# Balance laws

Mathematical models for continuum mechanics rely upon governing laws that ensure the balance of mass and momentum in the system being modeled.
Below we provide the equations for balance of mass and linear momentum used in the material methods section.

## Lagrangian view

| Law                         | Formulation                                                                           |
|-----------------------------|---------------------------------------------------------------------------------------|
| Balance of Mass             | $\rho J - \rho_0 = 0$                                                                 |
| Balance of Linear Momentum  | $\rho_0 \frac{\partial}{\partial t} \bm{V} - \nabla_X \cdot \bm{P} - \rho_0 \bm{g} = 0$ |

$\bm{P}$ is the *first Piola-Kirchhoff stress tensor*, $\bm{V}$ is the Lagrangian velocity defined in {eq}`vel-acc-lagrangian`, and $\bm{g}$ is the gravitational acceleration.
$\rho$ and $\rho_0$ denote the mass density and initial mass density, respectively.

## Eulerian view

| Law                         | Formulation                                                                  |
|-----------------------------|------------------------------------------------------------------------------|
| Balance of Mass             | $\frac{\partial}{\partial t} \rho + \nabla \cdot (\rho \bm{v}) = 0$          |
| Balance of Linear Momentum  | $\rho \frac{D}{D t} \bm{v} - \nabla \cdot \bm{\sigma} - \rho \bm{g} = 0$     |

where $\bm{v}$ is the Eulerian velocity {eq}`vel-acc-eulerian`.

:::{note}
The *balance of linear momentum* can be written in the equivalent form

$$
  \frac{\partial}{\partial t} \left(\rho \bm{v} \right) + \nabla \cdot \left( \rho \bm{v} \otimes \bm{v} \right) - \nabla \cdot \bm{\sigma} - \rho \bm{g} = 0.
$$ (equivalent-balance-linear-momentum)

by expanding $\rho \frac{D}{D t} \bm{v}$ using {eq}`acc-eulerian` as

$$
  \begin{aligned}
    \rho \frac{D}{D t} \bm{v} &= \rho \frac{\partial \bm{v}}{\partial t} + \rho \left( \nabla \bm{v} \right) \bm{v} = \frac{\partial \left( \rho \bm{v} \right)}{\partial t} - \frac{\partial \rho}{\partial t} \bm{v} + \left( \nabla \bm{v} \right) \left(\rho \bm{v} \right) \\
                              &= \frac{\partial \left( \rho \bm{v} \right)}{\partial t} + \bm{v} \nabla \cdot \left( \rho \bm{v} \right) + \left( \nabla \bm{v} \right) \left(\rho \bm{v} \right) \\
                              &= \frac{\partial \left( \rho \bm{v} \right)}{\partial t} + \nabla \cdot \left( \rho \bm{v} \otimes \bm{v} \right)
  \end{aligned}
$$

where we have used the identity $\nabla \cdot \left( \bm{u} \otimes \bm{v} \right) = \bm{u} \nabla \cdot \bm{v} + \left( \nabla \bm{u} \right) \bm{v}$.

:::