Units in the Navier-Stokes equation

[ailchenko23]

First post; I am starting to read the official problem description of the http://www.claymath.org/millennium/Navier-Stokes_Equations/navierstokes.pdf" and am having trouble understanding the units involved in the first equation
The equation, verbatim, is
\begin{equation}
\frac{\partial}{\partial t}u_i + \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j}
= \nu \Delta u_i - \frac{\partial p}{\partial x_i} + f_i(x,t) ,
\end{equation}
where $u$ is $u(x,t)$, the velocity vector, $\nu$ is the viscosity, $p$ is the pressure, and $f_i(x,t)$ are the components of a given, externally applied force. So, from right to left: I expect the units for $f_i(x,t)$ to be (kg m/s), however, I would also accept force per unit volume, kg/(s$^{2}$ m$^{2}$). Units for $\frac{\partial p}{\partial x_i}$ are kg/(s$^{2}$ m$^{2}$). Units for $\nu$ are (Pascal s), so units for $\nu \Delta u_i$ are, again, kg/(s$^{2}$ m$^{2}$). Then, on the left side, unit for $\frac{\partial u_i}{\partial x_j}$ is (1/s); so, $\sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j}$ is expressed in (m/s$^{2}$) as is $\frac{\partial}{\partial t}u_i$. My question is why do we have units of force per unit volume on the right side and only units for acceleration on the left side? Is there some understood conversion factor implicit in the equation? What is the advantage, if any, in using force per unit volume instead of simply force?

 

[timthereaper]

To be honest with you, I think he's missing a term on the right side. It should read:
$$\frac{\partial}{\partial t} + \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j} = \nu \Delta u_i - \frac{1}{\rho} \frac{\partial p}{\partial x} + f_i (x,t)$$

I say this because the Navier-Stokes equation I remember from fluid dynamics are:

$$\rho \left( \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z} \right) = - \frac{\partial p}{\partial x} + \rho g_x + \mu \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \right)$$
$$\rho \left( \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} + w \frac{\partial v}{\partial z} \right) = - \frac{\partial p}{\partial y} + \rho g_y + \mu \left( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} + \frac{\partial^2 v}{\partial z^2} \right)$$
$$\rho \left( \frac{\partial w}{\partial t} + u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y} + w \frac{\partial w}{\partial z} \right) = - \frac{\partial p}{\partial z} + \rho g_z + \mu \left( \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial z^2} \right)$$

To answer the question, your unit assumptions are a bit off. $$\nu$$ has units of $$m^2/s$$ because $$\nu = \frac{\mu}{\rho}$$ and in this case, $$f_i (x,t)$$ is just the gravity term in the ith direction. You can rearrange my version of Navier-Stokes to get the equation the paper gave if you just include the $$\frac{1}{\rho}$$ in the pressure term. Using the proper units for everything, you'll get units of acceleration across the board.

I don't think that the missing term changes any of the conclusions in the paper, though.