UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 7 Prof. Steven Errede LECTURE NOTES 7

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1 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede LECTURE NOTES 7 LAPLACE S EQUATION As we have seen in previous lectures, very often the priary task in an electrostatics proble is e.g. to deterine the electric field E( r) of a given stationary/static charge distribution e.g. via Coulob s Law: Charge density ρ ( r ) ẑ Field Point Source r = r r P Point(s) S r v Volue r eleent dτ Ο ŷ in volue v rˆ E( r) = ρ ( r ) dτ πε r 4 o v ˆx r r r r = r r rˆ = = r r r r = r r = x x + y y + z z is coplicated, and analytic calculation of E( r) ( p s) ( p s) ( p s) Oftenties ρ ( r ) is painful / tedious (or just plain hard). (Nuerical integration on a coputer is likely faster/easier... ) Oftenties it is easier to first calculate the potential V ( r ), and then use E( r) = V ( r) Here: V ( r) = ρ ( r ) dτ 4πε o v r But even doing this integral analytically often can be very challenging... ρ r ay not apriori (i.e. beforehand) Furtherore, often in probles involving conductors, be known! Charge is free to ove around, and often only the total free charge Q free is controlled / known in the proble. In such cases, it is usually better to recast the proble in DIFFERENTIAL for, using Poisson s equation: ρ ( r ) ie( r) = i V ( r) = V ( r) = ε o ρ ( r ) Or: V ( r) = Poisson s Equation ε o 5-8. All rights reserved.

2 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Poisson s equation, together with the boundary conditions associated with the value(s) allowed V r e.g. on various conducting surfaces, or at r =, etc. enables one to uniquely deterine for V ( r ) (we ll see how / why shortly...). The Poisson equation is an inhoogeneous second-order differential equation its solution consists of a particular solution for the inhoogeneous ter (RHS of Poisson s Equation) plus the general solution for the hoogeneous second-order differential equation: V r = Laplace s Equation coensurate with the boundary conditions for the specific proble at hand. Very often, in fact, we are interested in finding the potential containing no electric charge, i.e. where ρ ( r ) =. If ρ ( r ) =, then V ( r ) useless! = V r and the TRIVIAL solution is V ( r) = r, We seek physically eaningful / non-trivial solutions V ( r) boundary conditions on V ( r ) for a given physical proble. in a charge-free region, which is boring / that satisfy V ( r) = and the Now, before we go any further on this discussion, let s back up a bit and take a (very) broad generalized MATHEMATICAL view (or approach) to find V ( r ). First, let s siplify the discussion, by talking about one-diensional probles: If ρ ( x) =, Laplace s Equation in one-diension becoes (in rectangular/cartesian coordinates): V r = dv x dx Integrating this equation (both sides) once, we have: = Note the total (not partial) derivative with regards to x. d dv dv x dv x dv x st dx = dx = d = = dx = = constant of integration d x dx dx dx dx dv ( x) Then: dx = dx = dx dx nd Or: dv ( x) = V ( x) = x + b constant of integration So: V( x) = b+ x (equation for a straight line) is the general solution for y-intercept slope dv x =. dx 5-8. All rights reserved.

3 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Depending on the boundary conditions for the proble, e.g. suppose V ( x= 5) = Volts and V( x= ) = 4Volts, then together, these two boundary conditions uniquely specify what b and ust be we have two equations, and two unknowns ( & b) solve siultaneously: V ( x) = b+ x equation for a straight line y-intercept slope 5 5 V x= 5 = = b+ 5 b= 5 V x= = 4 = b+ 4 = 5+ = 4 V x = x or: = and b = 5. V x = xis the equation of a straight line for this proble. 5 b = 5 is y-intercept V(x) 4 3 slope = x x a x x+a 5-8. All rights reserved. 3

