# Question about the General Constitutive Law (mathy)

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1. Sep 9, 2015

### Hercuflea

I have attached a question about the general constitutive law for Newtonian fluids (it contains a lot of math, which is easier for me to type and send to PDF than put it on here.)

Basically, I do not like the einstein notation and I decided to write everything out loud and clear, since my professor had written it as

τ = μ[∇u+(∇u)T]

Which I think is not a very good notation because you cannot take the gradient of a vector valued function! You must resort to using a Jacobian matrix.

My main concern with this document is whether I got it right with regards to a time-dependent flow. For a time independent flow this should be fine, but I wasn't sure how to handle the Jacobian matrix for a flow u which depends on (x,y,z) and time (which is only defined on the set of nonnegative real numbers)

Please let me know whether you see any errors in this document or any comments you may have. This is simply for my own understanding of the material.

Thanks

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2. Sep 9, 2015

### Geofleur

The math looks alright to me, but I think there are also terms involving the pressure $p$ and the other Lame constant, $\lambda$, no? And that would be for an isotropic fluid. The general constitutive relation I am used to seeing as

$\tau_{ij} = -p \delta_{ij} + \frac{D_{ijkl}}{2} \left( \frac{\partial u_k}{\partial x_l}+\frac{\partial u_l}{\partial x_k}\right)$,

where repeated indices are summed over. For an isotropic fluid this becomes

$\tau_{ij} = -p\delta_{ij} + \lambda\delta_{ij}\frac{\partial u_k}{\partial x_k} + \mu\left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$,

where the repeated $k$'s are summed over. Your equation seems to contain only the last term of this.

3. Sep 10, 2015

### Staff: Mentor

Who says you cannot take the gradient of a vector valued function. The gradient of a vector valued function is a 2nd order tensor. In your case, it is the velocity gradient tensor.
As Geofleur has pointed out, the equation you wrote only includes the portion of the stress tensor from viscous stresses. The overall form is given by Geofleur's last equation. This is the most general form of the tensorial relationship between stress and deformation rate under the assumption that the stress tensor is a linear and isotropic in the rate of deformation tensor. So, unlike what you said in your write-up, the Newtonian constitutive law is based on a very sound mathematical theory.

Chet

4. Sep 11, 2015

### vanhees71

It's also indeed very bad notation, and you should get used to LaTeX (also for writing up scientific text in general, not only here in the forum; Word is simply ugly and very hard to read; usually, I don't read papers written in Word). In index-free notation, I'd write it as a formal tensor product like $\vec{\nabla} \otimes \vec{V}$, where $\vec{V}$ is an arbitrary vector field. In terms of the index (Ricci) calculus in Cartesian components its the tensor field $\partial_j V_k$.

The "consitutive" law for the stress-strain relation in isotropic fluids can be argued in terms of the irreducible representations of the rotation group. First of all we note that the equations of motion are nothing else than the momentum balance equation,
$$\partial_t (\rho v_j)=-\partial_i \Pi_{i j},$$
where $\Pi_{ij}=\Pi_{ji}$ is the momentum-flux tensor. It can be systematically derived from kinetic theory, but it's also possible to argue in a completely phenomenological way. First of all one can separate off the ideal-fluid part and write
$$\Pi_{ij}=p \delta_{ij} + \rho v_i v_j-\sigma_{ij},$$
where $\sigma_{ij}$ is the viscous stress tensor. It occurs in the kinetic derivation when you take into account the deviations from local thermal equilibrium (which describes the ideal fluid) and describes dissipation. Now you can assume that, if this deviation from local equilibrium is small that it is linear in the gradients of the velocity.

