2.7 Conservation of momentum

The law of conservation of momentum can be written4

|-----------------| Du--= r ☐☐☐ + b | --Dt--------------| \relax \special {t4ht=
(2.19)
where eqn represents any body force per unit mass. Body forces represent any force which does not act at a bounding surface, including those that act at a distance, such as gravitational force.

The equation is derived by considering the time rate of change of the momentum of a mass of particles. We consider a volume of material of fixed mass moving through space and therefore present rate of change by the material derivative eqn.

PICT\relax \special {t4ht=

Applying Gauss’s theorem and Eq. (2.16 ), the surface force is

Z Z Z Z t dS = n ☐☐☐ dS = (dS ☐☐☐) = r ☐☐☐ dV : S S S V \relax \special {t4ht=
(2.20)
Equating the rate of change of momentum to the forces and, noting mass is fixed so eqn is constant in time, gives
Z Du-- ☐ V Dt r ☐☐ b dV = 0: \relax \special {t4ht=
(2.21)
The integrand must equal 0, resulting in Eq. (2.19).

Divergence was described in section 2.4 as the flux across a surface per unit volume as eqn. The divergence of stress similarly represents the stress flux across the surface, i.e. force, per unit volume as eqn, given by

 Z r ☐☐☐ = lim --1- (dS ☐☐☐): V!0 V S \relax \special {t4ht=
(2.22)
Eq. (2.19 ) specifically relates to linear momentum. Instead, conservation of angular momentum5, in the absence of any “couple stresses” that generate a moment field, is given by
☐☐☐ = ☐☐☐T (☐☐☐ is symmetric): \relax \special {t4ht=
(2.23)
The derivation of Eq. (2.23 ) is fairly complex so is omitted here.

Tensor symmetry

Symmetry in a tensor eqn refers to components being symmetric about the diagonal, i.e. eqn, eqn and eqn. The transpose of a tensor, denoted by the ‘eqn’ superscript, switches components across the diagonal such that:

 0 1 Txx Tyx Tzx TT = B@ Txy Tyy Tzy CA T T T xz yz zz \relax \special {t4ht=
(2.24)
eqn is therefore symmetric if eqn. A skew (anti-symmetric) tensor has eqn. A tensor can be decomposed into symmetric and skew parts by:
 1 T 1 T T = 2-(T + T )+ 2(T T ) = sym T + skw T |------------{z------------} |------------{z------------} symmetric skew \relax \special {t4ht=
(2.25)

4Cauchy’s first law of motion (1827)
5Cauchy’s second law of motion (1827)

Notes on CFD: General Principles - 2.7 Conservation of momentum