# Pressure

Pressure ($$p$$) is the force per unit area applied on a surface, in a direction perpendicular to that surface, i.e. the scalar part of the stress tensor under equilibrium/hydrosatic conditions.

## Thermodynamics

In thermodynamics the pressure is given by

$p = - \left.\frac{\partial A}{\partial V} \right\vert_{T,N} = k_BT \left.\frac{\partial \ln Q}{\partial V} \right\vert_{T,N}$

where $$A$$ is the Helmholtz energy function, $$V$$ is the volume, $$k_B$$ is the Boltzmann constant, $$T$$ is the temperature and $$Q (N,V,T)$$ is the canonical ensemble partition function.

## Units

The SI units for pressure are Pascals (Pa), 1 Pa being 1 N/m2, or 1 J/m3. Other frequently encountered units are bars and millibars (mbar); 1 mbar = 100 Pa = 1 hPa, 1 hectopascal. 1 bar is 105 Pa by definition. This is very close to the standard atmosphere (atm), approximately equal to typical air pressure at earth mean sea level: atm, standard atmosphere = 101325 Pa = 101.325 kPa = 1013.25 hPa = 1.01325 bar

## Stress

The stress is given by

${\mathbf F} = \sigma_{ij} {\mathbf A}$

where $${\mathbf F}$$ is the force, $${\mathbf A}$$ is the area, and $$\sigma_{ij}$$ is the stress tensor, given by

$\sigma_{ij} \equiv \left[{\begin{matrix} \sigma _x & \tau _{xy} & \tau _{xz} \\ \tau _{yx} & \sigma _y & \tau _{yz} \\ \tau _{zx} & \tau _{zy} & \sigma _z \\ \end{matrix}}\right]$

where where $$\ \sigma_{x}$$, $$\ \sigma_{y}$$, and $$\ \sigma_{z}$$ are normal stresses, and $$\ \tau_{xy}$$, $$\ \tau_{xz}$$, $$\ \tau_{yx}$$, $$\ \tau_{yz}$$, $$\ \tau_{zx}$$, and $$\ \tau_{zy}$$ are shear stresses.

## Virial pressure

The virial pressure is commonly used to obtain the pressure from a general simulation. It is particularly well suited to molecular dynamics, since forces are evaluated and readily available. For pair interactions, one has (Eq. 2 in ):

$p = \frac{ k_B T N}{V} + \frac{ 1 }{ d V } \overline{ \sum_{i<j} {\mathbf f}_{ij} {\mathbf r}_{ij} },$

where $$p$$ is the pressure, $$T$$ is the temperature, $$V$$ is the volume and $$k_B$$ is the Boltzmann constant. In this equation one can recognize an ideal gas contribution, and a second term due to the virial. The overline is an average, which would be a time average in molecular dynamics, or an ensemble average in Monte Carlo; $$d$$ is the dimension of the system (3 in the "real" world). $${\mathbf f}_{ij}$$ is the force on particle $$i$$ exerted by particle $$j$$, and $${\mathbf r}_{ij}$$ is the vector going from $$i$$ to $$j$$: $${\mathbf r}_{ij} = {\mathbf r}_j - {\mathbf r}_i$$.

This relationship is readily obtained by writing the partition function in "reduced coordinates", i.e. $$x^*=x/L$$, etc, then considering a "blow-up" of the system by changing the value of $$L$$. This would apply to a simple cubic system, but the same ideas can also be applied to obtain expressions for the stress tensor and the surface tension, and are also used in constant-pressure Monte Carlo.

If the interaction is central, the force is given by

${\mathbf f}_{ij} = - \dfrac{\mathbf {r}_{ij}}{ r_{ij}} f(r_{ij}) ,$

where $$f(r)$$ the force corresponding to the intermolecular potential $$\Phi(r)$$:

$-\partial \Phi(r)/\partial r.$

For example, for the Lennard-Jones potential, $$f(r)=24\epsilon(2(\sigma/r)^{12}- (\sigma/r)^6 )/r$$. Hence, the expression reduces to $p = \frac{ k_B T N}{V} + \frac{ 1 }{ d V } \overline{ \sum_{i<j} f(r_{ij}) r_{ij} }.$

Notice that most realistic potentials are attractive at long ranges; hence the first correction to the ideal pressure will be a negative contribution: the second virial coefficient. On the other hand, contributions from purely repulsive potentials, such as hard spheres, are always positive.

## Pressure equation

For particles acting through two-body central forces alone one may use the thermodynamic relation

$p = -\left. \frac{\partial A}{\partial V}\right\vert_T$

Using this relation, along with the Helmholtz energy function and the canonical partition function, one arrives at the so-called pressure equation (also known as the virial equation): $p^*=\frac{\beta p}{\rho}= \frac{pV}{Nk_BT} = 1 - \beta \frac{2}{3} \pi \rho \int_0^{\infty} \left( \frac{{\rm d}\Phi(r)} {{\rm d}r}~r \right)~{\rm g}(r)r^2~{\rm d}r$

where $$\beta := 1/k_BT$$, $$\Phi(r)$$ is a central potential and $${\rm g}(r)$$ is the pair distribution function.