4 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede General features of -D Laplace s Equation V ( x) = and potential V ( x ) :. Fro above one-diensional case V ( x) = b+ x (general solution = straight line eqn.) we can see that: V ( x) is the average of V ( x+ a) and V ( x a) i.e. V ( x) = { V ( x+ a) + V ( x a) } Laplace s Equation is a kind of averaging instruction The solutions of V ( x ) are as boring as possible, but fit the endpoints (boundary conditions) properly. This ay be obvious in one-diension, but it is also true / also holds in -D and 3-D cases of V r =.. V ( r) i.e. V ( r ) tolerates / allows NO local axia or inia extrea ust occur at endpoints = requires the second spatial derivative(s) of V ( r ) to be zero. - Not a proof, because e.g. fcns ( x) where the second derivative vanishes other than at 4 endpoints - e.g. f ( x) = x (has a iniu at x = ). Laplace s Equation in Two Diensions (in Rectangular/Cartesian Coordinates) If V V ( x, y) = then V V x y V = + = n.b. now have partial derivatives of V ( r ). Because V = = + V ( x, y) now contains partial derivatives, the general solution x y does not contain just two arbitrary constants or any finite nuber - an infinite nuber of possible solutions (in general) the ost general solution is a linear cobination of haronic functions (sine and cosine functions of x and y in rectangular coordinates and other functions (Bessel Functions) in cylindrical coordinates). Nevertheless, V ( x, y ) will still wind up being the average value of V around a point ( x, y ) within a circle of radius R centered on the point ( x, y ) All rights reserved.

5 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede The Method of Relaxation - Iterative Coputer Algorith for Finding V ( x, y ): = π R (, ) V x y circle C of radius R centered on (x,y) V r dl - Start with V ( x, y) as specified on boundary (fixed) - Choose reasonable interpolated values of V ( x, y ) (fro boundary) on interior ( x, y ) points away fro the boundaries., - st pass reassigns V ( x y ) = average value at interior point (, ) neighbors. - nd pass repeats this process rd pass repeats this process... - etc.. After few iterations, V ( x, y ) of n th iteration settles down, e.g. when: x y of its nearest iteration n iteration n (, ) (, ) (, ) V x y = V x y V x y n n tolerance then QUIT iterating, V ( x, y ) is deterined after n th iteration is good enough. V ( x, y) again will have no local axia or inia all extrea will occur on boundaries. V ( x, y) = has solution V ( x, y ) which is the ost featureless function as sooth as possible All rights reserved. 5

6 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Laplace s Equation in Three Diensions Can t draw this on -D sheet of paper (because now this is a 4 diensional proble!), but: (,, ) V ( r) V x y z = = average value of V over a spherical surface of radius R centered on r. i.e. V ( r) Again V ( r ) = Vda 4π R sphere at r of radius R will have no local axia or inia - all extrea ust occur at boundaries of proble (see work-through proof in Griffiths, p. 4) - The average potential produced by a collection of charges, averaged over a sphere of radius R is equal to the value of the potential at the center of that sphere! Boundary Conditions on the Potential V ( r ) Dirichlet Boundary Conditions on V ( r ) : V ( r ) itself is specified (soewhere) on the boundary - i.e. the value of V ( r ) (soewhere) on the boundary. is specified Neuann Boundary Conditions on V ( r ) : The noral derivative of V ( r ) V( r ) nˆ = E ( r ) is specified soewhere on the boundary - i.e. i is specified soewhere on the boundary All rights reserved.

7 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Uniqueness Theore(s): Suppose we have two solutions of Laplace s equation, V( r) and V( r) sae boundary condition(s), i.e. the potentials V ( r) and V ( r), each satisfying the are specified on the boundaries. We assert that the two solutions can at ost differ by a constant. (n.b. Only differences in the V r are iportant / physically eaningful!) scalar potential Proof: Consider a closed region of space with volue v which is exterior to n charged conducting surfaces S, S, S 3... S n that are responsible for generating the potential V. The volue v is bounded (outside) by the surface S. S Charged Conducting Surfaces S S 3 S 4 vs, closed region of volue v, bounded by enclosing surface S. = Suppose we have two solutions V (r) and V (r) both satisfying V ( r) and V ( r ) = in the charge-free region(s) of the volue v. = i.e. V ( r) V (r) and V (r) satisfy either Dirichlet boundary conditions or satisfy Neuann boundary conditions V r i n on the surfaces S, S, S 3... S n. We also deand that V(r) be finite at r =. ˆ = Let us define: V ( r) V ( r) V ( r) = Since both V ( r) = and V ( r) difference in the two potential solutions at the point r. then: V ( r) = ( V( r) V( r) ) = V( r) V( r) = separately separately = = Note that: V ( r ) = i V ( r ) The potentials V i =, are uniquely specified on charged (equipotential) surfaces S, S, S 3,... S n in the volue v. Now apply the divergence theore to the quantity ( V V ) E r V r i V V dτ = V V ida= V E i da ( ) ( ) ( ) v S+ S+ S+ S+ S3+... Sn S+ S+ S3+... Sn Volue integral over enclosing volue v Surface integral over ALL surfaces in v ; we also define: 5-8. All rights reserved. 7

