1 Introduction. Green s function notes 2018

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1 Green s function notes 8 Introduction Back in the "formal" notes, we derived the potential in terms of the Green s function. Dirichlet problem: Equation (7) in "formal" notes is Φ () Z ( ) ( ) 3 Z Φ ( ) ( ) 4 4 () which is also Jackson eqn..44. The Dirichlet Green s function is symmetric in and (see Lea C.7. for the proof). ( ) ( ) R Neumann problem: Writing Φ Φ theaveragevalueof the potential over the surface equation (9) in "formal" notes is Φ () Φ 4 Z ( ) ( ) Z Φ () In order to make use of these expressions, we need a form for the Green s function that is relatively easy to integrate. That suggests that we expand the Green s function in orthogonal functions. The Dirichlet Green s function is symmetric in the two sets of coordinates and (See Lea pg 54 and J problem.4.) In the Neumann case it is possible to impose the symmetry as an additional condition. (In problems.4 and 3.7 you can see why this does not affect the computed potential). So it does not matter whether we compute ( ) or ( ) I find it easier to use as the variable and as fixed while finding to reduce the number of primes I have to write. Then the differential equation satisfied by is ("formal" notes eqn 3): ( ) 4 ( ) (3) This means that represents the potential at due to a unit point charge at multiplied by 4 This shows that the physical dimensions of are /length. The boundary conditions are Dirichlet case: ( ) for on (4) and Neumann case: ˆ 4 for on (5) where is the total area of the surface bounding the volume and, as usual, ˆ is the outward normal. In dimensions it is a line charge.

2 Division of region method The -step method for finding is outlined in Lea (pg 55). We begin by (Step ) drawing the region under consideration and (Step ) choog a coordinate system so that the boundaries are represented by constant values of one or more of the coordinates. In this method (Step 3) we place the unit point charge at an arbitrary point within the volume, and (Step 5) use this point to divide the region into two separate regions, I and II, with on the boundary between them. Then satisifies the simpler differential equation ( ) within each of regions I and II, (but not on the boundary between them). This means that we can (Step 4) make use of the eigenfunctions of Laplace s equation.. Dirichlet Green s Function in Spherical Coordinates Suppose our volume is the interior of a sphere of radius We place a unit point charge at an arbitrary (but, for the moment, fixed) point in the region with coordinates This point then divides the volume into two regions: Region I: Region II:

3 (Step 6) Then in region I the appropriate solution to Laplace s equation is X+ ( ) ( ) (6) where we have used only the functions and excluded (+) because the potential should be finite at The dependence on the coordinates and is contained in the coefficient Region II contains neither nor so we need both functions of : ( ) X+ µ + + ( ) However, here we have to deal with the boundary at where has to be zero: X+ µ ( on ) + + ( ) We make use of the orthogonality of the to argue that each term must separately equal zero: Then ( ) X+ µ + + ( ) (7) We still have two sets of unknown constants: the and But we have one more boundary to consider at (Step 7) Thefirst boundary condition we need is continuity of the potential () at ( ) X+ X+ ( ) ( ) Ã! ( ) + ( ) + ( ) ( ) Again we make use of the orthogonality of the to argue that Ã! ( ) ( ) + ( ) + for each Thus ( ) ( ) X+ X+ Ã! + ( ) + ( ) µ ( )

4 We d like to display the symmetry in and more obviously, and we can do that by relabeling ( ) (Remember that for the moment is a constant.) Then or ( ) ( ) ( ) X+ X+ X+ ( ) Ã ( ) µ + +! + ( ) + ( ) ( ) µ ( ) + + ( ) (8) where max( ) and the same expression (8) holds in both regions. We still need to find the andtodothis(step8) wemakeuseofthe differential equation (3). First we insert expression (8) for and evaluate the derivatives in and We express the delta function in terms of the spherical coordinates ug the result of Jackson Problem.. µ + ang X+ X X+ X+ ( ) µ + µ ( ) + + ( ) + ( ) µ ( ) + + ( +) ( ) 4 ( ) ( ) Now we multiply both sides of the equation by ( ) and integrate over the whole solid angle of the sphere. On the LHS we use the orthogonality of the OntheRHSweusethesiftingproperty. Ã!# " + ( ) ( ) Ã! + + ( +) + 4 ( ) (9) Now we can drop the primes on and Thus each contains a factor 4

