Problem 1 – Hint 4

When solving this problem, you will be needing an expression for the added mass of a thin circular disk of radius R. You can surely find a suitable expression by a web search, but using it without a proof does not adhere to the rules of the Physics Cup – you need to derive it. You can also find sources where it is derived, but most (if not all) of those sources are mathematically fairly advances. In what follows, a mathematically simple approach is outlined. NB! Once you have obtained the desired expression, do not ignore the facts that (a) our cylinder is not a thin disk, and (b) it is only plunging into water and, hence, not immersed entirely (the added mass you obtain assumes that the disk is entirely in the water). So, you need a smart way of using this added mass.

To begin with, it is convenient to switch from calculating the added mass to calculating the magnetic polarizability of a superconducting ellipsoid that is an ideal diamagnetic with magnetic susceptibility \chi=-1 (because then, we can use our knowledge about magnetic fields). This can be done, because (a) the field equation are identical (both the velocity and magnetic fields are curl-free, i.e. potential, and divergence-free, i.e. incompressible) , and (b) the boundary conditions are exactly the same: both fluid velocity and magnetic field are tangent to the surface of the ellipsoid. Added mass describes the kinetic energy of water that is matched (with a suitable conversion constant) to the energy of the magnetic field.

So, we need to calculate the energy of the magnetic field created by a polarized ellipsoid. When a superconducting body is placed into an homogeneous magnetic field, the surface currents adjust so that the magnetic field created by the surface currents cancels out the field inside the body. This means that if we exclude the external homogeneous field \mathbf{B}_0(\mathbf{r}), the field \mathbf{B}_s(\mathbf{r}) created by the surface currents is homogeneous inside the superconductor, and inhomogeneous outside. What we need to know is the energy of \mathbf{B}_s(\mathbf{r}), but only outside the body. Indeed, according to our mapping, the total field \mathbf{B}_0(\mathbf{r})+\mathbf{B}_s(\mathbf{r}) is matched to the velocity field \mathbf{v}_{\mathrm{cm}}(\mathbf{r}) as seen in a frame co-moving with the body, and \mathbf{B}_s(\mathbf{r}) is matched to the velocity field \mathbf{v}_r(\mathbf{r}) as seen in the rest frame where the fluid velocity vanishes at infinity. It is the kinetic energy of the fluid in the rest frame what we need. Pay attention to the fact that this kinetic energy excludes the region occupied by the body.

One might wonder, why we consider the disk as an infinitely thin ellipsoid, and e.g. not as an infinitely short cylinder. The answer lies in the fact that ellipsoid is the only body shape that, when homogeneously polarized (i.e. with the polarization \mathbf{M} being constant over the entire volume of the body), creates an homogeneously magnetic field inside the body (the same holds for electric field); see the final paragraph of this hint. All the other body shapes (sphere, infinite plate, infinite cylinder) with the same property can be considered as special cases of ellipsoids. This means that the only things we need to do is to consider an ellipsoid of homogeneous polarization \mathbf{M}, calculate the magnetic field energy of that ellipsoid, and in the resulting formula, substitute the polarization \mathbf{M} in terms of the magnetic field \mathbf{B}_0 inside the ellipsoid.

The magnetic field energy can be conveniently calculated as \frac 12\mathbf{m}\cdot\mathbf{B}_0, where \mathbf{m} denotes the full magnetic dipole moment of the ellipsoid. This conclusion can be obtained if we imagine the process of obtaining the polarized ellipsoid as turning individually the tiny molecular dipoles \mathbf{m}_0; the work done by turning each of those is \mathbf{m_0}\cdot\mathbf{B}(t), where \mathbf{B}(t) denotes the current value of the magnetic field inside the ellipsoid that is proportional to the current total dipole moment of the ellipsoid.

In principle, we should subtract the magnetic field energy inside the ellipsoid, but that is not needed in our case, because we use the limit case of infinitely thin ellipsoid of zero volume.

As a final closure, we need to express \mathbf{B}_0 in terms of \mathbf{m}. With the magnetic field being homogeneous inside the ellipsoid, It makes sense to calculate the magnetic field at that point inside the cylinder where the task is the easiest: at the centre of the infinitely thin ellipsoid. This can be done by integrating over the contributions made by circular currents around that point, the only thing we need is the current density as a function of distance from the symmetry axis. That expression can be obtained easily if we consider the surface density of the magnetization J\equiv Md, where d=d(r) is the (vanishingly small) thickness of the ellipsoid at a distance r from the axis.

Finally, to show that homogeneously polarized ellipsoid creates an homogeneous field inside itself, it is more convenient to consider electrical polarization. When going from magnetic fields to electric fields, notice that electric and magnetic fields created by electric and magnetic dipoles are equivalent as long as we consider points outside the dipoles. In this case, however, we consider points inside the polarized ellipsoid which are surrounded by molecular dipoles. We can imagine drilling a very narrow channel parallel to the magnetization axis: in this way, our point of interest A will be inside the channel and thus outside the dipoles – there, electric and magnetic fields are equivalent. The magnetic fields at A (inside the channel) and at an immediately neighbouring point A' outside the channel differ by J\mathbf M\mu_0, which is a constant vector. So, if the electric field is homogeneous inside a homogeneously polarized ellipsoid, the magnetic field is homogeneous, too. Now, let us start with a spherical charged shell of constant thickness. It is clear from Gauss theorem that the field inside such a shell must be zero. However, there is a different way of showing this, that idea was used by I. Newton. For an arbitrary point P inside the spherical shell, we draw a narrow cone with the tip at P; it is easy to see that the fields created by two opposing pieces of the shell, cut out by two such cones, cancel pairwise out. Now, let us make an affine transformation with our spherical shell, so that it becomes an ellipsoidal shell. The Newton’s method can still be used, proving that the field inside an ellipsoidal charged shell is zero, too. Now, we consider a homogeneously charged ellipsoid for which we want to find electric potential as a function of coordinates. Let us consider a tiny ellipsoid of the same shape, at the centre of this big ellipsoid. Since the new ellipsoid is very small, we can use only the quadratic terms of the Taylor’s expansion to express the potential inside the tiny ellipsoid as a quadratic polynomial. If the potential is a quadratic polynomial, the electric field is homogeneous. As shown before, the charges between the two ellipsoids have no contribution to the electric field and as long as we are interested in the electric field inside the tiny ellipsoid only, we can remove them. As a result, we can conclude that the tiny ellipsoid creates a perfectly homogeneous electric field inside itself.

Please submit the solution to this problem via e-mail to physcs.cup@gmail.com. The Hint No 4 will be published on 26th November, 13:00 GMT.