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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?

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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.

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  • $\begingroup$ 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. $\endgroup$
    – eimrek
    Commented Jan 5, 2017 at 20:28
  • $\begingroup$ @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 $\partial \rho /\partial t = 0$. $\endgroup$
    – tpg2114
    Commented Jan 5, 2017 at 21:14
  • $\begingroup$ Isn't it $\frac{D\rho}{Dt} = \frac{\partial\rho}{\partial t} + \vec{v}\cdot\nabla\rho=\frac{\partial\rho}{\partial t} + \nabla\cdot (\rho\vec{v}) - \rho\nabla\cdot\vec{v}$? $\endgroup$
    – eimrek
    Commented Jan 5, 2017 at 22:05

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