Mathematical Methods For Physics 2

8/21/2019 Mathematical Methods For Physics 2

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Maths Methods 2

Spring 2013

1 Useful Mathematical Results

1.1 Trigonometric Identities

We recall the following useful trigonometric identities:

Figure 1:

tan = sin

cos , cosec=

1

sin , sec =

1

cos , cot =

1

tan =

cos

sin .

From the well known identitycos2 + sin2 = 1 (1.1)

we have1 + tan2 = sec2

andcot2 + 1 = cosec2.

Addition and subtraction formulae:

sin( ) = sin cos cos sin and

cos( ) = cos cos sin sin If = then from the above addition and subtraction formulae we have

sin2= 2 sin cos

andcos2= cos2 sin2 . (1.2)

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Using (1.1) in (1.2) we havecos2= 1 2sin2

sin2 = 12(1

cos2) . (1.3)

Alternatively we have

cos2= 2 cos2 1

cos2 = 12(1 + cos 2) . (1.4)The formulas (1.3) and (1.4) are useful when trying to integrate cos 2 or sin2 .

1.2 Trigonometric functions

In Figure 2 we display graphs of the trigonometric functions sin , cos , tan , csc , sec and cot .

Figure 2:

1.3 Determinants of Matrices

For a 2 2 matrix we haveA=

a bc d

det A = |A| =ad bc.

For a 3 3 matrix we have

A=

a b cd e fg h i

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det A = |A| =ae fh i

b

d fg i

+ c

d eg h

= a (ei hf) b (di gf) + c (dh ge) .

1.4 Vectors

Vectors have magnitude and direction.The vector i has magnitude 1 and points in the increasing x direction. The vector j has magnitude 1and points in the increasing y direction. The vector k has magnitude 1 and points in the increasing zdirection.The vectors i, j and k are unit vectors. All unitvectors have magnitude 1 and usually they are writtenwith hats, e.g. u or v.

A standard vector

F= Fxi + Fyj + Fzkhas 3components,Fx, Fy andFz that are all scalars. Fx is thex component ofF, Fy is they componentofF and Fz is the z component ofF.

Themagnitudeof the vector F is denoted by|F| and is defined by

|F| =

F2x + F2

y + F2

z .

A unit vector in the direction ofF has magnitude 1 and points in the same direction as F, it is denotedby

F= 1

|F|F= 1

|F| (Fxi + Fyj + Fzk) .

Example 1.1 For the vectorG= 3i + 2j

find the three componentsGx, Gy andGz, then determine the magnitude ofG and the unit vector thatpoints in the same direction asG.

Solution:The three componentsGx, Gy andGz are

Gx = 3, Gy = 2 and Gz = 0.

Hence the magnitude ofG is

|G| = G2x+ G2y+ G2z = 32 + 22 + 02 = 13and the unit vector that points in the same direction as G is

G= 1

|G|G= 1

13(3i + 2j) .

Example 1.2 For the vectorF= 2yi 3zj + 9z2k

find the three componentsFx, Fy andFz and then determine the magnitude ofF.

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Solution:The three componentsFx, Fy andFz are

Fx = 2y, Fy = 3z, and Fz = 9z2

and hence the magnitude ofF is

|F| =

(2y)2

+ (3z)2 + (9z2)2

=

4y2 + 9z2 + 81z4.

1.4.1 The scalar (or dot) product

Thescalar productof 2 vectors A and B is given by

A B= |A| |B| cos where is the angle between the two vectors.

Since cos 0 = 1 and cos 2 = 0 we have

i i= |i| |i| cos0 = 1 1 1 = 1, j j= 1 and k k= 1.Furthermore

i j= |i| |j| cos2

= 1 1 0 = 0, i k= 0 and j k= 0.

IfA= Axi + Ayj + Azkand B = Bxi + Byj + Bzk

A B = (Axi + Ayj + Azk) (Bxi + Byj + Bzk)= AxiBxi+AxiByi+AxiBzi

+Ay iBxi+AyiBy i+Ay iBzi+AziBxi+AziByi+AziBzi

= AxBx(i i) + AxBy(i j) + AxBz(i k)+Ay Bx(j i) + AyBy(j j) + AyBz(j k)+AzBx(k i) + AzBy(k j) + AzBz(k k)

= AxBx+ AyBy+ AzBz.

Example 1.3 ForF= 3yi 2xk andG= i + 2zj 3k

calculateF G.

Solution: We have

F G = 3y 1 + 0 2z+ (2x) (3)= 3y+ 6x.

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1.4.2 The vector (or cross) product

The vector product of 2 vectors A and B is given by

AB= |A| |B| sin nwhere n is a unit vector perpendicular to A and B, pointing in the direction given by the right handscrew rule (i.e. the direction in which a screw would advance if it were turned fromA through the angle toB. The cross product can be evaluated by calculating the determinant of the 3 3 matrix

A B =

i j kAx Ay AzBx By Bz

= i

Ay AzBy Bzj Ax AzBx Bz

+ k Ax AyBx By

= i (Ay Bz ByAz)j (AxBz BxAz) + k (AxBy BxAy) .Example 1.4 For

A= 3xi 4k andB= 2i 9xj + kevaluateA B.Solution: We have

A B =

i j k3x 0 42 9x 1

=i 0 49x 1

j 3x 42 1+ k 3x 02 9x

= i ((0) (1) (9x) (4))j ((3x)(1) (2)(4)) + k ((3x) (9x) (2) (0))= i (36x)j (3x + 8) + k 27x2= 36xi (8 + 3x)j 27x2k.

Remark: AB =B A

1.5 Differentiation

Given a function f(x), (or f(t)) of a single variable x (or t), we denote the derivative off(x) byf (x)or df

dx (or the derivative off(t) by f(t) or df

dt) The following table gives the derivatives of some common

functions.

f(x) f (x)xn nxn1

sin x cos x

cos x sin xtan x sec2 xln |x| 1

x

ex ex

1.5.1 The product rule

Ifu = u(x) and v = v(x) then

d (uv)

dx =

du

dxv+ u

dv

dx=uv+ uv.

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1.5.2 The quotient rule

Ifu = u(x) and v = v(x) thend

dx u

v =

uv uvv2

.

1.5.3 The chain rule

Iff=f(u(x)) then

df

dx=

df

du

du

dx.

Example 1.5 Forf(x) = x3 cos x

calculatef(x).

Solution: Sincef(x) = x3 cos x we set

u= x3 andv = cos x

and hencedu

dx= 3x2 and

dv

dx= sin x

so from the product rule we have

df

dx =

d (uv)

dx =

du

dxv+ u

dv

dx

= 3x2 cos x + x3( sin x)= 3x2 cos x

x3 sin x.

Example 1.6 For

f(x) = ex

sin x

calculatef(x).

Solution: Settingu= ex and v= sin x

we havedu

dx=ex and

dv

dx= cos x.

Hence from the quotient rule we have

df

dx =

d

uv

dx

=dudx

v u dvdx

v2

= ex sin x ex cos x

sin2 x.

Example 1.7 Forf(x) = cos3x2

calculatef(x).

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Solution: Settingu = 3x2 we have f(x) = cos 3x2 = cos u and hence

du

dx= 6x and

df

du=

d cos u

du = sin u.

Thus from the chain rule we have

df

dx=

df

du

du

dx= sin u 6x= 6x sin u= 6x sin3x2.

RemarksWe note the following useful identities

ea+b =eaeb, eab = (ea)b

and for a,b >0

ln(ab) = ln a + ln b, lna

b

= ln a ln b and ln ba =a ln b.

1.6 IntegrationThe following table gives the integrals of some common functions.

f(x)

f(x)dx

kxn kn+1

xn+1 + c, n = 1sin kx - 1

kcos kx + c

cos kx 1k

sin kx + c

ekx 1k

ekx + c

kx

k ln |x|+ c

1.7 Integration by parts

Recalling the product rule we have

d

dx(fg) = fg+ f g

and hence by integration we obtain ba

d

dx(f g) dx=

ba

fgdx + b

a

f gdx

[f g]ba = b

a

fgdx + b

a

f gdx.

Manipulating the above equation gives the standard form ofintegration by parts formula ba

fgdx = [fg]ba b

a

f gdx.

Example 1.8 Use integration by parts to integrate 21

xexdx.

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Solution: We have 21

xexdx = [xex]21

21

exdx, g= x g = 1, f= ex f=ex

= [xex

]2

1 [ex

]2

1

= 2e2 e e2 1= e2.

1.7.1 Integration using substitution

ba

f(x)dx=

u(b)u(a)

f(u)dudx

du=

u(b)u(a)

f(u)

dx

du

du.

Example 1.9 Use integration using substitution to evaluate the following integral

1

0 x 1 + x23 dx.Solution: Settingu = 1 + x2 we have

du

dx= 2x dx

du=

1

2x, u(0) = 1 + 02 = 1 and u (1) = 1 + 12 = 2

and hence 10

x

1 + x23

dx =

u(1)u(0)

xu3dx

dudu,

=

2

1

xu3 1

2xdu

= 1

2 2

1

u3du

= 1

2

u4

4

21

= 1

8

24 14

= 15

8 .

1.8 Polar coordinates

There are two unit vectors associated with polar coordinates: e and e. They have magnitude 1 andpoint in the directions of increasing and respectively, also they are perpendicular so that

e e.

Remark 1 We note thate ande are position dependent unlikei, j andk (i.e. the direction in whichthey point depends on where they are in the(x, y) plane).

