I'm writing this question here because it is a mathematics issue not a physics problem in particular.

Here I was trying to write an extended version of Euler-Bernoulli (EB) equation for compressible and viscous flow when I realised it is actually the equation of conservation of linear momentum (Navier-Stokes NS) in Lagrangian form alongside the stream. As it has been mentioned in the book "Transport Phenomena" by Bird second edition page 86, to drive EB from NS we should multiply both sides of the NS equation by $\vec{v}\cdot$ and then change the $\vec{v}\cdot\vec{\nabla}$ parameters to $\left|\vec{v} \right|\frac{\partial}{\partial s}$. So let's start with the most general form of linear momentum equation for a steady-state flow:

$$\rho \check{v} \check{\nabla}^T \check{v} =\check{\nabla} \check{\sigma}\tag{1}$$

$$\rho \left(\boldsymbol{v} \cdot \boldsymbol{\nabla}\right) \boldsymbol{v} = \rho \boldsymbol{v} \cdot \left( \boldsymbol{\nabla} \boldsymbol{v} \right) =\boldsymbol{\nabla} \cdot \boldsymbol{\sigma}\tag{1 tensor form}$$

Where $\check{\sigma}$ is the matrix representation of Cauchy stress tensor $\boldsymbol{\sigma}$

$$\check{\sigma}= \check{\tau} -P\check{I}\tag{2}$$

Assuming Newtonian fluid

$$\check{\tau}=\eta\left( \check{\nabla}^T \check{v}+ \left( \check{\nabla}^T \check{v} \right)^T \right) +\lambda \left(\check{\nabla}\check{v}^T\right) \check{I}\tag{3}$$

$$\boldsymbol{\tau}=\eta\left( \boldsymbol{\nabla} \boldsymbol{v}+ \left( \boldsymbol{\nabla} \boldsymbol{v} \right)^T \right) +\lambda \left(\boldsymbol{\nabla}\cdot\boldsymbol{v}\right) \boldsymbol{I}\tag{3 tensor form}$$

Assuming the viscosity is constant:

$$\rho \check{v} \check{\nabla}^T \check{v} = \eta\left( \check{\nabla}\check{\nabla}^T \check{v}+ \check{\nabla}\left( \check{\nabla}^T \check{v} \right)^T \right) + \check{\nabla} \left(\lambda\check{\nabla}\check{v}^T-P\right) \tag{4}$$

Now if I have calculated correctly

$$\check{\nabla}\left(\check{\nabla}^T \check{v}\right)^T=\check{\nabla}\left(\check{\nabla} \check{v}^T\right)\tag{5}$$

$$\boldsymbol{\nabla}\cdot\left(\boldsymbol{\nabla} \boldsymbol{v}\right)^T=\boldsymbol{\nabla}\left(\boldsymbol{\nabla} \cdot \boldsymbol{v}\right)\tag{5 tensor form}$$

and we can write the equation in vector form:

$$ \rho \vec{v} \cdot \left(\vec{\nabla}\otimes \vec{v}\right) = \eta\left( \left( \vec{\nabla} \cdot \vec{\nabla} \right) \vec{v}+ \vec{\nabla}\left( \vec{\nabla} \cdot \vec{v} \right) \right) + \vec{\nabla} \left(\lambda\vec{\nabla}\cdot\vec{v}-P\right) \tag{6}$$

Expanding the left side as discussed here leads to:

$$ \vec{v} \cdot \left(\vec{\nabla}\otimes \vec{v}\right)= \left(\vec{v} \cdot \vec{\nabla}\right) \vec{v} = \frac{1}{2}\vec{\nabla}\left(\vec{v} \cdot\vec{v} \right)-\vec{v} \times \left( \vec{\nabla} \times\vec{v} \right) \tag{7}$$

