Paper 4, Section II, B

Dynamics and Relativity | Part IA, 2011

A particle with mass mm and position r(t)\mathbf{r}(t) is subject to a force

F=A(r)+r˙×B(r)\mathbf{F}=\mathbf{A}(\mathbf{r})+\dot{\mathbf{r}} \times \mathbf{B}(\mathbf{r})

(a) Suppose that A=ϕ\mathbf{A}=-\nabla \phi. Show that

E=12mr˙2+ϕ(r)E=\frac{1}{2} m \dot{\mathbf{r}}^{2}+\phi(\mathbf{r})

is constant, and interpret this result, explaining why the field B\mathbf{B} plays no role.

(b) Suppose, in addition, that B=ψ\mathbf{B}=-\nabla \psi and that both ϕ\phi and ψ\psi depend only on r=rr=|\mathbf{r}|. Show that

L=mr×r˙ψr\mathbf{L}=m \mathbf{r} \times \dot{\mathbf{r}}-\psi \mathbf{r}

is independent of time if ψ(r)=μ/r\psi(r)=\mu / r, for any constant μ\mu.

(c) Now specialise further to the case ψ=0\psi=0. Explain why the result in (b) implies that the motion of the particle is confined to a plane. Show also that

K=L×r˙ϕr\mathbf{K}=\mathbf{L} \times \dot{\mathbf{r}}-\phi \mathbf{r}

is constant provided ϕ(r)\phi(r) takes a certain form, to be determined.

[ Recall that rr˙=rr˙\mathbf{r} \cdot \dot{\mathbf{r}}=r \dot{r} and that if ff depends only on r=rr=|\mathbf{r}| then f=f(r)r^.]\left.\boldsymbol{\nabla} f=f^{\prime}(r) \hat{\mathbf{r}} .\right]

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