LAPLACES EQUATION IN SPHERICAL COORDINATES With Applications to Electrodynamics
We have seen that Laplaces equation is one of the most significant equations in physics. It is the solution to problems in a wide variety of fields including thermodynamics and electrodynamics. In your careers as physics students and scientists, you will encounter this equation in a variety of contexts. It is important to know how to solve Laplaces equation in various coordinate systems. The coordinate systems you will encounter most frequently are Cartesian, cylindrical and spherical polar. We investigated Laplaces equation in Cartesian coordinates in class and just began investigating its solution in spherical coordinates. Lets expand that discussion here. We begin with Laplaces equation: 2 V = 0 (1) We can write the Laplacian in spherical coordinates as: sin )(
2
V=1 rr
2 2 222
( r
2
V rr ) + 1 sin
(sin
V
)+ r 1
V
2
(2) where is the polar angle measured down from the north pole, and is the azimuthal angle, analogous to longitude in earth measuring coordinates. (In terms of earth measuring coordinates, the polar angle is 90 minus the latitude, often termed the co- latitude.) To make our initial calculations a little simpler, lets assume azimuthal symmetry; that means that our parameter V does not vary in the direction. In other words, V 0/ = , so we can write the Laplacian in (2) a bit more simply. Assuming azimuthal symmetry, eq. (2) becomes:
2
V=1 rr
2
( r
2
V rr ) + 1
2
sin
(sin
V
) (3) This is the form of Laplaces equation we have to solve if we want to find the electric potential in spherical coordinates. First, lets apply the method of separable variables to this equation to obtain
a general solution of Laplaces equation, and then we will use our general solution to solve a few different problems. To solve Laplaces equation in spherical coordinates, we write:
1
2V=1
rr
2
( r
2
V rr ) +
2
sin
(sin
V
0) = (4) First Step: The Trial Solution The first step in solving partial differential equations using separable variables is to assume a solution of the form: rRV = )()( (5) where R(r) is a function only of r, and () is a function only of . This means that we can set: V r = rR );()( V
= rR )()( (6) Substituting the relationships in (6) into (4) produces: 2V= )( rr
2
))(( rRr
2
+ rR )( r
2
sin
(sin
0))( = (7) If we multiply each term in (7) by r2 and then divide each term by V = R(r) (), we obtain: 2 V = rR 1 )( dr d ))(( rRr 2+ sin)( 1
d
d (sin
0))( = (8) Notice that the derivates in (8) are no longer partial derivatives. This is because the method of separable variables has produced two terms; one is solely a function of r and the other is solely a function of . Second Step: Separating Variables Equation (8) allows us to separate Laplaces equation into two separate ordinary differential equations; one being a function of r and the other a function of . As we have discussed in class, we realize that each term on the right hand side of (8) is equal to a constant. This means we can separate (8) into: 1 d rR )( dr )1())(( llrRr 2 = + and 1 sin)(
d
d (sin
))( = ll )1( + (9) We now have two different ordinary differential equations which we will solve. We realize that the product of solutions will allow us to use eq. (5) (along with appropriate boundary conditions) to determine the solution to Laplaces equation. You may wonder we we choose to write the separation constant as something as non-obvious as l(l+1).