8 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Then: V E da = V E da V E da V E da V E da ( i ) ( i S) ( i S ) ( i ) S i Sn... S S S S Sn S + S + S + Sn Recognizing that:. The conducting surfaces S, S, S 3,... S n, are equipotentials. Thus: V ( r) = V( r) V( r) (= a constant on surfaces S, S, S 3,... S n ) ust = at/on those surfaces!!!. The volue v is arbitrary, so let s choose volue v, and thus surface area S as well. 3. da i i =ΦE = electric flux through i th surface. E Si i 4. V ( r ) V( r ) V( r ) because V ( r ) = V ( r ) = (= constant on surface S ) ust =. i V V d = V E ida V E ida V E ida V E ida S S S Sn τ... S S S Sn v = S = S = S = Sn all space all space S S S =Φ n S E =Φ E =ΦE =Φ E Thus: i ( V V) dτ = v all space = V i V However, using the identity i ( V V ) = V ( V ) + ( V ) Then: i ( V V ) dτ V ( = V) τ v all space The only way ( V ) V atheatically v all space d + V dτ = = v all space atheatically v all space = ( V) dτ = ( Vi V) dτ = can be dτ = = = If ( V( r) ) ( V( r) V( r) ) ( ie.. Ar Ar = Ar ) v all space is iff (i.e. if and only if) the integrand i, then: V ( r) i for all points itself ust be = r in volue v. If V ( r) = for all points r in volue v, then V ( r) volue v. V ( r) = V ( r) V ( r) = = i =. ( V r ) ( V( r) V( r) ) = (sae) constant at all points in constant at all points in volue v All rights reserved.

9 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Dirichlet Boundary Conditions (V specified on surfaces S, S, S 3,..., S n ) If V (r) and V (r) are specified on the surfaces S, S, S 3,..., S n in the volue v enclosed by surface S (Dirichlet boundary conditions), then: V ( r) = V( r) V( r) = (i.e. the proble is over-deterined). ust =! V = ( r) throughout the volue v and V ( r) = V ( r) i.e. the two solutions V (r) and V (r) for V ( r) throughout the volue v. = are identical there is only one unique solution. Neuann Boundary Conditions ( E specified on surfaces S, S, S 3,..., S n ) If V ˆ i n= E and V ˆ i n= E are specified on the surfaces S, S, S 3,..., S n in the volue v enclosed by surface S (Neuann boundary conditions), then V ( r) = V( r) V( r) = at all points in volue v and V i nˆ =. Then V ( r) V ( r) V ( r) = = constant, but is not necessarily =!!! Here, solutions V (r) and V (r) can differ, but only by a constant V o. e.g. V( r) = V ( r) + V o proble is NOT over-deterined for V ( r ). ( E( r) Physical Exaple: is over-deterined / unique, but not V ( r ) ). The Parallel Plate Capacitor: E = V/d = V/ + V E V d = Or: E = V/d = V/ +5 V E +4 V d = V = V in both cases thus E-field is sae/identical in both cases! 5-8. All rights reserved. 9

10 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede If we instead specify the charge densit(ies) ρ ( r) within the volue v (see figure below), then we also have a uniqueness theore for the electric field associated with Poisson s equation i E r = V r = ρ r ε. ( o ) Q Charged Conducting Surfaces closed region of volue v, bounded S by enclosing surface S. S Q Q 3 S 3 Q 4 S 4 vs, ρ ( r) specified Suppose there are two electric fields E ( r) and E ( r), both satisfying all of the boundary conditions of this proble. Both obey Gauss law in differential and integral for everywhere within the volue v: E r = ρ r ε i E r = ρ r ε i and: o i encl E da = Qi and: th ε i conducting o th i conducting surface, Si surface, Si At the outer boundary (enclosing surface S) we also have: o E i da = Qi ε o encl encl E i da= Qtot and: S ε S o E i da= ε encl Qtot o We define the difference in electric fields: E ( r) E( r) E( r) the conductors, obeys E ( r) E ( r) E ( r) ( r) ( r) which, in the region between i = i i = ρ εo ρ εo =, and obeys encl encl E da = E da i i Ei da = Qi Qi ε ε = over each boundary surface S i. Si Si Si o o Even though we do not know how the charge Q i on the i th conducing surface S i is distributed, we do know that each surface S i is an equipotential, hence the scalar potential V V V on each surface is at least a constant on each surface S i (n.b. V ay not necessarily be =, since in general V ay not in general be equal to V on each/every surface S i ) All rights reserved.

