# Differential form of Conservation of Momentum related to the Navier-stokes equation

In the context of the Navier-Stokes equation the Conservation of Momentum is given in my textbook as $$\frac{d}{dt}\int_{W_t} \rho \mathbf{u}dV = -\int_{W_t}\rho \cdot \mathbf{n} - \mathbf{\sigma}(\mathbf{x},t)dA$$ where $$\sigma(\mathbf{x},t)$$ is called the $$\textit{stress tensor}$$ and is written as $$\sigma(\mathbf{x},t) = 2\mu[D - \frac{1}{3}(div(u))Id]+\xi (div(u))Id$$ where $$\mu$$ is called the first coefficient of viscosity and $$\xi = \lambda + \frac{2}{3}\mu$$ is called the second coefficient of viscosity, and $$D$$ called the deformation densor.

Plugging in the above $$\sigma$$ into the integral version of the conservation of momentum apparently yields the differential form of the conservation of momentum below:

$$\rho \frac{D \mathbf{u}}{Dt} = -\nabla \rho + (\lambda + \mu )\nabla (div(\mathbf{u}))+\mu\Delta \mathbf{u}$$

However, I cannot get the same result. Could someone please show me how this is done? Here $$\frac{D}{Dt} = \partial_t + u\cdot \nabla$$ denotes the material derivative. I can plug in $$\sigma$$, but how are the integrals removed?

• There is a lot going on here. You need to correct your errors in the original momentum equation in integral form. Also you need to be familiar with tools like the divergence theorem, Reynolds transport theorem, etc. to really understand how Navier-Stokes equations are derived.
– RRL
Mar 5, 2021 at 4:44

First, you need to recognize that $$W_t$$ represents a material control volume that is time-dependent as it moves and deforms with the fluid. Then you pass the derivative under the integral on the right-hand side by applying Reynolds transport theorem to obtain,

$$\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{\partial W_t}\rho \mathbf{u} \mathbf{u} \cdot \mathbf{n} \, dS,$$

where $$\mathbf{n}$$ is the outer unit normal vector field at the surface $$\partial W_t$$ bounding the control volume and the integral on the far right-hand side is a surface integral.

By the divergence theorem, it follows that

$$\tag{1}\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{W_t}\nabla \cdot(\rho \mathbf{u} \mathbf{u}) \,dV = \int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV$$

The right-hand side you have written is incorrect. This should represent the net force acting on the control volume due to pressure and stress at the surface $$\partial W_t$$. Correctly written it is

$$\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS,$$

where $$p$$ is the pressure field and $$\sigma$$ is the (deviatoric) stress tensor. Applying the divergence theorem, we get

$$\tag{2}\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$

Equating (1) and (2), we have

$$\int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$

Since this is true for any control volume $$W_t$$ we can eqaute the integrands, to obtain

$$\tag{3}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) =- \nabla p + \nabla \cdot \sigma$$

Now it is just a matter of simplifying both sides.

For the left-hand side we have

$$\tag{4}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right] + \left[\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) \right]\mathbf{u}$$

Conservation of mass is expressed in terms of the equation of continuity

$$\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) =0,$$

which can be derived in a similar way. Substituting into (4) , we get

$$\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right]= \rho \frac{D\mathbf{u}}{Dt},$$

and substituting into (3) with this result,

$$\rho \frac{D\mathbf{u}}{Dt} = -\nabla p + \nabla \cdot \sigma$$

It remains for you to expand $$\nabla \cdot \sigma$$ using the expression you gave for $$\sigma$$ in terms of the deformation tensor which should be $$D = \nabla \mathbf{u} + [\nabla \mathbf{u}]^T.$$

• Thank you for the detailed answer. I had in fact been looking at the Transport theorem, but my own self-doubts had confused me. I should have just worked through it as if I had done so I would have came to the same conclusion you did. However, I do have a single point of confusion lingering. Both you and my text utilizing equating the integrands, but I do not understand why being "true for any control volute $W_t$ allows us to do this equating. Could you please elaborate on this? Mar 6, 2021 at 22:29
• @ozeraozera: We can combine integrals in $\int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$ to get $\int_{W_t} \mathbf{f}(\mathbf{x}) \, dV = 0$, where $\mathbf{f}(\mathbf{x}) = \frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) + \nabla p - \nabla \cdot \sigma)$.
– RRL
Mar 7, 2021 at 3:30
• The usual physical assumption is that the velocity, pressure, and stress fields are smooth functions, i.e., continuous with continuous derivatives. Hence $\mathbf{f}$ is continuous and since $\int_{W_t} \mathbf{f}(\mathbf{x}) \, dV = 0$ for any region $W_t$, the integrand must be identically $0$.
– RRL
Mar 7, 2021 at 3:35
• Recall how to prove this. Assume at some point $\mathbf{x}$ and time $t$, $\mathbf{f}(\mathbf{x}) > 0$, then there is a neighborhood $\Omega_t$ where $\mathbf{f}(\mathbf{y}) > 0$ for all $\mathbf{y} \in \Omega_t$ and $\int_{\Omega_t} \mathbf{f}(\mathbf{y}) \, dV > 0$, a contradiction. Similarly if $\mathbf{f}(\mathbf{x}) < 0$ ...
– RRL
Mar 7, 2021 at 3:42