# How to derive Bernoulli's theorem for an elastic fluid from Lamb's equation

Here I state Lamb's version of the Euler's equation for an elastic fluid: $$\partial_t \vec v + \vec \Omega \times \vec v + \vec \nabla \left( \frac{\|\vec v\|^2}{2} + \frac{p}{\rho} + \psi + \phi_b\right) = 0$$ where $$\vec\Omega$$ is the vorticity of the fluid, $$p$$ the pressure, $$\rho$$ the density, $$\vec b$$ is a potential body force such that $$\vec b = \vec \nabla \phi_b$$ and $$\psi$$ is Helmholt's density of energy that in our course has been defined as the fuction $$\psi = \psi(\rho)$$ such that $$p(\rho) = \rho^2 \frac{\partial \psi}{\partial \rho}$$

Using Lamb's equation we stated this version of Bernoulli's theorem:

Theorem: Consider an elastic eulerian fluid, then in steady flow the quantity $$\frac{\|\vec v\|^2}{2} + \frac{p}{\rho} + \psi + \phi_b$$ is constant along the flow lines. (Usually I've been told that the quantity $$w :=\frac{p}{\rho} + \psi$$ is called Gibbs' free energy)

Proof: It sufficies to show that $$\vec v \cdot \vec \nabla \left( \frac{\|\vec v\|^2}{2} + w + \phi_b\right) = 0$$ but this is obvious from Lamb's equation.

My questions are:

• Why It sufficies to show that $$\vec v \cdot \vec \nabla \left( \frac{\|\vec v\|}{2} + w + \phi_b\right) = 0$$ to prove that theorem? (The obviousness of the next passage is clear to me)

• How this theorem can be linked to the "classical" Bernoulli statement: $$p = -\rho \frac{\|\vec v \|^2}{2} + k$$ where $$k$$ is a constant?

• I add this as a comment and not as an edit because it is a personal comment about the subject: "please excuse me for the naming of the equations and functions. With my surprise it's not so easy to find things about fluid-dynamics on internet and everyone seems to have different notations and names for the same things." Commented Mar 26, 2021 at 15:44
• Random internet search is not the best approach to learn this subject as you are discovering. There are many books of course, but the problem is that while some are very good in terms of the physics they do not present the mathematical details in a precise and rigorous way. An Introduction to Theoretical Fluid Dynamics by Stephen Childress offers a good balance.
– RRL
Commented Mar 26, 2021 at 21:02
• Excuse me but I've seen this comment only now. I know that this is not the best method but I'm already using Childress and my professor notes but under some aspects both of these sources sometimes assume things that seems obvious to their eyes but not to mine. Here internet comes in. Commented Mar 27, 2021 at 14:02
• No reason not to use the internet as a resource, of course, but then you inevitably run into quality control issues and lack of consistency in notation. There is no perfect book, so in my experience it is good to look at a number of references like Hydrodynamics by Lamb, Introduction to Fluid Dynamics by Batchelor, Theoretical Hydrodyanmics by Milne-Thomson, Vortex Dynamics by Saffman, etc.
– RRL
Commented Mar 27, 2021 at 17:24

A Streamline (flow line) is a curve with parametric representation $$\mathbf{x}_S(s)$$ at some fixed time $$t$$ such that

$$\mathbf{x}_S(s) \times \mathbf{v}(\mathbf{x}_S(s),t) = 0$$

The velocity field is, therefore, tangential to the streamline at any point, and

$$\frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t)$$

Defining the quantity $$H (\mathbf{x},t) = \frac{\| \mathbf{v}\|^2}{2} + w + \phi_b$$, we have

$$\mathbf{v}(\mathbf{x}_S(s),t) \cdot \nabla H(\mathbf{x}_S(s),t) = 0,$$

and, by the chain rule,

$$\frac{\partial}{\partial s} H(\mathbf{x}_S(s),t) = \nabla H(\mathbf{x}_S(s),t)\cdot \frac{d \mathbf{x}_S}{ds}(s) = \alpha(s) \mathbf{v}(\mathbf{x}_S(s),t)\cdot \nabla H(\mathbf{x}_S(s),t) = 0$$

Therefore, $$H$$ is constant along a streamline, where value of this constant can be different for each streamline.

