Fluid Dynamics

A recurring operator we'll use is the Material Derivative defined as

\begin{equation} \frac{D}{D t} = \vec{v}\cdot\vec{\nabla} + \frac{\partial}{\partial t}. \end{equation}

We can think of it as translating a time derivative in the frame of a given fluid packet to the lab frame.

The conservation of mass induces the Continuity Equation:

\begin{equation} \partial_{t}\rho + \vec{\nabla}\cdot(\rho\vec{v}) = 0. \end{equation}

Note we didn't consider viscosity in deriving this equation, we just used Newton's second law modified for fluid packets (to give us the material derivative of velocity as the "acceleration"). Since we didn't consider viscosity, there is no heat exchange between different parts of the fluid, nor between the fluid and its environment.

L&L note in adiabatic motion, the entropy of any particle of fluid remains constant as that particle moves around in space. So if \(s\) denotes the entropy per unit mass, we can express the adabiatic condition as

\begin{equation} \frac{\D s}{\D t} = \partial_{t}s + (\vec{v}\cdot\vec\nabla)s = 0. \end{equation}

L&L refer to the quantity \(\rho s\vec{v}\) as the Entropy Flux Density because we have the "equation of continuity" for entropy in terms of it:

\begin{equation} \partial_{t}(\rho s) + \vec{\nabla}\cdot(\rho s\vec{v}) = 0. \end{equation}

If, "as usually happens", the entropy is constant throughout the volume of the fluid at some initial instant, it retains everywhere the same constant value at all times and for any subsequent motion of the fluid. In this case we can write the adiabatic condition as

\begin{equation} s = \mbox{constant} \end{equation}

and L&L usually does so. In this case, such a motion is said to be Isentropic.

This requires some boundary condition. For an ideal fluid, this boundary condition is simply "the fluid cannot penetrate a surface solid". I.e., the component of the fluid velocity normal to the bounding surface must vanish if that surface is at rest:

\begin{equation} v_{\vec{n}} = 0. \end{equation}

In general, for the case when the surface is moving, \(v_{n}\) must be equal to the corrsponding component of the velocity of the surface.

The next steps are to study this equation when the fluid is at rest \(\vec{v}=0\) (hydrostatics) and when \(\partial_{t}\vec{v}=0\) is constant (Bernoulli's law).

When we have \(\rho\) be constant throughout the fluid, we can solve the simplified Euler's equation. Taking \(z\) as the vertical axis, so \(\vec{g} = -g\widehat{z}\), then we have

\begin{equation} \partial_{x}p = \partial_{y}p = 0 \end{equation}

and

\begin{equation} \partial_{z}p = -\rho g. \end{equation}

Hence the general solution for this system of equations is

\begin{equation} p = -\rho g z + \mbox{constant}. \end{equation}

If further the fluid is at rest has a free surface at height \(h\), to which an external pressure \(p_{0}\) (the same at every point) is applied, this surface must be the horizontal plane \(z = h\). From the condition \(p = p_{0}\) for \(z = h\), we find the constant is \(p_{0} + \rho gh\), so in this case

\begin{equation} p = p_{0} + \rho g(h - z). \end{equation}

But again, this is the special case of a special case: when density is constant.

When our fluid is a gas, or we have large amounts of our liquid, then we cannot approximate density as constant. This is especially true for gases like the atmosphere.

But we may suppose the fluid is in mechanical and thermal equilibrium. So the temperature is the same at every point. We use the thermal relation

\begin{equation} \D\Phi = -s\,\D T + V\,\D p \end{equation}

where \(\Phi\) is the thermodynamic potential (Gibbs free energy) per unit mass. For constant temperature, this becomes

\begin{equation} \D\Phi = V\,\D p = \frac{\D p}{\rho}. \end{equation}

This let's us write

\begin{equation} \frac{\vec{\nabla}p}{\rho} = \vec{\nabla}\Phi. \end{equation}

Euler's equation then becomes

\begin{equation} \vec{\nabla}\Phi = \vec{g}. \end{equation}

For a constant vector \(\vec{g}\) directed along the negative \(z\)-axis, we have

\begin{equation} \vec{g}:=-\vec{\nabla}(gz) \end{equation}

hence

\begin{equation} \vec{\nabla}(\Phi + gz) = 0, \end{equation}

or equivalently

\begin{equation} \Phi + gz = \mbox{constant}. \end{equation}

Particularly useful for me, when considering a fluid like the atmosphere, which is in mechanical equilibrium in a gravitational field, the pressure in it can be a function only of the altitude/"height" \(z\) (otherwise, if the pressure were different at different points with the same altitude, motion would follow).

