I think the answer to the question is that there is no mechanism by which angular momentum is transferred to the water from outside, after you stop the initial swirling. Rather, the vortex takes angular momentum from the draining water and
transfers it to that remaining in the bottle. I give a fuller explanation below -
it talks about water draining from a bath but the principles are the same.
Three basic physical principles apply:
First, angular momentum is conserved: at any time during the draining of the
bath the total angular momentum of the water (inside and outside the bath)
stays the same.
Second, the only energy input to the system is that available from gravity
as the water drains.
Third, the water is viscous, so that any rotating volume of water within the
bath will tend to transfer its angular momentum to the body of water as a whole.
The presence of the vortex tube with a water/air surface shows that the force
experienced by the water at that interface points radially outward and must be balanced by the normal pressure of the water in the bath. As water exits the bottom of the vortex, pressure from the body of bath water outside pushes more water inward towards the vortex centre.
Conservation of angular momentum means that as this water moves in, its angular
velocity and consequent rotational energy increase. At the same time viscosity
acts to decrease the shear in water velocities, transferring angular momentum
from the fast rotating water to the main body of water away from the vortex.
The two effects combine to build rotational energy near the vortex but
transfer angular momentum away from it.
What appears to the watcher to be the adding of angular momentum to the water is,
if this model is right, rather its transfer from one part of the water volume to
another. The liquid that falls out of the bottom end of the vortex has high angular energy but low angular momentum. Most of its original angular momentum remains in the liquid still in the vessel; as the amount of liquid in the vessel decreases so its angular speed increases.
The increase in angular energy is fuelled by the decrease in gravitational potential energy as the water falls down the vortex.
Some numbers. If the air tube at the centre of the vortex in a draining sink has a radius of 3mm, and the gradient of the vortex tube surface is 9, then the angular velocity of the water must be:
$\omega^2 \cdot 3\times 10^{-3} = 9g$
that is, $\omega=171$ rad/s; quite rapid. Conversely, a speck of water rotating with the same angular momentum at a distance of 20cm from the vortex centre would have
$\omega=3.9\times 10^{-2}$ rad/s (2.2 deg/s); almost too slow to notice. Though the two specks
of water have the same angular momentum, their rotational energies differ by a
factor of 4325.
Empirical evidence:
First of all, to fix our ideas, I assume that you are talking about an tangential wind profile that looks something like this:
$\phantom{texttexttexttexttexttexttexttexttextte}$
The plot shows the (azimuthally averaged, i.e. averaged along circles) tangential wind profile as a function of radius at a fixed height. Looking at this picture one is naturally reminded of a Rakine vortex, which consists of a forced vortex core surrounded by a free vortex. You said:
[...] a forced vortex has a velocity profile u∝r (r is radial distance from centre of vortex), concluding at some outer boundary r=R to avoid fluid particles travelling at infinite speed. At this outer boundary it requires an external torque to be constantly supplied to keep going.
The misconception here is that in a tropical cyclone, there is no solid (nor an effectively solid) boundary at the point where the "forced" vortex transitions into the "free" part. There is no wall that imparts momentum to the fluid by wall friction as in the cylinder.
So why does the air spin in a cyclone?
Because it is a low-pressure system. To good approximation (assuming we can neglect friction), there is a force balance between the pressure gradient force, the Coriolis force and the centrifugal force:
$$ -\frac{1}{\rho}\frac{\partial p}{\partial r}+fv+\frac{v^2}{r} = 0, \tag 1$$
where I denote by $v$ the tangential velocity and $f$ is the Coriolis parameter. It may be interesting to you that the approximate balance (1) follows from a scale analysis of the radial Navier-Stokes equation in cylindrical coordinates. The balance (1) is often called "gradient wind balance". From (1) it is clear that in a low-pressure system ($\frac{\partial p}{\partial r}<0$), there will be cyclonic motion ($v>0$) to sustain the balance, hence the name 'cyclone' for low-pressure systems.
So, plainly speaking, the tropical cyclone should not (or cannot) be compared to water in a rotating cylindrical tank which is maintained in solid body rotation by an external torque on the cylinder. Rather, the air spins because of the pressure deficit between environment and cyclone eye. Hence the question of what provides the torque on the "outer boundary" to maintain solid body rotation is misleading because no such boundary exists in a cyclone.
Why the tropical cyclone wind profile is more complicated than a Rankine vortex:
The Rankine vortex model (just like free and forced vortex models individually) describes an axis-symmetric vortex in a two-dimensional fluid (all motion occurs in the horizontal plane and motion is independent of z).