So let's look at the tensor $\partial_i v_j$. First you can split it in an antisymmetric and a symmetric part,
$$\Omega_{ij}=\frac{1}{2} (\partial_i v_j-\partial_j v_i).$$
It is equivalent to an axial vector via
$$\Omega_{ij}=\epsilon_{ijk} \omega_k.$$
It describes the rotation of a fluid element as a whole and does not contribute to the dissipative part of the momentum-flow tensor. Thus it must be symmetric. In linear order of the velocity gradients it can be split further into two irreducible parts of the corresponding tensor representation of the rotation group, because obviously the trace $\partial_j v_j=\vec{\nabla} \cdot \vec{v}=\mathrm{div} \vec{v}$ is a scalar. So separating the trace part and the trace-less part, you have
$$\partial_i v_j + \partial_j v_i=\frac{2}{3} \partial_k v_k \delta_{ij} + \left (\partial_i v_j + \partial_{j} v_i -\frac{2}{3} \partial_k v_k \delta_{ij} \right ).$$
Now the leading-order dissipative part of the momentum-flow tensor, linear in the velocity gradients, is consistent with the isotropy of the fluid if you use independent scalars in front of these two irreducible parts, i.e., you make the ansatz
$$\sigma_{ij}'=\zeta \partial_k v_k \delta_{ij} + \eta \left (\partial_i v_j + \partial_{j} v_i -\frac{2}{3} \partial_k v_k \delta_{ij} \right ).$$
Now $\partial_k v_k$ discribes the volume-changing part of the deformation of the fluid cell, and that's why one calls $\zeta$ the bulk viscosity. The traceless piece describes volume-conserving deformations, i.e., shears of the fluid element, and that's why $\eta$ is called shear viscosity. The viscosities are socalled transport coefficients and can be evaluated from kinetic theory using various methods to approximate the Boltzmann equation for the case of small deviations from equilibrium. The most common methods are the Chapman-Enskog method and Grad's method of moments, but of course one can take the constituent equations just as phenomenological equations with the viscosities determined empirically.

Assuming that the viscosities are varying only slowly with space, you obtain the Navier-Stokes equation, which reads (in vector notation)
$$\rho (\partial_t \vec{v}+(\vec{v} \cdot \vec{\nabla}) \vec{v})=-\vec{\nabla} p + \eta \Delta \vec{v} + \left (\zeta+\frac{\eta}{3} \right ) \vec{\nabla}(\vec{\nabla} \cdot \vec{v}).$$

5. Sep 12, 2015

### Hercuflea

Sorry I haven't responded in a few days. I've attached another document with responses to Geofleur and vanhees71. I was going to try and type it in here, but this one contains a motherload of a matrix which would have been an absolute nightmare to typeset myself (I use an automatic typesetting software). This pdf will require that you have a pdf viewer with a good zoom function, because I had to print it to poster size. Thanks for the insightful posts.

I've attached a document in response.

May be being pedantic here, but I think that the gradient operator can only be applied to functions which map to R, in other words - scalars. In the case of a vector valued function mapping to Rⁿ you would need to use the Jacobian operator - which is basically what you said. I just don't like the notation ∇ being applied to a vector since it is reserved for scalars. It gets confusing.

I understand that the mathematics behind it is consistent. What I meant in my previous write up is that, like other physical laws (Maxwell's equations, etc...), the General Constitutive Law is basically an experimentally observed phenomenon. From what I can tell (maybe wrong) it isn't something which can be derived by just starting with a set of mathematical axioms and working your way to it. I can't include any mathematical proof because as far as I know there isn't one, I can only try to make a statement of the GCL in a (hopefully) mathematically precise way.

What is a bad notation - the Einstein notation or the notation in my write-up?

I use a software called Scientific Workplace to type up my mathematics. It is basically a front-end to LateX which does not require you to do any actual coding. It's hotkey-based and I can type relatively quickly. I've gotten fast enough that I can take notes in math classes without ever using a pencil.

Unfortunately, when I tried to copy and paste the TeX code into the LateX editor in PhysicsForums, it did not compile. I think SWP has some proprietary content in their version.

More response to your post in the document.

#### Attached Files:

• ###### pdfresponse_09122015.pdf
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6. Sep 12, 2015

### Staff: Mentor

Who says it is reserved for scalars? See this current PF thread on how the del operator is applied to vectors to obtain a second order tensor:

Chet

7. Sep 13, 2015

### vanhees71

Concerning the notation I meant not the typography, which isn't so bad, and if it's even LaTeX it's the best one can have in the digital world for math/science typesetting. What I meant was the notation. It should be clear what object one has already in the notation.

Further, it is clear that the nabla-operator (or more generally the covariant-derivative operator) can be applied in various ways to all kinds of fields. As you see, in continuum mechanics you also need tensors and formal tensor products.

What I've written down is already the general case of the isotropic fluid. If you have anisotropic fluids, the viscosities become tensors themselves.