11 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede i i i, then: Using Griffith s product rule # 5: ( fa) = f ( A) + A ( f ) = i( VE ) V ( i E) + E i ( V ) However, in the region between conductors, we have shown (above) that and E V i VE = Ei V = Ei E = E., hence: i E r =, If we integrate this relation over the entire volue v (with associated enclosing surface S): i VE dτ = VEida= Edτ ( ) v all S v Note that the surface integral covers all boundaries of the region in question the enclosing outer surface S and all of the S i inner surfaces associated with the i conductors. Since V is a constant on each surface, it can be pulled outside of the surface integral (n.b. if the outer surface S is at infinity, then for localized sources of charge, V ( r = ) = ). Thus: V E da E dτ all S i = v But since we have shown above that E da i = for each surface S i, then E da S i all S i =. Therefore: Edτ v =. Note that the integrand E( r) = E( r) i E( r) is always non-negative. Hence, in general, the only way that this integral can vanish is if E ( r) E( r) E( r) = E r = E r. everywhere, thus, we ust have 5-8. All rights reserved.

12 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede ( = ) Solving Laplace s Equation V ( r) In general, when solving the potential V ( r ) in 3-D, -D and -D Situations probles in 3 (or less) diensions, first note the syetries associated with the proble. Then, if you have: Rectangular Solve Rectangular Cylindrical Syetry Proble Cylindrical Coordinates Spherical Using Spherical In -D and 3-D probles, the general solutions to V ( r) = are the haronic functions (an - series solution, in principle) e.g. of sines and cosines, Bessel functions, or Legendre Polynoials and/or Spherical Haronics. The boundary conditions / syetries will select a subset of the -solutions. We will now work through derivations of finding solutions to Laplace s Equations in 3-diensions in rectangular (i.e. Cartesian) coordinates, cylindrical coordinates, and spherical coordinates. We will also use / show the ethod of separation of variables. Laplace s Equation V( x y z),, = and Potential Probles with Rectangular Syetry (Rectangular / Cartesian coordinates) In Three Diensions: Solve Laplace s equation in rectangular / Cartesian coordinates: V V V,,,, x y z x y z V ( x y z) = + + V ( x y z) = + + = The solutions of V = in rectangular coordinates are known as haronic functions (i.e. sines and cosines) ( Fourier Series Solutions). It is usually (but not always) possible to find a solution to the Laplace Equation, V = which also satisfies the boundary conditions, via separation of variables technique, i.e. try a product solution of the for: (,, ) = X ( x) Y( y) Z( z) V x y z where: X (x) x Y (y) are functions only of y respectively. Z (z) z Then: V ( x y z) (,, ) (,, ) (,, ) V x y z V x y z V x y z,, = + + = x y z But: V ( x, y, z) = X ( x) Y( y) Z( z) 5-8. All rights reserved.

13 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Thus: X x Y y Z z X x Y y Z z X x Y y Z z x y z + + = X x Y y Z z = Y( y) Z( z) + X x Z z + X x Y y = x y z Now divide both sides of the above equation by X ( xy ) ( yz ) ( z ) : Then: But: X x Y y Z z = + + = X x x Y y y Z z z X x Y y Z z = + + = X ( x) x Y( y) y Z( z) z fcn(x) only fcn(y) only fcn(z) only independent of y, z independent of x, z independent of x, y = C = C = C3 i.e. C+ C + C3 = True for all points (x, y, z) in volue v of proble. The only way the above equation can be true for all points (x, y, z) in volue v is if: d X ( x) X x = constant C CX ( x) = X x x dx d Y( y) Y y = constant C CY ( y) = Y y y dy d Z( z) Z z = constant C 3 CZ 3 ( z) = #3 Z( z) z dz Note total derivatives now!!! Subject to the constraint: C + C + C 3 = Can now solve 3 ORDINARY -D differential equations, # 3, which are subject to C + C + C 3 =, PLUS the specific Dirichlet / Neuann boundary conditions for the proble on either V (x, y, z) or V( x, y, z) i nˆ at surfaces for this 3-D proble. Essentially, we have replaced the 3-D proble with three -D probles, and the constraint: C + C + C 3 =. # # 5-8. All rights reserved. 3