A real tropical cyclone is neither exactly symmetric and, more importantly, it is very much a three-dimensional flow structure. Before I go on, let me just say a few words about what that structure looks like:
$\phantom{texttettexttextte}$
Due to the low pressure in the centre of the cyclone, the air is set in circling motion according to (1). However, there is a layer close to the surface called 'planetary boundary layer' where the balance (1) is broken by frictional effects. Here, the flow is no longer along constant radius but has a component pointing from high to low pressure (i.e. a radial component). The air masses converge in the centre of the cyclone and hence are forced (by mass conservation) to move upward in the so-called eyewall. As the air rises, water vapour condenses releasing latent heat and giving the air parcels further buoyancy so they can keep rising up to the tropopause. this illustrates the fact that the three-dimensionality along with thermodynamic effects make the cyclone a more complicated problem than the theoretical forced or free vortices.
A first (attempted) explanation:
Nonetheless, we can explain why the velocity profile looks like it does.
To begin with, we remember that (even approximately) conserved quantities are useful when trying to understand physics problems so let's see if we can find one here. The azimuthal component of the NS-equations in cylindrical polars reads:
$$\frac{1}{r}\frac{D(vr)}{Dt}+fu = -\frac{1}{\rho r}\frac{\partial p}{\partial \theta} + F \tag{ 2},$$
where $D/Dt$ is the material derivative, $(u,v)$ are radial and azimuthal wind respectively, $\theta$ is the azimuthal angle and F is friction. Defining $M=rv+\frac{1}{2}fr^2$, the absolute angular momentum and treating $f$ as a constant, it follows from (2) that
$$\frac{DM}{Dt} = -\frac{1}{\rho}\frac{\partial p}{\partial \theta}+ Fr \tag{3}$$
Now if frictional effects are negligible and the flow is axisymmetric, the RHS can be set to 0 and absolute angular momentum is materially conserved.
Now, keeping this in mind, consider an air parcel in the ambient atmosphere which is "sucked" into the cyclone. As the air parcel travels inward, approximately conserving its absolute angular momentum, its tangential velocity $v$ has to increase. When $M$ is dominated by the $rv$ term (which is likely because $f$ is very small), then it needs to increase approximately as $v\approx \frac{1}{r}$, i.e. like a free vortex (isn't this pretty ^^). However, the free vortex cannot extend all the way to $r=0$, since the tangential velocity must go to zero at the origin due to frictional effects. Hence we expect that there will be a maximum at some finite radius. The only question remaining is why is the fluid inside the eye in solid body rotation, or equivalently:
Why is the vertical component of vorticity homogeneous inside the cyclone eye?
Assuming axisymmetry, this question is equivalent to your original question because $v=\omega r \Rightarrow \zeta = \frac{1}{r}\frac{\partial(rv)}{\partial r} =2\omega = \textrm{const.}$ for $\omega=\textrm{const.} $ I will offer two explanations: the first one is more hand-wavy (and possibly not entirely correct), the other more mathematical, but maybe less transparent for you, unless you have time to go through the maths.
- Since the velocity needs to go to zero between the radius of maximum wind and $r=0$, there is a significant shear (meaning large velocity gradient). Such a shear flow in the eye may fuel turbulence according to the turbulence kinetic energy equation, look at the "shear production" term. Turbulent diffusion in turn is a mechanism by which vorticity might be homogenised inside the eye.
This argument is hand-wavy because although the shear production term in the TKE equation is (empirically) almost always positive in the atmosphere, it is impossible to tell whether this will be the case here since we don't know the sign of the eddy correlation term. Also, there are many other terms in the TKE equation that could potentially cancel out the effects of shear production.
- More formally, we can do the following: in addition to the gradient wind balance (1) which holds in the horizontal, there is (again to good approximation) a hydrostatic balance in the vertical between gravity and the vertical pressure gradient:
$$ -\frac{1}{\rho}\frac{\partial p}{\partial z}-g=0 \tag 4$$.
Based on these two balances, a form of the 1st law of thermodynamics and two conservation laws, Emanuel (1986) (it's a very famous paper in meteorology) derived (on page 3 and top of p.4) an equation relating M and the distribution of specific entropy $s$ in the cyclone, his equation (13).
When doing my project at Imperial, I found out that the specific entropy at fixed height very closely follows a Gaussian distribution:
$$s(r) = \Delta s e^{-r^2/\left(2\lambda^2\right)}+s_\textrm{env}, \tag 5$$
with amplitude $\Delta_s$, offset $s_{env}$ and width $\lambda$. Why exactly it has this functional form has not yet been explained theoretically, but the fact that it has a maximum at/near the center makes sense because most of the diabatic heating due to condensation occurs in the eyewall and entropy is closely related to heat (given that temperature doesn't change much).
(Aside: in fact, the entropy needs to be monotonically decreasing with radius, otherwise (13) from Emanuels paper implies that the vortex is "inertially unstable").