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Figure 3:

1.9 3-D coordinate systems

In 3-d there are three commonly used coordinate systems:

coordinate system variables unit vectorsCartesian (x,y,z) (i, j, k)Cylindrical (,,z) (e, e, ez)Spherical (r,,) (er, e, e)

Figure 4:

1.9.1 Cylindrical polar coordinates

A pointP(x,y,z) can be written in terms of the cylindrical polar coordinates , , zsuch thatP(x,y,z) =P(,,z). One can think ofP(,,z) as being a point on the surface of a cylinder with radius , seeFigure 4.

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Remark 2 The unit vectorez points in the increasingz -direction (and is in factk !).

Figure 5:

1.9.2 Spherical polar coordinates

A pointP(x,y,z) can be written in terms of the spherical polar coordinates r, , such that P(x,y,z) =P(r,,). One can think ofP(r,,) as being a point on the surface of a sphere with radiusr, see Figure

5.

Remark 3 The unit vectors are such that er points in the increasing r-direction, e points in the in-creasing-direction, e points in the increasing-direction.

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2 2D Surface Integrals

2.1 Cartesian coordinates

The areaA under a graph y = f(x) between two pointsx = a and x = b (see Figure 6) can be calculated

by evaluating the integral

A=

ba

f(x)dx.

The area A can be approximated by the sum of the areas ofNthin strips Si centred at xi with widthx and height f(xi) (see Figure 6)) hence we have

A

area of strips Si =

Ni=1

f(xi)x.

In the limit as the x 0 i.e. as the number of strips tends to infinity we haveN

i=1

f(xi)x ba

f(x)dx= A.

Alternatively we could divide each strip into blocks with centre (xi, yj ), width x and height y

A N

i=1

(area ofSi)

where

area ofSi =

Mij=1

(area ofBj ).

Hence

A N

i=1

(area ofSi)

=N

i=1

Mij=1

(area ofBj )

=

Ni=1

Mij=1

yx area ofBj

.

Note that the number of blocks Mi, depends on the height ofSi, i.e. the value off(xi). In the limit as

y=f(x)

a b

A

f(x i)

xi

ba

y=f(x)

Figure 6:

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the number of blocks tends to infinity and x and y tend to zero we have

N

i=1Mi

j=1yx

area ofBj b

a

f(x)0

dydx= A.

In general the area of a 2D surface S can be approximated by

A =

(area of the surface elements)

=

dA.

In the limit as the area of the surface element tends to zero we have

A=

S

dA.

Here the integral Sis a double integral i.e. , and the dA is a double integrand i.e. dxdy

(1,0)

(0,1)

(0,0)

T

Figure 7:

Example 2.1 Calculate the area of the triangleT bounded by the linesx = 0, y= 0 andy= 1 x.Solution:First we draw the triangle T, see Figure 7. We have that

area ofT =

T

dA=

dy

dx

The inner integral relates to the area of a strip with width x. The base of this strip is aty = 0,the topvaries depending on the position that the strip is along the x-axis (i.e. it is a function ofx).

From Figure 7 we see that the top of the strip is 1 x, so the limits on the inner integral are

area ofT =

1x0

dy

dx.

For the limits on the outer integral we recall that this integral is associated with the sum of the strips.The left hand strip is centred atx = 0 and the right hand one is at x = 1, so we have

area ofT =

10

1x0

dydx.

To evaluate a double integral we first evaluate the inner integral and then we evaluate the resultingintegral

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A=

10

1x0

dy

dx =

10

y1x

0 dx

= 10

[(1 x) 0] dx

=

10

(1 x)dx=

x x2

2

10

= 1

2.

2.1.1 Polar coordinates

Figure 8:

If we want to find the area of a circular or part circular surface we use surface elements derived frompolar coordinates.

The area of a surface element can be approximated by (see Figure 8)

A .Thus the area of the surface

A

(area of the surface elements)

.

In the limit as the area of each surface element tends to zero we have

A=

S

dA=

S

dd dA

.

The limits on the integrals depend on the surface that you are integrating over.

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Example 2.2 Calculate the area of the annular regionS by0 /2 and1 2.Solution: First we draw the annular region S, see Figure 9.

(1,0) (2,0)

(0,2)

(0,1) S

Figure 9:

We see that

area ofS=

S

dA=

d

d

From Figure 9 we see that the limits for the integral are 1 and 2, while the limits for the integral are/2 and /2. Thus

area ofS = 2

0

21

dd

=

2

0

21

d

d

=

2

0

2

2

21

d

=

2

0

2 1

2

d

=

2

0

3

2d

= 32

[]

2

0

= 3

2( 0)

= 3

4 .

2.2 Calculating mass and charge density

The techniques that we have used to calculate the area of surfaces can be applied to calculate the massof a 2D object or the charge on a 2D object.

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If the density of the rectangular plate S bounded by 0 x 2 and 0 y 1 is given by f(x, y) =(1 +x2)(1 +y) kgm2 we can approximate the mass ofSby dividing it up into small surface elementsand summing the masses of each of these elements.Since for an object with constant density, mass = area x density, we can approximate the mass of a

surface element by its area multiplied by the density at its mid-point

i.e. (mass of S.E.) xyf(xi, yj )Thus we have

S=

(mass of S.E.) =

i,j

xyf(xi, yj )

f(x, y)dydx.

Adding the required limits for x and y gives

Mass ofS =

20

10

(1 + x2)(1 + y)dydx

=

20

10

(1 + x2)(1 + y)dy

dx

= 20

(1 + x2) 10

(1 + y)dy dx=

20

(1 + x2)

y+

y 2

2

10

dx

=

20

(1 + x2)

1 +

1

2

0

dx

=

20

3

2(1 + x2)dx

= 3

2

x +

x3

3

20

= 7 kg.

Example 2.3 If the charge density on the semi circular disc S in Figure 10 is f(, ) = 3 Cm2,calculate the total electric charge onS.

Solution: We havetotal charge onS=

i

charge on S.E.

charge on S.E. area of S.E. charge density at the mid-point of S.E.(i, j )= f(i, j ).

So

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(0,1)

(0,1)

(1,0)

S

Figure 10:

total charge onS =

2

2

10

f(, )dd

=

2

2

10

3dd

=

2

2

10

32d

d

=

2

2

31

0d

= 2

2

d

= []2

2

= C.

2.3 General Formulas

We recall the following useful formulas.

Area ofS= S

dA,

Mass ofS=

S

f dA wheref= density,

charge on S=

S

f dA wheref= charge density .

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2.4 Double integrals from a mathematical point of view

So far we have been looking at things from a physical point of view. We could have asked the previousquestions in the following mathematical ways:

Example 2.4 Evaluate the integral of the functionf(x, y) = (1 +x2)(1 +y) over the two dimensionalrectangular region with0 x 2, 0 y 1.

Solution: 20

10

f(x, y)dydx=

20

10

(1 + x2)(1 + y)dydx= 7.

Example 2.5 Find the integral off(, ) = 3 over the semi-circular region0 1, 2

2.

Solution:

2

2

10

3dd= .

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3 Surface Areas in 3D

To evaluate the area of a 3D surface we use

A= S dA.As in the case of 2D surface integrals

S

is a double integral i.e.

, and dA is a double integrand i.e.dxdy.

3.1 Cartesian coordinates

If we wish to evaluate the surface area (or the mass, or the charge) of the hollow block 0 x 7,0 y 5 and 1 z 3, we need to break the block up into six 2D surfaces, i.e. its six faces.

On the top and bottom faces of the block x and y vary, and z is constant, so we have

area of top = 5

0

70

dxdyand area of bottom = 5

0

70

dxdy.

On the left and right hand faces x and z vary and y is constant, so we have

area of left hand face =

31

70

dxdzand area of right hand face =

31

70

dxdz.

On the remaining two faces y andz vary, and x is constant, so

area of front = 3

1 5

0

dydzand area of back = 3

1 5

0

dydz.

Example 3.1 Find the integral off(x,y,z) = 1 + x + y+ z over the surface of the hollow blockS.

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Solution: We haveS

f dA =

top

f dA +

bottom

f dA +

left side

f dA +

right side

fdA +

front

f dA +

back

f dA

= 50

70

1 + x + y+ z(=1)

dxdy+ 50

70

1 + x + y+ z(=3)

dxdy+

31

70

1 + x + y

(=0)

+ z

dxdz+

31

70

1 + x + y

(=5)

+ z

dxdz

+

31

50

1 + x

(=0)+ y+ z

dydz+

31

50

1 + x

(=7)+ y+ z

dydz

=

50

70

(1 + x + y+ 1) dxdy+

50

70

(1 + x + y+ 3) dxdy

+

31

70

(1 + x + z) dxdz+

31

70

(1 + x + 5 + z) dxdz

+ 31

50

(1 + y+ z) dydz+ 31

50

(1 + 7 + y+ z) dydz

=

50

70

(6 + 2x + 2y) dxdy+

31

70

(7 + 2x + 2z) dxdz+

31

50

(9 + 2y+ 2z) dydz

=

50

6x + x2 + 2yx

70

dy+

31

7x + x2 + 2zx

70

dz+

31

9y+ y2 + 2zy

50

dz

=

50

(91 + 14y) dy+

31

(98 + 14z) dz+

31

(70 + 10z) dz

=

91y+ 7y25

0+

91z+ 7z23

1+

70z+ 5z23

1

= 630 + 336 98 + 255 75 = 1048.

