MAGNITUDE OF THE POLFLUCHTKRAFT

One can derive the Polfluchtkraft by considering the potential energy associated with an iceberg located at a given latitude compared with the situation where the water making up the iceberg is molten. This latter state defines potential energy zero. The excess in potential energy associated with the iceberg is obtained by starting in the molten state and raising each water molecule in the combined gravitational and centrifugal force field of the earth against the lines of force such as to build-up the iceberg (residual movements perpendicular to the lines of force do not further change the potential energy). By this procedure, one can obtain the excess potential in principle exactly from measurable quantities; namely, one needs to know the gravitational force exerted by the earth down to typical depths and up to typical heights reached by icebergs and can then evaluate the excess potential by taking the appropriate integral over the space filled by the iceberg.

For the purpose of the present treatment, the gravitational force will be modeled by a suitably accurate parametrization as described below. Before entering into the details, however, it should be noted that there are two independent small parameters in this problem:

• • The oblateness ε of the earth, defined by parametrizing the earth radius R as a function of latitude as

  (1) $$R(\vartheta ) = R_0 \left( {1 - {\rm \it \varepsilon} \left( {\mathop {\cos} \nolimits^2 \vartheta - \displaystyle{1 \over 3}} \right)} \right)$$
  i.e. as an expansion in Legendre polynomials, truncated at the quadrupole term. Latitude is measured in terms of the angle $\vartheta $ with $\vartheta= 0$ at either the North or South Pole depending on the hemisphere under consideration. For the earth, ε ≈ 1/300.

• • The ratio δ = d/R ₀, where d denotes the separation (along a line perpendicular to the ocean surface) between the center of mass of the iceberg and the center of mass of the displaced water (center of buoyancy) as shown in Fig. 1; R ₀ is defined by the above parametrization of the earth radius.

The Polfluchtkraft will turn out to be an effect of order O(δε), and it is necessary to carefully keep all terms up to this order. Note that δ ≪  ε  ≪ 1, i.e. one can neglect terms of order O(δ ²) with respect to terms of order O(δε).

Consider now successively building up the iceberg by raising water molecules against lines of the combined gravitational and centrifugal force $\vec f$ , as described above. The lines of force are characterized by the tangential unit vector $\vec e_f = \vec f/\vert\vec f\vert$ , and if one transports a water molecule by an arc length s ₂ − s ₁, it acquires the potential energy

(2) $$\eqalign{ V(s_2, s_1 ) & = {\int}_{s_1} ^{s_2} {\rm d}s^{\prime}\vec e_f (s^{\prime})\vec f\,(s^{\prime}) \cr & = \int_{s_1} ^{s_2} {\rm d}s^{\prime}\left \vert {\vec f\,(s^{\prime})} \right \vert} $$

Expanding $\vert\vec f(s')\vert$ in a Taylor series in the arc length s′ around the point s′ = 0 where the line of force pierces the ocean surface, it is clear that the leading contribution to V is given by

(3) $$V(s_2, s_1 ) = (s_2 - s_1 )\left \vert {\vec f\,(s^{\prime} = 0)} \right \vert $$

because (s ₂ − s ₁) is of the order of d. Analogously expanding $\vec e_f (s')$ , the relation

(4) $$\eqalign{ \vec r(s_2 ) - \vec r(s_1 ) & = - \int_{s_1} ^{s_2} {\rm d}s^{\prime}\vec e_f (s^{\prime}) \cr & = - \vec e_f (s^{\prime} = 0)\int_{s_1} ^{s_2} {\rm d}s^{\prime} + O(d^2 )} $$

implies that, to leading order in d, the line of force can be regarded as straight and perpendicular to the ocean surface, and (s ₂ − s ₁) is nothing but the vertical displacement of the water molecule.