It follows then that the density

\begin{equation} \rho = \frac{-1}{g}\frac{\D p}{\D z} \end{equation}

is also a function of \(z\) only. The pressure and density determine the temperature, which is therefore again a function of \(z\) only. So in mechanical equilibrium in a gravitational field, the pressure, density, and temperature distributions depend only on the altitude.

Conversely, if, say, the temperature is different at different points with the same altitude, then mechanical equilibrium is impossible. (Thus this fact is useful in contrapositive form, too.)

When we have a large mass of fluid, say a star, whose separate parts are held together by gravitational attraction. Let \(\phi\) be the Newtonian gravitational potential of the field due to the fluid. It satisfies the differential equation

\begin{equation} \nabla^{2}\phi = 4\pi G\rho \end{equation}

where \(G\) is the Newtonian gravitational constant. The gravitational acceleration is \(-\vec{\nabla}\phi\), and the force on a mass \(\rho\) is \(-\rho\vec{\nabla}\phi\). The condition of equilibrium is therefore

\begin{equation} \vec{\nabla}p = -\rho\vec{\nabla}\phi. \end{equation}

Dividing both sides by \(\rho\) then taking the divergence of both sides gives us

\begin{equation} \nabla\cdot\left(\frac{1}{\rho}\vec{\nabla}p\right) = -4\pi G\rho. \end{equation}

This seems like a great deal more difficult. But we did not assume thermodynamic equilibrium, and in exchange we have a condition for mechanical equilibrium.

If we further assume the body is not rotating, then it will enjoy a spherical symmetry when in equilibrium, and the density & pressure distributions will be spherically symmetric. We have in spherical coordinates our condition become

\begin{equation} \frac{1}{r^{2}}\frac{\D}{\D r}\left(\frac{r^{2}}{\rho}\frac{\D p}{\D r}\right) = -4\pi G\rho. \end{equation}

Again, this fails to hold for a rotating body (like the Earth).

In a steady flow, the isentropic Euler equations becomes

\begin{equation} \frac{1}{2}\vec{\nabla}(v^{2}) - \vec{v}\times(\curl\vec{v}) = -\vec{\nabla}w. \end{equation}

Streamlines are lines such that the tangent to a streamline at any point gives the direction of the velocity of that point. They are determined by the system of differential equations

\begin{equation} \frac{\D x}{v_{x}} = \frac{\D y}{v_{y}} = \frac{\D z}{v_{z}}. \end{equation}

In steady flow, the streamlines do not vary with time (i.e., the choice of \(t\) does not vary the solution of \(\varphi\)), and coincide with the paths of the fluid particles.

BUT for non-steady flows, this coincidence no longer occurs: the tangents to the streamlines give the directions of the velocities of fluid particles at various points in space at a given instant, where the tangents to the paths give the directions of the velocities of given fluid particles at various times.

Question: for non-steady flows, are streamlines still a valuable concept?

Taking the equation for steady flows, then forming the scalar product with the unit vector tangent to the streamline at each point (call this unit vector \(\vec{\ell}\)). The projection of the gradient on any direction is the derivative in that direction (duh, the directional derivative). So

\begin{equation} \vec{\ell}\cdot(\vec{\nabla}w) = \partial w/\partial\ell. \end{equation}

The vector \(\vec{v}\times(\curl\vec{v})\) is orthogonal to \(\vec{v}\), hence its dot product with \(\vec{\ell}\) vanishes.

Thus we get the simplified equation

\begin{equation} \frac{\partial}{\partial\ell}\left(\frac{v^{2}}{2} + w\right) = 0. \end{equation}

It follows that \(w + v^{2}/2\) is constant along the streamline, i.e.,

\begin{equation} \frac{1}{2}v^{2} + w = \mbox{constant}. \end{equation}

This is precisely Bernoulli's Equation.

In general, the constant \(w + v^{2}/2\) takes different values for different streamlines.

So, if our fluid flows in a gravitational field, the acceleration \(\vec{g}\) due to gravity must be added to the right-hand side of the isentropic Euler equations for steady flows:

\begin{equation} \frac{1}{2}\vec{\nabla}(v^{2}) - \vec{v}\times(\curl\vec{v}) = -\vec{\nabla}w + \vec{g}. \end{equation}

Let us take the direction of gravity as the \(z\)-axis, with \(z\) increasing upwards. This choice let's us relate the cosine of the angle between \(\vec{g}\) and \(\vec{\ell}\) to be equal to the derivative \(-\D z/\D\ell\), so the projection of \(\vec{g}\) on \(\vec{\ell}\) is \(-g\,\D z/\D\ell\).

