Solving Laplace's Equation in Spherical Coordinates: Electric Potential & Applications, Summaries of Differential Equations

Download Solving Laplace's Equation in Spherical Coordinates: Electric Potential & Applications and more Summaries Differential Equations in PDF only on Docsity! LAPLACE’S EQUATION IN SPHERICAL COORDINATES With Applications to Electrodynamics We have seen that Laplace’s 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 Laplace’s equation in various coordinate systems. The coordinate systems you will encounter most frequently are Cartesian, cylindrical and spherical polar. We investigated Laplace’s equation in Cartesian coordinates in class and just began investigating its solution in spherical coordinates. Let’s expand that discussion here. We begin with Laplace’s equation: 02 =∇ V (1) We can write the Laplacian in spherical coordinates as: )( sin 1)(sin sin 1)(1 2 2 222 2 2 2 φθθ θ θθ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ =∇ V r V rr Vr rr V (2) where θ is the polar angle measured down from the north pole, and φ 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, let’s assume azimuthal symmetry; that means that our parameter V does not vary in the φ direction. In other words, 0/ =∂∂ φV , so we can write the Laplacian in (2) a bit more simply. Assuming azimuthal symmetry, eq. (2) becomes: )(sin sin 1)(1 2 2 2 2 θ θ θθ ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ =∇ V rr Vr rr V (3) This is the form of Laplace’s equation we have to solve if we want to find the electric potential in spherical coordinates. First, let’s apply the method of separable variables to this equation to obtain a general solution of Laplace’s equation, and then we will use our general solution to solve a few different problems. To solve Laplace’s equation in spherical coordinates, we write: 0)(sin sin 1)(1 2 2 2 2 = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ =∇ θ θ θθ V rr Vr rr V (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: )()( θΘ= rRV (5) where R(r) is a function only of r, and Θ(θ) is a function only of θ. This means that we can set: )()();()( θ θ θ Θ′= ∂ ∂ Θ′= ∂ ∂ rRVrR r V (6) Substituting the relationships in (6) into (4) produces: 0))((sin sin )())(()( 2 2 2 2 =Θ′ ∂ ∂ +′ ∂ ∂Θ =∇ θθ θθ θ r rRrRr rr V (7) If we multiply each term in (7) by r2 and then divide each term by V = R(r) Θ(θ), we obtain: 0))((sin sin)( 1))(( )( 1 22 =Θ′ Θ +′=∇ θθ θθθ d drRr dr d rR V (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 θ. Second Step: Separating Variables Equation (8) allows us to separate Laplace’s equation into two separate ordinary differential equations; one being a function of r and the other a function of θ. 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())((sin sin)( 1)1())(( )( 1 2 +−=Θ′ Θ +=′ ll d dandllrRr dr d rR θθ θθθ (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 Laplace’s equation. You may wonder we we choose to write the separation constant as something as non-obvious as l(l+1). 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: 1cos0100)(cos),( 0 <<== ∑ ∞ = θθθ forPaArV l l l 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 (-π, π); 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: ∑ ∞ = = 0 )()( l ll xPcxf (17) where the coefficients, cl are determined by: ∫ − + = 1 1 )()( 2 12 dxxPxflc ll (18) We can see that equation (17) applies to eq. (16) with f(x) = 100, and cl = Al al, or l l l a c A = (19) All we have to do now is determine the values of the coefficients cl from (18), substitute these values into (19) and then use those values of Al in (15) to determine the complete solution to the potential inside the sphere. We can determine several of the coefficients cl 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, θ): ...])(cos 16 7)(cos 4 3)(cos)( 2 1[100),( 3 3 10 0 +⎟ ⎠ ⎞ ⎜ ⎝ ⎛−+= θθθθ P a rP a rP a rrV (20) and you can expand the various Legendre polynomials explicitly in terms of cosθ 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: θθ 2 0 cos),( VaV = (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 Al must go to zero since r can get very large, allowing us to simplify (14) as: )(cos),( )1( 0 θθ l l l l PrBrV +− ∞ = ∑= (22) Now, we apply the boundary condition (21) and obtain: ∑ ∞ = +− == 0 2 0 )1( cos)(cos),( l l l l VPaBaV θθθ (23) This is just another form of eq. (17). Here, the function f(x) is V0cos2 θ, and the coefficient BBl a stands in the place of c-(l+1) l. So, our task now is very familiar: compute the coefficients cl using (18), use these to determine the values of BlB , and substitute these values of BBl into (22) to find our complete solution. Let’s begin by finding the coefficients cl. We can set x=cos θ; since θ varies from 0 to π¸ x then 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 cl from: ∫ ∫ ∫ − − − + ⋅= + = + = 1 1 1 1 1 1 2 0 2 0 )( 2 12)( 2 12)()( 2 12 dxxPxlVdxxPxVldxxPxflc llll (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 let’s 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 cl for odd l vanish since the integrand in (24) becomes the product of an even function (x2) and an odd function (Pl(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. Let’s compute coefficients: ∫ ∫ ∫ ∫ − − − − =−== === 1 1 1 1 0 2 2 02 2 02 1 1 1 1 0 2 00 2 00 3 2 ) 2 1 2 3( 2 5)( 2 5 3 1 2 1)( 2 1 V dxxxVdxxPxVc VdxxVdxxPxVc (25) You will find that all higher index coefficients vanish; does it make sense that this function is expressible in terms of only P0(x) and P2(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 cl to find the values of BBl that substitute back into (22); eq. (23) tells us that: 3 2 ; 3 , 3 03 22 01 00 1 aV acB aV acBsoacB l ll ===== + (26) Using these values of BBl in our general solution (22) gives us the complete answer to this problem: )](cos)(2)[( 3 )(cos)(cos),( 2 30 2 3 20 1 0 θθθθ P r a r aV PrBPrBrV +=+= −− (27) Third Example Let’s say now that we want to find the potential outside a sphere of radius a whose surface is held at a potential given by )3cos(0 θV . 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 BBl . The process we follow is identical to the example immediately above, except now f(x) = )3cos(0 θV rather than . We saw in the example above how we could simplify our calculations by realizing we could set x = cos θ; we would like to express our current f(x) in terms of x=cos θ, but we will have to do a little trig and algebra manipulation to accomplish this. θ0 cosV 2 2 2 Let’s start by writing cos(3θ) = cos(2θ+θ). We now expand this as: θθθθθθθθθθθ cossin2cos)sin(cossin2sincos2cos)2cos( 222 −−=−=+ = [ ] θθθθθθ cos3cos4)cos1(2)}cos1({coscos 3222 −=−−−− = (28) xx 34 3 − where x= cos θ. 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) = V0(4x3 – 3x), and solve for cl: