lecture 19 spherical polar coordinates remember phils problems and your notes = everything come to...

Lecture 19Lecture 19

Spherical Polar CoordinatesSpherical Polar Coordinates

http://www.hep.shef.ac.uk/Phil/PHY226.htmRemember Phils Problems and your notes = everything

Come to see me before the end of term• I’ve put more sample questions and answers in Phils Problems• Past exam papers•Have a look at homework 2 (due in on 15/12/08)

Hints for Homework 2Hints for Homework 2

You know that

mxmx meedx

d

mxmx meedx

d and that

mxmx meedx

d33

and thatmxmx mee

dx

d

33

So if you had

mxedx

d3 you must write mxmx mee

dx

d

33

In questions that ask you to prove an expression is a solution of an equation, simply stick the expression into the equation starting at the heart of the equation working outwards and show that the LHS = RHS.

(see Q25 of tutorial questions and corresponding answer at Phils Problems)

Polar Coordinate SystemsPolar Coordinate Systems

1. Spherical Polar Coordinates

Spherical polars are the coordinate system of choice in almost all 3D problems. This is because most 3D objects are shaped more like spheres than cubes, e.g. atoms, nuclei, planets, etc. And many potentials (Coulomb, gravitational, etc.) depend on radius.

Physicists define r, as shown in the figure. They are related to Cartesian coordinates by:

sin cos , sin sin , cosx r y r z r

. 222 zyxr

2. 3D Integrals in Spherical Polars2 sindV r drd d

.20,0,0 r

The volume element is (given on data sheet).

To cover over all space, we take

Example 1 Show that a sphere of radius R has volume 4R3/3.So

R

sphere

drrdddddrrdVV0

2

0

2

0

2

sinsin 3

4

3

3

0

3

020

RrR

cos


Example 2 Find the Fourier transform of a screened Coulomb potential,

As before we have the 3D Fourier transform spaceall

i dVerfF k.r232

1k )(

)()(

/

In this case f(r) is a function only of the magnitude of r and not its direction and so has perfect radial symmetry.

2 sindV r drd d Again the volume element is

We therefore have spaceall

i ddrdrerfF

sin)()(

)(/

2k.r232

1k

There is a standard ‘trick’ which is to chose the direction of k to be parallel to the polar

(z) axis for the integral. Then k.r becomes . cosrkk.r

Now clearly the whole integral is a function only of the magnitude of k, not its direction, i.e. F(k) becomes F(k):

spaceall

ikrr

dddrrer

ekF

sin4)2(

1)( 2cos

02/3

r

erf

r

04

)(


We therefore write

The integral over is trivial: it just gives a factor of 2.

But note that the factor involves r and . Which integral should we do next?

spaceall

ikrr

dddrrer

ekF

sin4)2(

1)( 2cos

02/3

00

2

0023 4

1

2

1sin

)()( cos

/ikrr edredrdkFso

cosikre

The presence of the together with the makes integration by

substitution the obvious choice:

cosikresin

dikrdAsoikrA sincos

sinsinsin cos

ikr

dAede Aikr

)(sin

sincos krkr

krkri

ikree

ikre

ikrdA

ikr

e ikrikrikrA

sinc22

2111

0

let Rewrite

So


kr

krredrkF r sin2

24

1

)2(

1)(

002/3

00

23

1

2

1drekr

kkF r

)(sin

)()(

/

We are then left with the integral over r:

)(sinc2sin2

sin2111

sin0

0coscos kr

kr

krkri

ikree

ikre

ikrde ikrikrikrikr

From previous page:

This type of integral was met earlier in the tutorial question exercises on Fourier transforms. It’s best to write the sine in terms of complex exponentials:

22

0

)()(

0 0

11

2

1

2

1

2

1)(sin

k

k

ikikidree

idreee

idrekr ikrikrrikrikrr

This gives the final result: )(

1

)2(

1)(

220

2/3 kkF


3. 2 in Spherical Polars: Spherical Solutions

As given on the data sheet, 2

2 22 2 2 2 2

1 1 1sin

sin sinr

r r r r r

(Spherically symmetric’ means that V is a function of r but not of or .)

