Introductory ﬂuid mechanics JF207

Lecture#3

Returning to our continuumdescription of ﬂuid motion,we will distinguish between two kinds

of forces exerted on an inﬁnitesimal continuum ﬂuid element (recall that such an element has

to be large compared to the relevant microscopic dimensions - λ

MFP

for dilute gases - for a

continuum description to be valid).These are the body forces and the surface forces.The

former refer to long-ranged forces such as gravity or that arising from an electromagnetic

ﬁeld,and so on.Such forces,as the name implies,vary slowly with the distance between

interacting elements;for instance,the gravitational force per unit mass acting on any object is

very well represented by the constant vector g (|g| = 9.8m/s

2

) for distances from the earth’s

surface that are less than the order of the earth’s radius (∼ 6.4 x10

6

m).A consequence

of this slow variation is that such forces act equally on all the matter within a suﬃciently

small (inﬁnitesimal) element of our ﬂuid continuum,and thus,long-ranged forces manifest as

body forces.Assuming the body force per unit mass at a point x in the ﬂuid,and at time t,

to be F(x,t),the total force on an inﬁnitesimal element around this point with volume dV

is given by ρF(x,t)dV,where ρF is now the force per unit volume.For gravity,this force is

simply the weight of the element,ρgdV;for a conducting ﬂuid in an magnetic ﬁeld(B),for

instance,there would be an additional Lorentz force (F

L

∝ i∧B,i being the current density.

The second group of forces are short-ranged forces that have a direct molecular origin,and

as a result,are characterized by length scales relevant to the microscopic dynamics.This

would mean that such forces decay extremely rapidly with distance on the macroscopic

scale.Again,considering the example of an dilute gas,one expects such short-ranged forces

to decay on distances of the order of a mean free path.Since any inﬁnitesimal element in

a continuum description is,by deﬁnition,much larger than all microscopic scales,the eﬀect

of such forces would be negligible unless two interacting elements are directly in contact;

and then,one would expect these forces to act across the contact surface

1

.In other words,

the short-ranged forces manifest as surface forces,in a continuum description,that act to

transport momentum across the boundaries of an inﬁnitesimal element.In dilute gases,

the momentum transport occurs due to molecules randomly crossing the boundary under

consideration and thereby carrying momentum across in the appropriate direction(often

referred to as the kinetic contribution to the stress);in liquids,the tranport of momentum

can occur without physical translation of molecules via short-ranged forces acting between

pairs of molecules on either side of the boundary and separated by a distance comparable

to the range of the inter-molecular potential (referred to as the potential contribution to

the stress).It must now be clear that the total eﬀect of short-ranged forces acting on a

diﬀerential element is decided by its surface area rather than the volume.Hence,it is more

appropriate to consider a plane surface element in the ﬂuid and specify the local short-ranged

force as the total force exerted by the ﬂuid on one side of the element on the other side.If the

area of the element is δA,this may be written as Σ(n,x,t)δA,where Σ is the force per unit

1

Strictly speaking,such forces would be contribute to momentum exchange in a layer of O(λ

MFP

) that

encompasses the contact surface,but since λ

MFP

/V

1

3

→ 0,this layer may be regarded as coincident with

the contact surface itself in a continuum description.

1

n

-n

A

(x,n,t)

(x,-n,t)

Surface force acting

on the element

A

Figure 1:A surface force,Σ(x,n,t) acting on a surface element,δA,with unit normal n.

An equal and opposite force,Σ(x,−n,t) = −Σ(x,n,t) acts on the opposite face.

area (or the stress vector) and n is the unit normal to the element.One has a dependence on

n since one expects the surface force to depend on the orientation of the plane element,and

the latter is determined by n.We will adopt a convention that would amount to a tension

being interpreted as a positive force.Thus,in the above expression for the surface force,

Σ(n,x,t) is the force exerted by the ﬂuid,on the side to which n points,on the ﬂuid on the

side which n points away from.