Then we have our projected equation become

\begin{equation} \frac{\partial}{\partial\ell}\left(\frac{v^{2}}{2} + w + gz\right) = 0. \end{equation}

Hence Bernoulli's equation states that along a streamline we have

\begin{equation} \left(\frac{v^{2}}{2} + w\right) = \mbox{constant}. \end{equation}

For our fluid, the energy per unit volume of fluid is \(\rho v^{2}/2 + \rho\varepsilon\) the sum of its kinetic energy and internal energy, where \(\varepsilon\) is internel energy per unit mass.

The change in this energy is given by its partial derivative with respect to time.

This follows by computing \(\partial_{t}(\rho v^{2}/2 + \rho\varepsilon)\), the change in energy with respect to time. The first and second terms are rather involved, and delegated to "lemmas" below. But the lemmas then give us

\begin{equation} \partial_{t}\left(\frac{\rho v^{2}}{2} + \rho\varepsilon\right) = -\left(\frac{v^{2}}{2} + w\right)\vec{\nabla}\cdot(\rho\vec{v}) -\rho\vec{v}\cdot\vec{\nabla}(w + v^{2}/2) \end{equation}

The right-hand side can be rewritten to give the final result.

We have an abstract equation we assert is correct through a long string of equations. Believe me? Well, let's see what it means. We need to integrate it over some volume \(V\):

\begin{equation} \partial_{t}\int\left(\frac{\rho v^{2}}{2} + \rho\varepsilon\right)\,\D V = -\int\vec{\nabla}\cdot\left[\rho\vec{v}\left(\frac{v^{2}}{2}+w\right)\right]\,\D V \end{equation}

We convert the integral on the right to a surface integral

\begin{equation} \partial_{t}\int\left(\frac{\rho v^{2}}{2} + \rho\varepsilon\right)\D V = -\oint\left[\rho\vec{v}\left(\frac{v^{2}}{2}+w\right)\right]\cdot \D\vec{f}. \end{equation}

• The LHS is the rate of change of the energy of the fluid in some volume.
• The RHS is the amount of energy flowing out of this volume in unit time.
• We may call \(\rho\vec{v}(w + v^{2}/2)\) the Energy Flux Density. Its magnitude is the amount of energy passing in unit time through unit area perpendicular to the direction of the velocity. Nifty.

The momentum of unit volume is \(\rho\vec{v}\). What is its rate of change with respect to time \(\partial_{t}(\rho\vec{v})\)? We will switch to index notation for ease of computation.

We can use the product rule to find

\begin{equation} \frac{\partial}{\partial t}(\rho v_{i}) = \rho\frac{\partial v_{i}}{\partial t} + \frac{\partial\rho}{\partial t}v_{i} \end{equation}

We can use the continuity equation in the form of

\begin{equation} \partial_{t}\rho = -\sum_{k}\frac{\partial(\rho v_{k})}{\partial x_{k}} \end{equation}

and Euler's equations in the form of

\begin{equation} \partial_{t}v_{i} = -\frac{1}{\rho}\frac{\partial p}{\partial x_{i}} - \sum_{k}v_{k}\frac{\partial v_{i}}{\partial x_{k}} \end{equation}

to obtain

\begin{equation} \partial_{t}(\rho v_{i}) = \rho\left(-\frac{1}{\rho}\frac{\partial p}{\partial x_{i}} - \sum_{k}v_{k}\frac{\partial v_{i}}{\partial x_{k}}\right) + \left(-\sum_{k}\frac{\partial(\rho v_{k})}{\partial x_{k}}\right) v_{i} \end{equation}

which simplifies to

\begin{equation} \partial_{t}(\rho v_{i}) = -\frac{\partial p}{\partial x_{i}} - \sum_{k}\frac{\partial}{\partial x_{k}}(\rho v_{i}v_{k}). \end{equation}

The first term on the right hand side can be rewritten using index gymnastics as

\begin{equation} \frac{\partial p}{\partial x_{i}} = \sum_{k}\delta_{ik}\frac{\partial p}{\partial x_{k}}. \end{equation}

Then we have

\begin{equation} \partial_{t}(\rho v_{i}) = -\sum_{k}\frac{\partial\Pi_{ik}}{\partial x_{k}} \end{equation}

where the tensor \(\Pi_{ik}\) is defined as

\begin{equation} \Pi_{ik} = p\delta_{ik} + \rho v_{i}v_{k}. \end{equation}

Since \(\Pi_{ik}\) is the sum of two symmetric tensors, it is thus a symmetric tensor.