Example 3 Find spherically symmetric solutions of Laplace’s Equation 2V(r) = 0.

Therefore we can say 0)(1

)( 22

2

rV

dr

dr

dr

d

rrV

Really useful bit!!!!

If (as in the homework) we were given an expression for V(r) and had to prove that it was a solution to the Laplace equation, then we’d just stick it here and start working outwards until we found the LHS was zero.

If on the other hand we have to find V(r) then we have to integrate out the expression.


0)(1

)( 22

2

rV

dr

dr

dr

d

rrV

0)(2

rV

dr

dr

dr

d

ArVdr

dr )(2

2)(r

ArV

dr

d

Multiplying both sides by r2 gives

Integrating both sides gives where A is a constant.

This rearranges to and so ….

Integrating we get the general solution: Br

ArV )(

.

We’ve just done Q3(i) of the homework backwards!!! (see earlier note in red)


4. The Wave Equation in polar coordinates

Let’s only look for spherically symmetric solutions (r,t), so the equation can be written

22 2

2 2 2

1 ( , ) 1 ( , )( , )

r t r tr t r

r r r c t

The wave equation is 2

2

22 1

tc

( , ) ( ) ( )r t R r T t

2

2

22

2

)()(1)()(1

dt

tTdrR

cdr

rdRtTr

dr

d

r

2

2

22

2

)(

)(

1)(

)(

1

dt

tTd

tTcdr

rdRr

dr

d

rrR

As usual we look for solutions of the form

As usual substitute this back in

As usual separate the variables

We equate both sides to a constant and since we expect LHO solutions this is -ve

2c

22

2

1

cdr

rdRr

dr

d

rrR

)(

)(

2

2

2

2

1

cdt

tTd

tTc

)(

)(

To make maths easier let this be


22

2

1

cdr

rdRr

dr

d

rrR

)(

)(

2

2

2

2

1

cdt

tTd

tTc

)(

)(

)()(

tTdt

tTd 22

2

( ) ~ i tT t e The equation for T(t) is easy to solve giving

)()(

rRcdr

rdRr

dr

d

r

22

2

1

Now we need to solve

There is a standard trick which is to define ( )

( )u r

R rr

, solve for u(r) and thus find R(r).

This is tricky to solve …..

Start by differentiating R(r) with respect to r using the product rule. 2

1)(

)(1

rru

dr

rdu

rdr

dR

Multiply both sides of the expression above by r2 gives )()(2 ru

dr

rdur

dr

dRr

Now differentiate again. 2

2

2

2

dr

rudr

dr

rdu

dr

rdu

dr

rudrru

dr

rdur

dr

d )()()()()(

)(

(*)


Therefore 2

22

2

)(1)(1

dr

rud

rdr

rdRr

dr

d

r

)()(

rRcdr

rdRr

dr

d

r

22

2

1

Remember we originally needed to solve

r

ru

cdr

rud

r

)()(2

2

21

So equation (*) becomes:

(*)

so )()()(

rukrucdr

rud 22

2

2

ikrikr BeAeru )(

r

Be

r

AerR

ikrikr

)(

tiikrikr

er

BeAetTrRtr

)()(),(

Thus we have solutions of the form:

( )( )

u rR r

rRemember and so

And from the start of this example

2

2

2

2

dr

rudr

dr

rdu

dr

rdu

dr

rudrru

dr

rdur

dr

d )()()()()(

)(


tiikrikr

er

BeAetTrRtr

)()(),(

Yet again I’ve made a mistake in the notes !!!!


These are spherical waves moving in and out from the origin.

tiikrikr

er

BeAetTrRtr

)()(),(

Note the factor of 1/r. Intensity is related to amplitude squared.

Our solution gives intensity as

2

22

r

Ae

r

Aee

r

Aetr ti

ikrti

ikr

),(

For waves moving out from the central point (origin). ti

ikr

er

AetTrRtr )()(),(

This is the well known inverse square law.

lecture 19 spherical polar coordinates remember phils problems and your notes = everything come to...

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