lecture 19 spherical polar coordinates remember phils problems and your notes = everything come to...
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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
Polar Coordinate SystemsPolar Coordinate Systems
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
)(
Polar Coordinate SystemsPolar Coordinate Systems
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
Polar Coordinate SystemsPolar Coordinate Systems
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
Polar Coordinate SystemsPolar Coordinate Systems
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.
Polar Coordinate SystemsPolar Coordinate Systems
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)
Polar Coordinate SystemsPolar Coordinate Systems
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
Polar Coordinate SystemsPolar Coordinate Systems
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 )()()()()(
)(
(*)
Polar Coordinate SystemsPolar Coordinate Systems
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 )()()()()(
)(
Polar Coordinate SystemsPolar Coordinate Systems
tiikrikr
er
BeAetTrRtr
)()(),(
Yet again I’ve made a mistake in the notes !!!!
Polar Coordinate SystemsPolar Coordinate Systems
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.
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