The reason is that we have read ahead in the script, and know that by writing the separation constant in this way we will produce a well known differential equation whose solution we already know. Notice that separation constant is positive in one equation (the radial part) and negative in the other (the angular part); this is necessary so that the sum of equations is zero as required by Laplaces equation. The radial equation Lets start by solving the radial equation of eq. (9). We multiply through by R(r) and expand the derivate to find: r 2 Rd
2
dr
2
+2 r dR dr Rll 0)1( + = (10) This is a fairly simple example of a Frobenius (see http://www.luc.edu/faculty/dslavsk/courses/other/classnotes/frobeniusexample.pdf )differential equation. This is also an example of an Euler (or Cauchy) differential equation. See http://www.luc.edu/faculty/dslavsk/courses/other/classnotes/euler.pdf for more details about solving Eulers equation. Using either the method of Frobenius or methods of Eulers equations, we can find the solution to equation (10): ArrR )( = l + Br )1( l + (11) where A and B are constants which will be determined once we apply specific boundary equations. The angular equation We solve the angular portion of equation (9) by multiplying through by () and expanding the derivative to obtain: d
2
+ cos sin
dd + ll 0)1( + = (12) This is actually a differential equation you are very familiar with, although perhaps not in this exact form. Refer back to the solutions for the very first homework set of the semester, http://www.luc.edu/faculty/dslavsk/courses/phys301/homework/phys301hw1s.pdf and review again the solutions to problems 4 and 5. You will see that the equation you have derived in (12) is just the well known Legendre equation. We know that the solutions to the Legendre equation are the Legendre polynomials, P l (cos ). Third Step: Constructing the complete solution
Having separated Laplaces equation into two ordinary differential equations, we can use the results above to substitute into eq. (5) to realize that the general solution to Laplaces equation in spherical coordinates will be constructed of a sum of solutions of the form: rV (),( = Ar l + Br )1( l + )(cos) P l (13) From our experience with Laplaces equation in Cartesian coordinates, we know that the full solution will be constructed by taking a sum of solutions of the form of (13); in other words, our general solution to Laplaces equation in spherical coordinates is: ),( =
( + )1(
)(cos) (14) =
0
Now, all we need are boundary conditions to determine the values of the coefficients A l and B l rV rBrA l
l l+l Pll
. Applying Boundary Conditions First Example (Boas pp. 647-649) Lets see how we can use (14) as the starting point to determine a solution to Laplaces equation with specific boundary conditions. For this purpose, lets use the example in Boas pp. 647-649. Without any loss of meaning, we can use talk about finding the potential inside a sphere rather than the temperature inside a sphere. So, lets assume there is a sphere of radius a, and the potential of the upper half of the sphere is kept at a constant +100, and the potential of the lower half of the sphere is held at 0. How can we determine the potential at any point inside the sphere? First, lets write the boundary conditions as: V = 0,100
<
< cos02/ or < < 1 2/,0
< < or cos1 < < 0 Remember that the north pole of the sphere corresponds to =0, and = /2 in the equatorial plane. Now, lets look a bit more closely at (14). We are asked to find the potential at any point inside the sphere. This regime includes, of course, the point r=0, and we can look at (14) and realize that the solution diverges at r=0 unless B l =0. Applying the necessity for a meaningful physical solution to this problem allows us to set all coefficients B
l
to zero, so that (14) simplifies to: rV ),(
PrA l )(cos =
0 l l
(15) l
Now, we use the boundary condition for the surface of the sphere. When r = a, we know that V = 100 in the upper half sphere and V = 0 in the lower half sphere. This means we can write (15) as: rV ),(
PaA l 100)(cos cos0 1 =
0 l
l = for < < l (16) The expression in (16) should look familiar to us: we are seeking to write a function (in this case the function equals the constant 100) in terms of an infinite series. We have seen how to do this using both Fourier series and Legendre Polynomials. We know that our function can be expanded in a series if and only if we can expand that function in terms of a complete set of orthogonal functions. Fourier series are possible because sin and cos represent a complete set of orthogonal functions on (-, ); expansion in terms of Legendre polynomials is possible since we have learned that Legendre polynomials are a complete set of orthogonal functions on (-1, 1). Thus, we can expand any function f(x) on (-1, 1) as: xf )( =
xPc )( (17)
l = 0
where the coefficients, c
l ll
are determined by: c
l
= 12
l +2
1
dxxPxf )()( l
1
(18) We can see that equation (17) applies to eq. (16) with f(x) = 100, and c
l
al, or
l
=A
l
Al =cla (19) All we have to do now is determine the values of the coefficients c
l
from (18), substitute these values into (19) and then use those values of A
l
in (15) to determine the complete solution to the potential inside the sphere. We can determine several of the coefficients c
l
easily by direct integration; in fact this is done in Boas on p. 581. Using these Legendre coefficients with f(x) = 100 and substituting into (16) we obtain an explicit expansion of our solution for V(r, ): rV [100),( =12 )( a r
0P0
)(cos + 3 4 ar
P 1 )(cos 16 7 ar
3
P
3
...])(cos + (20) and you can expand the various Legendre polynomials explicitly in terms of cos if you wish, but there is really no need to go beyond the expression as it is written in (20).
