For this derivation, we use three principles. These are:
- As so far as water waves are concerned, water behaves in an irrotational manner (more on this below).
- It also acts as if it were incompressible, and
- The motions of the parcels of water are governed by Newton's second law of motion F = ma .
We shall derive all equations in two dimensions, in the x and y coordinates, assuming everything is constant in the z direction. This allows us to calculate those long ocean waves, but not waves on a choppy sea.
Irrotational flow and the velocity potential
Fluid dynamics experts know that most common fluid phenomena take place in the irrotational regime. That is to say, that if we look at all the blobs (or parcels) of water in the system, that none of them are spinning or have a net circulating current flowing around their perimeter. This is a little confusing because in a wave, the blobs actually move in circular paths, but at the same time they are not spinning. This is similar to the riders of a ferris wheel, traveling in a circular path but not turning head-over-heals. This has to do with the viscosity of water being a fairly minor force in water wave dynamics. It usually takes viscosity and shear forces to cause spinning of the water parcels. We will revisit this later, but for now, it allow us to invent a mathematical construct, a potential, called the velocity potential, defined such that the velocity v at any point is given by minus the gradient of this velocity potential φ:
where we assume that everything is constant in the z direction. The velocity potential is a scalar, i.e. is not a vector and has no directional components. The advantage of using a velocity potential is that this one variable replaces the two velocity components and is therefore less work to manipulate. Furthermore by using it, we automatically will have only irrotational flow and need not worry further about this aspect. In fact, creation of the scalar velocity potential is only possible for fluid flows that are irrotational. Below we show animations featuring irrotational flow and the velocity potential of water waves.
Incompressible flow, divergence equation
Next, we must insure that the flow in a wave behaves as though the fluid is incompressible, that the flow moving into any fixed volume of space in the region of the fluid must equal that coming out the other side of the same volume. Mathematically, we insure this by setting its divergence equal to zero:
for our two dimensional wave, where we can ignore z since nothing depends on z.
|Pierre-Simon Laplace, 1749-1827. Wikipedia|
We can now substitute the velocity potential (1a) and (1b) above in for the two velocity components in (2b) to get:
Written out component by component this is:
This equation is called Laplace's Equation after Pierre-Simon Laplace who first studied it. This equation occurs in many continuum mechanics problems, e.g. problems involving electromagnetic fields, fluid flow, and deformation of solids.
Separation of variables, product solution
One common way to solve Laplace's Equation is by separation of variables.
In our case we are looking for a φ(x,y,t) that satisfies (3). The separation of variable methods has us look for product solutions of the form
where X(x) is only a function of x, Y(y) is only a function of y and T(t) is only a function of t. When we substitute (4) into (3), cancel out similar factors, divide by XYT and rearrange, we have:
We now have a situation where the left hand side is only a function of x and the right hand side is only a function of y. Now we can argue that because of this, the equation must be constant. We do this by reminding ourselves that the two sides are supposedly equal for the entire x-y space of interest. Pick one point. The two sides are equal here (as they are everywhere in the space). Now vary only x keeping y constant. That is, move in a direction parallel to the x axis in the space. Y(y) is only a function of y so the right side of the equation must remain constant, since we are not varying y. Since the two side are supposed to stay equal, the left side must also remain constant, which means X(x) must be constant even though x is varying. Thus the left side is a constant, independent of x. And since the left side is constant and the right side supposedly equals the left side, the right side therefore must also be constant, i.e. equal to the same constant as the left hand side.
Armed with this knowledge, we set both sides equal to a constant. This constant is called a separation constant. We will call it −κ2, since in the problem at hand, it will turn out to be minus a wave number squared. We can set the left side of (5) equal to −κ2 and also the right side of (5) equal to −κ2:
These can be rearranged into the more standard differential equation form as:
Thus, we started with one differential equation of two variables and now have two differential equations of only one variable each. Furthermore, these are very standard differential equations with well known solutions. At the same time, we should remember that we have limited our search of solutions to only those that can be expressed as product solutions of the form given by (4) above. It will turn out that this has forced us to only arrive at nicely sinusoidal functions and related exponential functions.
|Fig. 5. The coordinate system used in this derivation.|
SolutionsWe are looking for wave-like solutions in the horizontal or x direction. Luckily for us, the most common solutions to (6a) are sin κx, cos κx, exp(iκx), and exp(−iκx). All of these are wave-like solutions and mathematically we could use any combination of these to form our desired solution. Perhaps we'll use the last one, the X(x) = exp(−iκx), and work in the complex representation for our water waves. This solution will be correct for traveling waves moving in the positive x direction. (A good exercise for students is to plug each of the solution types into (6a) and confirm that the left side of (6a) equals the right side for each of these substitutions, i.e. that they are indeed solutions.)
Equation (6b) is almost the same as (6a) except for the lack of the minus sign. This sign means that the solutions are sinh κy, cosh κy, exp(κy), and exp(−κy). If we pick the y coordinate system as shown to the right, with the bottom of the body of water corresponding to y = 0, then we need a solution such that vy = 0 at y = 0 so that the water isn't allowed to move in and out of the solid bottom. Using (1b) we see that this means that:
|Fig. 6. Four possible solutions to Equation (6b).|
Since we don't want X(x) or T(t) to be zero and wipe out our wave, φ(x,y,t) = X(x) Y(y) T(t), it must be that the derivative factor is zero at y = 0, i.e.
Thus we are looking for a Y(y) that has a derivative equal to zero at y = 0. The four solutions mentioned above are plotted at the left in Fig. 6. Looking at the graph, we see that only Y(y) = cosh κy has a derivative equal to zero at y = 0, so this is the solution to use.
So far we have
X(x) = exp(−iκx) (7a) and Y(y) = cosh κy (7b) .
T(t) is still yet to be discovered.
At this point, we will simply assume that T(t) is cyclic, i.e.
T(t) = exp(iωt) (8)
with an unknown angular frequency ω. Our solution becomes:
φ(x,y,t) = X(x) Y(y) T(t) = A exp(−iκx) cosh(κy) exp(iωt)
= A cosh(κy) exp i(ωt−κx) = A cosh(κy) ei(ωt−κx) , (9)
where A is a constant amplitude.
Velocity and displacement equationsWe still don't have a relationship between ω and κ, but more on that in a bit. First, let's see what (9) looks like. Looking at (1), i.e.
we see that we can differentiate φ to get the x and y velocities of water in the wave:
These give the velocities of water parcels as a function of x, y, t for two dimensional traveling water waves propagating in the positive x direction. We can integrate these velocities with respect to time to get the x and y displacements from the equilibrium position as a function of x and y. We will label the displacements as Δy and Δy. The integrations yield:
where we have ignored the integration constant because we are interested in the time varying part (the wave part). Eqs. (12a) and (12b) are the circular and elliptical paths discussed in the last posting (and shown in Figs. 1 and 2 above) and are also responsible for the cycloidal surface of a water wave. I might point out, that (12a) and (12b) are approximations (the first so far) that assume that the velocity of a parcel (even when it has moved away from its equilibrium position) is given by the velocity of water at the parcel's equilibrium position. This approximation is valid for small amplitude waves in which parcels do not move very far from the equilibrium positions, but becomes less so for larger amplitude waves.
NEXT: Water waves derivation 2: dynamics of the free surface PREVIOUS TOPIC: Water waves
|Good references on WAVES||Good general references on resonators, waves, and fields|