Section 3.3 Momentum balance

With mass conservation now handled, we turn our eyes towards a law that governs the conservation of momentum. This is Newton’s second law: the rate of change of momentum of a body is equal to the sum of all forces acting on the body.

Our task in this section is to prove the following equation for the conservation of momentum of a fluid:

Theorem 3.3.1 Momentum equation.

The differential form of the law of conservation of momentum is

\begin{equation} \rho \DD{\bu}{t} = \rho \left(\pd{\bu}{t} + (\bu \cdot \nabla)\bu\right) = - \nabla p + \rho \bg,\tag{3.3.1} \end{equation}

where \(\rho = \rho(\bx,t)\) is the density of the fluid, \(\bu = \bu(\bx, t)\) is the velocity of the fluid, \(p = p(\bx, t)\) is the pressure, and \(\bg\) is the acceleration due to a body force (typically gravity).

Note that the term \((\bu \cdot \nabla)\bu\) above can also be written as \(\bu \cdot \nabla\bu\text{,}\) using the notation \(\nabla \bu\) for the gradient of a vector. This is explained in Remark 2.1.6.

Subsection 3.3.1 Proof of the momentum equation for an inviscid fluid

Consider again a material volume \(V(t)\) of the fluid. Then the rate of change of net momentum of this volume is equal to

\begin{equation} \dd{}{t} \iiint_{V(t)} \rho \bu(\bx, t) \, \de{V}.\tag{3.3.2} \end{equation}

It is helpful to remember that the mass of a volume element is \(\rho \de{V}\) and the acceleration of a fluid particle is \(\DD{\bu}{t}\text{,}\) so the above is similar to mass times acceleration. However, we work with the more general form above to allow for changes in the density throughout the fluid, and also for variable fluid elements, \(V(t)\text{.}\)

We must equate the above to the sum of all surface and body forces applied to the fluid element.

An example of a body force is the force of gravity. For a small volume element of mass \(\rho \de{V}\text{,}\) the force of gravity is equal to \((\rho \de{V})\bg\text{.}\) Therefore, the total external force due to gravity on the volume is equal to

\begin{equation} \iiint_{V(t)} \rho \bg \, \de{V}.\tag{3.3.3} \end{equation}

There may be other external body forces. For example, if your fluid is electrically conductive (like a plasma), there may be electromagnetic forces that must be considered. In any case, \(\bg\) can be considered as the analogous body force per unit mass.

The final type of forces we should consider are surface forces, which are applied to the boundary of the fluid element, denoted \(\partial V\text{.}\) Let us assume that at each point on the boundary, there is a surface force per unit area, \(\bF\text{,}\) that decomposes into a component normal to the boundary, \(F_n \bn\text{,}\) and a component tangential to the boundary, \(F_t \bt\text{.}\) So the surface forces, summed over the boundary, will be

\begin{equation} \iint_{\partial V} \left[ F_n \bn + F_t \bt\right] \, \de{S}.\tag{3.3.4} \end{equation}

In the above, the interpretation is that the force on a small patch of surface area \(\de{S}\) is equal to the per-area force, say \(F_n\text{,}\) multiplied by the area, then directed into the normal direction, \(\bn\text{.}\)

At this point, we make a key assumption that is applied in the particular case of inviscid fluids.

Theorem 3.3.2 Forces in inviscid fluids.

In the case of inviscid fluids, we assume that the surface force exerted on (interior) volume elements is accounted for solely by a pressure, \(p\text{,}\) which acts in the inward normal direction at each point, with

\begin{equation*} F_n \bn = -p \bn. \end{equation*}

Consequently, the surface force, given by (3.3.4), is

\begin{equation} \iint_{\partial V} (-p) \bn \, \de{S} = \iiint_V (-\nabla p) \, \de{V}.\tag{3.3.5} \end{equation}

In particular, for the case of inviscid fluids, we ignore tangential internal forces.

Proof.

The divergence theorem in its usual form applies to the flux of a vector field, \(\iint_{\partial V} \boldsymbol{F} \cdot \bn \, \de{S} = \iiint_V \nabla \cdot \boldsymbol{F} \, \de{V}\text{,}\) whereas the left-hand side of (3.3.5) is a vector. We therefore take its dot product with an arbitrary constant vector \(\boldsymbol{c}\text{,}\) which reduces the problem to scalars.

Since \(\boldsymbol{c}\) is constant, it may be moved inside the integral, giving

\begin{equation*} \boldsymbol{c} \cdot \iint_{\partial V} (-p) \bn \, \de{S} = \iint_{\partial V} (-p \boldsymbol{c}) \cdot \bn \, \de{S}. \end{equation*}

Now apply the divergence theorem with \(\boldsymbol{F} = -p\boldsymbol{c}\text{.}\) Again using that \(\boldsymbol{c}\) is constant, \(\nabla \cdot (-p \boldsymbol{c}) = -\boldsymbol{c} \cdot \nabla p\text{,}\) and hence

\begin{equation*} \iint_{\partial V} (-p \boldsymbol{c}) \cdot \bn \, \de{S} = \iiint_V (-\boldsymbol{c} \cdot \nabla p) \, \de{V} = \boldsymbol{c} \cdot \iiint_V (-\nabla p) \, \de{V}. \end{equation*}

Combining the two results,

\begin{equation*} \boldsymbol{c} \cdot \left[ \iint_{\partial V} (-p) \bn \, \de{S} - \iiint_V (-\nabla p) \, \de{V} \right] = 0. \end{equation*}

This holds for every constant vector \(\boldsymbol{c}\text{.}\) Choosing \(\boldsymbol{c} = \be_1, \be_2, \be_3\) in turn shows that each component of the bracketed vector is zero, so the vector itself is zero, which is (3.3.5).

Finally, summing the forces above, Newton’s second law gives

\begin{equation*} \dd{}{t} \iiint_{V(t)} \rho \bu(\bx, t) \, \de{V} = -\iiint_{V(t)} \nabla p \, \de{V} + \iiint_{V(t)} \rho \bg \, \de{V}. \end{equation*}

We can now apply Corollary 3.2.4 to each component of momentum, taking \(h = u, v, w\) in turn (i.e. \(f = \rho u, \rho v, \rho w\)). This allows us to conclude that

\begin{equation*} \iiint_{V(t)} \left( \rho \DD{\bu}{t} + \nabla p - \rho \bg\right) \, \de{V} = 0. \end{equation*}

Again, the above holds for all material volumes \(V(t)\) and therefore it must follow that the integrand is zero. We conclude thus with the following result as stated in Theorem 3.3.1:

\begin{equation*} \rho \DD{\bu}{t} = - \nabla p + \rho \bg. \end{equation*}

In Exercise 3.6.3, you will practise the derivation of the mass and momentum equations.

Prev Top Next