As a direct consequence, the excess potential energy of the entire iceberg at latitude $\vartheta $ can be calculated to leading order in d from the combined gravitational and centrifugal force (Fig. 1) at the ocean surface,

(5) $$V(\vartheta ) = d{\kern 1pt} \vert \vec f \vert _{r = R(\vartheta )} $$

It should be noted that this argument strictly speaking is valid only for horizontal iceberg extensions of the order of d. However, icebergs can be orders of magnitude larger in the horizontal direction, such that the $\vartheta $ -dependence of $\vert\vec f\vert_{r = R(\vartheta )} $ should not be passed over without comment. Corrections due to this dependence are indeed suppressed by a factor of similar smallness as effects subleading in δ; namely, by a factor εγ, where γ denotes the ratio between the horizontal extension of the iceberg and the earth radius.

To proceed, it is necessary to introduce a model for the gravitational and centrifugal force field at the ocean surface. This can be constructed as follows: Consistent with (1), the gravitational potential v _G can be written as a rapidly converging expansion in Legendre polynomials, truncated at the quadrupole term,

(6) $$v_{\rm G} (r,\vartheta ) = v_0 (r) + {\it \varepsilon} v_2 (r)\left( {\mathop {\cos} \nolimits^2 \vartheta - \displaystyle{1 \over 3}} \right)$$

Also, this potential at (or, more precisely, anywhere above) the ocean surface must solve the Laplace equation Δv _G = 0, which implies that

(7) $$v_{\rm G} (r,\vartheta ) = - GM\left( {\displaystyle{1 \over r} - \displaystyle{{{\varepsilon} B} \over {r^3}} \left( {\mathop {\cos} \nolimits^2 \vartheta - \displaystyle{1 \over 3}} \right)} \right)$$

with integration constants GM and B. Furthermore, in the rotating frame of reference of the earth, the iceberg experiences an centrifugal force, which can be described as originating from an additional effective potential

(8) $$v_{\rm C} (r,\vartheta ) = - \displaystyle{1 \over 2}\omega ^2 r^2 \mathop {\sin} \nolimits^2 \vartheta $$

and the combined gravitational and centrifugal field of the earth is described by the sum v = v _G + v _C. Note that for the present discussion of the excess potential energy associated with an iceberg, the Coriolis force arising in the rotating frame of reference is irrelevant.

Before proceeding, it is useful to eliminate some of the diverse constants. On the one hand, the angular frequency ω in v _C and the oblateness ε are related via the Earth's planetary dynamics. Empirically, for the Earth the relation

(9) $$\omega ^2 R_0^2 = {\it \varepsilon} \displaystyle{{GM} \over {R_0}} $$

is satisfied to a good approximation (Lang Reference Lang1980); note that if the earth were a homogeneous fluid ellipsoid, the right-hand side of this relation would appear multiplied by a factor 4/5. On the other hand, $R(\vartheta )$ represents an equipotential surface of v,

(10) $$0 = \displaystyle{{\rm d} \over {{\rm d}\vartheta}} v(R(\vartheta ),\vartheta ) = \left. {\displaystyle{{\partial v} \over {\partial \vartheta}}} \right \vert _{r = R(\vartheta )} + \left. {\displaystyle{{\partial v} \over {{\rm d}r}}} \right \vert _{r = R(\vartheta )} \displaystyle{{{\rm d}R} \over {{\rm d}\vartheta}} $$

whence, by solving with respect to ${\rm d}R/{\rm d}\vartheta $ and inserting the representations (1) and (7), (8) one obtains the relation

(11) $$B = R_0^2 \left( {1 - \displaystyle{{\omega ^2 R_0^3} \over {2{\it \varepsilon} GM}}} \right)$$

to leading order in ε, where it should be observed, cf. (9) that ω ² is a quantity of order O(ε).

Using (9) and (11) to eliminate the integration constants in v, one obtains

(12) $$\eqalign{ v(r,\vartheta ) & = - \displaystyle{{\omega ^2 R_0^3} \over {{\rm \varepsilon} r}} + \displaystyle{{\omega ^2 R_0^5} \over {2r^3}} \left( {\mathop {\cos} \nolimits^2 \vartheta - \displaystyle{1 \over 3}} \right) \cr & \quad - \displaystyle{1 \over 2}\omega ^2 r^2 \mathop {\sin} \nolimits^2 \vartheta} $$

Consequently, the excess potential energy associated with an iceberg of mass m can be evaluated by inserting $\vert\vec f\vert = m\vert\vec \nabla v\vert$ into (5),

