Unit 9: Physical examples of fields.Potential energy function.

Unit 10: Gradient of potential.Gradient as directional derivative.

Units 9 & 10 are on overlapping topics. We shall develop them together.

PCD-08

Equipotentials and Field Lines of an electric dipolePCD-08

Primary, elementary entities in Physics:

Physical Universe is made up of Particles and Fieldsand their mutual interactions

Particles: Material world

Fields: ‘Action at a distance’

Fields PotentialsIntegration / Constant of IntegrationBoundary Value problem

DifferentiationPCD-08

Scalar/ Vector Fields: ‘Point’ function

( ) ( , , ) ( , , ) ( , , )

( ) ( , , ) ( , , ) ( , , )

r x y z r z

A A r A x y z A r A z

ψ ψ ψ ψ θ ϕ ψ ρ ϕ

θ ϕ ρ ϕ

= = = =

= = = =

r

ur ur ur ur urr

Examples of Scalar Point Functions

• temperature• gravitational/electrostatic

potentials• pressure in a liquid column

Examples of Vector Point Functions

• velocity field• electric field• magnetic field

PCD-08

Scalar/ Vector Fields: ‘Point’ function

Functions of ‘space’ and ‘time’

( , ) ( , , , ) ( , , , ) ( , , , )

( , ) ( , , , ) ( , , , ) ( , , , )

r t x y z t r t z t

A A r t A x y z t A r t A z t

ψ ψ ψ ψ θ ϕ ψ ρ ϕ

θ ϕ ρ ϕ

= = = =

= = = =

r

ur ur ur ur urr

PCD-08

0 0

The rate of change of with distance is a scalar given by

( ) ( ) ( )lim lims s

s

d r r r rds s sδ δ

ψ

ψ δψ ψ δ ψδ δ→ →

+ −= =

r r r r

DIRECTIONAL DERIVATIVE is a SCALAR QUANTITY

which has a DIRECTIONAL ATTRIBUTE.

This ‘rate’ (‘slope’) depends on the direction in which the displacement is considered.

r

PCD-08

δ r

Ratio of two scalar quantities.

It has a ‘directional attribute’

0 0

=

( ) (

) lim lim

=

s s

x y zx y z

z

d r

z

r

d s

r

r rs sδ δ

ψ ψ ψδ δ δ

ψ ψ ψδρ δϕ δρ ϕ

ψ ψ ψδ δ

ψ δψ ψ δ ψδ δ

ϕ

δψ

θ δϕθ

→ →

∂ ∂ ∂+ +

∂ ∂ ∂

∂ ∂ ∂+ +

∂ ∂ ∂

∂ ∂ ∂= + +∂ ∂

+ −= =

∂

r r r

Cartesian Coordinate System

Cylindrical Polar Coordinate System

Spherical PolarCoordinate System

(r)= ( , , )

(r)= ( , , )

(r)= ( , , )

x y z

z

r

ψ ψ

ψ ψ ρ ϕ

ψ ψ θ ϕ

r

r

r

PCD-08

0

=

The directional derivative( ) ( ) lim

=

s

dx dy dzx ds y ds z ds

d d dzds ds z ds

dr d dr ds

d

d

r r rds

s ds

sδ

ψ ψ ψ

ψ ρ ψ ϕ ψρ ϕ

ψ ψ δ ψ

ψ ψ θ ψϕ

δ

ϕθ

→

∂ ∂ ∂+ +

∂ ∂ ∂

∂ ∂ ∂

+ −=

+ +∂ ∂ ∂

∂ ∂ ∂= + +∂ ∂ ∂

r r r

PCD-08

Cartesian Coordinate System

Cylindrical Polar Coordinate System

Spherical PolarCoordinate System

These expressions can be written very nicely by using the expressions for the displacement in various coordinate systems.dr

uur

(r)= ( , , )

(r)= ( , , )

(r)= ( , , )

x y z

z

r

ψ ψ

ψ ψ ρ ϕ

ψ ψ θ ϕ

r

r

r

PCD-08

ˆ ˆ ˆr =e e e zd d d dzρ ϕρ ρ ϕ+ +r

ˆ ˆ ˆr =e e ex y zd dx dy dz+ +r

ˆ ˆ ˆr = e e ex y zx y z+ +r

ˆ ˆr = e e zzρρ +r

ˆr =re rr ˆ ˆ( ) ( )r rdr dr e r de= +

r

ˆ ˆˆ( ) r rr

e edr dr e r d dθ ϕθ ϕ

⎡ ⎤∂ ∂= + +⎢ ⎥∂ ∂⎣ ⎦

r

ˆ ˆ ˆsinrdr dre rd e r d eθ ϕθ θ ϕ= + +r

ˆ ˆ ˆr =( )e ( e ) ( )e zd d d dzρ ρρ ρ+ +r

ˆ ˆ ˆr =e e ex y zd dx dy dz+ +r

ˆ ˆ ˆ sinrdr e dr e rd e r dθ ϕθ θ ϕ= + +r

Position and Displacement vectors in various coordinate systems

In order to avoid making careless mistakes, always try to write unit vectors first, differential elements last!