Second Example Consider a sphere of radius a that has a potential on its surface given by: aV ),( = V 0 cos 2 (21) and we are asked to find the potential at points exterior to the sphere. We go back to eq. (14) and begin to apply boundary conditions. First, we realize that A
l
must go to zero since r can get very large, allowing us to simplify (14) as: rV ),(
=
PrB l )1( )(cos 0
l +l
(22)
l =
Now, we apply the boundary condition (21) and obtain: aV ),(
PaB l =
)1( l +l
)(cos = V 0 cos 2 (23)
l 0
This is coefficient just another B
l
a -(l+1)
form of eq. (17). Here, the function f(x) is V
0
cos2 , and the stands in the place of c l . So, our task now is very familiar: compute the coefficients c
l
using (18), use these to determine the values of B
l
, and substitute these values of B l into (22) to find our complete solution. Lets begin by finding the coefficients c
l
. We can set x=cos ; since varies from 0 to then x varies from -1 to 1, which is very convenient in calculating Legendre coefficients since the Legendre polynomials are a complete, orthogonal set on (-1, 1). With this substitution, we will calculate our coefficients c
l
from: c
l
= 12 l +2 1 dxxPxf )()( l 1 = 12 l +2 1 VdxxPxV
0 2
l )( 1 =
0
12 l +
1
dxxPx
2l 1
(24) The final integral on the right is pretty easy to do; Legendre polynomials are, well, polynomials, and multiplying them by x2 just produces another polynomial which is easy to integrate between these limits. But lets think a bit more and make our lives even easier. We recall that Legendre polynomials are even functions for even values of l, and are odd functions for odd values of l. This means that all c
l
2 )( for odd l vanish since the integrand in (24) becomes the product of an even function (x2) and an odd function (P
l
(x) for an odd l). This means the integrand in (24) is odd whenever l is odd, and the integral of an odd function between limits symmetric with respect to the origin vanishes. Lets compute coefficients:
1c 0
= 12 dxxPxV 0 1
1 2 0
)( =12 VdxxV 0
1 2
1 dx V
=3
0
(25) c
2
= 52 dxxPxV 0 1
1 2 2
)( =52 xV 0
1 1
You will find that all higher index coefficients vanish; does it make sense that this function is expressible in terms of only P
0 2
( 3 2x
2
12 ) =2
0
(x) and P
2
(x)? There are only two terms which will contribute to the series expansion of V, namely the l=0 and l=2 terms. We remember from before that we use our values of c
l
to find the values of Bl that substitute back into (22); eq. (23) tells us that: acBsoacB l =l3 ;
2 3 2023 3l +1 0 =01
= aV 0 acB = = aV
(26) Using these values of B l , in our general solution (22) gives us the complete answer to this problem: rV ),( = PrB 0
1 0
)(cos + PrB 2 3 2 )(cos = aV
30
)(2)[( r + (cos )] (27) Third Example Lets say now that we want to find the potential outside a sphere of radius a whose surface is held at a potential given by 0 )3cos( ar
3P
2 V . We know that since we are dealing with exterior points our solution will be of the form of eq. (22), and that we will have to find the coefficients B l . The process we follow is identical to the example immediately above, except now f(x) = V 0 )3cos( rather than V 0 cos 2
. We saw in the example above how we could simplify our calculations by realizing we could set x 2 = cos 2 ; we would like to express our current f(x) in terms of x=cos , but we will have to do a little trig and algebra manipulation to accomplish this.
Lets start by writing cos(3) = cos(2+). We now expand this as: cos2cos)2cos( + = sin2sin = (cos 2 cos)sin 2 sin2 2 cos = cos [ {cos 2 cos1( 2 cos4)cos1(2)} 2 ] = 3 cos3 = 34 xx 3 (28) where x= cos . Now it is a fairly straightforward task to find the necessary coefficients to solve our problem. We follow the example of eq. (24), now with f(x) = V
0
(4x3 3x), and solve for c
l
11=
2 )()34( 1 =32 ))(34( 1 dxxPxxVc
13 03
dxxxxV
=53V 31 0 0 1 1 1=
==1c
38
72 dxxPxxV 0 )()34(
3 3
72 xxV 0 )[34(
3
12 )35( Vdxxx 3 5 These results tell us that: 5
0
3 acB 1 =
12
= aV
05 2
; acB
3 =34
=8 aV
04
and we use the form of (24) to write the final answer as: rV ),( = PrB 0
2 1
)(cos + PrB 3 4 3 )(cos = V 5 0 )(3[ a r
2P1
)(8)(cos + a r
4
P
3
(cos )]
