# Continuity Equation for Momentum

Momentum is a conserved quantity, which makes me wonder if we can write an equation for the local conservation of momentum in the form of a continuity equation. If we're considering a system of discrete objects, momentum isn't really a vector field. It is a quantity intrinsic to each object in the space. So I'm not sure if a continuity equation makes sense in this case.

If there is such an equation, it should be the mathematical statement that the total momentum of a system of bodies plus any changes in momentum from forces (or potential) originating outside of the system is zero. Can we write such an equation?

Of course we can. First off, when discussing conversation laws, continuity typically refers to the conservation of mass specifically:

$$\frac{D \rho}{D t} = 0$$

in a Lagrangian frame or:

$$\frac{\partial \rho}{\partial t} + \frac{\partial \rho u_i}{\partial x_i} = 0$$

Now, for momentum. This equation holds for both fluids and solids and is an expression of Newton's Second Law. In a Lagrangian frame:

$$\frac{D \rho u_i}{D t} + \frac{\partial \sigma_{ij}}{\partial x_j} = 0$$

where $\sigma_{ij}$ is the Cauchy stress tensor. And likewise, for an Eulerian frame, this is:

$$\frac{\partial \rho u_i}{\partial t} + \frac{\partial \rho u_i u_j}{\partial x_j} + \frac{\partial \sigma_{ij}}{\partial x_j} = 0$$

The form of the Cauchy stress tensor depends on what is being studied of course. It is also possible to write relativistic forms of this equation (and this is probably a better link).

When this is specialized for a fluid in a gravitational field, we get:

$$\frac{\partial \rho u_i}{\partial t} + \frac{\partial \rho u_i u_j}{\partial x_j} = - \frac{\partial p}{\partial x_i} + \mu \frac{\partial^2 u_i}{\partial x_i x_j} + \rho g_i$$

where the Cauchy stress tensor is split into the pressure and deviatoric stress components. A virtually identical equation set can be written for solids as well.

• Shouldn't your first equation be $\frac{D\rho}{Dt}+\rho\nabla\cdot \vec{v} =0$? $\frac{D\rho}{Dt} =0$ should be the condition for incompressibility. Commented Jan 5, 2017 at 20:28
• @Shepherd No, the notation here is $D/Dt$ is the Lagrangian derivative, also called the substantial or material derivative. It is the derivative following the infinitesimal fluid volume. That gets expanded out into $\partial \rho/\partial t + \rho \nabla \cdot \vec{v}$ when transformed into an Eulerian reference frame, which is the second equation. In the second equation, incompressibility will show up as $\partial \rho /\partial t = 0$. Commented Jan 5, 2017 at 21:14
• Isn't it $\frac{D\rho}{Dt} = \frac{\partial\rho}{\partial t} + \vec{v}\cdot\nabla\rho=\frac{\partial\rho}{\partial t} + \nabla\cdot (\rho\vec{v}) - \rho\nabla\cdot\vec{v}$? Commented Jan 5, 2017 at 22:05