Example:

0

( ) ( )lim

dx dy dz = + +

x ds ds ds

s

d r r rds s

dds y z

δ

ψ ψ δ ψδ

ψ ψ ψ ψ

→

+ −=

∂ ∂ ∂∂ ∂ ∂

r r r

ˆ ˆ ˆ=e e e

ˆ ˆ ˆ= e e e

x y z

x y z

x y z

x y zψ ψ

∂ ∂ ∂∇ + +

∂ ∂ ∂

⎡ ⎤∂ ∂ ∂∇ + +⎢ ⎥∂ ∂ ∂⎣ ⎦

ur

ur

s 0

ˆ

ˆ lim

, tiny

increament

, differential

increament

d uds

r drus ds

s r

ds dr

δ

ψ ψ

δδ

δ δ

→

= •∇

= =

=

=

r

uur uur

uur

uur

ˆ ˆ ˆr =e e ex y zd dx dy dz+ +r

Gradient in the Cartesian Coordinate System

PCD-08

( ) 1ˆ ˆ ˆ ˆ ˆ ˆe e e e e ez zzzρ ϕ ρ ϕ

δψ

δρ ρδϕ δ ψρ ρ ϕ

=

⎛ ⎞∂ ∂ ∂+ + • + +⎜ ⎟∂ ∂ ∂⎝ ⎠

:

ˆ ˆ ˆr =e e e

:

z

How should we definesuch that

r

zwhere

ρ ϕ

ψ

δψ δ

ρ ϕ

ψ

δ δ ρδ δ+

∇

= • ∇

+

ur

uur ur

r

1ˆ ˆ ˆ=e e

e

!

z

Following form of thegradient operator will work

zρ ϕρ ρ ϕ∂ ∂ ∂+ +

∂ ∂∇

∂

ur

( )

ˆ ˆ ˆe e e z

z

r

z

z

ρ ϕ

δ

δψψ ψ ψ

ψ

δρ ρδϕ δ

δρ δϕ δρ ϕ

ψ

= •

=∂ ∂ ∂

= + +∂ ∂

∇

+ + • ∇

∂uur ur

ur

1Note how the

cancelsρρ

PCD-08

Cylindrical Polar Coordinate System

( ) ( ) ( ) 1 1ˆ ˆ ˆ ˆ ˆ ˆe e e sin e e esinr rr r r

r r rθ ϕ θ ϕ

δψ

δ δθ θ δϕ ψθ θ ϕ

=

⎛ ⎞∂ ∂ ∂⎡ ⎤+ + • + +⎜ ⎟⎣ ⎦ ∂ ∂ ∂⎝ ⎠

( ) ( ) ( )ˆ ˆ ˆr =e e e

s

:

:

in r

How should we definesuch that

r

rwher

re

rθ ϕδ δ δθ θ δ

δ ψ

ϕ

ψ

ψ δ

+ +

∇

= • ∇

ur

uur

r

ur

1 1ˆ ˆ ˆ=e e es

i

!

nr

Following form of thegradient operator wil

r r

l wo k

r

r

θ ϕθ θ ϕ∂ ∂ ∂+ +

∂∇

∂ ∂

ur

( ) ( ) ( )