3.2 Cylindrical polar coordinates

a

x

y

4

0

Figure 11:

What about if we wish to find the mass of a hollow cylinder S in Figure 11 whose surface density isf(,,z) = 1 + z2?

We use Mass=S

f dA, but what is dA? First we split the surface of the cylinder into three parts; the

top, the base and the curved surface (C.S.), so

mass=

top

f dA +

base

f dA +

C.S.

fdA.

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The top and the base are 2D surfaces, so as in Section 2 we have (see Figure 12)top

f dA=

20

a0

fdd and

base

f dA=

20

a0

fdd

with z constant on both surfaces.

On the curved surface we have is constant, and in fact = a where a is the radius of the cylinder.

Figure 12:

We know that dA is related to the area of a surface element on the curved surface of the cylinder seefigure 12.

Soarea= az and hence dA = addz.

Thus C.S.

f dA=

40

20

faddz.

So

mass=

20

a0

fdd +

20

a0

fdd +

40

20

faddz.

Sincef= 1 + z2 we have

mass =

20

a0

1 + z2

(=16)

dd +

20

a0

1 + z2

(=0)

dd +

40

20

1 + z2

addz

=

20

a0

17dd +

20

a0

dd +

40

20

1 + z2

addz

=

20

172 a

2d +

20

a2

2d +

40

2a

1 + z2

dz

= 17a2 + a2 + 2a

z+

z 3

3

40

= 18a2 + 50 23 a.

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3.3 Spherical polar coordinates

Figure 13:

If we want to evaluate the integral off(r,,) over the surface of a sphere with radius R we note thatsincer is constant (= R) and and vary on the surface of the sphere we have (see Figure 13)

area of S.E. = R R sin = R2 sin

and hence dA= R2 sin dd.

Remark 4 The angle only varies from0 to , while the angle varies from 0 to 2 (for the wholesurface of a sphere).

Example 3.2 Given a hollow sphere with centre at the origin, radius 2, with charge densityf(R,,) =sin2 + cos2 , calculate the total charge on the surface of the sphere.

Solution: We have

total charge =

S

f dA (f is charge density)

=

0

20

sin2 + cos2

R2 sin dd

=

0

20

122 sin dd

=

0

20

4sin []20 d

= 8

0

sin d

= 16.

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3.4 Scalar and vector fields

A scalar (vector) field is a distribution of scalar values (vector values) on/in a specified surface/region inspace, such that there is a unique scalar (vector) associated with each point on/in the surface/region.

Examples ofscalar fields are temperature in a room or pressure in a room.Examples of vector fields are magnetic flux around a bar magnet or water velocity on the surface of ariver.

3.5 Flux of a vector field

The flux of a vector field F = Fxi+Fyj+Fzk (F= Fe+ Fe+ Fzez, F= Frer+ Fe+ Fe),through a surface S is equal to the surface integral of the component ofF normal to the surface.

Mathematically we have that

Flux = =

SFndA= S

F ndA

whereFn denotes the component ofF that points in the direction ofn.Ifn points out of the surface we have outward flux, ifn point in to the surface we have the inwardflux.

Example 3.3 Evaluate the outward flux of F= 3e 2ez through the base of a cylinder centred at theorigin, with heightHand radiusR.

Solution:On the base of the cylinder n = ez and hence

F n= (3e 2ez) (ez)

= (3)(0) + (0)(0) + (2)(1) = 2Since

Flux = =

S

F ndA

we have

=

S

2dA.

But what is dA? On the base of a cylinder and vary and we have dA = dd so,

=

20

R0

2dd=

20

2

2

2

R0

d= 2R2.

Alternatively, since we know that the area ofS= SdA, we have = 2 area ofS= 2R2.If we want to calculate the outward flux ofG = ze+ cos()e ez through the curved surface of thecylinder in Example 3.3, we have n = e and so

G n = (2ze+ cos()e ez) e= (2z)(1) + cos()(0) + (1)(0)= 2z

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and hence the outward flux

=

S

2zdA

= H2H

2

20

2zRddz

=

H2

H2

20

2RzRddz

=

H2

H2

20

2R2zddz

=

H2

H2

2R2z

20

dz

=

H2

H2

4R2zdz

= 4R2z2

2

H2H

2

= 0.

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4 Volume Integrals

4.1 Cartesian, cylindrical polar and spherical polar coordinates

Figure 14:

The integral of a function g(x,y,z), or g(,,z) org (r,,), over a 3D object is given byobject

gdV

wheredVis the limit of the volume of a small volume element in the object.

In Cartesian coordinates dV =dxdydz.

In cylindrical polar coordinatesdV =dddz (see Figure 14).

In spherical polar coordinatesdV =r2 sin()dddr (see Figure 15).

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Figure 15:

Volume integrals can be used to calculate the volume of 3D objects, or the mass or charge of/in a 3Dobject. Also they can be used to calculate the moment of inertia of a 3D object. For the objects in Figure16 we have:

Figure 16:

In cartesian coordinates V

gdV =

c0

b0

a0

gdxdydz.

In cylindrical polar coordinates V

gdV =

H0

20

a0

gdddz.

In spherical polar coordinates

VgdV =

2

0

0 R

0

gr2 sin()drdd.

Example 4.1 Evaluate the volume integral ofg(r,,) = 4r cos(/8)over a sphere centred at the originwith radius2.

Solution: We have sphere

f(,,z)dV =

20

0

20

g(r,,)r2 sin()drdd

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with g(r,,) = 4r cos(/8). Thussphere

f(,,z)dV =

20

0

20

4r3 cos(/8)sin()drdd

= 20

cos(/8)d 0

sin()d 20

4r3dr= [8sin(/8)]

20 [ cos()]0

r42

0

= 8

1

2 0

((1 1)) (16 0)

= 256

2.

4.2 Moment of Inertia

The moment of inertia of a mass m about the z-axis is given by I = m2, where =

x2 + y2 is thedistance from the z-axis.

The moment of inertia of a solid objectV about the z-axis is equal to the sum of the moments of inertiaof the individual volume elements in the object.

Since the volume elements are small we can approximate their mass by their volume multiplied by thedensity at their mid pointspi. Also we can approximate their distance from the z-axis by the distance oftheir mid points pi from the z-axis.

Hence the moment of inertia of a volume element is approximated by

Ii = 2Mass f(pi)Vi2i .

Thus we have that the moment of inertia Iis such that

IN

i=1Ii=

N

i=1f(pi)Vi

2i object

f 2dV.

Example 4.2 Calculate the moment of inertia of a cylinder with radius1, heightH, centre the originand densityf=z2.

Solution: We have

I =

H2

H2

20

10

2z2dddz

=

H2

H2

20

10

4z2dddz

= H2

H2 2

0 1

0 5

5 1

0

z2ddz

=

H2

H2

20

1

5z2ddz

=

H2

H2

2

5 z2dz

=

2

15z3H

2

H2

= H3

120.

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5 Line Integrals

5.1 Work done by a force

Figure 17:

Line integrals are used to calculate the work doneby a forceF (or a vector field) F = Fxi+Fyj+Fzk, on

a particle in moving it along a curve C=AB see Figure 17.

The work done by aconstant force F on a body undergoinglinear displacementis given by the distance|AB| multiplied by the component ofF in the direction AB , i,e, in the direction S, see Figure 18.

Figure 18:

Thus we have

W =F S= |F| |S| cos()and since

F= FsS + FsSwith

Fs = |F| cos()we have

W =Fs |S| .For a non-constant force F acting on a particle moving along a non-linear path C we can approximatethe work done by approximating the path Cby small linear path segments, and then on each of these

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Figure 19:

segments we can approximate the force F by the value ofF at the start of each linear segment (i.e. by aconstant value).On the ith segment along the curve that has start point at ri = xii + yij+ zik and at end pointri+1 = xi+1i +yi+1j +zi+1k (see figure 19), we have that the work done by the constant force F(ri) inmoving a particle along the segment is

Wi = F(ri) ri.So

W

i

Wi=

i

F(ri) ri.

In the limit as the size of the line segments tends to zero, we have

W = C

F(r) dr.

SinceF = Fxi+Fyj+Fzkand r = xi + yj + zk we have

F dr = (Fxi+Fyj+Fzk) (dxi + dyj + dzk)= Fxdx + Fydy+ Fzdz.

Hence

W=

C

Fxdx + Fy dy+ Fzdz. (5.5)

We need to write (5.5) as a definite integral involving a single variable, say x,y, z,t or .

5.2 Line integrals in the x-y plane

Example 5.1 Calculate the work done byF = 3yi 5xj+100xyk on a particle moving along the pathAB shown in Figure 20.

Solution: On the path AB we have y = 12 x and z = 3 and hence dy

dx= 12 and

dzdx

= 0.We can write (5.5) as a definite integral in x:

W =

C

Fxdx + Fydy+ Fzdz=

XBXA

(Fx(x)dx

dx+ Fy (x)

dy

dx+ Fz(x)

dz

dx)dx

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Figure 20:

whereXA is thex coordinate at the start point A and XB is the x coordinate at the end B .Thus XA = 0 and XB = 4. Furthermore since

Fx(x) = 3y=3

2x, Fy(x) = 5x, Fz(x) = 100xy = 50x

anddx

dx= 1,

dy

dx=

1

2 and

dz

dx= 0

we have

W =

40

3

2x

(1) (5x)

1

2

+ (50x)(0)

dx

=

4

0

xdx=x

2

2

40

= 8.