5 and eqn (8) becomes ( ) X+ µ ( ) + + ( ) Nowwehavesymmetryinallthecoordinates,andtheremainingsetofconstants should be independent of all coordinates. The remaining equation (9) takes the form ( ) µ + ( ) µ ( +) 4 ( ) (Step 9) To make use of this equation we multiply both sides by,thenintegrate across the internal boundary from to +. Z 4 + Z + ½ ( ) µ + ( ) µ + ( ) + + ¾ ( +) The integral on the RHS equals. The second term on the LHS as because the integrand is continuous by construction and has no gularities in the range of integration. Thus we are left with lim ( ) µ At the upper limit, + is so At the lower limit, is so Thus ( Ã ) ( ) + + "!# + ( ) + 4 " Ã!# ( ) + ( ) +( +) + ( ) + ( ) + ( ) + " # ( ) + ( +) ( + ) ( ) is independent of all coordinates, as expected, and thus ( ) X+ 4 ( ) µ ( ) () 5

6 (Step ) has dimensions of /length, as required, and displays the necessary symmetry. It is also positive, ce throughout the region. In the limit we get X+ X+ X+ 4 ( ) ( ) ( ) ( ) ( ) which is the correct result for a point charge in infinite space.. Use of the result Suppose we have a sphere of radius with potential on one half and on the other half. Inside the sphere, a line charge of length ( )with uniform line charge density runs along the diameter of the dividing plane and is centered at the center of the sphere. Find the potential inside. Firstwehavetochoosecoordinates. Nomatterwhatwedo,wedonothave azimuthal symmetry. I m going to put the polar axis along the line charge, givingachargedensity () [ ( ) + ( +)] ( ) We can check this by finding the total charge on a differential piece of the line that runs from to + with Then Z Z + [ ( ) + ( +)] () X (Do you understand the factor of? Draw a picture and see!) The potential on the surface is then Thus the potential is (eqn ) Φ + if if Φ () 4 4 Z Z + Z Z Z + ( ) ( ) [ ( ) + ( +)] ( ) ( ) Φ ( ) ( ) () 6

7 Now the field point is fixed and is variable. The first integral gives potential Φ where 4 Φ () Z Z + Z X+ 4 ( ) µ ( ) ( ) [ ( ) + ( +)] ( ) ( ) By orthogonality of the the integral over gives zero unless in whichcaseweget so we have Z + Z 4 ( ) µ () ( )[( ) + ( +)] ( ) where p ( +)4 Next we use the sifting property of the deltafunctions to do the integral over Z ( ) µ + + ()[ () + ( )] ( ) + Since ( ) ( ) the result is zero unless is even, when the square bracket equals. Finally we do the integral over The integrand is zero for so we have 4 Φ () Z µ + ( ) + + () () even Looking at the integral over by itself, we note that if we have to split the integral into two parts: and. Z µ ( ) + Z µ + ( ) + Z Ã! + + ( ) + ( ) + µ µ µ ( +) ( +) + + for + Note that this fails for because of the in the denominator. For we get Z µ Z ³ + Z + ln ( ) ³ + ln 7

8 Including the factor of 4 the first term in the potential for is ½ µ Φ () + ln ()+ 4 + ( +) + ( +) + + () + even ½µ +ln (3) µ ( +) + ( +) () even The log term in (3) is expected near a line charge, and even indicates the reflection symmetry about the equatorial ( )plane. If then throughout the range of integration, and the integral over is Z µ µ ( ) for and the potential (first term) is Φ () + µ () (4) even Separating out the term, we have Φ () µ 4 + even µ () The first term is the potential due to a point charge at the center of a grounded sphere of radius Again this is what we would expect. Finally we note that the potential Φ is continuous at We still need to evaluate the second term in (). The outward normal from thevolumeatthesurface is ˆ +ˆ so ˆ Thus we have Z Z + 4Φ () Φ ( ) ( ) µz Z Z + X+ ( ) 8 4 ( ) µ

9 On the outer surface, so and we have X+ µz 4 Z Z + + ( ) à ( ) + Doing the integral over first, we get zero if and for 6 µz Z! + ( ) + ( ( ) ) So we get zero for even and 4 for odd. Since the sum over now starts at Thus the integral has reduced to X+, odd 6 + ( ) à + + ( ) + ( )! Z + ( ) () is an even function of if + is even and an odd function if + is odd. Thus for the integral over to be non-zero, we need + to be even, and thus odd. Let s call the integral Thederivativeis à + + ( ) + ( )! ( +) + + ( +) + and thus we have 4Φ () X+ odd, odd ³ 6 ( +) + ( ) Finally we should tidy this up by combining the positive and negative terms. Remember that ( ) and so ( ) That makes ( ) Thus ( ) + () + () () 9