ˆ ˆ ˆe e e sin r

r

r r

rr

rθ ϕ

δ ψ

δ δθ

δψψ ψ ψδ δθ

θ δϕ

δϕϕ

ψ

θ

= •

=∂ ∂ ∂

= + +

∇

⎡ ⎤+ + • ∇⎣

∂ ∂ ∂

⎦

uur ur

ur

PCD-08

Note the cancellation of the factors that are circled

Spherical Polar Coordinate System

0

( ) ( )lim

=

s

d r r rds s

d d d dzds ds dss zd

δ

ρ ψ

ψ ψ δ

ϕ ψ ψρ ϕ

ψδ

ψ

→

∂ ∂ ∂+ +

∂

+ −=

∂ ∂

r r r

s 0

ˆ

ˆ lim

, tiny

increament

, differential

increament

d uds

r drus ds

s r

ds dr

δ

ψ ψ

δδ

δ δ

→

= •∇

= =

=

=

r

uur uur

uur

uur

Directional Derivative in the Cylindrical Polar Coordinate System

ˆ ˆ ˆr =e e ezd d d dzρ ϕρ ρ ϕ+ +r

1ˆ ˆ ˆ=e e e

1ˆ ˆ ˆ= e e e

z

z

z

z

ρ ϕ

ρ ϕ

ρ ρ ϕ

ψ ψρ ρ ϕ

∂ ∂ ∂∇ + +

∂ ∂ ∂

⎡ ⎤∂ ∂ ∂∇ + +⎢ ⎥∂ ∂ ∂⎣ ⎦

ur

ur

PCD-08

0

( ) ( ) lim

=

s

dr d

d r r rds s

d dr ds dds s ds

δ

ψ ψ δ ψδ

ψ ψ ψ θ ψ ϕθ ϕ

→

+

∂ ∂ ∂= + +∂ ∂ ∂

−=

r r r

s 0

ˆ

ˆ lim

, tiny

increament

, differential

increament

d uds

r drus ds

s r

ds dr

δ

ψ ψ

δδ

δ δ

→

= •∇

= =

=

=

r

uur uur

uur

uur

ˆ ˆ ˆ sinrdr e dr e rd e r dθ ϕθ θ ϕ= + +uur

1 1ˆ ˆ ˆ=e e esin

1 1ˆ ˆ ˆ= e e esin

r

r

r r r

r r r

θ ϕ

θ ϕ

θ θ ϕ

ψ ψθ θ ϕ

∂ ∂ ∂∇ + +

∂ ∂ ∂

⎡ ⎤∂ ∂ ∂∇ + +⎢ ⎥∂ ∂ ∂⎣ ⎦

ur

ur

PCD-08

Directional Derivative in the Spherical Polar Coordinate System

1 1ˆ ˆ ˆ=e e esinr r r rθ ϕθ θ ϕ

∂ ∂ ∂∇ + +

∂ ∂ ∂

ur

1ˆ ˆ ˆ=e e ez zρ ϕρ ρ ϕ∂ ∂ ∂

∇ + +∂ ∂ ∂

ur

ˆ ˆ ˆ=e e ex y zx y z∂ ∂ ∂

∇ + +∂ ∂ ∂

ur

s 0

ˆ

ˆ lim

, tiny

increament

, differential

increament

d uds

r drus ds

s r

ds dr

δ

ψ ψ

δδ

δ δ

→

= •∇

= =

=

=

r

uur uur

uur

uur

PCD-08

ˆ ˆx yT e e∇ = +r

T = 20 + x + y

T = 100 + x y

This picture shows the value of scalar function T;

Brighter regions have larger values of T

ˆ ˆy xT xe ye∇ = +r

T = 100 + x y

Example:

Map of a scalar field

fields -- gradient

This picture shows the value of scalar function T;

Brighter regions have larger values of T

2 2

Equipotential surf1

c00

a eT x y= =

2 2 400T x y= =

fields -- gradient

We can visualize several equipotential surfaces for different values of Te.g. countours

fields -- gradient

Example

Φ(x,y,z) = x2 +y2+z2

level surface Φ =c ( a positive constant) is a sphere of radius √c centered at the origin.

∇ Φ = 2x ex + 2y ey+ 2 zez = 2 r, along er

gradient at a point with position vector r is perpendicular to the equipotential at r

Divide gradient a point by its magnitude to get the unit vector

The shades of grey represent the intensity of the scalar field; the blacker, the more intense.The blue vectors indicate the directions of the gradient.

Example 1 Example 2 : uniform gradient

a) T = 20 + x + y

b) T= 100+xy

c) T = x2y2

Examples(calculation of gradient and visualization of the fields)

ˆ ˆx yT e e∇ = +r

ˆ ˆy xT xe ye∇ = +r

2 2ˆ ˆ2 2x yT xy e x ye∇ = +r

fields -- gradient

Example

Φ(x,y,z) = x2 +y2+z2

level surface Φ =c ( a positive constant) is a sphere of radius √c centered at the origin.