14 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede * If one has a -D rectangular coordinate proble ( V( x, y) ) d X ( x) d X x = C CX ( x) = X x dx dx dy( y) dy y = C CY ( y) = Y y dy dy Subject to the constraint: C + C =, i.e. C = C. Plus BC s: either on V (x, y) or V( x, y) i nˆ for the -D proble. * If one has a -D rectangular coordinate proble V ( x) =, then: V (x, y) = X(x)Y(y) (only). ( ) dv x d X x d X x = = = = C dx dx X ( x) dx d X x =, then: V(x) = X(x) (only). = X ( x) = V( x) = ax+ b is the -D general solution. dx ( ) For -D proble V ( x) =, only need to solve one ordinary differential equation subject to the constraint C = and BC s on either V(x) or dv ( x) dx All rights reserved.

15 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede The General Solution V ( x, y, z) = X ( x) Y( y) Z( z) for V( x y z) in Rectangular Coordinates,, = Since we have the constraint C + C + C 3 =, at least one of the C i s (i =, or 3) ust be less than zero. Let us choose C = α, C = β, C = γ 3 Then: C + C + C 3 = α β + γ = or: α + β = γ The boundary conditions on the surfaces will define α and β, and hence define γ. IMPORTANT NOTE: The geoetry (x y z) of the proble and the boundary conditions dictate whether: C > or C < C > or C < C 3 > or C 3 < x x i.e. have sine / cosine type solutions vs. sinh / cosh (or e, e ) type solutions for x, y, z. Then the General Solution is (for above choice of C = α, C = β, C = γ ): (,, ) sin ( α ) sin ( β ) sinh ( γ ) n n n,n= could be could be could be cos cos cosh = αn+ β 3 V x y z = A x y z so we also have the additional series solutions:,n= ( α ) ( β ) ( γ ) + Bn cos nx cos y sinh nz,n= = αn+ β ( α ) ( β ) ( γ ) + Cn sin nx sin y cosh nz,n= = αn+ β ( α ) ( β ) ( γ ) + Dn cos nx cos y cosh nz = αn+ β n.b. ix ix ix ix sin( x) = ( e e ) cos( x) = ( e + e ) i i n.b. ix e = cos( x) + isin( x) ix e = cos( x) isin( x) x x n.b. cosh( x) ( e e x x = + ) sinh( x) = ( e e ) x x n.b. e = cosh( x) + sinh( x) e = cosh( x) sinh( x) The BC s and syetries will deterine which of the coefficients A n, B n, C n, D n = All rights reserved. 5

16 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede We solve for the non-zero coefficients A pq, B pq, C pq and D pq by taking inner products. i.e. we ultiply V( x, y, z) = ( stuff ) by e.g. sin( α px)sin( β qy) to project out the p-q th coponent (i.e. we use the orthogonality properties of the individual ters in sin and cos Fourier Series.) and then integrate over the relevant intervals in x and y: e.g. xo yo V( x, y)sin( α x)sin( β y) dxdy p q xo y o = sinsinsinh*sinsin A n αnx β y γnz αpx βq y n, = = constant here + Bn cos( αnx)cos( β y)sinh( γnz)*sin( αpx)sin( βq y) n, = = constant here + Cn sin( αnx)sin( β y)cosh( γnz)*sin( αpx)sin( βq y) n, = = constant here + Dn cos( αnx)cos( β y)cosh( γnz)*sin( αpx)sin( βq y) dxdy n, = = constant here Fourier Functions: orthonorality properties of sin and cos : sin( αnx)sin( αpx) dx= δ np = x o x o ( nx) ( p ) = for n = p ( for n p ) soe constant cos α sin α x dx= = for n = p Kroenecker δ -function: δ np ( = for n p ) So all ters in above Σ s vanish, except for a single ter (in each su) that for the A pq /B pq /C pq /D pq coefficient!!! The BC s will e.g. kill off 3 out of reaining 4 non-zero ters, thus only one ter survives Suppose only the A pq coefficient survives. Its analytic for is now known for all integers p and q. Then the analytic for of 3-D potential V ( x, y, z) is now known it is an infinite series solution of the for: = n n n,n= = αn+ β (,, ) sin ( α ) sin ( β ) sinh ( γ ) V x y z A x y z All rights reserved.