We grasp the meaning of the momentum flux density tensor by looking at its integral over some volume

\begin{equation} \partial_{t}\int \rho v_{i}\,\D V = -\int\sum_{k}\frac{\partial\Pi_{ik}}{\partial x_{k}}\,\D V. \end{equation}

We can apply Green's theorem to the right hand side, obtaining

\begin{equation} \partial_{t}\int \rho v_{i}\,\D V = -\sum_{k}\oint\Pi_{ik}\,\D f_{k}. \end{equation}

• The left-hand side is the rate of change of the \(i\)-th component of momentum contained in the volume (i.e., the domain of integration).
• The RHS is thus the amount of momentum flowing out through the bounding surface in unit time.

So \(\Pi_{ik}\,\D f^{k}\) is the \(i\)-th component of momentum flowing through the surface element \(\D f\). If we write \(\D f_{k}\) in the form of \(n_{k}\,\D f\), where \(\D f\) is the area of the surface element and \(\vec{n}\) is a unit vector along the outward normal, we find that \(\Pi_{ik}n^{k}\) is the flux of the \(i\)-th component of momentum through unit surface area. Note that

\begin{equation} \Pi_{ik}n^{k} = p n_{i} + \rho v_{i}v_{k}n^{k}. \end{equation}

or in vector form

\begin{equation} \Pi\vec{n} = p\vec{n} + \rho\vec{v}(\vec{v}\cdot\vec{n}). \end{equation}

We may be excited and hope to base several arguments upon this non-steady flow result. For example…

• Consider steady flow past some body. Let the incident flow be uniform at infinity. Its velocity \(\vec{v}\) is a constant (so \(\curl\vec{v}=0\) on all streamlines). Hence we conclude that \(\curl\vec{v}=0\) along the whole of every streamline (i.e., all of space).
• Suppose at some instant we have irrotational flow throughout the volume of the fluid. Then the velocity circulation around any closed contour in the fluid is zero.
  • Simplifying Assumption. (For simplicity, we assume the fluid occupies a simply-connected region of space; for a multiply-connected region, we need restrictions on the choice of contours for this derivation to work.)
  • Thus we then conclude this will hold at any future instant, i.e., we should find that — if there is potential flow at some instant, then there is potential flow at all subsequent instants. (In particular, any flow for which the fluid is initially at rest must be a potential flow.)
  • This is in accordance with the fact if \(\curl\vec{v}=0\), then the isentropic Euler equations hold automatically.

Alas, these conclusions are of limited validity. Why? The proof given above that \(\curl\vec{v} = 0\) all along a streamline is, strictly speaking, invalid for a line which lies in the surface of a solid body past which the flow takes place, since the present of this surface makes it impossible to draw a closed contour in the fluid encircling such a streamline.

The equations of motion for an ideal fluid admit solutions for which separation occurs at the surface of the boundary: the streamlines follow the surface, until the surface ends, and the streamlines become separated and continue into the fluid. The resulting flow pattern is characterized by the presence of a Surface of Discontinuity proceeding from the body. On this surface, the fluid velocity — which is everywhere tangential to the surface — has a discontinuity. In other words, at this surface one layer of fluid "slides" on another. (Mathematically speaking, the discontinuity in the tangential velocity components corresponds to a surface on which the curl of the velocity is nonzero.)

When we include discontinuous flows, there are infinitely many solutions to the equations of motion for an ideal fluid. The tangential discontinuities are absolutely unstable and become turbulent and, well, all hell breaks loose.

But when we, e.g., spray an object with water, there is some flow, i.e., empirically there is a solution. What's going on here? Well, water isn't an ideal fluid, there's some viscosity. Even if we had some hypothetical ideal fluid, we would expect similar results. Why? Because the properties of the flow in the boundary layer determines the choice of solution.

In the general case of flow past bodies of arbitrary form, solutions with separation must be taken, which in turn results in turbulence.

For certain special shapes, the flow differs little from irrotational flow. Well, it will be irrotational flow except in a thin layer of fluid at the surface of the body and in a relatively narrow "wake" behind the body. Nifty.

Another important case of potential flow occurs for small oscillations of a body immersed in fluid. (Isn't the Earth immersed in its atmosphere?) We can show that, if the amplitude \(a\) of the oscillations is small in comparison with the linear dimension \(\ell\) of the body (\(a\ll \ell\)), then the flow past the body will be potential flow. How to prove it?