Remark 5

1. The line integral of a vector fieldF along a pathP =P1+ P2 can be written asP

F dr=

P1

F dr+

P2

F dr.

2. BA

F dr= A

B

F dr.

Example 5.2 Calculate the work done by F = 3xi + 5xj2k on a particle moving along the pathAc= AB+ BCshown in Figure 21.Solution: We have C

A

F dr = B

A

F dr+ C

B

F dr

=

BA

Fxdx + Fy dy+ Fzdz+

CB

Fxdx + Fydy+ Fzdz.

On path P1 we have

x= 1, z = 0 dxdy

= 0 and dz

dy = 0

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Figure 21:

and on path P2 we have

x= 1, y = 2 dxdz

= 0 and dy

dz = 0.

Thus we have

BA

F dr = YB

YA

(Fx(y)dx

dy+ Fy(y)

dy

dy+ Fz(y)

dz

dy)dy C

B

F dr = ZC

ZB

(Fx(z)dx

dz + Fy (z)

dy

dz+ Fz(z)

dz

dz)dz

B

A

F dr = 2

0

((3x)(0) + (5x)(1) (2)(0))dy = 2

0

5dy= 10

C

B

F dr = 1

0

((3x)(0) + (5x)(0) (2)(1))dz = 1

0

2dz = 2

and hence CA

F dr= 10 2 = 12.

Example 5.3 Calculate the work done byF = yi + xj on moving a particle along the pathAB shownin Figure 22.

Solution: The work done is given by

W = B

A

F

dr= B

A

Fx

dx + Fy

dy= XB

XA

(Fx

(x)dx

dx+ F

y(x)

dy

dx)dx

and since

y= (4 x2) 12 dydx

=1

2(4 x2)(2x) = x

(4 x2) 12we have

W =

XBXA

(4 x2) 12 (1) + (x)

x(4 x2) 12

dx

=

20

(4 x2) 12 x2(4 x2) 12

dx.

This is not an easy integral to solve, so hopefully there is a better way of calculating W.

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Figure 22:

Figure 23:

5.3 Parametric Equations

Any path in 3D may be expressed in terms of three parametric equations x(t),y(t) andz(t) that involvea parameter t that varies from t1 tot2 such that

x(t1) = x1, y(t1) = y1, z(t1) = z1, x(t2) = x2, y(t2) = y2 and z(t2) = z2.

We have (see Figure 23) P2P1

F dr= t2

t1

(Fx(t)dx

dt + Fy(t)

dy

dt + Fz(t)

dz

dt)dt.

Example 5.3 (continued)Using polar coordinates we can write the curve C=AB in terms of three parametric equations

x = 2 cos , y= 2 sin , z= constant

dxd

= 2sin , dyd

= 2 cos , dz

d= 0.

When = 0 x= 2, y = 0 and z = constant and hence we are at point A. While when = 2 x= 0,y= 2 and z = constanthence we are at point B .Thus we have

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BA

F dr = B

A

(y dxd

+ xdy

d)d

= 20

[(2sin )(2sin ) + (2cos )(2cos )] d

=

2

0

(4sin2 + 4 cos2 )d

=

2

0

4d= 2.

Example 5.4 Evaluate the integral ofF= 2xi + 3zj 5k along the curve given parametrically by

x(t) = t, y(t) = 2t, z(t) = t2

wheret varies from0 to1.

Solution: Since

x(t) = t, y(t) = 2t, z(t) = t2

dxdt

= 1, dy

dt = 2,

dz

dt = 2t

we have C

F dr = t2

t1

(Fx(t)dx

dt + Fy (t)

dy

dt + Fz(t)

dz

dt)dt

=

10

[(2x)(1) + (3z)(2) + (5)(2t)] dt

= 10

(2t 6t2 + 10t)dt

=

10

(12t 6t2)dt

=

6t2 2t310

= 4.

5.4 Path dependence/independence

A line integral of a field/force F from a point A to a point B is said to be path dependent if it dependson the path taken to get from A to B , otherwise it is said to be path independent. If the line integral ofF is path dependent F is non conservative.

Example 5.5 By considering two paths that start fromA = (2, 0)and end atB = (0, 2)(see Figure 24),show that the line integral of the vector fieldF = yi + xj is path dependent.

Solution: From Example 5.3 we have that P2

F dr= 2.

While on P2 we haveP2

F dr= B

A

Fxdx + Fydy=

XBXA

(Fx(x)dx

dx+ Fy(x)

dy

dx)dx.

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Figure 24:

Since the path P2 is defined by y = 2 x we have dydx = 1 and hence

P2

F dr = 0

2

[(y)(1) + (x)(1)] dx

=

02

[(2 x) x] dx

=

02

2dx= [2x]02 = 4.

So

P1F dr =

P2F dr and hence the integral of F is path dependent which tells us that F is non-

conservative.

5.5 Closed curves (loops)

Figure 25:

For a closed curveCthat comprises of the pathC1 from rA torB (see Figure 25) and the pathC2 fromrB torA, we have that the line integral ofF around the closed curve (loop) Cis given by

C

F dr=

C1

F dr+

C2

F dr.

The circle around the integral ofC indicates that Cis a closed curve.

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5.6 Conservative fields (or forces)

Remark 6 1. The line integral of a conservative field from a pointA to a pointB is path independent,i.e. it only depends on the initial pointA and the end pointB .

2. The line integral of a conservative field around any closed loop is always equal to zero, i.e.C

F dr=0 (5.6)

for a conservative fieldF and for any closed curveC.

3. If

CG dr =0 thenG is non conservative.

4. If

CG dr=0 thenG is mightbe conservative.

Example 5.6 Show that a conservative forceF satisfies (5.6).

Solution:

Since CF dr= C1 F dr+ C2 F dr and BA F dr= AB F dr, and for a conservative field C1 F dr=

C2F dr, we have the following for any conservative field.

C

F dr =

C1

F dr+

C2

F dr

=

C2

F dr+

C2

F dr= 0.

Figure 26:

Example 5.7 Calculate the line integral of F = x2i+yj along the two pathsC1 = AP+P B andC2shown in Figure 26. Say if the fieldF could be a conservative field.

Solution:The path C1 consists of the two straight line paths AP and P B. On the first path AP, y = 0 and xvaries from 0 to 1 dy

dx= 0.

On the second path P B, x = 1 andy varies from 0 to 1 dxdy

= 0.Hence we have

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C1

F dr = P

A

F dr+ B

P

F dr

= XPXA

(Fx(x) dxdx

+ Fy(x) dydx

)dx + YBYP

(Fx(y) dxdy

+ Fy (y) dydy

)dy

=

10

(x2)(1) + (y)(0)

dx +

10

(x2)(0) + (y)(1)

dy

=

10

x2dx +

10

ydy

=

x3

3

10

+

y2

2

10

=1

3+

1

2=

5

6.

On C2, y = x dydx = 1

C2

F dr = AB

Fxdx + Fydy

=

XAXB

(Fx(x)dx

dx+ Fy(x)

dy

dx)dx

=

01

(x2)(1) + (y)(1)

dx

=

01

(x2 + x)dx

=

x3

3 +

x2

2

01

= 56

.

Since the path C1+ C2 is a closed curve withC1+C2

F dr =

C1

F dr+

C2

F dr

= 5

6 5

6= 0

it follows thatF may be conservative.

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6 Partial derivatives

Figure 27:

The partial derivative of a functionF(x,y,z,t,...) with respect to x is defined by Fx

and is evaluated bytreating the other variables as constant and differentiating with respect to x. Consider a function of twovariables F(x, t), the partial derivatives F

x and F

tat a given point (x, t) of a functionF(x, t), represent

the slopes of the 3D graph y = F(x, t) in the direction of the x-axis and the t-axis at the given point(x, t) (see figure 27). We define

F

x = lim

h0F(x + h, t) F(x, t)

h (6.7)

andF

t = lim

h0F(x, t + h) F(x, t)

h . (6.8)

Example 6.1 Given thatf(x,y,t) = 3x2tey find fx

, ft

and fy

.

Solution: We have

f

x = 3tey

d

dxx2 = (3tey)(2x) = 6teyx

f

t = 3x2ey

d

dtt= 3x2ey

f

x = 3x2t

d

dyey = 3x2tey.

6.1 The chain rule for partial derivatives

For a function f(u(x,y,t)) we have

xf(u(x,y,t)) =

df

du

u

x,

yf(u(x,y,t)) =

df

du

u

y and

tf(u(x,y,t)) =

df

du

u

t. (6.9)

Example 6.2 Forf= cos(3x2y) find fx

.

Solution: Settingu = 3x2y we have

cos(3x2y) = cos u with df

du= sin u, and u

x= 6xy.

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Hence from the chain rule (6.9) we have

f

x = 6xy sin(3x2y).

6.2 Higher order partial derivatives

From the example above we see that partial derivatives of a function F(x,y,t) are also functions ofx, yandt, and so can be partially differentiated themselves.

Second order partial derivatives take the following form

2F

x2 =

x

F

x2F

yx =

y

F

x

2F

y 2 =

y

F

y

2F

xy =

x

F

y.

Example 6.3 ForF(x, y) = 3ex cos2y calculate

F

x,

F

y,

2F

x2,

2F

yx,

2F

y 2 and

2F

xy.