(13) $$V(\vartheta ) = dm\left. {\left( {\displaystyle{{\omega ^2 R_0^3} \over {{\it \varepsilon} r^2}} - \displaystyle{3 \over 2}\displaystyle{{\omega ^2 R_0^5} \over {r^4}} \left( {\mathop {\cos} \nolimits^2 \vartheta - \displaystyle{1 \over 3}} \right) - \omega ^2 r\mathop {\sin} \nolimits^2 \vartheta} \right)} \right \vert _{r = R(\vartheta )} $$

(14) $$= dm\omega ^2 R_0 \left( {\displaystyle{1 \over {\it \varepsilon}} - \displaystyle{7 \over 6} + \displaystyle{3 \over 2}\mathop {\cos} \nolimits^2 \vartheta} \right)$$

to linear order in ε. Note that, to this order, $\vert\vec f\vert = m\partial v/\partial r$ . From this, the Polfluchtkraft F _p is simply given in terms of the gradient of V in the azimuthal direction. Specifically, a shift in latitude ${\rm d}\vartheta $ implies a change in potential

(15) $${\rm d}V = \displaystyle{{{\rm d}V} \over {{\rm d}\vartheta}} {\rm d}\vartheta = - \displaystyle{3 \over 2}dm\omega ^2 R_0 \sin 2\vartheta {\kern 1pt} d\vartheta $$

whereas the distance induced by this shift is $(R_0 + O({\varepsilon} )){\rm d}\vartheta $ . Thus, one obtains for the Polfluchtkraft

(16) $$F_{\rm p} = - \displaystyle{1 \over {R_0}} \displaystyle{{{\rm d}V} \over {{\rm d}\vartheta}} = \displaystyle{3 \over 2}dm\omega ^2 \sin 2\vartheta $$

This result coincides with the one obtained by Epstein, seemingly by an accidental cancellation of errors, as discussed in the Appendix.

In terms of an operation on the model (7), (8) for the earth's combined gravitational and centrifugal potential, the Polfluchtkraft generally takes the form

(17) $$\eqalign{ F_{\rm p} & = - \displaystyle{1 \over {R_0}} \displaystyle{{\rm d} \over {{\rm d}\vartheta}} V(R(\vartheta ),\vartheta ) \cr & = - \displaystyle{1 \over {R_0}} \left. {\left( {\displaystyle{{\partial V} \over {\partial \vartheta}} + \displaystyle{{\partial V} \over {\partial r}}\displaystyle{{{\rm d}R} \over {{\rm d}\vartheta}}} \right)} \right \vert _{r = R(\vartheta )} = \displaystyle{{dm} \over {R_0 v_{\rm r}}} (v_{{\rm rr}} v_\vartheta - v_{{\rm r}\vartheta} v_{\rm r} )} $$

where the subscripts in the last expression denote partial derivatives. Here, $V = d\vert\vec f\vert = dmv_{\rm r} $ to linear order in ε was used, and ${\rm d}R/{\rm d}\vartheta $ was expressed in terms of partial derivatives of v by using (10).

MOVEMENT OF FLOATING ICEBERGS

How does a floating iceberg move under the influence of the Polfluchtkraft? If there were no friction of whatever kind, no Coriolis force and no ocean currents, the answer would be given in this ideal case of physics directly by the excess potential $V(\vartheta )$ and energy conservation.

An iceberg starting at latitude $\varphi _1 = 90{\rm ^\circ} - \vartheta _1 $ with its center of gravity on the potential surface effectively sinks to another surface with a potential difference of

(18) $$\delta L = L^{\prime}_2 - L^{\prime}_1 = \displaystyle{3 \over 2}d\omega ^2 R(\mathop {\sin} \nolimits^2 \vartheta _2 - \mathop {\sin} \nolimits^2 \vartheta _1 )$$

while it reaches the latitude φ ₂ and acquires the speed corresponding to this difference:

(19) $$v = \sqrt {3d\omega ^2 R(\mathop {\sin} \nolimits^2 \vartheta _2 - \mathop {\sin} \nolimits^2 \vartheta _1 )} $$

The potential (18) is formally identical to that of a mathematical pendulum (a mass suspended by a string of length l), which reads:

