## Event horizons in water flow: the math!

In a previous post, I discussed recent research which demonstrated the creation of an artificial ‘event horizon’ in a fiber optic cable. In that post, I described how a river speeding up as it goes towards a waterfall has an event horizon: waves that are created past the horizon have no possibility of escape. This was illustrated by the figure below:

As you can see, I’ve drawn the wavefronts created by rocks dropped in the water as ellipses, which seems like the obvious solution: waves will be ‘stretched out’ along the flow of the river, while they will spread normally perpendicular to the flow. Being a nitpicky sort of guy, though, I wanted to demonstrate that this is the case mathematically, which I do below the fold… (warning: algebra and calculus follow!)

To start, we draw a picture, illustrating and defining the problem at hand:

The origin of the coordinate system is taken to be at the center of the river, right on the ‘event horizon’. The velocity of the river increases with increasing x; the simplest form for such a behavior (indeed the easiest one to solve for) is:

$v_{\rm river}(x)= v_0 +\alpha x$,

where $v_0$ is the speed of water waves and α is a constant of proportionality; larger α indicates a more rapid increase of river speed.

The source of the wave is taken to be at $(x_1,y_1)$; we imagine a rock is dropped in the water at time $t=0$. In still water, a circular wave would be generated which spreads at speed $v_0$; we can parameterize this wave by its x and y-components and the angle θ,

$v_x=v_0 \cos\theta$,

$v_y=v_0\sin\theta$.

With the problem set up, we now want to determine the shape of the wavefront as it travels in the river. The velocity of the wavefront in the x-direction is increased by the local velocity of the river, i.e.

$v_x=v_0\cos\theta+v_{\rm river}(x)$.

We rewrite this using our definition of $v_{\rm river}$ in the form,

$v_x=q+\alpha x$,

where $q\equiv v_0(1+\cos\theta)$. We are now in a position to solve for the x-behavior of the wavefront! The x-component of velocity is

$v_x=\frac{dx}{dt}$,

so we have

$\frac{dx}{dt}= \alpha x+q$.

This is a first-order linear ordinary differential equation, whose solution can be readily found,

$x=c_0e^{\alpha t}-q/\alpha$.

Here $c_0$ is a constant determined by our initial condition. We require that, at $t=0$, $x(0)=x_1$. The constant $c_0$ is found to be

$c_0= x_1+q/\alpha$.

The equation for x is therefore

$x(t)=(x_1+q/\alpha)e^{\alpha t}-q/\alpha$.

The equation for y can be found by straightforward integration of $v_y$,

$y(t)=y_1+v_0\sin\theta t$.

For any moment of time t, we can trace out the shape of the wavefront by evaluating this pair of equations for all values of θ. We can do better than that, however, and actually find the equation that defines the shape of the wavefront. To do this, we rewrite our equations for x and y so that the entire angular dependence is on the right-hand side:

$(x-x_1e^{\alpha t}-\frac{v_0}{\alpha}e^{\alpha t}+\frac{v_0}{\alpha})/(\frac{v_0}{\alpha}(e^{\alpha t}-1))=\cos\theta$,

$(y-y_1)/(v_0 t)=\sin\theta$.

If we square these two equations and add them together, we can use the familiar trigonometric identity,

$\cos^2\theta+\sin^2\theta =1$,

to write a single equation for the wavefront independent of θ:

$(x-x_1e^{\alpha t}-\frac{v_0}{\alpha}e^{\alpha t}+\frac{v_0}{\alpha})^2/(\frac{v_0}{\alpha}(e^{\alpha t}-1))^2+(y-y_1)^2/(v_0 t)^2=1$.

This is the equation for an ellipse! The standard form of an ellipse is:

$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$,

where a is the semimajor axis and b is the semiminor axis, and $(x_0,y_0)$ is the center of the ellipse:

Thus our initial guess is correct! The equation for the ellipse is quite complicated, however. We note that it is unrealistic for the river velocity to increase linearly for a long distance; we therefore expect that the ‘elongation’ will only happen over a short period of time t. We can write the ellipse equation in a more understandable form by making a Taylor series approximation of the exponential:

$e^{\alpha t}\approx 1+\alpha t+\frac{1}{2}(\alpha t)^2$.

The equation for the ellipse therefore becomes:

$(x-x_1-v't-\frac{1}{2}a't^2)^2/(v_0(t+1/2\alpha t^2))^2+(y-y_1)^2/(v_0 t)^2=1$,

where

$v'\equiv \alpha x_1+v_0$,

$a'\equiv \alpha^2 x_1+\alpha v_0$.

The center of the ellipse propagates downstream with a velocity v‘, and with an effective acceleration a‘. This shows, at least partially, that a velocity-dependent river acts like an ‘effective force’ on objects traveling in the river. The size of the major axis increases quadratically, while the minor axis increases linearly.

The behavior is illustrated in the figure below:

Here I used $v_0=1$, $\alpha =0.1$, and $t=1,2,3,4$ (all in dimensionless units). The wave is emitted at (-1,0) for the first picture, (0,0) for the second and fourth, and (1,0) for the third. When the wave is emitted before the event horizon, we end up elliptical wavefronts whose spacing is compressed upstream and stretched downstream. For waves emitted at the event horizon, the wavefronts are infinitely compressed on the left side, resulting in ellipses which meet on the left. For waves emitted past the event horizon, the ellipses cross each other. The wavefronts for a still water river are shown for comparison.

So my initial picture was correct for the ‘before’ and ‘at’ parts, but incorrect for the ‘past’ part! Hopefully this post will serve as a nice example as how to formulate and solve a simple mathematical physics problem from scratch.

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