We estimate the order of magntidue of the various terms in Euler's equation

\begin{equation} \partial_{t}\vec{v} + (\vec{v}\cdot\vec{\nabla})\vec{v} = -\vec{\nabla}w. \end{equation}

The velocity \(\vec{v}\) changes markedly over a distance of the order of dimension \(\ell\) of the body by an amount of the same order as the velocity \(\vec{u}_{\text{osc}}\) of the oscillating body. Hence:

• the derivatives of \(\vec{v}\) with respect to the coordinates \(\nabla\vec{v}\sim\bigOh(u/\ell)\);
• the order of \(\vec{v}\) at fairly small distances from the body is \(\vec{v}\sim\bigOh(\vec{u})\)
• hence \((\vec{v}\cdot\vec{\nabla})\vec{v}\sim\bigOh(u^{2}/\ell)\)
• the derivative \(\partial_{t}\vec{v}\sim\omega u\) where \(\omega\) is the frequency of oscillations
  • since \(\omega\sim u/a\) we have \(\partial_{t}\vec{v} = u^{2}/a\)
• since \(a\ll\ell\), the term \((\vec{v}\cdot\vec{\nabla})\vec{v}\) is small compared to \(\partial_{t}\vec{v}\)
  • hence we may neglect \((\vec{v}\cdot\vec{\nabla})\vec{v}\)

• the equation of motion becomes

  \begin{equation} \partial_{t}\vec{v} = -\vec{\nabla}w \end{equation}
• taking the curl of both sides, we obtain \(\partial_{t}(\curl\vec{v})=0\)
  • hence \(\curl\vec{v} = \mbox{constant}\).
  • In oscillatory motion, the time average of the velocity vanishes; hence \(\curl\vec{v} = 0\).
• Thus the motion of a fluid executing small oscillations is potential flow, to first approximation.

• Only isentropic flow can be potential flow (otherwise, for non-isentropic flow, if we have potential flow at some instant, the vorticity will be nonzero at subsequent instants, leading to non-potential ["rotational"] flow)
• In potential flow, the velocity circulation along any closed contour vanishes \(\oint \vec{v}\cdot\D\ell = \iint\curl\vec{v}\cdot\D\vec{f}=0\)
  • Hence closed streamlines cannot exist in potential flow.
• In rotational flow, the velocity circulation is not zero in general.
  • There may be closed streamlines in rotational flow, but the presence of closed streamlines is not necessary for rotational flow.

• We could write the velocity as the gradient of a Velocity Potential and denote it \(\phi\),

  \begin{equation} \vec{v} = \vec{\nabla}\phi \end{equation}

  • The isentropic Euler equations become

    \begin{equation} \vec{\nabla}(\partial_{t}\phi + \frac{v^{2}}{2} + w) = 0. \end{equation}

  • The first integral of this equation looks like

    \begin{equation} \partial_{t}\phi + \frac{v^{2}}{2} + w = f(t) \end{equation}

    where \(f(t)\) is "an arbitrary function of time". We may set it equal to zero without loss of generality (because the velocity potential is not uniquely defined: it's defined such that \(\vec{v}\) is a spatial derivative of \(\phi\), so adding any function of time to \(\phi\) will produce the same results)

  • For steady flows, \(\partial_{t}\phi = 0\), \(f(t)\) is constant, and the first integral becomes Bernoulli's equation

    \begin{equation*} 0 + \frac{v^{2}}{2} + w = \mbox{constant}. \end{equation*}
  • In general, the "constant" on the right hand side is a constant along any given streamline, but varies for different streamlines.
  • In a potential flow, it is constant throughout the fluid. This enhances the importance of Bernoulli's equation in the study of potential flow.

The equations of motion take advantage of the identity

\begin{equation} \frac{1}{2}\vec{\nabla}v^{2} = \vec{v}\times(\curl\vec{v}) + (\vec{v}\cdot\vec{\nabla})\vec{v} \end{equation}

and we can rewrite the equations of motion using this, and the continuity equation, to be

\begin{equation} \partial_{t}(\curl\vec{v}) = \curl(\vec{v}\times(\curl\vec{v})). \end{equation}

We don't need isentropy, we have the same equations of motion though.