Solution: We haveF

x = 3ex cos2y,

F

y = 6ex sin2y

and

2F

x2 =

x

F

x =

x(3ex cos2y) = 3ex cos2y

2F

yx =

y

F

x =

y(3ex cos2y) = 6ex sin2y

2F

y 2 =

y

F

y =

y(6ex sin2y) = 12ex cos2y

2F

xy =

x

F

y =

x(6ex sin2y) = 6ex sin2y.

Remark 7 It is always the case that

2

Fxy

= 2

Fyx

.

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7 Directional derivatives

Figure 28:

The spatial rate of change of a scalar field in a specified direction is known as a directional derivative.

Example 7.1

Consider a 40 cm thick wall that has a temperature of 0C on one side and 20C on the other, with thetemperature varying uniformly within the wall (see Figure 28).The rate of change of temperature (T) in the direction specified byu is given by

change inT

distance in specified direction=

(10 5)CP Q

=5cos

10

Ccm

= 0.5cos C

cm. (7.10)

SinceP R= P Q cos we haveP Q=

P R

cos =

10

cos cm.The equation (7.10) is an example of a directional derivative. At any point P there is an infinity ofdirectional derivatives.

A vector at Pwith magnitude equal to the largest directional derivative, and pointing in the directionin which this largest directional derivative occurs, is known as the gradient vectoror more commonly asgrad.

Remark 8 The directional derivative atPin the directionucan be obtained by taking the scalar productofu with the gradient vector.Example 7.2 What is gradT atPin the previous example? Also show that

u gradT = 0.5

Ccm

.

Solution: Since grad T has magnitude equal to the largest directional derivative ofTwe see that gradT atPis when cos = 1 i.e. when = 0 and grad T points in the direction ofn. So from (7.10) we have

gradT= 0.5nCcm

.

Since u n= |u| |n| cos = cos we have that u gradT= 0.5u n C

cm = 0.5cos

Ccm

.

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7.1 Directional derivative at a point

Figure 29:

The directional derivative of a scalar field at a point at r in the directionu(see Figure 29) is given byu(r) = lim

r0(r + ru) (r)

r .

If we take the limit as r tends to zero we have the directional derivative of at the point at r in thedirectionu.The directional derivative of at a point atr = (x,y,z) in the direction of the x-axis, i.e. in the directionofi is

i(r) = limx0

(x + x,y,z) (x,y,z)x

.

This is the partial derivative of with respect to x. Hence

i(r) =(r)

xSimilarly we have

j(r) =(r)

y andk(r) =

(r)

z

7.2 The gradient vector, grad

We recall that any directional derivative in the direction u can be calculated by evaluating

u =

u grad.

So we havei= i grad, j = j grad, k = k grad

Since gradis a vector we can write it as

grad = ai + bj + ck

and hence

i = i (ai + bj + ck) = a =

x

j = j (ai + bj + ck) = b =

y

k = k (ai + bj + ck) = c =

z.

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Thus we have

grad =

xi +

yj +

zk. (7.11)

Example 7.3

1. Find the gradient of the scaler fieldU= 12 (x2 + y2 + z2).

2. Evaluate the magnitude of gradUat the pointP = (1, 2, 3).

3. Calculate the directional derivative ofU atP in the directions= i +j.

Solution:

1. Using (7.11) we have

gradU=U

xi +

U

yj +

U

zk =xi + yj + zk.

2. |gradU(P)| = 12 + 22 + 32 = 14.3. The directional derivative ofU atPin the direction s = i +jis given by

Us(P) = grad U(P) s= (i + 2j + 3k) i +j2

= 1

2+

22

= 3

2.

7.3 Gradient and physical laws

The following physical laws involve the gradient vector.

F= gradU F - conservative force field U - potential energy field (7.12)

E= gradV E- electric field V - electric potential (7.13)

h= k grad T h - heat flow T - temperature k - thermal conductivity (7.14)

Example 7.4 The electric potentialVat a point is given byV(x,y,z) = x2ey(z+1). Find the magnitudeand direction of the electric field vectorE at the point(1, 0, 0).

Solution: Since

E= gradV = Vx

i Vy

j Vz

k

we havegradV = 2xey (z+ 1)i x2ey(z+ 1)j x2eyk

and hence the direction of the electric field E at the point (1, 0, 0) is

E(1, 0, 0) = gradV(1, 0, 0) = 2ij k

and its magnitude is|E(1, 0, 0)| =

(2)2 + (1)2 + (1)2 =

6.

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7.4 Differential operators

The differential operator x

is an example of a scaler differential operator,it operates on a scalar field T

to give another scalar field Tx

.

The differential operator i x

is an example of a vector differential operator, it operates on a scalar field

Tto give a vector field i Tx

.The expression for the gradient of a scalar field involves a differential operator.

gradf = f

xi +

f

yj +

f

zk

=

i x

+ j

y+ k

z

(1)

f.

Here (1) =

i x

+ j y

+ k z

is a vector differential operator, it acts on the scalar field f to give a

vector field gradf

The differential operator (1) is commonly known as del ornabla, and is denoted by the symbol

gradf= f= fx

i + f

yj +

f

zk.

We can write the physical laws (7.12)-(7.14) in terms of differential operator:

h= kT, E= V, F= U.

7.5 Other coordinate systems

We have seen that it is often easier to work in coordinate systems other than rectangular Cartesian. Wewill, in fact, concentrate on two others, cylindrical polars and spherical polars. We will indicate here,however, a general approach for finding differential operators in other systems.

Suppose the new system (q1, q2, q3) has orthogonal vectors e1, e2, e3 at each point ( eg e, e, ez). Thenwe can write a small distance ds as ds2 = dx2 +dy2 +dz2 = h21dq

21 +h

22dq

22 +h

23dq

23. For example,

in 2-d, we have ds2 = d2 +2d2. The coefficients hi are called the metric, and generally in 3-dspace they will form a 3x3 matrix. They are a key component of Einsteins theory of General Rel-ativity, in which the metric of 4-dimensional space-time is determined by the distribution of matter.Famously, the metric for the space-time outside a point mass Mis called the Schwarzschild metric, givenbyds2 = (1 2GM

rc2 )dt2 + (1 2GM

rc2 )1dr2 +r2d2 +r2 sin2 d2. Notice that for large r , this reduces

tods2 = dt2 + dr2 + r2d2 + r2 sin2 d2, which is the (Minkowski) metric for flat space-time, in whichwe can see embedded the familiar 3-d spherical polar distance.

We now use the hi coefficients to find a general definition of the gradient operator. We can write thesmall change in any quantity asd =

q1dq1+

q2

dq2+ q3

dq3, but we also have that

d= .(h1e1dq1+ h2e2dq2+ h3e3dq3). This shows that h1.e1 =

q1 etc, and hence

= 1h1

q1

e1+ 1h2

q2

e2+ 1h3

q3

e3.

Our two important cases give

i) Cylindrical Polars=

e+ 1

e+

zez, and

ii) Spherical Polars= r

er+ 1r

e+ 1r sin

e.

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8 The divergence of a vector field

So far we have only looked at the spatial variations of scalar fields, what about the spatial variations ofvector fields?

There are two fields that are used to describe spatial variations of vectors fields F, one is a scalar fieldknown as the divergence ofF, and the other is a vector field known as thecurl ofF.

8.1 Mathematical approach to divergence

The divergence of a vector field F is the scalar field

div F = F=

i

x+ j

y+ k

z

(Fxi+Fyj+Fzk)

= Fx

x +

Fyy

+ Fz

z .

Example 8.1 Calculate the divergence of the vector fieldF = 3xeyi+2yj+z2kand evaluateF(2, 0, 4).Solution: We have

divF = F= (3xey)

x +

(2y)

y +

(z2)

z

= 3ey + 2 + 2z

F(2, 0, 4) = 3 + 2 + 8 = 13.

8.2 Physical approach to divergence

Physically the divergence of a vector field F is related to the flow of energy or matter, e.g. flow of water

in a pipe or flow of heat through some conducting material.

Consider the case of heat flow in a rod. Heat generated (by fission of uranium nuclei say) at a rate (r),flows via conduction to the cooler outer edge of the rod. If we define the heat flow rate by h(r) then thetwo fields (r) and h(r) are related by conservation of energy.

Consider a very small regionR of the rod with volume Vand surfaceS, such thatR contains the pointPwith position vectorrp. SinceR is small we can approximate the total rate at which heat is generatedinR by (rp)V. Since the temperature of the region does not change we must have that the amount ofheat generated in R must be equal to the rate of flow of heat out ofR through its surface S.

Thus we have

net outward heat flow rate across S (rp)V net outward heat flow rate across S

V (rp).

If we let Vtend to zero then the left hand side describes the net outward flow rate per unit volume atthe point P. We define this to be the divergence ofh at the point P.Thus

div h(P) = limV0

net outward heat flow

V

= (rp).

Since this equation holds for all points P we have

div h = .