Multiplying the sides of the equation with $\vec{v}\cdot$ and if my presumptions of

$$\vec{v}\left(\vec{\nabla}\cdot\vec{\nabla} \right)\vec{v}=\left(\vec{v}\cdot\vec{\nabla}\right)\left(\vec{\nabla}\cdot\vec{v}\right)\tag{8}$$


$$\vec{v}\cdot\left( \vec{v} \times \left( \vec{\nabla} \times\vec{v} \right)\right)=0\tag{9}$$

are correct we can write:

$$ \frac{1}{2}\rho\frac{\partial}{\partial s} \left(\vec{v} \cdot\vec{v} \right) = \frac{\partial}{\partial s} \left(\left(2\eta+\lambda\right)\left(\vec{\nabla}\cdot\vec{v}\right)-P\right) \tag{10}$$

Which is mesmerizingly elegant and for incompressible flow $\vec{\nabla}\cdot\vec{v}=0$ or inviscid fluids $\eta,\lambda=0$ it reduces to EB equation. However the issue is that the standard EB equation is valid for inviscid and incompressible and and irrotational flow, not just inviscid or incompressible. There must be terms for inviscid-compressible and viscous-incompressible which does not exist in my result. Presumably I have done some calculation mistakes.

I would appreciate if you could help me know where are my mistakes and how I can solve them.

  • 1
    $\begingroup$ So, what exactly does "Lagrangian form alongside the stream" mean? Does it mean that you want to put the equation in the form $$\frac{\partial}{\partial s}A = \frac{\partial}{\partial s} B$$ where $A$ and $B$ are scalars? $\endgroup$ Aug 4, 2023 at 21:39

2 Answers 2


As you suspect, the mechanical energy equation you derive is missing terms that account for viscous effects. The other terms you derive are correct.

The RHS should be $\nabla \cdot \mathbf{\sigma}$, the divergence of the stress tensor, and you have edited accordingly.

This clearly is the correct term to include because the rate of increase of momentum in a fixed volume $V$ of fluid should be related to the viscous stress force of the surrounding fluid acting on the bounding surface of the volume $\partial V$,

$$\text{rate of momentum change } = \int_{\partial V}\mathbf{\sigma} \cdot \mathbf{n} \, dS = \int_V \nabla \cdot \mathbf{\sigma} \, dV.$$

Your error stems from incorrectly forming the dot product of the velocity and the viscous term $\eta\,\nabla \cdot [\nabla \mathbf{v} + \nabla \mathbf{v}^T] = \eta \, \nabla^2 \mathbf{v} + \eta \, \nabla (\nabla \cdot \mathbf{v})$.

In particular, you are missing $ \mathbf{v} \cdot [\eta\,\nabla \cdot \nabla \mathbf{v}]$ which should give rise to the viscous dissipation term, representing the action of viscous forces in converting useful mechanical energy into heat.

The correct treatment of that term is

$$\tag{*} \eta\,\mathbf{v} \cdot\,\nabla \cdot [\nabla \mathbf{v} + \nabla \mathbf{v}^T] = \underbrace{-\eta \nabla \mathbf{v} \,\mathbf{:}\, \nabla \mathbf{v}}_{\text{viscous dissipation}} + \frac{\eta}{2}\nabla^2(|\mathbf{v}|^2) + \eta \, \mathbf{v} \cdot \nabla (\nabla \cdot \mathbf{v}), $$

where the double dot product represents $\sum_{i=1}^3 \sum_{j=1}^3 |\frac{\partial v_i}{\partial x_j}|^2$, a norm of the velocity gradient, and $|\mathbf{v}|$ is the usual Euclidean norm of the velocity vector.

For an easy derivation, we can expand $\mathbf{v} \cdot\,\nabla \cdot [\nabla \mathbf{v}+ \nabla \mathbf{v}^T] $ using Cartesian coordinates and the Einstein convention (repeated index implies summation over that index). First we obtain

$$\nabla \cdot [\nabla \mathbf{v}+ \nabla \mathbf{v}^T]_{i} = \partial_j(\partial_j v_i + \partial_iv _j) = \partial_j(\partial_j v_i) + \partial_i(\partial_jv_j),$$

and forming the dot product we get

$$ \mathbf{v} \cdot\,\nabla \cdot [\nabla \mathbf{v} + \nabla \mathbf{v}^T] = v_i\,\partial_j(\partial_j v_i) + v_i \,\partial_i(\partial_jv_j) \\ = -\partial_jv_i \, \partial_j v_i + \partial_j[v_i(\partial_j v_i)] + v_i \,\partial_i(\partial_jv_j) \\ = -\partial_jv_i \, \partial_j v_i + \frac{1}{2}\partial_j\,\partial_j (v_iv_i) + v_i \,\partial_i(\partial_jv_j), $$

which is the component form of the coordinate-free term on the RHS of (*).