Now,we proceed to show that it is a stress tensor rather than the stress vector Σ that is

a more fundamental quantity,characterizing the ﬂuid response to an imposed deformation.

In order to do this,we need to extract the n-dependence of Σ.Note that while Σ(n,x,t)

is the force exerted by the ﬂuid on the side to which n points,that exerted by the ﬂuid

on the opposite side should be equal and opposite,since there can be no net force acting

on a surface element (which has zero mass).On the other hand,in accordance with our

notation,this same force can also be written in terms of the unit normal pointing in the

opposite direction(−n) as Σ(−n,x,t).Thus,we must have that Σ(−n,x,t) = −Σ(n,x,t),

implying that the stress vector is an odd function of n.In order to deduce the speciﬁc

dependence on n,we consider the tetrahedral volume element shown in ﬁgure 2.The three

orthogonal faces have areas δA

1

,δA

2

and δA

3

,and unit (outward) normals −a,−b and −c,

respectively;the fourth inclined face has an area δA and unit normal n.In accordance with

our notation above,the total surface force acting on the tetrahedral element is:

Σ(n,x,t)δA+Σ(−a,x,t)δA

1

+Σ(−b,x,t)δA

2

+Σ(−c,x,t)δA

3

,(1)

where we have assumed the volume element to be small enough(speciﬁcally,much smaller

than the characteristic length scales on which Σ varies) that the position coordinate corre-

sponding to all of the faces of the tetrahedron may be approximated as x.Since only the

2

unit normal changes from one term to another in (1),we suppress the dependence on the

arguments x and t in what follows.Further,using Σ(n) = −Σ(−n),and the fact that both

the total volume force and the acceleration of the element must be O(δl)

3

,where δl is a

characteristic linear dimension of the element,we must have that the total surface force on

the given element mush vanish as O(δl) for δl →0.Thus,in the limit,one has

Σ(n)δA−Σ(a)δA

1

−Σ(b)δA

2

−Σ(c)δA

3

= 0.(2)

That is to say,the surface forces,acting on a suﬃciently small element,must exist in balance.

Note that the diﬀerences between the surface forces acting on the various faces,arising from

the O(δl) diﬀerence in the position vectors x specifying the diﬀerent faces of the tetrahedron,

is O(δl)

3

,and may be balanced by a combination of the body forces and the acceleration of

the element.

Now,the volume of a tetrahedron is again proportional to the scalar triple product (discussed

in an earlier lecture on Cartesian tensors),although,unlike a parallelopiped,there is a

prefactor of 1/6;that is,V

tetrahedron

=

1

6

t

1

t

2

∧t

3

,where t

1

,t

2

and t

3

are vectors emanating

from a common vertex.Notwithstanding the proportionality factor of 1/6,this may be used

to write the volume of the tetrahedron in ﬁgure 2 as δAn δl

2

b,choosing B as the common

vertex;here,we have taken δAn to replace t

2

∧ t

3

in the general formula above,and t

1

to

correspond to the edge of length δl

2

along b.Alternately,taking δA

2

b to correspond to t

2

∧t

3

,

we have the volume being equal to δA

2

δl

2

.Clearly,the two expressions must be equal,which

is only possible if δA(n b) = δA

2

.Similarly,we have the other relations δA(n a) = δA

1

and

δA(n c) = δA

3

from alternate expressions for the tetrahedral volume based on A and C as

the common vertices.Using these relations in (2),we have

Σ(n) = [Σ(a)a +Σ(b)b +Σ(c)c] n,(3)

after cancelling δA on both sides;or,in index notation:

Σ

i

(n) = {Σ

i

(a)a

j

+Σ

i

(b)b

j

+Σ

i

(c)c

j

} n

j

,(4)

Since the vectors n and Σ do not in any way depend on the choice of the axes of refer-

ence ([a,b,c]) the expression within brackets in (4) that relates Σ and n must likewise be

independent of the particular choice of axes.In other words,for given i and j,this expres-

sion must correspond to the ij

th

component of a second-order tensor which we denote by σ.