∇ Φ = 2x ex + 2y ey+ 2 zez = 2 r, along er

T = 15 + ρ cos φ

Geostationary Orbits ⊂ Geosynchronous

Orbits Geostationary Orbit:The satellite orbits the earth at exactly the same speed as the earth turns and

at zero latitude (equator). It appears at the same zenith point from the equatorial point underneath it.

Geosynchronous Orbit:The satellite’s orbit is synchronized with the earth's rotation. The plane of the

satellite’s orbit may be tilted with respect to the equatorial plane. The satellite will then appear to move around along a latitude even if it will remain at the same longitude.

Many advantages! We need detailed maps of ‘potentials’ and ‘fields’ to launch such objects in the sky.

WMAP: Wilkinson Microwave Anisotropy ProbeOrbits around the Lagrange point L2 of the Sun-Earth system.

The Lagrange points are analogous to the points where geosynchronous orbits can be launched, but with regard to a 3-body problem.

At the Lagrange points, the resultant gravitational pull of the sun and the earth on an object, such as a satellite, will precisely equal the centripetal force that is required for that object to rotate with them. At the Lagrange points, net force ~ zero!Lagrange found five such points (‘LAGRANGE POINTS’) for the sun-earth system using the PRINCIPLE OF LEAST ACTION.

SOHO

Solar and HeliosphericObservatory

WMAP

WilkinsonMicrowaveAnisotropy Probe

The L1 point of the Earth-Sun system provides uninterrupted view of the sun.

At L2: Sun, Earth, & Moon are

always behind the instrument's field of view.

http://map.gsfc.nasa.gov/mission/index.html

http://en.wikipedia.org/wiki/Lagrangian_point

http://map.gsfc.nasa.gov/media/990533/index.html

Colonizing space?

At Lagrange points?

Wilkinson Microwave Anisotropy Probe: WMAP

The average temperature is 2.725 Kelvin (degrees above absolute zero; equivalent to -270 C or -455 F), and the colors represent the tiny temperature fluctuations, as in a weather map.

Red regions are warmer and blue regions are colder by

about 0.0002 degrees.

The cosmic microwave temperature fluctuations from the 5-year WMAP satellite data seen over the full sky.

http://map.gsfc.nasa.gov/news/index.html

MEASURING COSMIC ASYMMETRY?

http://www.aerith.net/comet/catalog/1998J1/pictures.html

Roughly eighty-five percent of the SOHO discoveries, and also this one, are fragments from a once great comet that split apart in a death plunge around the Sun, probably many centuries ago. The fragments are known as the Kreutz group and now pass within 1.5 million kilometres of the Sun's surface when they return from deep space.

The Kreutz-Group comet SOHO-1500 was spotted on June 25th 2008 in images taken by the LASCO C2 coronagraph.

http://soho.nascom.nasa.gov/hotshots/2008_06_23/1500thcomet.tif

Most successful comet catcher in history. SOHO has discovered its 1500th comet, making it more successful than all the other discoverers of comets throughout history put together.

SOHO: spacecraft that was designed as a solar physics mission. SOHO's history-making discovery was made on June 25th 2008 by US-based amateur astronomer Rob Matson. This is Rob's 76th SOHO comet find.

SOlar and Heliospheric Observatory has one big advantage over everybody else: its location.

Situated between the Sun and the Earth, it has a privileged view of a region of space that can rarely be seen from Earth.

From the surface of the planet, the space inside our orbit is largely obscured because of the daytime sky and so we only clearly see close to the Sun during an eclipse.

http://www.physics.montana.edu/faculty/cornish/lagrange.html

[1] WMAP uses the moon to get a gravity assisted boost on to the L2 point. This saves a lot of fuel that can be used later to extend WMAP's operational life. This animation shows the flight path around the Moon. Credit: NASA / WMAP Science Team http://map.gsfc.nasa.gov/media/990534/990534_320.mp4

[2] WMAP Flight to L2 WMAP orbits around the L2 Lagrange point, one million miles beyond Earth. The sun shield/solar panels always protect it from the radiation generated by the Sun. This animation Shows the flight path from Earth to L2 http://map.gsfc.nasa.gov/media/990533/990533_320.mp4[3] SOHO’s orbit around the sun and around L1 http://www.youtube.com/watch?v=cGm-nrgAIzM&feature=related

We shall now show three video clips, two from NASA, and one from youtube

Understanding gradient

Electric field and potential inside a conductor

Conductors have `free electrons’

Consider charges to be placed on a conducting solid/hollow sphere

Electrostatic field inside a conductor is zero

Any net charge resides on the surface only, not inside

A conductor is an equipotential

For any two points and inside the conductor,

( ) ( ) . 0 and hence ( ) ( )b

a

a b

U a U b E dl U a U b− = − = =∫rr

The field lines just outside the conductor are perpendicularto the surface of the conductor