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8.3 Equivalence of the two definitions of divergence

Figure 30:

If we take the small region R in the heated rod to be a cube with centre C= (X+ L2 , Y , Z ) and sides oflengthL, then we can calculate the net outward heat flow through R by using

Flux= =

R

h ndA=

R

hndA

=

6i=1

face i

h ndA

wherehn denotes the component ofh pointing in the direction ofn andn is the outward pointing normaltoR.If we set (1) to be the face of the cube lying in the plane x = X(see Figure 30) and (2) to be the face ofthe cube lying in the plane x = X+ L, then the outer unit normal to (1) is i and the outer unit normalto (2) isi. AlsoQ = (X , Y , Z ) is the mid point of (1) and Q= (X+ L,Y,Z) is the mid point of (2) andthe volume V =L3.Since the cube is small we can approximate the heat flow vector h on each face of the cube by its valueat the mid points, hence on face (1) we can approximate h by h(Q) = h(X , Y , Z ) and on face (2) we canapproximateh by h(Q) = h(X+ L,Y,Z).Hence on (1) we have h n= h (i) = h i= hx(X , Y , Z ) (i.e. minus the x component ofh at Q),and on (2) we have h n= h i= hx(X+ L,Y,Z).So

face 1

h ndA

face 1

hx(X , Y , Z )dA= hx(X , Y , Z )

face 1

dA= hx(X , Y , Z )L2

face 2

h ndA face 2

hx(X+ L,Y,Z)dA= hx(X+ L,Y,Z)L2.

So

net outward heat flow across (1)+(2)

V =

L2 [hx(X+ L,Y,Z) hx(X , Y , Z )]L3

= hx(X+ L,Y,Z) hx(X , Y , Z )

L

limV0


V

= lim

L0

hx(X+ L,Y,Z) hx(X , Y , Z )

L

=

hxx

.

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Similarly

limV0


V

=

hyy

and

limV0

net outward heat flow across (5)+(6)V

= hzz

.

Thus

div h = limV0

net outward heat flow across S

V

=

hxx

+ hy

y +

hzz

.

8.4 Interpretation of divergence in terms of sources and sinks

Figure 31:

We have seen that the divergence of a steady state heat flow field at any point is the heat source densityat that point. Similar ideas can be applied to other flow fields.

We can regard positive electric charge as being the source of an electric field E

div E =

0 (one of Maxwells equations).

Here is the electric charge density and 0 is the permittivity of free space.

The interpretation of divergence in terms of sources can be illustrated in field line diagrams.For example the field lines in the region outside a positively charged sphere are regularly spaced continuous(i.e. unbroken) radial lines directed outwards see Figure 31. In 3D the density of these lines representsthe magnitude of the field.You can only make the association of field magnitude and density of continuous field lines in regionswhere the divergence of the vector field is zero.

Field lines have sources (i.e. starting point) in regions ofpositive divergence, and sinks(end points) inregions ofnegative divergence.

If a vector field has zero divergence everywhere, then there are no sources or sinks anywhere and the fieldlines are closed loops.Magnetic fields are examples of divergence free fields see Figure 32.

div B= 0 Maxwells equation, B - magnetic field.

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Figure 32:

8.5 The divergence and physical laws

The following are examples of physical laws that involve the divergence of vector fields:

divh= , h steady state heat flow, - heat source density,

divE=

0, E electric field, - electric charge density, 0 - permittivity of free space,

anddivB= 0, B magnetic field.

Example 8.2 Calculate the divergence of the vector fieldF = y cos2 xi+yz2j+ezxkat the point(, 0, 2).

Solution: We have

div F = F= Fxx

+ Fy

y +

Fzz

= 2y cos x sin x + z2 + ezx

and hencediv F(, 0, 2) = 2(0)(cos sin ) + 22 + e2= 4 + e2.

Example 8.3 Could either of the following vector fieldsF orG in principle represent a magnetic field

F= xy2i + y3j + (x2y

4y2z)k or G= 2yi

3xyj

(9 + 5z)k?

Solution: Since

div F =Fx

x +

Fyy

+ Fz

z =y2 + 3y2 4y2 = 0

and

div G =Gx

x +

Gyy

+ Gz

z = 3x 5

it follows thatF could in principle represent a magnetic field and G could not.

Example 8.4 If the scalar fieldfis equal to the divergence of the vector fieldG = x cos yi+x2eyj+z2yk,calculateH = gradf.

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Solution: We have

f = divG=Gxx

+ Gy

y +

Gzz

= cos y+ x2ey + 2zy

and hence

H = grad f=f

xi +

f

yj +

f

zk

= 2xeyi + ( sin y+ x2ey + 2z)j + 2yk= 2xeyi + (x2ey + 2z sin y)j + 2yk.

RemarkIn operator terms the vector field H in the above example takes the form

H= f= ( G).

Example 8.5 Iff= 3x2

y+ y cos z+ y2

calculate divF, whereF= gradf.Solution: We have

F= grad f=f

xi +

f

yj +

f

zk

and henceF= 6xyi + (3x2 + cos z+ 2y)j y sin zk

div F = Fxx

+ Fy

y +

Fzz

= 6y+ 2 y cos z.


Using the same notation as earlier, we can find a general expression for the divergence of a vector fieldfrom the definition above

divh(P) = limV0

net outward heat flow

V

.

We have V = h1h2h3dq1dq2dq3, and we find, for example, that the flux of a vector field u across twoopposite facesq1=constant is given by

q1

(u1h2h3)dq1dq2dq3. Then, taking the limit, we get

divu = u= 1h1h2h3

(

q1(u1h2h3) +

q2(u2h1h3) +

q3(u3h1h2)).

This gives

i) Cylindrical polars

F = 1

(F) +

1

F

+ Fz

z

= F

+

F

+1

F

+ Fz

z

ii) Spherical polars

F= 1r2

r(r2Fr) +

1

r sin

(Fsin ) +

1

r sin

F

.

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8.7 The scalar operator del squared

In operator terms we have div (grad f) = f= 2f, here 2 is the scalar operatordel squared, suchthat

2 = 2x2

+ 2y 2

+ 2z 2

with

2f= 2f

x2+

2f

y2 +

2f

z 2.

We can also have del squared acting on a vector field, such that

2F = 2 (Fxi+Fyj+Fzk)= 2Fxi+2Fyj+2Fzk.

We can use the results from the previous sections to find the expression for the Laplacian operator in

other coordinate systems. We find

2= 1h1h2h3

q1

h2h3h1

q1

+

q2

h3h1h2

q2

+

q3

h1h2h3

q3

.

This gives

i) Cylindrical polars

2 = 1

+

1

22

2 +

2

z 2


2= 1r2 sin

sin

r

r2

r

+

sin

+

1

sin

2

2

.

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9 The curl of a vector field

9.1 Mathematical definition of curl

The mathematical definition of the curlof a vector field F is obtained by taking the cross product of del

(nabla) with F i.e.

curl F = F=

i j k

x

y

z

Fx Fy Fz

= i

Fzy Fy

z

j

Fzx Fx

z

+ k

Fyx Fx

y

.

Example 9.1 Calculate the curl of the vector fieldF= x2yi + y2zj + z2xk and then evaluate curlF atP = (1, 2, 3)

Solution: We have

curlF =

i j k

x

y

z

x2y y2z z2x

= i

y

z2x

z

y2z

j x

z2x

z

x2y

+ k

x

y2z

y

x2y

= y2i z2j x2k.

HencecurlF(1, 2, 3) = 4i 9j k.

9.2 Physical definition of curl

The physical definition of curl F is given in terms of the work done by F during one complete traversalof a closed loop. The work done depends on the orientation of the loop. Using the right hand screw rulewe can specify the orientation of the loop in terms of a unit vector n at right angles to the plane of theloop. If we consider a loop that lies in the (x, y) plane like in (Figure 34) then an anti-clockwise traversalof the loop is associated with n = k, whereas a clockwise traversal is associated with n = k.Assume that we have calculated W =

L

F dr then as the loop gets smaller and smaller so does W.However, if we divide W byA=area enclosed by L, then we have a quantity that remains finite as theloop gets smaller. If we define

C= limA0

W

A

thenCis in fact the component of curl F in the direction specified by n

i.e. C= curl F n.In our example n = k and hence Cis the z-component of curl F withC= curl F k.

9.3 Local rotation

The curl of a vector field F (or rot F as it is sometimes called) can be thought of in terms of the localrotation caused by F.

Consider the case of the water velocity on the surface of a river. In fast flowing rivers small floatingobjects rotate as they flow downstream (and this rotation is faster nearer the banks than it is near themiddle of the river.) This is because the velocity of the river is greater in the middle of the river than atthe edges (see figure 33). The magnitude of these rotations give the z-component of curl v.

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Figure 33:

H

H

B

R

S

Q

CTD

A

P

x

y

Figure 34:

9.4 Equivalence of the physical and mathematical definitions of curl

We recall that curl F n= limA0 WA whereW is the work done by F during one complete traversal ofa closed loop L that encloses an area A. We take L to be a square loop in the (x, y) plane with centreP = (X , Y , Z ) (see Figure 34)). For the loop L we have

W =

L

F dr=

AB

F dr+

BC

F dr+

CD

F dr+

DA

F dr.

Furthermore

AB F dr= AB Fxdx + Fydy+ Fzdz= x(B)

x(A) Fx dxdx + Fy dydx + Fz dzdx dxwherex(A) andx(B) respectively denote the x-coordinates of the points A and B . Sincey = Y H2 andz = constant on AB we have dy

dx= dz

dx= 0 and hence

AB

F dr= X+H

2

XH2

Fxdx.

Since His small we can approximate the x component ofF on AB by its value at the midpoint S andhence we have

AB

F dr= X+H

2

XH2

Fxdx HFx(S)

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Similarly CD

F dr = x(D)

x(C)

Fx

dx

dx+ Fy

dy

dx+ Fz

dz

dx

dx=

XH2

X+H2

Fxdx HFx(T)

BC

F dr = y(C)y(B)

Fx

dx

dy + Fy

dy

dy+ Fz

dz

dy

dy =

Y+H2YH

2

Fydy HFy (Q)DA

F dr = y(A)

y(D)

Fx

dx

dy+ Fy

dy

dy+ Fz

dz

dy

dy=

YH2

Y+H2

Fy dy HFy(R).

