Bernoulli’s Equation Revisited and Magnus Effect

Recently, one of my friends was intrigued by a physical phenomenon known as Magnus effect, in which a moving object’s path is curved due to its rotation. While I was trying to explain this effect using Bernoulli’s principle, I found that my previous derivation of this principle was incorrect. Specifically, I was too vague when dealing with PDEs. So this time I will derive it in a more rigorous manner, and use it to explain Magnus effect.

Bernoulli’s Equation

We will start from Newton’s second law.

Assume that the fluid is light and has no viscosity, i.e. the only force acting on a particle in the fluid is the pressure gradient force. Then, apply the second law on the little cube above, we have \begin{alignat*}{4} & \ve{F} && && = m && \ve{a} \\ & - \Delta p && \cdot \ve{S} && = \rho V && \ve{a} \\ & - \ve{\nabla} p dr && \cdot dr^2 && = \rho dr^3 && \ve{a} \\ & - \ve{\nabla} p && && = \rho && \ve{a}, \tag{$\ast$} \label{n2l} \end{alignat*} where $\ve{S}$ is the surface normal vector and $\nabla$ is the gradient operator. So we get the same PDE as before, but this time, instead of recklessly treating it as if it is an ODE, we want to be as careful as possibly at each step.

First of all, let’s write the the acceleration $\ve{a}$ in terms of partial derivatives. \begin{alignat*}{4} \ve{a} & = \frac{d \ve{v}}{dt} \\ & = \frac{\partial \ve{v}}{\partial t} && + \frac{\partial \ve{v}}{\partial x} \frac{\partial x}{\partial t} && + \frac{\partial \ve{v}}{\partial y} \frac{\partial y}{\partial t} && + \frac{\partial \ve{v}}{\partial z} \frac{\partial z}{\partial t} \\ & = \frac{\partial \ve{v}}{\partial t} && + \frac{\partial \ve{v}}{\partial x} v_x && + \frac{\partial \ve{v}}{\partial y} v_y && + \frac{\partial \ve{v}}{\partial z} v_z, \end{alignat*} where $x$, $y$, $z$ are the coordinates of the particle, and $v_x$, $v_y$, $v_z$ are the three components of its velocity $\ve{v}$. We now make another assumption, in addition to the the fluid being inviscid, that it is steadily flowing, so that the term $\partial \ve{v} / \partial t$ vanishes, and thus \[ \ve{a} = \frac{\partial \ve{v}}{\partial x} v_x + \frac{\partial \ve{v}}{\partial y} v_y + \frac{\partial \ve{v}}{\partial z} v_z. \]

Note that the three terms look so similar, and in order to address this similarity, as well as to simplify our notation, we could write \[ \ve{a} = \sum_{j = 1}^{3} v_j \frac{\partial \ve{v}}{\partial x_j}, \] where $x_j$ is the $j$’th component of the coordinate basis, and $v_j$ is the $j$’th component of the velocity vector. And following the same idea, when we plug the acceleration back into \eqref{n2l}, we may write \[ - \frac{\partial p}{\partial x_i} = \rho v_j \frac{\partial v_i}{\partial x_j}, \tag{$\dagger$} \label{eulers-equation} \] omitting the $\sum$ sign as it is implied in the repeating use of index $j$. For more information, check out the Wikipedia article on Einstein notation.

We now start to integrate the PDE. Having observed that the right hand side is essentially a directional derivative along vector $\ve{v}$, we can turn the left hand side into one along the same direction if we multiply both sides by $v_i$: \[ - v_i \frac{\partial p}{\partial x_i} = \rho v_j \frac{\partial v_i}{\partial x_j} v_i, \] i.e. \[ v_i \frac{\partial}{\partial x_i} (-p) = v_j \frac{\partial}{\partial x_j} (\frac{1}{2} \rho v^2). \] Integrating along a line tangent to the flow velocity at every point (known as a streamline) gives \[ p + \frac{1}{2} \rho v^2 + C = 0, \tag{$\ddagger$} \label{bernoullis-equation} \] where $C$ is an arbitrary constant.