10 and so the second term in the potential is Φ () X+ odd, odd X+ odd, odd X+ odd, odd ³ () ³ + ( )! 4 ( + )! ³ () µ + ( )! ( + )! ()(5) The total potential is the sum of the two terms (3 or 4 and 5): µ Φ () +ln + + X+ odd, odd even " Ã ( +) ( +) + + µ!# + ( +) ( )! ³ ( + )! ()... () and Φ () + µ + X+ odd, odd even µ () ( +) ( )! ³ ( + )! ()... Compare the second term, in method and result, with pages in Lea. Let s evaluate the first few terms in our result. Thus for Φ () Z ½µ 5 +ln + ( ) ³ ( ) + ( µ +ln ¾ µ3 3 3cos ) 3

11 and for µ Φ () + 3 µ ½ + µ 5 3cos ¾ + 3 Lea 8.67 gives the value of for larger odd values of and Dirichlet Green s function in Cylindrical coordinates Here we will find the Green s function for the interior of an infinitely long tube of radius Because we are going to use the result to find a potential that does depend on, we need the three-dimensional Green s function. (Step : see Lea Figure C.6) (Step ) We choose cylindrical coordinates with -axis along the axis of the tube, and (Step 3) place our unit point charge at with (fixed) coordinates ( ) (Step 4) The solutions of Laplace s equation in this coordinate system are or ½ () () ½ () () ¾ ± ¾½ cos The sets of functions { times () or ()} and { times or cos } form complete orthogonal sets of functions, so we may divide our space in either or (Step 5) Let s divide in (The other choice is in Lea pg 5-55.) (Step 6) TheninregionI, we need a function that is finite at which is Here we have no boundaries at finite so we have no way to determine specific values for So we have ( ) + ¾ () () (6) In region II, we have both modified Bessel functions. ( ) At so + + ()[ ()+ ()] ()[ ()+ ()] By orthogonality of the in and we can equate each term separately to zero, so we get () ()

12 and thus ( ) + () () () () () (Step 7) Now we use continuity of the potential at to get + + () ( ) () ( ) () () ( ) (7) and again by orthogonality we have () () () ( ) ( ) () Thus (6) becomes + ( ) () () ( ) ( () ) () (8) We may combine the two results (8) and (7) to get where and ( ) + () ( ) (9) () ( ) ( () () ) () () () () () () (Step 8) Next we make use of the differential equation, expresg the delta function on the right in cylindrical coordinates (cf Problem.): ()+ + 4 ( ) ( ) + ½ () () 4 ( ) ( ) ¾

13 Now we multiply both sides by and integrate from to in On the RHS we use the sifting property, and on the left hand side, we use orthogonality of the Z ( ) () () ( ) 4 ( ) ( ) Now we can drop the prime on Similarly we multiply by and integrate over all We use Lea eqn 6.6 to get ½ ¾ () ( ) () ( ) ½ ¾ ( ) () ( ) ( ) Drop the primes on the and relabel, remembering that and are fixed for the moment. () () (Step 9) Finally we multiply both sides by and integrate across the boundary at Z + ½ ¾ Z () () + ( ) Making use of the fact that and are well behaved in the range of integration, we get lim () () + AttheupperlimitweareinregionIIwhere (), and at the lower limit we are in region I where (), so ½ () ( ) () () ( ) () ( ¾ ) ( ( ) ) () ½ () () () ( )+ () ( ) ( ¾ ) () ( ) Thus () () ( ) ()[ ( ) ( ) ( ) ( )] The denominator is the Wronskian ( ) of the modified Bessel differential equation. To evaluate it, let s use the large argument form of the functions. We can do this because () isthesamefunctionforall. (Also see ) r () ' ; () ' 3

14 () () () () () r µ 3 r r µ 3 () Thus and then (8) becomes () () ( ) () ( ) + for and (7) ( ) () ( ) ( ) () () ( ) ( ) () () ( ) ( ) [ () ( ) () ( )] () () ( ) for or, in both cases, (9) + () ( ) ( ) () () () () () ( ) ( ) () ( ) [ () () () ()] ( ) ( ) ( ) () [ () ( ) () ( )] ( ) ( ) (3) (Step ) The Green s function has dimensions /length (from the integral over ) as expected. It also displays the necessary symmetry. As () and so + ( ) ( ) ( ) ( ) (Jackson 3.48), as required. Now let s solve the same problem that is in the math phys book (pg 54) with on the wall between and + and the rest of the cylinder 4

