Friday, June 02, 2006

Shiraz Minwalla: fisherman and pole

Kyriakos Papadodimas has defended his PhD in the morning, and he did so with the help of videoconferencing. Congratulations! Prof. Shiraz Minwalla was sitting at the other side of the impressive high-quality IP videoconnection, in India, and smiled happily for many good reasons - the biggest among which is his baby boy.

The committee has convinced Kyriakos to answer many questions about the phase transitions of Yang-Mills theory on a sphere - and on a K3 - in depth. The most difficult calculation they did involved three-loop vacuum diagrams in dimensionally reduced gauge theory. If the expectation values of the Wilson loops are zero in the confining phase, there must be a phase transition. Their interesting work might be described sometimes later.

But what I want to share with you now is the last problem that Shiraz asked Kyriakos at the end. You will see that a PhD defense may be non-trivial and string theorists have to know not only string theory but also the fine art of fishing. ;-)

The problem

Imagine a fisherman who wants to attach his boat to the land. He takes the rope and winds it around a pole "w" times. Why does he do so and how does the ability of the rope to hold the boat still depend on "w"? Be careful, the rest of the article contains spoilers.


Let me try to offer you what we think is the correct solution. It's a refined version of what we just re-checked with Suvrat Raju. :-)

At every point "x" of the rope (here "x" is a linear coordinate around the rope), there is tension "T(x)" in the rope; the tension is a kind of force whose unit is 1 Newton. If there were no friction, "T(x)" would have to be x-independent, i.e. constant. The fisherman would need exactly the same force in his hand as the force exerted by the boat on the other side of the rope.

However, there exists friction between the rope and the pole. The rope will only move if the "driving force" of the motion exceeds a critical bound related to the friction for every "x" where the rope is in contact with the pole.

Consider an infinitesimal piece of the rope whose length is "dx" that touches the pole. If there is tension "T(x)" at this point, it will act as a normal force "F(x)" in the radial direction. What is their relation? The force is "dE / dr" where "E" is the energy. Imagine that the infinitesimal motion "dr" is changing the radius of the circle around which the rope is wound. Then clearly,
  • F = dE / dr
  • T = dE / d(2.pi.r)
because the tension "T(x)" inside the rope changes the circumference while the normal force "F(x)" changes the radius. Consequently, the radial force is
  • F(x) = 2.pi.T(x)
We must divide this force uniformly around the circle of radius "r". That means that the radial force per infinitesimal length "dx" of the rope equals
  • dF(x) = 2 pi T(x) dx / 2 pi r = T(x) dx / r
As you know, the tangential friction force "FR(x)" is proportional to the normal force with the dimensionless friction coefficient "mu". Infinitesimally,
  • d FR(x) = mu dF(x) = mu T(x) dx / r.
The driving force trying to move the rope is the derivative
  • T(x+dx) - T(x) = dx T '(x).
The rope will only move if the driving force is sufficiently large at every point, i.e.
  • dx T '(x) >= mu T(x) dx / r
  • T ' (x) >= mu T(x) / r
Incidentally, if you had the opposite sign of "T '(x)", it could move in the opposite direction:
  • -T ' (x) >= mu T(x) / r
Let's take the first inequality. If you know the tension "T0" in the rope between the pole and the fisherman's hand,
  • T(x) = T0 ... for all negative "x" (the side of the fisherman, for "x=0", the rope touches the pole for the first time, and for some very large positive "x", you find the boat),
you can easily integrate the differential inequality - assuming that "T(x)" has a uniform sign - to obtain
  • T(x) >= exp(x mu / r) T0.
This inequality must be satisfied for the rope to move. You can see that your tiny force "T0" is enough to keep the boat still as long as the force from the boat is less than
  • Tmax = exp(L mu / r) T0
where "L" is the total length of the rope wrapped around the pole, which equals
  • L = 2 pi r w
which is why
  • Tmax = exp(2 pi w mu) T0
where "w" is the winding number i.e. the number of loops. You see that you only need an exponentially (in "w") small force "T0" to be able to hold the boat the boat.

The punch line is that the fishermen secretly know the same know-how as inflationary cosmologists such as Alan Guth, inflationary economists such as Alan Greenspan, and string theorists such as Allan Adams and others, namely the secret of the exponential function. :-)

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