What you refer to as the "classical" Bernoulli statement,

$$\tag{1} p + \frac{1}{2} \rho \|\mathbf{v}\|^2 = k,$$

where $$k$$ is constant at all points in the domain is valid for steady inviscid, incompressible and irrotational flow.

For a proof, begin with the Euler equations governing steady, inviscid flow,

$$\tag{2}\rho (\mathbf{v} \cdot \nabla) \mathbf{v} = - \nabla p,$$

along with the incompressibility condition $$\nabla \cdot \mathbf{v} = 0$$.

By the general vector identity

$$\nabla (\mathbf{a} \cdot \mathbf{b}) = (\mathbf{a} \cdot \nabla) \mathbf{b} + (\mathbf{b} \cdot \nabla) \mathbf{a} + \mathbf{a} \times (\nabla \times \mathbf{b}) + \mathbf{b} \times (\nabla \times \mathbf{a}),$$

we get with $$\mathbf{a} = \mathbf{b} = \mathbf{v}$$,

$$\nabla (\|\mathbf{v}\|^2) = \nabla (\mathbf{v} \cdot \mathbf{v}) = 2(\mathbf{v} \cdot \nabla) \mathbf{v} + 2\mathbf{v} \times (\nabla \times \mathbf{v})$$

For irrotational flow, where $$\nabla \times \mathbf{v} = 0$$, this reduces to

$$\tag{3}(\mathbf{v} \cdot \nabla) \mathbf{v} = \nabla (\frac{1}{2}\|\mathbf{v}\|^2)$$

Since $$\rho$$ is constant we obtain after substituting (3) into (2),

$$\nabla (\frac{1}{2}\rho\|\mathbf{v}\|^2) = -\nabla p,$$

whence $$\nabla (p + \frac{1}{2}\rho\|\mathbf{v}\|^2) = 0$$ and (1) must hold where $$k$$ is a global constant.

• Thank you for your answer but I still can't get the link between the two statements of the Bernoulli theorem. How can we deduce the "classical" form of the theorem from the theorem we proved? Commented Mar 27, 2021 at 10:15
• @Gabrielek: Let's put aside the irrelevant details pertaining to "elastic fluid" and free energy. First if we have an incompressible and inviscid flow (and are given nothing else) then Bernoulli's principle (as derived above) is that the quantity $H= p + \frac{1}{2} \rho \|\mathbf{v}\|^2 = C(\psi)$. (Note that body force term if there is one is absorbed into $p$.) What this says is $H$ is constant along any streamline and the value of the constant $C(\psi)$ may vary from one streamline to another. This will apply to any incompressible, inviscid flow even if there is non-zero vorticity.
– RRL
Commented Mar 27, 2021 at 17:11
• The result you probably have seen, $H= p + \frac{1}{2} \rho \|\mathbf{v}\|^2 = K$ where $K$ is a global constant, is true for flows that are also irrotational. An incompressible, inviscid and irrotational flow is necessarily a potential flow where the velocity is the gradient of a potential: $\mathbf{v} = \nabla \phi$.
– RRL
Commented Mar 27, 2021 at 17:13
• I can show you how to derive $H= p + \frac{1}{2} \rho \|\mathbf{v}\|^2 = K$ when $\nabla \times \mathbf{v} =$ and $\mathbf{v} = \nabla \phi$ if you like.
– RRL
Commented Mar 27, 2021 at 17:31
• If you would be so kind to add that derivation in your answer I would thank you very much and of course I would accept the answer as well! Commented Mar 28, 2021 at 9:09

I show an intuitive argument. The velocity field defines some integral curves. These are referred to as flow lines and have the property of having the velocity field as tangents. Clearly any quantity F with a gradient always orthogonal to the velocity field will have the gradient orthogonal to the integral curves as well. In such a situation F must be constant along the integral curves.