 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s7/1/2/solution

The Euler-Lagrange equations in cylindrical coordinates are

\ddot{R} = R \dot{ϕ}^{2} - \frac{C z ˙}{R}, \frac{d}{d t} (R^{2} \dot{ϕ}) = 0, \overset{z}{¨} = \frac{C R ˙}{R} .

The last two follow because

ϕ

and

z

ℓ = R^{2} \dot{ϕ}, k = \overset{z}{˙} - C lo g R .

Since the Lagrangian is autonomous, its energy function is conserved as well:

E = \dot{R} \frac{\partial L}{\partial R ˙} + \dot{ϕ} \frac{\partial L}{\partial ϕ ˙} + \overset{z}{˙} \frac{\partial L}{\partial z ˙} - L = \frac{1}{2} (\dot{R}^{2} + R^{2} \dot{ϕ}^{2} + \overset{z}{˙}^{2}) .

The velocity-linear term cancels from this energy, but it still changes the equations of motion. A dimensionless logarithm means

lo g (R / R_{0})

for some fixed reference length; replacing the printed

lo g R

by that expression changes the Lagrangian only by a total time derivative and shifts the numerical value of

k

, not the trajectory.

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