Bernoulli's equation simplifies for an incompressible fluid. We simply replace the heat function \(w\) by \(p/\rho\):

\begin{equation} \frac{1}{2}v^{2} + \frac{p}{\rho} + gz = \mbox{constant}. \end{equation}

For an incompressible fluid, we also write \(p/\rho\) in place of \(w\) for the energy flux as

\begin{equation} \rho\vec{v}\left(\frac{1}{2}v^{2} + \frac{p}{\rho}\right). \end{equation}

Did you notice that the equations of motion are satisfied if

\begin{equation} \curl\vec{v} = 0. \end{equation}

Indeed, \(\vec{v} = \vec{\nabla}\phi\) substituted into the continuity equation gives us

\begin{equation} \nabla^{2}\phi = 0. \end{equation}

We need boundary conditions at the surfaces where the fluid meets solid bodies. At fixed surfaces, the fluid velocity component \(v_{n}\) normal to the surface must be zero, whereas for moving surfaces it must be equal to the normal components of the velocity of the surface (i.e., a given function of time). The velocity \(v_{n}\) is equal to the normal derivative of the potential \(\phi\):

\begin{equation} v_{n} = \frac{\partial\phi}{\partial n}. \end{equation}

Thus the general boudnary conditions are that \(\partial\phi/\partial n\) is a given function of coordinates and time at the boundaries.

For potential flow, the velocity is related to the pressure by results from the previous section (§9). In an incompressible fluid, we can replace \(w\) in this equation by \(p/\rho\)

\begin{equation} \partial_{t}\phi + \frac{v^{2}}{2} + \frac{p}{\rho} = f(t). \end{equation}

We may notice here the following important property of potential flow of an incompressible flow:

Suppose some solid body is moving through the fluid. If the result is potential flow, it depends at any instant only on the velocity of the moving body at that instant, and not (e.g.) on its acceleration. Since \(\nabla^{2}\phi=0\) does not explicitly contain time, which enter the solution only through the boundary conditions, and these contain only the velocity of the moving body.

Consider a fluid flowing past a body, as diagrammed in the next figure. For Bernoulli's equation \(v^{2}/2 + p/\rho\) equal to some constant, we see in steady flow of an incompressible fluid (not in a gravitational field), the greatest pressure occurs where velocity is zero. Such a point usually occurs on the surface of a body past which the fluid is flowing and is called a Stagnation Point. If \(\vec{u}\) is the velocity of the incident current (= the fluid velocity at infinity), and \(p_{0}\) is the pressure at infinity, then the pressure at the stagnation point is

\begin{equation} p_{\text{max}} = p_{0} + \frac{\rho u^{2}}{2}. \end{equation}

To solve problems of two-dimensional flow of an incompressible fluid, it is sometimes convenient to express the velocity in terms of what is called the Stream Function. From the equation of continuity

\[ \vec{\nabla}\cdot\vec{v}=\partial_{x}v_{x} +\partial_{y}v_{y}=0 \]

we see the velocity components can be written as derivatives

\begin{equation} v_{x} = \partial_{y}\psi,\quad\mbox{and}\quad v_{y} = -\partial_{x}\psi \end{equation}

of some function \(\psi(x,y)\) called the Stream Function. The equation of continuity is satisfied automatically, if we write the velocity in this manner (prove this to yourself).

The equation that must be satisfied now: the equation of motion. Simply plug in the definition of velocity as the derivative of the streamfunction into the equations of motion to find:

\begin{equation} \frac{\partial}{\partial t}\nabla^{2}\psi - \frac{\partial\psi}{\partial x}\frac{\partial}{\partial y}\nabla^{2}\psi + \frac{\partial\psi}{\partial y}\frac{\partial}{\partial x}\nabla^{2}\psi = 0. \end{equation}

Now, if we know the stream function, we can immediately determine the streamlines of steady flow. The differential equation for stream lines are

\begin{equation} \frac{\D x}{v_{x}}=\frac{\D y}{v_{y}}\implies v_{y}\,\D x - v_{x}\,\D y=0. \end{equation}

This expresses the fact that the direction of the tangent to a streamline is the direction of the velocity. Substituting the definition of velocity components as the derivative of the stream function

\begin{equation} \frac{\partial\psi}{\partial x}\D x + \frac{\partial\psi}{\partial y}\D y=\D\psi=0, \end{equation}

hence \(\psi\) is constant. Thus the streamlines are the family of curves obtained by putting the stream function \(\psi(x,y)\) equal to an arbitrary constant (aren't these the "level curves" we learned about back in calculus?).