Thus,the stress vector is a linear functional of n,and one may write:

Σ

i

(n) = σ

ij

n

j

,(5)

relating the stress vector and tensor.One may readily verify this assertion by substituting

(5) in (4) with n (successively) replaced by a,b and c,so as to obtain:

σ

ij

n

j

= (a

j

a

k

+b

j

b

k

+c

j

c

k

)σ

ik

n

j

,(6)

which reduces to an identity on noting that the quantity within brackets is just a represen-

tation of δ

jk

in the chosen coordinate system.

3

n

b

a

c

A

1

A

3

A

2

-a

-c

-b

O

A

B

C

Figure 2:The balance of surface forces on a tetrahedral surface element.

In a manner similar to that above,one may consider a balance of the moments of the various

forces acting on the ﬂuid within an inﬁnitesimal volume element.The moment of the surface

forces about a point in the interior of the volume is given by

ǫ

ijk

x

j

σ

kl

n

l

dA.Noting that

the moment arm is O(δl),we have that the moment of the surface forces is O(δl)

3

in the

limit δl →0.On the other hand,the moment of the body forces,given by

ǫ

ijk

x

j

(ρF

k

)dV is

O(δl

4

),and the angular momentum of the element,given by

ǫ

ijk

x

j

ρu

k

dV,is again at least

as small as O(δl)

4

;the angular momentum may,in fact,be reduced to a smaller order ((δl)

5

)

by taking moments about the center-of-mass so that

x

i

dV = 0,and the ﬁrst non-zero

contribution arises due to the variation of the linear momentum about the origin.In any

case,we note that,for δl → 0,the moment of the surfaces forces must be identically zero,

since it cannot be balanced by the other terms (moment of the body forces and angular

acceleration of the element) in the equation for the conservation of angular momentum.

Thus,for δl →0,the conservation of angular momentum implies

ǫ

ijk

x

j

σ

kl

n

l

dA = 0,(7)

and on applying the divergence theorem,one obtains

ǫ

ijk

∂

∂x

l

(x

j

σ

kl

dV =0,(8)

⇒

ǫ

ijk

x

l

∂σ

kl

∂x

l

+σ

kj

dV =0.(9)

Noting that the ﬁrst term is again O(δl) smaller,and that the volume dV is arbitrary,we

4

have that

ǫ

ijk

σ

kj

= 0.(10)

The above relation is satisﬁed provided the stress tensor,σ is symmetric;that is,σ

ij

= σ

ji

.

In other words,there are only six independent elements of the stress tensor (rather than nine

as is the case for the general second-order tensor).

There are two comments in order here.The ﬁrst is that we have obviously made a choice

of the point about which the moment of the forces above have been evaluated - this point

was taken as an interior point which made the moment arm of O(δl).The conclusions

with regard to the stress tensor above,however,are independent of this choice.An al-

ternate choice of a ﬁxed exterior point,say x

0

,could be used;in this case,the moment

arm may be written as the sum of a ﬁxed vector from the center-of-mass to the given

exterior point,and a vector of O(δl).The conservation of angular momentum would

then reduce to the form O(δl)

3

.

ǫ

ijk

(x

l

−x

(CM)l

)

∂σ

kl

∂x

l

+σ

kj

dV + O(δl)

2

(x

0

−x

CM

) ∧

[Total surface force on the element];leading to the same conclusions.

