{x} + 4\mu \dot{x^2} = \frac{d}{{dt}}(\frac{\partial L}{{\partial \dot{x}}}) \end{align}
(20)
\begin{align} -\frac{\partial H}{{\partial x}} = -(-2\mu\dot{x^2}+2\mu x g) = \frac{\partial L}{{\partial x}} \end{align}

And we already know that

(21)
\begin{align} \frac{d}{{dt}}(\frac{\partial L}{{\partial \dot{x}}}) = \frac{\partial L}{{\partial x}} \end{align}

QED

Scott agrees!

#### Discussion

What is the canonical Hamiltonian form? Is it

(22)
\begin{align} \dot{x} = \frac{\partial H}{\partial p_x}\\ \dot{p_x} = -\frac{\partial H}{\partial x}\\ \end{align}

Yep, that's it all right.

page revision: 38, last edited: 10 Apr 2007 17:58
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