If we draw a curve between two points \(A\) and \(B\) in the \(xy\)-plane, the mass flux \(Q\) across this curve is given by the differencein the values of the stream function at these two points regardless of the curve. If \(v_{n}\) is the component of the velocity normal to the curve at any point, we have

\begin{equation} Q = \rho \oint_{A}^{B}v_{n}\,\D\ell = \rho \oint^{B}_{A}(-v_{y}\,\D x + v_{x}\,\D y) = \rho\int^{B}_{A}\D\psi \end{equation}

or

\begin{equation} Q = \rho(\psi_{B} - \psi_{A}). \end{equation}

We can use complex analysis to solve a lot of problems of two-dimensional potential flow of incompressible fluids past bodies of various shapes. Look at the relation between the components of velocity and the stream function:

\begin{equation*} v_{x}=\partial_{x}\phi = \partial_{y}\psi, \end{equation*}

and

\begin{equation*} v_{y}=\partial_{y}\phi = -\partial_{x}\psi. \end{equation*}

These are the Cauchy–Riemann equations for the complex expression

\begin{equation} w = \phi + \I\psi \end{equation}

to be an analytic function of the complex argument \(z = x + \I y\). The function \(w\) is called the Complex Potential and \(\D w/\D z\) the Complex Velocity. The modulus of this complex velocity give the magnitude \(v\) of the velocity and the angle \(\theta\) between the direction of the velocity and the \(x\)-axis:

\begin{equation} \frac{\D w}{\D z} = v\E^{-\I\theta}. \end{equation}

At a solid surface past which the flow takes place, the velocity must be along the tangent; i.e., the profile contour of the surface must be a streamline (i.e., \(\psi\) is constant along it). The constant may be taken to be zero, then the problem of flow past a given contour reduces to determining an analytic function \(w(z)\) which takes real values on the contour.

The integral of an analytic function around any closed contour \(C\) is known to be \(2\pi\I\) times the sum of residues of the function at its simple poles inside \(C\). Hence

\begin{equation} \oint w'\,\D z = 2\pi\I\sum_{k}\mathrm{Res}(w', z_{k}) \end{equation}

where \(\mathrm{Res}(w', z_{k})\) are the residues of the complex velocity. We also have

\begin{align*} \oint w'\,\D z &= \oint(v_{x} - \I v_{y})(\D x + \I\,\D y)\\ &= \oint(v_{x}\,\D x + v_{y}\,\D y) + \I\oint(v_{x}\,\D y - v_{y}\,\D x). \end{align*}

The real part of this is just the velocity circulation \(\Gamma\) around the contour \(C\). The imaginary part, multipled by \(\rho\), is the mass flux across \(C\). If there are no sources of fluid within the contour, this flux is zero and we then have

\begin{equation} \Gamma = 2\pi\I\sum_{k}\mathrm{Res}(w', z_{k}) \end{equation}

all the residues are in this case purely imaginary.

Let us consider the conditions which the fluid may be regarded as incompressible. When the pressure changes adiabatically by \(\Delta p\), the density changes by \(\Delta\rho = (\partial\rho/\partial p)\Delta p\).

According to Bernoulli's equation, however, the order of \(\Delta p\sim\rho v^{2}\) in steady flow. (L&L promises to show in § 64 that the derivative \((\partial p/\partial\rho)\) is the square of the velocity of sound \(c\) in the fluid, so \(\Delta p\sim\rho v^{2}/c^{2}\).)

The fluid may be regarded as incompressible if

\begin{equation} \frac{\Delta\rho}{\rho}\ll 1. \end{equation}

We see a necessary condition for this is the fluid velocity may be regarded as small compared to that of sound:

\begin{equation} v\ll c. \end{equation}

But this is for a steady flow. What do we do for a non-steady flow? We need an additional condition.

• Let \(\tau\) and \(\ell\) be a time and length of the order of the times and distances over which the fluid velocity undergoes significant changes.
• If \(\partial_{t}\vec{v}\) and \((1/\rho)\vec{\nabla}p\) in Euler's equation are comparable, then we find the orders of magnitude \(v/\tau\sim\Delta p/\ell\rho\) or \(\Delta p\sim \ell\rho v/\tau\)
• The corresponding change in \(\rho\) is \(\Delta\rho\sim\ell\rho v/\tau c^{2}\).

• In the equation of continuity, compare the terms \(\partial_{t}\rho\) and \(\rho\vec{\nabla}\cdot\vec{v}\), we find that \(\partial_{t}\rho\) may be neglected if \(\Delta\rho/\tau\ll\rho v/\ell\), or

  \begin{equation} \tau\gg\ell/c. \end{equation}
• This has the meaning: the time \(\ell/c\) taken by a sound signal to traverse the distance \(\ell\) must be small compared to the time \(\tau\) during which the flow changes appreciably, so the propagation of interactions in the fluid may be regarded as instantaneous.
• If both this condition \(\tau\gg\ell/c\) and our previous condition \(v\ll c\) are fulfilled, the fluid may be regarded as incompressible.