The second comment is with regard to an assumed absence of body couples acting on the

ﬂuid element under consideration.There are substances such as ferroﬂuids that experience a

body couple under suitable conditions;a ferroﬂuid is essentially a suspension of tiny magnetic

particles,and since the microscopic constituents (magnetic dipoles) experience an aligning

torque when the ferroﬂuid is subject to an ambient magnetic ﬁeld,there will in general be

a second O(δl)

3

term in the angular momentum conservation,that depends on the total

magnetic couple exerted on the particles within the continuum element under consideration,

and that balances the torque due to the surface forces.In these cases,the stress tensor will

not be symmetric.Referring to ﬁgure 3,we note that the volume element under consideration

will have an additional term in the conservation equation due to the contribution from the

microscopic couples.If we denote the orientation of the α

th

magnetic dipole as p

α

,then

the total torque acting on a continuum element,containing N such dipoles,is given by

N

α=1

ǫ

ijk

(x

j

+sp

α

j

)F

α

k

+(x

j

−sp

α

j

)(−F

α

k

),where the forces constituting a microscopic couple

are assumed to be separated by a distance 2s.Assuming the dipole density to be n,so that

N = nδV,and further that a subset (ndV )Ω(p)dp have orientations in the range [p,p+dp],

where Ω(p) is a probability density deﬁned on the unit sphere

2

,one may write the body

couple contribution as

dV

dpnΩ(p)ǫ

ijk

p

j

(2sF

k

).Denoting

dpnΩ(p)p

j

(2sF

k

) as L

i

,the

angular momentum balance should now take the form ǫ

ijk

σ

kj

= −L

i

.

Restricting our consideration to simple ﬂuids,where body couples are absent,we conclude

that the surface forces are entirely characterized by a stress tensor that must be symmetric.

The diagonal elements of the stress tensor are known as normal stresses while the non-

diagonal elements constitute the tangential or shearing stresses.Furthermore,for a ﬂuid

at rest,and with an isotropic microstructure,the stress tensor is expected to be isotropic

since there can be no preferential direction in which the randommolecular motion transports

2

Here,the argument p is taken to denote a dependence on the angles θ and φ,and thus Ω(p) ≡ Ω(θ,φ)

is a probability density deﬁned on the unit sphere such that

π

0

2

0

πdθ sinθdφΩ(θ,φ) = 1

5

p

2s

-F

F

B

Figure 3:The torques on the microscopic constituents,in the presence of an applied magnetic

ﬁeld,leading to a couple per unit volume.

momentum.Thus,

σ

ij

= −pδ

ij

,(11)

where,in accordance with our convention,p denotes the magnitude of the (compressive) pressure

in a static ﬂuid.The compressive interpretation of the stress in a ﬂuid at rest is consistent

with the inability of simple ﬂuids,under normal conditions,to sustain tensile stresses.Thus,

(11) implies that,under conditions of static equilibrium,there can be no tangential stresses.

This also conforms to our original postulate of what a ﬂuid is - a state of matter that cannot

remain static under the action of shearing stresses.

Having characterized the form of the surface forces,and in particular,the simpliﬁed form

that they assume in the absence of ﬂuid motion,we now consider the force balance on a

continuum element under the action of both body and surface forces;this will allow us to

obtain the constraints required on the body forces in order for there to be no motion.This

force balance may be written in the form:

ρF

i

(x,t)dV +

σ

ij

n

j

dA = 0,(12)

and,on using (11),we have

ρF

i

dV −

pn

i

dA = 0.(13)

6

Further,applying the divergence theorem and the arbitrariness of the chosen volume,we

have

∂p

∂x

i

= ρF

i

.(14)

which must be satisﬁed at every point in the ﬂuid volume.Thus,ﬂuid motion is absent only

when the body force per unit volume,ρF is expressible as the gradient of a scalar.This is,

of course,not true in general,since the Helmholtz decomposition tells us that an arbitrary

vector is in general the sum of the gradient of a scalar potential (φ) and the curl of a vector

potential (Ψ);that is,any vector a may be written as a = ∇φ +∇∧Ψ.

In the simplest case of a constant density ﬂuid acted on by gravity,ρF = g,and (14) reduces

to the familiar statement of hydrostatic equilibrium viz.