15 grounded. The outward normal is ˆ +ˆ and at so Φ () Z Φ ( ) ( ) 4 4 Z Z Z + + () () ( ) ( ) [ () ( ) () ( )] The integral over gives zero unless expected. so Φ is independent of as Φ () 4 Z + Z () () ( ) [ () () () ()] Here again we have the Wronskian of the modified Bessel functions () (), so Φ () + Z + Z Z + () () Z + Z + Z + () () ( ) ( ) + () () () () () () cos The result is dimensionally correct. Since () () and so the integral converges. Let s see how this looks for Φ ( ) Z + Z Z Z + cos ( + )+ ( ) if (from the second delta function) if (from the first delta function) otherwise So we get back the correct boundary values. 5

16 Comparing with the result in the book (pg 55), Φ () X ( ) h ³ cosh ( ) we see that the main difference is that we have an integral instead of an infinite sum. We can evaluate the integral numerically. For we have R : + () cos() () R 3 : + (3) cos () R 5 : + (5) cos () 8 58 R 7 : + (7) cos () R 9 : + (9) cos () R For we have + () cos() () Compare with Lea Figure C.7. The results are the same. 3 Expansion in eigenfunctions without division. This method is outlined in Jackson section 3. and Lea sections C. and C.7.5. The result is a gle function, valid everywhere in the region, at the expense of an extra sum. We ll begin with a one-dimensional problem, and then extend the result to three dimensions. We start with the Sturm-Liouville equation µ () + () (4) valid for where we either have boundary conditions (Dirichlet or Neumann) that guarantee orthogonality of the solutions, or else () on the boundary. The solutions are the eigenfunctions () with eigenvalues Let s normalize the eigenfunctions (as we did with the )sothattheysatisfy the orthonormality relation Z i () () () (5) Now we look for a Green s functions that satisfies the differential equation µ ( ) () ( )+ () ( ) 4 ( ) (6) where the constant does not equal for any Since is expected to be well-behaved, we may expand ( ) in the eigenfunctions () : ( ) X ( ) () 6

17 and substitute into the differential equation (6). X ( ) µ () () X 4 ( ) ( ) ()+ () X ( ) () We use the eigenfunction equation (4) to express the first two terms in terms of : X ( )[ () ()] + () X ( ) () 4 ( ) Now we multiply both sides by () and integrate over the range to We use orthonormality (eqn 5) on the LHS and the sifting property on the RHS. ( ) ( ) 4 ( ) Thus ( ) 4 ( ) ( ) Now it is clear why cannot equal any Then ( )4 X ( ) () (7) To extend this result to potential problems in 3-d, we start with equation (3). The eigenvalue in this equation is zero, so we need eigenfunctions that satisfyanequationwithnon-zero and that is the Helmholtz equation. ()+ () (We have solved this equation before: See e.g waveguide notes pages 8 and 9.) Let s find the Dirichlet Green s function for the interior of a rectangular box measuring by by We put the origin at one corner, with Cartesian axes along the three sides, and solve by separation of variables: ( ) () () (). Then the eigenfunctions we want are the solutions of with () () () () and () () and and µ + + We still need to normalize the eigenfunctions. The orthogonality integral is Z 7 Then

18 so the normalized function is () r and so r r r ( ) Now we just stuff this result into (7) (extended to 3-d) with ( ) 4 3 Ãr! 8 X + ³ Again check that has dimensions of [/length]. The benefit of this approach is that we have a gle expression that we can use in the whole region. The disadvantage is that we have an extra sum. Suppose the box, with its walls remaining grounded, contains a sheet of charge with uniform surface charge density that extends from 4 to 34 and 4 to 34 at Find the potential inside the box. In this case we have the volume integral in (), but the surface integral is zero. The charge density is ³ () 4 34, 4 34 otherwise So Φ () 4 4 Z Z Z Z 34 Z 34 Z ( ) ( ) 3 ³ + Z Z

19 The result is zero if is even, so Φ () odd odd ( ) ( ) ( ) ( ) + cos ³ µ cos 3 4 cos 4 + cos 34 4 µ cos 3 cos 4 4 Now we combine the coes to get Φ () odd odd odd ( ) ( ) ³ + + ( ) ( )+( )+( ) ³ odd If + then µ ( +) cos 4 +cos 4 µ +cos ( )( ) if is odd ( ) if is even Thus replacing (odd) with +and similarly for and we get Φ () 6 4 where Check the dimensions! ³ (+) ( ) (+) ( +) + (+) ³ ³ integer part + integer part ( +) ( +)

20 At 4 we have Φ ( 4) 6 ( 4 ) ( ) +++ h ( +)( +) ( +) + (+) + (+) ( +) ( +) ( +) 4 The diagram shows the dimensionless potential (8), computed ug values of and from zero to five, and with i (8).5. z x.4 y.6.8