The general form of Laplace's equation is

\begin{equation} \phi=\frac{-a}{r}+\vec{A}\cdot\vec{\nabla}\frac{1}{r}+\bigOh(r^{-3}) \end{equation}

where \(a\), \(\vec{A}\) are independent of coordinates.

Since we are seeking the velocity at large distances, the terms of higher order than quadratic may be neglected, so we have

\begin{equation} \phi = \vec{A}\cdot\vec{\nabla}(1/r)=-\vec{A}\cdot\widehat{r}/r^{2} \end{equation}

and the velocity vector is

\begin{equation} \vec{v}=(A\cdot\vec{\nabla})\vec{\nabla}\frac{1}{r} =\frac{3(\vec{A}\cdot\widehat{r})\widehat{r}-\vec{A}}{r^{3}} \end{equation}

where \(\widehat{r}\) is a unit vector in the direction of \(\vec{r}\). At large distances the velocity diminishes as \(r^{-3}\). The vector \(\vec{A}\) depends on the shape and velocity of the body and can be determined only by solving the equation \(\nabla^{2}\phi=0\) at all distances, taking into account the appropriate boundary conditions at the surface of the moving body.

We can compute the total momentum \(\vec{P}\) for the fluid from the energy. How?

It follows from the definition of induced mass tensor that

\begin{equation} P_{i} = m_{ik}u^{k}. \end{equation}

And by comparing results we find

\begin{equation} \vec{P} = 4\pi\rho\vec{A} - \rho V_{0}\vec{u}. \end{equation}

The momentum transmitted to the fluid by the body in unit time is \(\D\vec{P}/\D t\). With the opposite sign it gives the reaction of the fluid (i.e., the force acting on the body):

\begin{equation} \vec{F} = -\frac{\D\vec{P}}{\D t} \end{equation}

The components of \(\vec{F}\) parallel to the velocity of the body is called the Drag Force and the perpendicular component is called the Lift Force.

Now, everything that has been said, we stress, has assumed an infinite volume for the fluid. If the fluid has a free surface, then a body moving uniformly parallel to this surface will experience a drag (Wave Drag) due to the occurrence of a "system of waves" propagated on the free surface, which continually remove energy to infinity.

We assume the velocity of the fluid varies slightly over distances of the order of the dimension of the body. Let \(\vec{v}\) be the fluid velocity at the position of the body would be if the body were absent; i.e., \(\vec{v}\) is the velocity of the unperturbed flow. By our given assumption, \(\vec{v}\) may be supposed constant throughout the volume occupied by the body. We therefore denote the velocity of the body by \(\vec{u}\) as before.

We can determine the force acting on the body and setting it in motion as follows:

• If the body were wholly carried along with the fluid (so \(\vec{u}=\vec{v}\)), then the force acting on it would be the same as the force that would be acting on the liquid in the same volume if the body were absent.
• The momentum of the fluid is \(\rho V_{0}\vec{v}\).
• The force acting on the fluid would therefore be \(\rho V_{0}\,\mathrm{d}\vec{v}/\mathrm{d}t\).
• But in reality, the body is not wholly carried along the fluid: there is a motion of the body relative to the fluid, and due to this the fluid itself acquires some additional motion.
• The additional momentum of the fluid is \(m_{ik}(u^{k}-v^{k})\) since we must now replace \(\vec{u}\) by the relative velocity \(\vec{u}-\vec{v}\) of the body relative to the fluid.
• The change of this momentum with time results in the appearance of an additional reaction force on the body of \(-m_{ik}\,\mathrm{d}(u^{k}-v^{k})/\mathrm{d}t\).
• Thus the total force on the body is \(\rho V_{0}\mathrm{d}v_{i}/\mathrm{d}t - m_{ik}\mathrm{d}(u^{k}-v^{k})/\mathrm{d}t\).
• We equate this force to the time derivative of the body momentum, thus obtaining the following equation of motion:

\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}(M u_{i}) = \rho V_{0}\frac{\mathrm{d}v_{i}}{\mathrm{d}t} - m_{ik}\frac{\mathrm{d}}{\mathrm{d}t}(u^{k}-v^{k}) \end{equation}

We integrate both sides with respect to time to obtain:

\begin{equation} (M\delta_{ik} + m_{ik})u^{k} = (m_{ik} + \rho V_{0}\delta_{ik})v^{k}. \end{equation}

We put the integration constant to zero, since the body's velocity \(\vec{u}\) must vanish when \(\vec{v}\) vanishes.

The relation determines the velocity of the body from that of the fluid.

As a sanity check, if the density of the body is equal to that of the fluid (so \(M = \rho V_{0}\)), we obtain \(\vec{u}=\vec{v}\) as we would expect.