Ref: Introduction to Electrodynamics, Griffiths

Understanding gradient

Field and potential due to a point dipole

2 20

3

3

2 cos=

1 sin=

1 =0sin

=

=

=

Field is the negative gradient

The elctrostatic potential due to a point dipole is cos 1( ,

of potential

, ) = where 4

Thus

U kpr rU kp

r rU

r

r

k r p kprU r kr r

E

E

E

θ

ϕ

θ

θθ

θ ϕ

θθ ϕπε

∂∂

∂∂

∂∂

−

−

−

⋅= =

r r

3 ˆ ˆ( , , ) ( 2cos sin )rkpE r e er θθ ϕ θ θ= +

r

pr

θcosp θsinp θ

z

Ref: Introduction toElectrodynamics, Griffiths3

ˆ ˆsince cos siˆ ˆ3( )

nr

r rpp e p e

e e pp

E kr

θθ θ=

⋅ −=

−rr

r

r

1 1ˆ ˆ ˆU= e e esinr U

r r rθ ϕθ θ ϕ⎡ ⎤∂ ∂ ∂

∇ + +⎢ ⎥∂ ∂ ∂⎣ ⎦

ur

Understanding gradient

Electric field and potential due to an infinitely long wire2ˆ where is the uniform charge density per unit lengthkE eρλ λρ

=r

ˆ ˆ ; comparing coefficients of and

2 ; 0

he 12 ln some conce nstant (indep. of )

. . 2 l

n

E U e e

U k

ci e U

U k

k

Uρ ϕ

λρ

λ

λρ

ρ

ρ

ϕ

ρ

= −∇

∂ ∂= − =

⎧ ⎫

∂

= +⎨ ⎬⎩ ⎭

⎧ ⎫= ⎨

⎭

∂

⎬⎩

r r

1ˆ ˆ ˆU= e e ez Uzρ ϕρ ρ ϕ

⎡ ⎤∂ ∂ ∂∇ + +⎢ ⎥∂ ∂ ∂⎣ ⎦

ur

Understanding gradient

Obtaining the field for a given potential

1 0 2 2

2 sinconsider an electrostatic potential tan2

aU Ua

π ρ ϕρ

−⎧ ⎫⎛ ⎞⎪ ⎪⎜ ⎟= +⎨ ⎬

⎜ ⎟−⎪ ⎪⎝ ⎠⎩ ⎭

0

2 2

2

2 sin1

024

1 sin2

0

tan at 2

2 sin2 sin

2 cos

a

aa

U U a

aUU U aE

aUE

ϕρ

ρ

ϕρ

ϕ

π ρ

ϕϕρ ρ ρ

ϕρ

⎛ ⎞−⎜ ⎟⎝ ⎠

+

⎧ ⎫= + >>⎨ ⎬⎩ ⎭

∂= − = =

∂ ⎧ ⎫⎨ ⎬⎩ ⎭

−=

1ˆ ˆ ˆU= e e ez Uzρ ϕρ ρ ϕ

⎡ ⎤∂ ∂ ∂∇ + +⎢ ⎥∂ ∂ ∂⎣ ⎦

ur

2 4( ) 2 3U x Ax Bx= −

Plot for A=B=1

Understanding gradient: Shapes of fields and potentials, equilibria

2

2

1ˆ ˆ ˆ

. . cos 2 U= cos 2 + ( , ) 2

This implies sin 2 .

Comparing now the corresponding terms of ,

: 0 ( )

Now ;

zU U UF e e e

zUi e f z

U f

Ufwe get f f z c

U f Uand zz z z

ρ ϕρ ρ ϕρρ ϕ ϕ ϕ

ρ

ρ ϕϕ ϕ

ϕ

∂ ∂ ∂= − − −

∂ ∂ ∂

∂= ⇒

∂∂ ∂

= − +∂ ∂

− ∇∂

= ⇒ = +∂

∂ ∂ ∂= = − ⇒

∂ ∂ ∂

ur

ur

2

2 2

2

( , , ) cos 22 2

zf

zU z kρρ ϕ ϕ

−=

∴ = − +

Given: Force experienced by a particle is

ˆ ˆ ˆ( , , ) cos 2 + sin 2 zzF z e e eρ ϕρ ϕ ρ ϕ ρ ϕ= − +ur

Obtaining the potential for a given field