Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 8E (Classical Dynamics)

Table of contents
- a 8E
  - Solution a
- b 8E
  - Solution b

a

Solution

Let

ξ_{i} (t) = \frac{\partial Q _{i} ( s , t )}{\partial s}_{s = 0}

be the infinitesimal generator of the continuous symmetry, and let

p_{i} = \frac{\partial L}{\partial q ˙ _{i}}

be the canonical momentum. Noether theorem states that along every trajectory satisfying the Euler-Lagrange equations,

J = i = 1 \sum n p_{i} ξ_{i}

is conserved.

Indeed, differentiate the invariance of

L

with respect to

s

at zero:

0 = \sum_{i} (\frac{\partial L}{\partial q _{i}} ξ_{i} + \frac{\partial L}{\partial q ˙ _{i}} \dot{ξ}_{i}) .

On an Euler-Lagrange trajectory,

\partial L / \partial q_{i} = d p_{i} / d t

, so

0 = \sum_{i} (\overset{p}{˙}_{i} ξ_{i} + p_{i} \dot{ξ}_{i}) = \frac{d}{d t} \sum_{i} p_{i} ξ_{i} .

This proves the Noether conserved quantity for a mechanical point symmetry.

b

Solution

In cylindrical coordinates,

L = \frac{m}{2} (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2} + \overset{z}{˙}^{2}) - f (r, ϕ - k z) .

The screw transformation

z \mapsto z + s, ϕ \mapsto ϕ + k s

leaves

r

and

ϕ - k z

fixed, so it is the helical symmetry of a potential. Its infinitesimal generator has

ξ_{z} = 1

and

ξ_{ϕ} = k

. Noether's theorem therefore gives the conserved quantity

p_{z} + k p_{ϕ} = m \overset{z}{˙} + km r^{2} \dot{ϕ} .

This is a linear combination of axial linear momentum and axial angular momentum.

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version