  • $\begingroup$ Thanks a lot for your response. If you don't mind here in the comments I will try to go through your points and ask question to better understand them. $\endgroup$
    – Foad
    Mar 11, 2018 at 19:14
  • $\begingroup$ 1. I'm not sure what you mean by my first equation is wrong because it is basically the NS equation in convective form omitting the time dependant part plus combined with continuity equation. I would appreciate if you could help me know what terms are missing from LHS or RHS $\endgroup$
    – Foad
    Mar 11, 2018 at 19:20
  • $\begingroup$ 2. I have tried to avoid the tensor notation because I find it inconsistent. The dyadic product of first rank tensors $\boldsymbol{a}\boldsymbol{b}$ and dot product $\boldsymbol{a} \cdot \boldsymbol{b}$ both can be expressed better if we use the matrix representation of the tensors, $a b^T$ and $a^T b$ where $a$ and $b$ are row matrices. or even the vector representation (outside the context of differential geometry) would be better IMHO. excuse my ignorance but tensors are very abstract entities and I have no idea what a transpose of a tensor would mean. $\endgroup$
    – Foad
    Mar 11, 2018 at 19:26
  • $\begingroup$ Your notation is unconventional and cumbersome and clearly has lead you astray. If you know what the transpose of a matrix is then you can you can understand the transpose of a tensor. A tensor has a matrix representation with respect to a basis. $\endgroup$
    – RRL
    Mar 11, 2018 at 19:36
  • $\begingroup$ @rrt sorry if my notation bothers you. I will use the conventional notation in this discussion from now on. $\endgroup$
    – Foad
    Mar 11, 2018 at 19:39

Thanks to RRL I found my mistake. Equation (8) is wrong. Del/Nabla operator is not a regular vector and it does not behave the same behaviour when distributed over vector operators (dot/cross/outer). So far the best reduction of NS equation in Lagrangian form is :

$$\rho \vec{v} \cdot \frac{\partial \vec{v}}{\partial s} = \frac{\partial}{\partial s} \left(\left(\eta+\lambda\right)\left(\vec{\nabla}\cdot\vec{v}\right)-P\right) +\eta\frac{\vec{v}}{\left|\vec{v}\right|}\cdot\left(\left(\vec{\nabla}\cdot\vec{\nabla} \right)\vec{v}\right) \tag{11}$$

Or considering this vector identity (ref)

$$\left(\vec{\nabla}\cdot \vec{\nabla} \right)\vec{v}= \vec{\nabla} \left(\vec{\nabla} \cdot \vec{v}\right)- \vec{\nabla} \times \left(\vec{\nabla} \times \vec{v}\right) \tag{12}$$

we have

$$\rho \vec{v} \cdot \frac{\partial \vec{v}}{\partial s} = \frac{\partial}{\partial s} \left(\left(2\eta+\lambda\right)\left(\vec{\nabla}\cdot\vec{v}\right)-P\right) -\eta\frac{\vec{v}}{\left|\vec{v}\right|}\cdot\left(\vec{\nabla} \times \left(\vec{\nabla} \times \vec{v}\right)\right) \tag{13}$$

this implies that the Bernoulli equation applies to a flow either if it is inviscid or it is incompressible and irrotational. The fluid doesn't have to be both inviscid and incompressible.

I'm not sure if it is possible to change the last part to a $\left(\vec{v}\cdot\vec{\nabla}\right)\left(scalar\right)$ required for the conversion. If I find any way to do it I will add it here to this post.


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