(20) $$L = - gl(1 - \cos \psi ) = - 2gl\mathop {\sin} \nolimits^2 \displaystyle{\psi \over 2}$$

One simply has to replace ψ/2 by φ and 2gl by (3/2)dω ² R and omits the here irrelevant constant terms. Owing to this identity, the floating iceberg would oscillate around the equator exactly like the pendulum swings around the vertical plane. If and only if latitudes are low, the force is elastic, proportional to φ and the oscillation becomes sinusoidal:

(21) $$\varphi = \varphi _0 \sin \Omega t$$

Since $v = R\dot \varphi $ the oscillation angular frequency Ω and the oscillation period T in days become:

(22) $$\Omega = \sqrt {\displaystyle{{3d\omega ^2} \over R}} ;\quad T = \sqrt {\displaystyle{R \over {3d}}} $$

For instance, with d = 100 m we obtain T = 112 days or 28 days for the trip from any low latitude to the equator. The speed is

(23) $$v = R\dot \varphi = \varphi _1 R\Omega \cos \Omega t$$

i.e. v  =  2 m s⁻¹ for φ ₁ = 30^° and d = 100 m.

This isochrony (independence of period on initial latitude) gets lost for higher latitudes, like with the pendulum for higher amplitudes. Then, the elementary expression for φ(t) must be replaced by elliptic integrals of the first kind. The speed can still be expressed as (19). Starting from 60^° our iceberg with d = 100 m would attain the equator with a speed v = 3.5 m s⁻¹ after 38 days.

The real movement of an iceberg is of course limited by frictional resistance in the ocean and by melting kinetics. As soon as a large iceberg breaks off from an iceshelf in Antarctica, it starts to move North under the influence of the Polfluchtkraft. It will reach a stationary speed v _st when the Polfluchtkraft and the Newton resistance force F _n in water are equal.

(24) $$F_{\rm p} = F_{\rm n} = \displaystyle{3 \over 2}dm\omega ^2 \sin 2\vartheta = \alpha A\rho v_{{\rm st}} ^2 $$

where α is a form factor, A is the frontal surface area, ρ is the density of water.

(25) $$v_{{\rm st}} = \sqrt {\displaystyle{3 \over 2}\displaystyle{{dm\omega ^2} \over {\alpha A\rho}} \sin 2\vartheta} $$

This stationary velocity oriented North gives rise to a Coriolis force F _c oriented perpendicular to the West.

(26) $$F_{\rm c} = 2m\omega v_{{\rm st}} \cos \vartheta $$

This Coriolis force is ~10 times smaller than the Polfluchtkraft, and since the area A′ of an elongated iceberg moving sideways is ~10 times larger, the stationary velocity to the West is also ~10 times smaller. In addition, the general West to East moving circumpolar ocean current counteracts the Coriolis force. This is consistent with the generally observed dominance of movement to the North. Concentrating therefore on the equation of motion for the northward velocity, v,

(27) $$\dot v = - R\ddot \varphi = \displaystyle{3 \over 2}\omega ^2 d\sin 2\varphi - \displaystyle{{\alpha A\rho} \over m}R^2 \dot \varphi ^2 $$

can be integrated substituting φ with −φ and

(28) $$q(\varphi ) = \dot \varphi ^2 (t)$$

furnishing

(29) $$q^{\prime} + 2cq = - k\sin 2\varphi ;\;k = \displaystyle{{3\omega ^2 d} \over R};\;c = \displaystyle{{\alpha A\rho R} \over m}$$

This first order differential equation is solved by

(30) $$\eqalign{ q = &- \displaystyle{k \over {2c^2 + 2}}[c\sin 2\varphi - \cos 2\varphi - e^{2c(\varphi _0 - \varphi )} (c\sin 2\varphi _0 \cr &- \cos 2\varphi _0 )]}$$

and the velocity is

(31) $$v = R\sqrt q $$

which also respects the initial condition q = 0 for $\varphi = \varphi _{{\kern 1pt} 0} $ . We look for the turning point of the trajectory, where v = 0. Putting x = 2c φ, it can be identified by the equation

(32) $$f\,(x) = e^x \left( {\sin \displaystyle{x \over c} - \displaystyle{1 \over c}\cos \displaystyle{x \over c}} \right) = f\,(x_0 )$$

As a plot of this function shows, under consideration of

(33) $$c = \displaystyle{{\alpha \rho R} \over {\rho ^{\prime}l}}{\rm \gg} 1$$

Equation (32) is solved by x ≈ 1, i.e.

