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with J given by expression (6.16).

This equation constitutes a conservation law for the probability density of a particle. Note

that equation (6.15) could be derived rigorously from the Schrodinger equation (6.13) so

that the conservation law (8.4) and the expression for the current J follow from the basic

equation of the system.

Problem b: Why is the source term on the right hand side of (8.4) equal to zero?

8.2. THE CONTINUITY EQUATION 75

8.2 The continuity equation

In this section we will consider the conservation of mass in a continuous medium such as a

uid or a solid. In that case, the quantity Q is the mass-density . If we assume that mass

is not created or destroyed, the source term vanishes: S = 0. The vector J is the mass

current, this quantity denotes the ow of mass per unit volume. Let us consider a small

volume V . The mass within this volume is equal to V . If the velocity of the medium

is denoted with v, the mass- ow is given by V v. Dividing this by the volume V one

obtains the mass ow per unit volume, this quantity is called the mass density current:

J= v : (8.5)

Using these results, the principle of the conservation of mass can be expressed as

@+ ( v) =0 : (8.6)

@t r

This expression plays a very important role in continuum mechanics and is called the

continuity equation.

Up to this point the reasoning was based on a volume V that did not change with time.

This means that our treatment was strictly Eulerian we considered the change of physical

properties at a xed location. As an alternative, a Lagrangian description of the same

process can be given. In such an approach one speci es how physical properties change as

they are moved along with the ow. In that approach one seeks an expression for the total

d

time derivative dt of physical properties rather than expressions for the partial derivative

@

@t . These two derivatives are related in the following way:

d = @ + (v : (8.7)

dt @t r)

Problem a: Show that the total derivative of the mass density is given by:

d + (r v) =0 : (8.8)

dt

Problem b: This Lagrangian expression gives the change of the density when one follows

the ow. Let us consider a in nitesimal volume V that is carried around with the

ow. The mass of this volume is given by m = V . The mass within that volume

is conserved (why?), so that m = 0. (The dot denotes the time derivative.) Use

_

this expression and equation (8.8) to show that (r v) is the rate of change of the

volume normalized by size of the volume:

_

V = (r v) : (8.9)

V

We have learned a new meaning of the divergence of the velocity eld, it equals the

relative change in volume per unit time.

CHAPTER 8. CONSERVATION LAWS

76

8.3 Conservation of momentum and energy

In the description of a point mass in classical mechanics, the conservation of momentum

and energy can be derived from Newton's third law. The same is true for a continuous

medium such as a uid or a solid. In order to formulate Newton's law for a continuous

medium we start with a Lagrangian point of view and consider a volume V that moves

with the ow. The mass of this volume is given by m = V . This mass is constant. Let

the force per unit volume be denoted by F, so that the total force acting on the volume

is F V . The force F contains both forces generated by external agents (such as gravity)

and internal agents such as the pressure force or the e ect of internal stresses (r )

;rp

(with being the stress tensor). Newton's law applied to the volume V takes the form:

d ( V v) = F V : (8.10)

dt

Since the mass m = V is constant with time it can be taken outside the derivative

in (8.10). Dividing the resulting expression by V leads to the Lagrangian form of the

equation of motion:

dv = F : (8.11)

dt

Note that the density appears outside the time derivative, despite the fact that the density

may vary with time. Using the prescription (8.7) one obtains the Eulerian form of Newton's

law for a continuous medium:

@v + v =F : (8.12)

@t rv

This equation is not yet in the general form (8.3) of conservation laws because in the rst

term on the left hand side we have the density times a time derivative, and because the

second term on the left hand side is not the divergence of some current.

Problem a: Use expression (8.12) and the continuity equation (8.6) to show that:

@( v) + ( vv) = F : (8.13)

@t r

This expression does take the form of a conservation law it expresses that the momen-

tum (density) v is conserved. (For brevity we will often not include the a x \density"

in the description of the di erent quantities, but remember that all quantities are given

per unit volume). The source of momentum is given by the force F, this re ects that

forces are the cause of changes in momentum. In addition there is a momentum current

J = vv that describes the transport of momentum by the ow. This momentum current

is not a simple vector, it is a dyad and hence is represented by a 3 3 matrix. This is not

surprising since the momentum is a vector with three components and each component

can be transported in three spatial directions.

You may nd the inner products of vectors and the in expressions such

r-operator

(8.12)-confusing, and indeed a notation such as v can be a source of error and confu-

rv

sion. Working with quantities like this is simpler by explicitly writing out the components

8.3. CONSERVATION OF MOMENTUM AND ENERGY 77

of all vectors or tensors and by using the Einstein summation convention. (In this con-

vention one sums over all indices that are repeated on one side of the equality sign.) This

P

notation implies the following identities: v = 3 vi @i Q = vi @i Q (where @i is an

2 = P3 vi vi = vi vi and an equation such as (8.12) is

i=1

rQ

abbreviated equation for @=@xi ), v i=1

written in this notation as:

@vi + v @ v = F : (8.14)

jji i

@t

Problem b: Rewrite the continuity equation (8.6) in component form and redo the

derivation of problem a with all equations in component form to arrive at the

conservation law of momentum in component form:

@( vi ) + @ ( v v ) = F : (8.15)

j ji i

@t

In order to derive the law of energy conservation we start by deriving the conservation

law for the kinetic energy (density)

EK = 2 v2 = 1 vi vi :

1 (8.16)

2

Problem c: Express the partial time-derivative @( v2 )=@t in the time derivatives @( vi )=@t

and @vi =@t, use the expressions (8.14) and (8.15) to eliminate these time derivatives

and write the nal results as:

@( 1 vi vi ) = 1

2 j 2 vi vi vj + vj Fj : (8.17)

@t ;@

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