# Help with understanding indices for Navier Stokes Equation

I am using Kundu and Cohen's textbook on Fluid Mechanics. I am not a mechanical engineering major, but I am trying to understand the indices for a project that I am doing. This is what I have so far:

We focus on finding the Navier-Stokes Equation for an incompressible fluid as our project involves water. An incompressible fluid is a fluid whose density does not change with pressure (kundu, 81). Here, $$g$$ represents gravity, $$\mu$$ represents viscosity, $$\rho$$ represents the density of the fluid, $$p$$ represents pressure, and u or $$\overrightarrow{u}$$ represents a velocity vector.

$$\rho \frac{Du_i}{Dt} = - \frac{\partial p}{\partial x_i} + \rho g_i + \underbrace{2\mu \frac{\partial e_{ij}}{\partial x_j}} - \frac{2 \mu}{3} \frac{\partial}{\partial x_i} (\nabla \cdot \overrightarrow{u}).$$

Using Einstein's notation, we have that:

$$\underbrace{2\mu \frac{\partial e_{ij}}{\partial x_j}} = 2\mu \left(\frac{\partial e_{i1}}{\partial x_1} + \frac{\partial e_{i2}}{\partial x_2} + \frac{\partial e_{i3}}{\partial x_3}\right).$$

Here, we note that the strain rate tensor is: $$e_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right).$$

Reapplying Einstein's notation to our strain rate tensor and then substituting, we get:

$$2\mu \frac{\partial e_{ij}}{\partial x_j} = 2\mu \left[\frac{\partial}{\partial x_1} \left(\frac{1}{2} \left(\frac{\partial u_i}{\partial x_1} + \frac{u_1}{x_i}\right)\right) + \frac{\partial}{\partial x_2}\left(\frac{1}{2} \left(\frac{\partial u_i}{\partial x_2} + \frac{u_2}{x_i}\right)\right) + \frac{\partial}{\partial x_3}\left(\frac{1}{2} \left(\frac{\partial u_i}{\partial x_3} + \frac{u_3}{x_i}\right)\right) \right].$$

Combining our partial derivatives and factoring out 2, our cleaner equation is now:

$$2 \mu \frac{\partial e_{ij}}{\partial x_j} = \mu \left[ \frac{\partial^2 u_i}{\partial x_1^2} + \frac{\partial^2 u_1}{\partial x_1 \partial x_i} + \frac{\partial^2 u_i}{\partial x_2^2} + \frac{\partial^2 u_2}{\partial x_2 \partial x_i} + \frac{\partial^2 u_i}{\partial x_3^2} + \frac{\partial^2 u_3}{\partial x_3 \partial x_i} \right].$$

Regrouping so that our $$u_i$$ are together we get:

$$2 \mu \frac{\partial e_{ij}}{\partial x_j} = \mu \left[ \left(\frac{\partial^2 u_i}{\partial x_1^2} + \frac{\partial^2 u_i}{\partial x_2^2} + \frac{\partial^2 u_i}{\partial x_3^2}) + (\frac{\partial^2 u_1}{\partial x_1 \partial x_i} + \frac{\partial^2 u_2}{\partial x_2 \partial x_i} + \frac{\partial^2 u_3}{\partial x_3 \partial x_i}\right)\right].$$

Here, we make the note that the Laplacian of $$u_i$$ is:

$$\nabla^2 u_i = \frac{\partial^2 u_i}{\partial x_j \partial x_j} = \frac{\partial^2 u_i}{\partial x_1^2} + \frac{\partial^2 u_i}{\partial x_2^2} + \frac{\partial^2 u_i}{\partial x_3^2}.$$

Making the above substitution, and factoring $$\frac{\partial}{\partial x_i}$$ we get:

$$2 \mu \frac{\partial e_{ij}}{\partial x_j} = \mu \left[ \nabla^2 u_i + \frac{\partial}{\partial x_i} \left(\frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3 }\right) \right].$$

Notice that we can rewrite the above to be:

$$2 \mu \frac{\partial e_{ij}}{\partial x_j} = \mu \big[ \nabla^2 u_i + \frac{\partial}{\partial x_i} (\nabla \cdot \overrightarrow{u} ))\big].$$

So, when we substitute the under-braced component below with the above, we get that: $$\rho \frac{Du_i}{Dt} = - \frac{\partial p}{\partial x_i} + \rho g_i + \underbrace{2\mu \frac{\partial e_{ij}}{\partial x_j}} - \frac{2 \mu}{3} \frac{\partial}{\partial x_i} (\nabla \cdot \overrightarrow{u}) = - \frac{\partial p}{\partial x_i} + \rho g_i + \mu \big[ \nabla^2 u_i + \frac{\partial}{\partial x_i} (\nabla \cdot \overrightarrow{u} )) \big] - \frac{2 \mu}{3} \frac{\partial}{\partial x_i} (\nabla \cdot \overrightarrow{u}).$$

We are close to achieving our goal of the Navier-Stokes Equation for an incompressible fluid!

We now make the final note that $$\nabla \cdot \overrightarrow{u} = 0$$, and using vector notation, we get:

$$\begin{array}{c} \displaystyle \rho \frac{D\mathbf{u}}{Dt} = - \nabla{p} + \rho \mathbf{g} + \mu \nabla^2 \mathbf{u}. \\ \\ \textrm{Navier-Stokes Equation for an Incompressible Fluid!} \end{array}$$

I apologize in advance if my question isn't clear. I am trying to understand the indices i and j. From what I gather when working on this, j varies from 1 through 3 and is representative of the 3 dimensions. But, what does the i index represent?

$$i$$ is a free index, and can take any value from 1 to 3. Note on the LHS, you have the acceleration of the fluid particle, which is a vector. Hence it has three components and you need an index $$i$$ to get one component.
• @mathnoob No, you are summing over $j$, so you still get equation about a vector (acceleration). Feb 15, 2023 at 21:46