Since we were integrating along a streamline, the constant $C$ will only necessarily be the same if two points are on the same streamline. And to be more sure about this, we can make a thought experiment: imagine a right cylindrical tube filled with a fluid of constant pressure, and the fluid is flowing through in a direction parallel to the wall. Then, it seems that different points on a cross-sectional surface could have different speeds, as long as the speed is constant along the streamline. In this case, pressure gradient is zero everywhere, and acceleration is zero everywhere as well, so physically it is possible; but as different points on a cross-sectional surface have possibly different flow speed and same pressure, their constants $C$’s must be of different values. Therefore, \eqref{bernoullis-equation} is only necessarily true for points on the same streamline.

We can now state the Bernoulli’s equation as follows:

For a light and inviscid fluid, if it is steadily flowing, then \[ p + \frac{1}{2} \rho v^2 + C = 0 \] along any streamline, where $p$ is the pressure at a point, $v$ is the flow speed, and $C$ is a fixed constant.

For instance, when the fluid is not steady, but being irrotational at the same time, one could obtain \[ p + \frac{1}{2} \rho v^2 + \frac{\partial \phi}{\partial t} + C = 0 \] along any streamline, where $\phi$ is the velocity potential, i.e. $\ve{v} = \nabla \phi$.

Lift Force

Now, let’s look at the lift force generated by the rotation of the object. Intuitively, the rolling object draws air in a certain direction, and the reaction force draws the object in the other direction, which is the lift.

But where comes Bernoulli’s principle? Well, since one side of the object rotates with the fluid, and the other against the fluid, the flow on the two side is either accelerated or decelerated by the rotation, and the difference in flow speed creates pressure difference, which creates the lift. Note that now we assume the same constant $C$ in \eqref{bernoullis-equation} at every point in the fluid, as at infinitely far away, all streamlines share the same flow speed and pressure.

There is actually a nice formula for calculating the lift of any airfoil under ideal conditions, known as Kutta-Joukowski theorem. It shows that for an inviscid, incompressible fluid irrotationally flowing, the lift force per unit width $L$ is given by \[ L = \rho V \Gamma, \] where $\rho$ is the density of the fluid, $V$ is the speed of the airfoil, and $\Gamma$ is the circulation, which is basically the rotation of the fluid generated by the airfoil. Since the fluid is irrotational, i.e. the curl of flow field is zero everywhere, it follows from Stokes’ theorem, that the integral \[ \oint \ve{v} \cdot d\ve{S} \] along any closed curve around the object (either clockwisely or anticlockwisely) should evaluate to the same value, which is the circulation $\Gamma$.

A rigorous derivation of this theorem can be found by using complex analysis, but only an intuitive one will be given below. We shall set the object fixed, so as to simplify the model. Imagine two streamlines, one above the object, and another below, are sufficiently away from the object so that the speed along both is approximately $V$. Choose a large volume of fluid between the two streamlines around the object.

The Wikipedia Article

Actually, if you read carefully the Wikipedia article on Magnus effect at the time of writing, you will find that the circulation around a rotating smooth cylinder is taken as $2 \pi r v$, where $v$ is the speed of its circumference relative to its axis. This result is quoted from an educational page by NASA, in which they argue that the flow field can be obtained by superimposing the flow field from an ideal vortex centered in the cylinder with a uniform free stream flow. But I suspect that the nonlinearity of \eqref{eulers-equation} in terms of flow field $\ve{v}$ prevents us from superimposing two arbitrary flow fields.

An earlier version of this article quoted another result from a German physics forum, which differs from the circulation given by NASA exactly by a factor of two. The idea here is also to superimpose two flow fields, with one of them being an ideal vortex. However, the other field is a constant, meaning that the fluid effectively ignores the cylinder, and flows through it!