Thus we have

W =

L

F dr= H Fx(S)HFx(T) + HFy(Q)HFy(R)= H(Fx(S) Fx(T)) + H(Fy (Q) Fy(R))

and so

curl F n = limA0

W

A = lim

H0(H(Fx(S) Fx(T)) + H(Fy (Q) Fy(R)))

H2

= limH0

Fx(X, Y H2, Z) Fx(X, Y + H2, Z)H

+ limH0

Fy(X+ H

2, Y , Z ) Fy(X H2, Y , Z )H

= Fxy

+ Fy

x .

As we have calculated the work done by traversing the loop in an anti-clockwise direction, from the righthand screw rule we have that n = k and hence we have that the z-component of curl F is given by

Fyx Fx

y

. For the x and y components of curl F we would need to consider loops in the y z and xz

planes.

9.5 Conservative fields

When the work done by a vector fields F for all possible loops in the domain ofF is equal to zero, thenthe curl ofF equals zero everywhere and we define F to be conservative.This gives us a test for conservative fields

curlF = 0 for any conservative field. (9.15)

The term conservative is used for any vector field that satisfies (9.15).

Example 9.2 Is the vector fieldF =

x2 35j + 2zk conservative?Solution: We have

curl F =

i j k

x

y

z

0 x2 35 2z

= i

y(2z)

z

x2 35j

x(2z)

z(0)

+ k

x

x2 35

y(0)

= 0i + 0j + 2xk.

Since curl F =0 for all values ofx, y,z F is non conservative.

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9.5.1 Maxwells equations in differential form

Maxwells equations involve div and curl:

curlE =

B

t Faradays law for induction

curl B = 0J + 00curl E

t Amperes law

div B = 0 Gausss law for magnetism

div E =

0Gausss law for electricity.

HereE - electric field, B - magnetic field, J - electric current density, 0 - permeability of free space, 0- permittivity of free space, - electric charge density.

9.5.2 Operator identities

Recall the following useful operator identities:

= del ornabla =i x

+ j

y+ k

z

f = grad f=i fx

+ jf

y+ k

f

z

div F = F= Fxx

+ Fy

y +

Fzz

curlF = F=

i j k

x

y

z

Fx Fy Fz

= i

Fzy Fy

z j

Fzx Fx

z + k

Fyx Fx

y div (gradf) = f= 2f=

2f

x2 +

2f

y 2 +

2f

z 2.

Also note that for all vector fields F we have

div (curlF) = (F) = 0curl (grad f) = f=0.

Example 9.3 Verify that div(curlF) = 0 for the vector fieldF= aexi z cos yj + 3xyzk.Solution: Set

G = curl F =

i j k

x

y

z

aex z cos y 3xyz

= i (3xz+ cos y)j (3yz 0) + k (0 0)= (3xz+ cos y)i 3yzj

and hence

divG = div(curlF) = G

x +

G

y +

G

z= 3z 3z+ 0 = 0.

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We can use Stokes Theorem ( see next section) to find an expression for the curl of a vector field in ageneral coordinate system. We find

curl F = F= 1h1h2h3

e1h1 e2h2 e3h3

q1

q2

q3

h1F1 h2F2 h3F3

So we geti) Cylindrical polars

curlF = F= 1

e e ez

z

F F Fz


curlF = F= 1r2 sin

er er e3h3

q1

q2

q3

h1F1 h2F2 h3F3

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10 Stokess and Gausss theorems

10.1 Stokess theorem

The surface integral (flux) of curl F across any simple surface Sis equal to the line integral ofF around

the closed boundary curveC. S

curl F ndA=

C

F dr.

The sense of the traversal of the line integral is related to the unit normal of the surface by the righthand screw rule.

10.2 Gausss theorem (divergence theorem)

The volume integral of div F over a region B is equal to the surface integral (outward flux) ofF acrossthe surface enclosing B.

B div FdV = S F ndA where n is the outward unit normal to S.

Figure 35:

Remark 9 In Stokess theorem we have thatS

curlF ndA=

C

F dr

whereSis the surface enclosed by the closed curveC. For any closed curveC there are infinitely manysurfacesS that are enclosed byC.IfCis the closed curvex2 + y2 = 1, z = 0, one possibleScould be the circular diskx2 + y2 = 1, z = 0(see Figure 35). Alternatively S could be the hemispherex2 +y2 +z2 = 1, z 0. Or Scould be thesurface of the cylinder = 1, 0

2, 0 < z

2 (i.e. the curved surface and top, but not the base).

Example 10.1 (of Gausss theorem)Verify that Gausss theorem holds for the vector field F = 3xyi 2zxk, and the brick B bounded by1 x 3, 0 y 2, 2 z 5.

Solution: We need to show that

Bdiv FdV =

S

F ndA where Sdenotes the surface of the brick B.For the right hand side we have

S

F ndA=6

i=1

(

face i

F ndA)

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such that

on face 1, x = 1, 0 y 2, 2 z 5 andn = ion face 2, x = 3, 0 y 2, 2 z 5 andn = i

on face 3, y = 0, 1 x 3, 2 z 5 andn = jon face 4, y = 2, 1 x 3, 2 z 5 and n = j

on face 5, z = 2, 1 x 3, 0 y 2 andn = kon face 6, z = 5, 1 x 3, 0 y 2 and n = k.

Hence we have face 1

F ndA =2

0

52

(3xyi 2zxk) (i)dzdy

=

2

05

2 3ydzdy=

20

3ydy

52

dz

=

3

2y22

0

[z]52 = 18

face 2

F ndA =2

0

52

(3xyi 2zxk) idzdy

=

20

52

3xydzdy

=

20

52

9ydzdy

=

9

2y22

0

[z]52 = 54

face 3F ndA =

3

15

2(3xyi 2zxk) (j)dzdx

=

31

52

0dzdx= 0

face 4

F ndA=3

1

52

(3xyi 2zxk) jdzdx= 0

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face 5

F ndA =3

1

20

(3xyi 2zxk) (k)dydx

=

31

20

2zxdydx

=

31

20

4xdydx

=

31

xdx

20

4dy

= 32and

face 6

F ndA =3

1

20

(3xyi 2zxk) kdydx

=

31

20

2zxdydx

=

31

20

10xdydx= 80.

So

S F ndA= 18 + 54 + 0 + 0 + 0 + 32 80 = 12.For the left hand side we have

B

div FdV =Fx

x +

Fyy

+ Fz

z = 3y 2x.

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So B

div FdV =

52

20

31

(3y 2x) dxdydz

=

52

20

3yx 2x23

1dydz

=

52

20

[(9y 9) (3y 1)] dydz

=

52

20

(6y 8)dydz

=

5

2 3y2 8y

2

0dz

=

52

(12 16) dz

=

52

4dz = 4 [z]52 = 12.

Since the left hand side equals the right hand side, the theorem is verified.

Example 10.2 (of Gausss theorem)Verify that Gausss theorem holds for the vector field F = r2er + r sin cos e and the sphere B withcentre the origin and radius1.


Bdiv FdV =

S

F ndA whereSdenotes the surface of the sphere B .For the right hand side we have

S

F ndA=

S

F erdA=

S

r2dA.

Sincer = 1 on the surface of the sphere and dA = R2 sin dd= sin ddwe haveS

F ndA=

S

r2dA =

20

0

sin dd

=

20

d

0

sin d

= []20 [

cos ]0

= (2 0)((1 1)) = 4.For the left hand side we note that

divF = F= 1r2

r(r2Fr) +

1

r sin

(Fsin ) +

1

r sin

F

and hence

div (r2er+ r sin cos e) = 1

r2

r(r2r2) +

1

r sin

(0sin ) +

1

r sin

(r sin cos )

= 1

r2

r(r4) +

1

r sin

(0) +

r sin

r sin

(cos )

= 4r sin .

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Thus sincedV =r2 sin drddwe haveB

divFdV =

B

(4r sin )dV

= 20

0 1

0(4r sin )r2 sin drdd

=

20

0

10

(4r3 r2 sin )sin drdd

=

20

0

r4 1

3r3 sin

10

sin dd

=

20

0

1 1

3sin

sin dd

=

20

1 1

3sin

0

sin d

d

= 2

0 1 13sin [ cos ]0d= 2

20

1 1

3sin

d

= 2

+

1

3cos

20

= 4.

Example 10.3 (of Stokess theorem)Verify that Stokess theorem holds for the vector fieldF = y zi xzj+zk and the closed curveC givenbyx2 + y2 = 1, z = 3.


S curlF ndA= C F drwhere n is related to the direction of traversal of the curve Cby the right hand screw rule, and S is asurface enclosed by C.We choose our surfaceSto be the disc x2 + y2 1, z= 3, and we take a clockwise traversal of the curveC, so thatn = k.For the right hand side we have

C

F dr =

C

Fxdx + Fydy+ Fzdz

=

21

(Fxdx

d+ Fy

dy

d+ Fz

dz

d)d

where1 = 2, 2 = 0, x= cos , y= sin , z= 3 dx

d = sin , dy

d = cos , dz

d = 0.Hence C

F dr =0

2

(yz sin xz cos ) d

=

02

3sin2 3cos2 d=

02

3d= [3]02 = 3 (0 2) = 6.