(34) $$\varphi _i = \displaystyle{1 \over {2c}};\quad R\varphi _i = \displaystyle{l \over 2\alpha} $$

The iceberg always shoots 1/α of its own length l beyond the equator.

It is also interesting to know, where the maximal speed occurs. Due to inertia, this is somewhat beyond the 45^° parallel, where the force is at a maximum. The derivative of (30) vanishes at

(35) $$2e^x \left( {c\cos \displaystyle{x \over c} + \sin \displaystyle{x \over c}} \right) = 2c \left[{- e^{x_0} \left( {c\sin \displaystyle{{x_0} \over c} - \cos \displaystyle{{x_0} \over c}} \right)}\right]$$

This occurs at 35^° for a typical iceberg of length l = 1000 km, at 43^° for l = 100 km, at 44.4^° for l = 10 km, and at 44.94^° for l = 1 km.

The computation of the time for such a trip leads again to elliptic integrals. In the stationary case

(36) $$v_{{\rm st}} = R\sqrt {\displaystyle{k \over {2c}}\sin 2\varphi} = \sqrt {\displaystyle{3 \over 2}\displaystyle{{\omega ^2 dl} \over \alpha}\sin 2\varphi} $$

Calculating the Polfluchtkraft for real icebergs is impossible for lack of sufficient data. Instead, we present some results for a wide range of model icebergs in order to get some insight into the order of magnitude for this important force acting on floating ice masses. Figure 2 shows the quasi-stationary velocity of three icebergs, 1, 10 and 100 km long. The transit time from a starting position at latitude 65^° is given in Fig. 3. If these icebergs are protected somehow from melting, they all reach the equator, the 1 km berg would theoretically get there in 18 years.

Fig. 2. Quasi-stationary velocity of icebergs 1, 10 and 100 km long and 300 m thick versus latitude starting at 65°S after reaching equilibrium between Polfluchtkraft and frictional force.

Fig. 3. Transit time of icebergs as in Fig. 2 starting at 65°S.

The relaxation time for obtaining the stationary speed is

(37) $$\tau = \displaystyle{{v_{{\rm st}}} \over {\dot v_{{\rm max}}}} = \sqrt {\displaystyle{2 \over {kc\sin 2\varphi}}} = \sqrt {\displaystyle{{2l} \over {3\alpha d\omega ^2\sin 2\varphi}}} $$

If the force would suddenly cease, within this time there would be an overshoot by a distance

(38) $$R\delta \varphi = R\dot \varphi \tau = R\displaystyle{1 \over c}$$

This is a good estimate for the overshoot beyond the equator as calculated above. Even in the stationary case, the time

(39) $$t = \int_{\varphi _1} ^{\varphi _2} \displaystyle{{{\rm d}\varphi} \over {\dot \varphi}} = \sqrt {\displaystyle{2c \over {k}}} \int_{\varphi _1} ^{\varphi _2} \displaystyle{{{\rm d}\varphi} \over {\sqrt {\sin 2\varphi}}} $$

cannot be expressed in elementary terms. Rather close to the equator, where φ ≪ 1

(40) $$t = \sqrt {\displaystyle{2c \over {k}}} \int_{\varphi _1}^{\varphi _2} \displaystyle{{{\rm d}\varphi } \over {\sqrt {2\varphi } }} = -2\left[{\displaystyle{R \over \omega }\sqrt {\displaystyle{\alpha \over {3ld}}} (\sqrt {\varphi _1} - \sqrt {\varphi _2} )}\right]$$

We have been discussing the trajectory of an iceberg whose mass and dimensions remain constant. Melting modifies this behavior by diminishing both the mass and the frontal surface of the iceberg. Melting is a complicated process, which can only be treated in a simplified manner. It greatly reduces the distances traveled by icebergs. Large icebergs often break-up into smaller pieces, which exposes them to more melting (Scambos and others, Reference Scambos, Sergienko, Sargent, MacAyeal and Fastook2005).