∂p

∂x

i

= ρg

i

,(15)

where g

i

is the acceleration vector due to gravity.Of course,(14) with F

i

= g

i

does not

require the density to be a constant.The ﬂuid will continue to be at rest provided the

density varies only in the direction of gravity.This may be seen by taking the curl of (14).

The left-hand-side equals zero,since (∇∧ ∇)p = 0,so that

ǫ

kji

∂

∂x

j

(ρg

i

) = 0,⇒ǫ

ijk

∂ρ

∂x

j

g

i

= 0.(16)

Since ∇ρ is the normal to the constant density surfaces,the above condition implies that

the constant density surfaces must be perpendicular to gravity;in other words,the ﬂuid

density can vary along gravity.A simple illustration of how ﬂuid motion would result on the

violation of this condition is when one tilts a container of water (here,the density variation

is restricted to the interfacial region between water and air,and ∇ρ can be regarded as a

vector perpendicular to the instantaneous interface);tilting causes the water interface to

become inclined to gravity.This immediately leads to ﬂuid motion that acts to restore the

original horizontal orientation

3

.

Restricting our consideration to gravity as the body force,but allowing for the variation of

density as a function of pressure,one may evaluate the pressure variation with height in a

static atmosphere.Since air may be approximated as an ideal gas,one has p = n

a

RT with

n

a

being the number of moles of air,or in terms of the air density,p = ρ

a

(R/M

a

)T where

R = 8314J/K.kmol is the molar gas constant with M

a

≈ 30kg/kmol (a value close to N

2

,

the dominant atmospheric component).Using this relation,and assuming g = −1

3

in (15),

we have

dp

a

dx

3

= −ρ

a

g,(17)

⇒

dp

a

dx

3

= −

gM

a

RT

p.(18)

3

The tilted interface leads to the non-coincidence of the constant pressure and constant density surfaces;

a term proportional to δp ∧ δρ,then acts as a (baroclinic) source of vorticity,inducing ﬂuid motion

7

Our analysis will be restricted to the troposphere which accounts for about 80%of the entire

atmospheric mass,and wherein the temperature only varies by about 20% of its value at the

earth’s surface.Thus,an isothermal atmosphere is a reasonable leading-order approximation.

Further,the height of the troposphere ranges from about 20km at the equator to less than

10km near the poles.These distances being considerably smaller than the radius of the

earth,g may also be taken as a constant

4

.With these assumptions,(18) may be easily be

solved to obtain:

p

a

= p

0

a

e

−

z

H

a

,(19)

where p

0

a

is the pressure at the earth’s surface (z = 0) and H

a

= RT/(M

a

g) is referred

to as the pressure-scale height;that is,the height over which the atmospheric pressure

varies by a signiﬁcant amount (speciﬁcally,by a factor 1/e).Using the values of R and M

a

above,with g ≈ 10m/s

2

,and T = 300K,one ﬁnds (RT/M

a

g) ≈ 8km.This is consistent

with our initial assumption of the pressure varying with height much more rapidly than the

temperature (which was taken as constant).

4

The troposphere is largely transparent to the incoming solar radiation whose intensity peaks at a fre-

quency in the mid-visible range.It is the earth’s surface that absorbs the solar radiation,and the troposphere

is,in eﬀect,heated from the bottom by the upwelling (infrared) radiation emitted by the surface.This heat-

ing leads to convection(vertical motion) and mixing - the troposphere being identiﬁed by the resulting

characteristic decrease in temperature with height.The tropopause separates the troposphere from the

stratosphere;in contrast to the troposphere,the temperature increases with height in the stratosphere due

to direct absorption of solar radiation(primarily by ozone).Since the surface heating is strongest at the

equator,the vertical motion is strongest in the equatorial regions and the tropopause is thus at the greatest

height close to the equator.

8

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