Paper 4, Section II, D

Variational Principles
Part IB, 2010

A function θ(ϕ)\theta(\phi) with given values of θ(ϕ1)\theta\left(\phi_{1}\right) and θ(ϕ2)\theta\left(\phi_{2}\right) makes the integral

I=ϕ1ϕ2L(θ,θ)dϕI=\int_{\phi_{1}}^{\phi_{2}} \mathcal{L}\left(\theta, \theta^{\prime}\right) d \phi

stationary with respect to small variations of θ\theta which vanish at ϕ1\phi_{1} and ϕ2\phi_{2}. Show that θ(ϕ)\theta(\phi) satisfies the first integral of the Euler-Lagrange equation,

L(θ,θ)θ(L/θ)=C,\mathcal{L}\left(\theta, \theta^{\prime}\right)-\theta^{\prime}\left(\partial \mathcal{L} / \partial \theta^{\prime}\right)=C,

for some constant CC. You may state the Euler-Lagrange equation without proof.

It is desired to tow an iceberg across open ocean from a point on the Antarctic coast (longitude ϕ1\phi_{1} ) to a place in Australia (longitude ϕ2\phi_{2} ), to provide fresh water for irrigation. The iceberg will melt at a rate proportional to the difference between its temperature (the constant T0T_{0}, measured in degrees Celsius and therefore negative) and the sea temperature T(θ)>T0T(\theta)>T_{0}, where θ\theta is the colatitude (the usual spherical polar coordinate θ)\left.\theta\right). Assume that the iceberg is towed at a constant speed along a path θ(ϕ)\theta(\phi), where ϕ\phi is the longitude. Given that the infinitesimal arc length on the unit sphere is (dθ2+sin2θdϕ2)1/2\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)^{1 / 2}, show that the total ice melted along the path from ϕ1\phi_{1} to ϕ2\phi_{2} is proportional to

I=ϕ1ϕ2(T(θ)T0)(θ2+sin2θ)1/2dϕI=\int_{\phi_{1}}^{\phi_{2}}\left(T(\theta)-T_{0}\right)\left(\theta^{\prime 2}+\sin ^{2} \theta\right)^{1 / 2} d \phi

Now suppose that in the relevant latitudes, the sea temperature may be approximated by T(θ)=T0(1+3tanθ)T(\theta)=T_{0}(1+3 \tan \theta). (Note that (1+3tanθ)(1+3 \tan \theta) is negative in the relevant latitudes.) Show that any smooth path θ(ϕ)\theta(\phi) which minimizes the total ice melted must satisfy

θ2=sin2θ(14k2tan2θsin2θ1),\theta^{\prime 2}=\sin ^{2} \theta\left(\frac{1}{4} k^{2} \tan ^{2} \theta \sin ^{2} \theta-1\right),

and hence that

sin2θ=21(1+k2)1/2sin2(ϕϕ0),\sin ^{2} \theta=\frac{2}{1-\left(1+k^{2}\right)^{1 / 2} \sin 2\left(\phi-\phi_{0}\right)},

where kk and ϕ0\phi_{0} are constants.

[Hint:

dxx(α2x4+x21)1/2=12arcsin[x22x2(1+4α2)1/2]\int \frac{d x}{x\left(\alpha^{2} x^{4}+x^{2}-1\right)^{1 / 2}}=\frac{1}{2} \arcsin \left[\frac{x^{2}-2}{x^{2}\left(1+4 \alpha^{2}\right)^{1 / 2}}\right]