Without going into the details of such a modeling exercise, we find that, under plausible assumptions about the melting kinetics, icebergs released at 65^° latitude with an initial thickness of 300 m and 100 km length as in Fig. 3 typically do not reach the equator, but only persist up to moderate latitudes.

Iceberg B-10, that originated as the floating tongue of Thwaites Glacier, broke off in 1992, ran aground for several years, was trapped and pushed in packice, and finally, in 1997, it escaped and started to travel North at an increasing speed helped by ocean currents and the Polfluchtkraft. The iceberg dimensions were 38 km × 78 km, the thickness is not well known. If all of its speed were due to the Polfluchtkraft, its thickness would have to be 1000 m. If only half of its speed were from Polfluchtkraft, its thickness would be 250 m. The fact that this iceberg survived such a long time indicates that it is very thick with the potential of traveling much further North. The importance of the Polfluchtkraft is also illustrated by a much smaller iceberg that broke off earlier from B-10. This iceberg was carried away by local ocean currents and wind towards the West, while the main iceberg B-10 took off in a northerly direction clearly assisted by the Polfluchtkraft against the prevailing ocean current.

INFLUENCE OF CURRENTS AND WINDS

Any ocean current carries of course the iceberg away with its speed $\vec w$ . The resultant speed is then the vector sum of $\vec w$ and the equatorward speed due to the Polfluchtkraft and relative to the water, which we just estimated. The eastward currents driven by the West winds between 40^° and 60^°S have an average speed $w \approx 0.2{\kern 1pt} \;{\rm m/s}$ , which is comparable with the Polflucht-drift speed of an iceberg ~100 km long. Such an iceberg ought to move in NE direction. The smaller an iceberg, the more eastward is its motion. The westward currents close to the Antarctic continent are somewhat faster. Moreover trapping in packice and stranding in shallow waters on the continental shelf can delay the progression of icebergs for several years.

The average wind velocity is some 20 times faster than the speed of ocean currents, but the wind hits only 1/7 as much surface as the current, and its density is 800 times lower. Thus, due to a force ρAv ², the wind influences the course of an iceberg 5–10 times less than the ocean current. It is not uncommon to observe smaller icebergs and sea ice to move in one direction propelled by strong surface winds and larger icebergs moving in the opposite direction following mostly the combined ocean currents and the Polfluchtkraft.

ORIENTATION OF AN OBLONG ICEBERG

From the $sin2\vartheta $ -dependence follows that the Polfluchtkraft is largest at 45^° latitude decreasing towards the equator and the pole. An oblong iceberg experiences a torque that tends to align the berg in N–S direction when traveling between the pole and 45^° and in E–W direction traveling from there to the equator.

How long would it take to rotate an iceberg into its equilibrium position? The torque is

(41) $$D = L^2 \displaystyle{{{\rm d}F_{\rm p}} \over {R{\rm d}\varphi}} \approx \displaystyle{{3L^2 md\omega ^2 cos2\varphi} \over R} \approx \displaystyle{{3L^3 WH^2\Delta \rho \omega ^2 cos2\varphi} \over R}$$

L, W, H is the length, width and height of the iceberg, Δρ is the difference of water and ice densities.

The opposing torque from friction in water is of the order

(42) $$D_{\rm r} = L^2 Hv^2 \rho = L^4 H\omega^{\prime 2} \rho $$

where ω′ is the angular velocity of rotation of the iceberg. The result is

(43) $$\omega ^{\prime} \approx \sqrt {\displaystyle{{WH\Delta \rho} \over {RL\rho}}} \omega $$

With W = H = 1 km and L = 10 km the rotation about an angle of order 1 rad would require ~2 years. It is observed that large tabular icebergs do not rotate much unless they break-up, melt, get grounded, or collide with other obstacles. But if the iceberg lingers around at high latitudes for long enough time it might be able to gain an orientation most favorable for traveling North. This is indeed true for iceberg B-10. At latitude 60^° South its long axis was oriented exactly N–S.