If you use this Gaussian distribution to solve (13) from Emanuel's paper, you can derive formulae for the wind profile, angular momentum and pressure distributions (cf. the following paper which I co-authored :)). If you want you can check the maths, it's simple algebra. In particular the velocity distribution at fixed height that one obtains is:
$$v(r) = \sqrt{2\Delta p \alpha}\sqrt{\frac{2\lambda^2}{r^2}\left(1-e^{-r^2/\left(2\lambda^2\right)}\right)-e^{-r^2/\left(2\lambda^2\right)}}-\frac{1}{2}fr^2,\tag 6$$
where $\Delta p$ is the pressure deficit between environment and cyclone center and $\alpha$ is specific volume. In the paper, we show that this profile is indeed a good fit for wind profiles simulated by solving the three-dimensional Navier-Stokes equations using a supercomputer.
Taylor expanding (6) for small radii and using that $f$ term much smaller than $\frac{\sqrt{\Delta p \alpha}}{\lambda}$ for typical parameter values, gives
$$v(r) = \left(\frac{\sqrt{\Delta p \alpha}}{\sqrt{2}\lambda}-\frac{1}{2}f\right) r + O(r^2) \approx \frac{\sqrt{\Delta p \alpha}}{\sqrt{2}\lambda}r,\tag 7$$
This is indeed a linear profile whose slope is governed by thermodynamic factors: $dW \equiv\Delta p \alpha$ is the work done by the an air parcel as it expands during the inflow from ambient pressure to the lower central pressure. In the the scope of model it may be shown that the radius of maximum wind depends (to good approximation) purely on $\lambda$, while the value of the maximum wind is governed by $dW$. Hence it is consistent that the slope of $v$ increases with increasing $dW$ for fixed $\lambda$.
The physical interpretation of the result however is not clear to me at this point.
Hope this helps, if you have any further questions, let me know and I'll try to improve my answer.
If anyone else has any ideas about the constant vorticity in the cyclone eye I'd be very interested.
EDIT 1: Breaking Rossby waves outside the eyewall
I forgot to mention in the foregoing answer another interesting dynamical aspect that contributes to the observed "sharpness" of the transition between the regions of the constant vorticity (in the eye) and the varying vorticity outside. The mean swirling flow in a tropical cyclone provides a background vorticity profile on which so-called "vortex Rossby waves (VRWs)" can propagate. Such VRWs can reach large amplitudes and "break" (the term has a specific definition in this context). This breaking of VRWs is very efficient at mixing vorticity (or more precisely "potential vorticity (PV)"). This leads to a steepening of the vorticity jump in the eye wall and a homogenisation just outside of it (in the so-called "surf-zone") because a breaking VRW mixes high PV air from the eye interior with lower PV air outside.
EDIT 2: Concentric eyewalls
It may be of interest to you that the velocity profile at the beginning of this post is by no means always present in TC. In fact there is a phenomenon called "concentric eyewall" or "secondary eyewall", where a second maximum occurs in the tangential wind profile. This affects the storm structure immensely, both in its horizontal extent and in its intensity. Therefore, it is an important issue to reproduce them correctly in supercomputer simulations (e.g. to make better forecasts). This is just to illustrate that there are many interesting phenomena associated with TCs :).
EDIT 3: Barotropic instability and potential vorticity mixing inside the eye
A dynamic explanation for why the vorticity is constant. I have learned just recently that, from a dynamical point of view, there is an answer to why vorticity in the TC eye is constant. The explanation goes something like this: consider some nonlinear velocity profile close to the origin, let's say $V\sim x^\alpha$, $\alpha\neq 1$.
If $\alpha<1$ then the vorticity is infinite at the origin, so let's exclude this case. If $\alpha >1$, then the profile will be be "U-shaped", i.e. the vorticity will be smallest in the centre of the eye, then it will be large in a region extending up to the radius of maximum wind, and it will be low outside the eye. In other words, there is a ring of high vorticity, rather than a disk.
Now the critical realisation, backed up by numerical experiments, is that such a ring of vorticity is prone to a process called barotropic instabilty which gives rise to disturbances (VRWs) growing in amplitude by taking energy out of a mean shear. These growing disturbances reach finite amplitudes and break inside the eye to mix (potential) vorticity, thus mixing and homogenising it. See Fig. 3 in this paper by Schubert for an illustration of what I mean. This paper is also cool because it gives a derivation of the equilibrium vortex configuration a maximum entropy method.
EDIT 4:
I found a paper by Glenn Shutts (now at MetOffice UK, written during his time at Imperial College) which explores the bathtub-hurricane analogy further. It seems to be not well known but it is a very interesting read.
Best Answer
It's all about velocity induction in ideally irrotational flows (irrotational except for an infinitely small region of the domain, like vortex tubes).
I'll let you think about it. I'll edit the answer further if need it. To get a in-depth knowledge of fluid dynamics is not so easy