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For the left hand side we have S

curlF ndA= 2

0

10

curl F ndd

with

curlF =

i j k

x

y

z

Fx Fy Fz

=

i j k

x

y

z

yz xz z

= i

z

y (xz)

z

j

z

x (yz)

z

+ k

(xz)

x (yz)

y

= xi + yj 2zk.

Hence

curlF n = (xi + yj 2zk) (k)= 2z = 6 (sincez = 3).

So S

curlF ndA = 2

0

10

6dd

=

10

6d

20

d

=

321

0[]

20

= 6.


Example 10.4 (of Stokess theorem)Verify that Stokess theorem holds for the vector fieldF = yzi xzj+zk and the closed curve C thatlies in the planex = 1 and consists of the three lines L1, L2 and L3 where on L1 we havez = 0 and

0 y 1, onL2 we havey= 1 and0 z 1 and onL3 we havez = y and0 y 1.Solution: We need to show that

S

curlF ndA=

C

F drwhere n is related to the direction of traversal of the curve Cby the right hand screw rule, and S is asurface enclosed by C.We choose our surface S to be the two dimensional triangular surface lying in the plane x = 1 that isbounded by the three linesL1,L2 andL3, and we take an anticlockwise traversal of the curve C, so thatn= i.For the right hand side we have

C F dr= L1 F dr + L2 F dr + L3 F dr.Sincex = 1 andz = 0 on L1 we have

dx

dy =

dz

dy = 0

and sincey varies from 0 to 1 we haveL1

F dr =

L1

yzdx xzdy+ zdz

=

10

yz

dx

dy xz dy

dy+ z

dz

dy

dy

= 1

0

xzdy = 1

0

0dy= 0.

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Sincex = 1 andy = 1 on L2 we havedx

dz =

dy

dz = 0

and sincez varies from 0 to 1 we haveL2

F dr = L2

yzdx xzdy+ zdz

=

10

yz

dx

dz xz dy

dz+ z

dz

dz

dz

=

10

zdz =

z2

2

10

= 1/2.

Sincex = 1 andz = y onL3 we have

dx

dy = 0 and

dz

dy = 1

and sincey varies from 0 to 1 we haveL3

F dr =

L3

yzdx xzdy+ zdz

=

10

yz

dx

dy xz dy

dy+ z

dz

dy

dy

=

10

(xz+ z)dy

=

10

(y+ y)dy= 1

0

0dy= 0.

Hence C

F dr= L1

F dr + L2

F dr + L3

F dr= 0 + 1/2 + 0 = 1/2.From the previous example we have that curlF= xi + yj 2zkand hence for the left hand side we have

S

curl F ndA=

S

curlF idA=

S

xdA.

SinceSlies in the plane x = 1 we haveS

xdA=

S

dA= area of S = 1/2.


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11 The calculus of variations

11.1 The shortest distance

Suppose we want to find the shortest distance between two points in 2-d space. If they are (x1, y1), (x2, y2),

then any distance between the two is given by

I=

x2x1

dx2 + dy2 =

x2x1

1 + y2dx

where we do the integration along the route y = y(x). We wish to find the route that minimizes theintegral.

Suppose that y(x) is the required answer, and that we write any other curve as Y(x) = y(x) + h(x),with h(x1) = h(x2) = 0. Thinking of I as I(), we want to find the minimum value of I, so we wantdId

= 0 when = 0.

We find dI

d

= x2

x1

1

2

1

1 + Y22Y

dY

d

dx. UsingY (x) = y(x) + h(x) gives

dY

d =h(x), so that

dI

d =0= x2

x1

y(x)h(x)1 + y2

dx= 0.

We now integrate by parts, to get dI

d =0= yh(x)

1 + y2

x2x1 x2

x1

h(x) d

dx

y1 + y2

dx.

The first term is zero, because h(x1) = h(x2) = 0, so to achieve a minimum we need ddx

y

1+y2

= 0.

This means that

y1+y2

= constant, soy = constant. Thus, the required curvey(x) is a straight line,

as expected.

11.2 Eulers Equation

Suppose we now generalize the problem, to ask which curve y(x) minimizes the integral I= x2x1 F(x,y,y)dx,whereFis a given function.As before, consider any curve Y(x) = y (x) +h(x), with h(x1) = h(x2) = 0. Thinking ofI as I(), wewant to find the minimum value ofI, so we want dI

d= 0 when = 0.

We find dI

d =

x2x1

FY

dY

d +

F

Y dY

d

dx. Using Y (x) = y (x) + h(x) again gives

dY

d =h(x), so that (

dI

d)=0 =

x2x1

Fy

h +F

y h

dx= 0.

Using parts

x2x1

F

y hdx=

Fy

hx2

x1 x2

x1

d

dx

Fy

hdx. The integrated term is zero, again because

of the boundary conditions, so we finally get

dId=0 = x2

x1 Fy ddx Fy hdx= 0.Sinceh is arbitrary, we have

F

y d

dx

F

y = 0,

which is the Euler ( sometimes known as Euler-Lagrange) equation.

11.3 The Brachistochrone Problem

This is a famous problem, in which we must find the shape of a smooth wire, joining the points (x1, y1)and (x2, y2), so that a bead will slide down the wire in the shortest time.Clearly, we need to minimize

dt.Taking initial speed v = 0, and the reference level for gravitational

potential energy as y = 0, then at the point (x, y) we have that the kinetic energy = 12 mv2 and the

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potential energy =mgy ( we are measuring y as positive in the down direction). Conservation ofenergy then gives 12 mv

2 mgy = 0, so thatv = 2gy.So we need to minimize the integral

dt=

ds

v =

ds

2gy =

12g

1 + y2

y dx.

It turns out that the easiest way (because Fdoes not explicitly contain x) to use the Euler equation is

by using ds =

dx2 + dy2 = dy

1 + x2, so that we try to minimize 1

2g

1 + x2

y dy. We now use

the alternate version of the Euler equation: F

x d

dy

F

x= 0, because the integration is with respect to

y, and this, when applied to they-integral, gives(

1+x2

y )

x d

dy

(

1+x2y

)

x = 0.

The first term is zero, since F has no explicit dependence on x, so we get d

dy

(

1+x2y

)

x = 0. This

integrates immediately to x1 + x2

y

= constant, which we can rearrange as dx

dy =

cy

1 cy . This

integrates, by separation of variables, to give x= yc y2 + 1

2ccos1(1 2cy) + c. If the curve passes

through the origin, then we have c = 0. This curve is in fact the equation of a cycloid, and it is moresimply written using a parameter , when we get

x= 1

2c( sin ), y= 1

2c(1 cos ).

A cycloid is well known as the locus of a point on the circumference of a circle as that circle rolls alonga level surface.

11.4 Hamiltons Principle

We may generalize our original question, of minimizingI=

x2

x1F(x,y,y)dx, to the case whereFdepends

on several variables y1, y2,..., each of which depend on x, so that we have I= x2x1 F(x, y1, y2,...,y1, y2,....)dx.Using the same method as before, we now get a set of Euler equations, of the form

F

yi d

dx

F

y i= 0, i= 1, 2,.....,n.

This is particularly useful in analyzing systems which depend on several coordinates. If we define theLagrangian L of a system as the difference of its kinetic and potential energies, so L = T V, thenHamiltons principle says that a system evolves in such a way that the time integral of the Lagrangian

is a minimum. This leads to the Lagrange equations L

xi d

dt

F

xi= 0, i= 1, 2,.....,n, where now the

variables are labeled xi, and they all depend on t.This formulation of mechanics is equivalent to the Newtonian one, but it involves only scalars, and it ismore amenable to choosing different coordinate systems to the usual Cartesian x, y,z. We can easily see

that in one dimension, it reduces to the usual form, since if we take T = 12 mx2 and potential V(x) ( sothat the force in the x-direction is F = dV

dx), we get (V)

x d

dtTx

= 0, or F =mx, which of course isjust Newtonss second law.

11.5 Noethers Theorem

Hamiltons principle can be used in non-mechanical contexts, such as electromagnetism, with an appro-priate definition of the Lagrangian. In general, in known theories, it simply reproduces results alreadyknow, albeit sometimes in a simpler and more elegant way. However, where a theory is incomplete, apostulated Lagrangian can be a useful way forward. Such an approach has been widely used in particlephysics and string theory.

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An example of this is Noethers Theorem, which states that the invariance of the Lagrangian with respectto a variable ( such as position or time) implies the conservation of a particular quantity (linear momentumand energy, respectively, for the two examples mentioned). We can see this in the very simplest example,with L = 12 mx

2 V(x). Applying Eulers equation gives Lx d

dtFx

= 0, or Lx

= ddt

Lx

.

Now consider dLdt =Lx

x+ Lxx+ Lt , using the chain rule of partial differentiation, and assuming

that L = L(x, x, t). If, in fact, L does not explicitly contain t, then Lt

= 0. Using the result in the

previous paragraph, we then get dL

dt =

d

dt(

L

x)x+

L

xx, which we can see is the result of the product

rule dL

dt =

d

dt(

L

xx).

Therefore we have d

dt(L L

xx) = 0, and so L L

xx= constant. But

L Lx

x = 12 mx2 V(x) mx2 = 12 mx2 V(x), so we have 12 mx2 +V(x) = constant ie the

conservation of energy.

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Mathematical Methods For Physics 2

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