17 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Laplace s Equation V ( ρϕz),, = And Potential Probles with Cylindrical Syetry (Cylindrical Coordinates) ẑ r r = ρ + z = ρρˆ + zzˆ r = ρ + z ˆx Ο ŷ V ( ρϕz),, = V V V ϕ ρ = ρ + + = ρ ρ ρ ρ ϕ z V V V V ˆϕ = = ρ ρ ρ ρ ϕ z ˆρ Again, we use the separation of variables technique: ( ρϕ ) ( ρ) ( ϕ) V,, z = R Q Z z V = yields 3 ordinary differential equations: Note(s): dz z dz dqϕ dϕ kz z = Z z = e + ν Q ϕ = Q ϕ = e dr ± kz ± iνϕ d R ρ ρ ν + + k R ( ρ ) = dρ ρ dρ ρ.) k is arbitrary without iposing boundary conditions..) k appears in both Z(z) and R(ρ) equations. Q ϕ = Q ϕ+ π ), ν ust be an integer! 3.) In order for Q(φ) to be single-valued (i.e. = α j j j= dr( x) d R x ν Let x kp Then: + + R ( x) = dx x dx x R( x) x a x Power Series Solution α = ± ν a 4 j j ( + α ) a j j for j =,,, 3,.... Bessel s Equation All odd powers of x j have vanishing coefficients, i.e. a = a 3 = a 5 = a j+ = 5-8. All rights reserved. 7

18 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Coefficients a j expressed in ters of a : where a a j ( α ) ( α ) j j j j ( α ) Γ + = a = ( + α ) j! Γ j+ + j! Γ j+ + = Γ α ( x) = Gaa Function Γ α + There exist TWO solutions of the Radial Equation (i.e. Bessel s Equation): They are: Bessel Functions of st kind, of order ± ν : ν j j x ( ) x J+ ν ( x) = j= j! Γ j+ ν + These series converge for J ν n.b. ( x) ν j ( ) ( j ν ) x x = j! Γ + all values of x. j= j If ν is not an integer (which is not the case here), then the J ( x) linearly independent solutions to the nd order Bessel s Equation: R x = A J x + A J x for ν integer ν ν ν ν ± ν for a pair of However, note that if ν = integer (which is the case for us here) then the Bessel functions Jν ( x) and J ν ( x) are NOT linearly independent!! If ν = =integer (,,, 3, ), then J ( ) x = J( x) We ust find another linearly independent solution for R(x) when ν = It is custoary to replace J ± ν ( x) by just Jν ( x) and another function Nν ( x) (called Neuann Functions) Where: N ( x) ν NOTE: N ( x) nd = Bessel Function of kind ν is divergent (i.e singular) at x cos( νπ ) ν sin ( νπ ) J x J x Coplex Bessel Functions = Bessel Functions of 3 rd kind = Hankel Functions Hankel Functions are coplex linear cobinations of Jν ( x) and N ( x) and nd kind respectively). They are defined as follows: ( ) ( ) H x J x + in x The Hankel Functions ν ν ν ν =integer ν (Bessel Functions of st H x and H x ν ν also for a H x J x in x fundaental set/basis of solutions to the Bessel equation. ν ν ν All rights reserved.

19 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede The General Solution for V ( ρϕz) ( ρϕ,, ) = ( ρ) ( ϕ) V z R Q Z z,, = in Cylindrical Coordinates: cosh(k n z) is also allowed ( ρ, ϕ, ) = ( ρ) sinh sin ( ϕ) + cos( ϕ) V z J kn knz An Bn n, = n, = cosh(k n z) is also allowed ( ρ ) sinh sin ( ϕ) cos( ϕ) + N kn knz Cn + Dn Apply ALL boundary conditions on surfaces (and also ipose for r =,that V(r = ) = finite! {If r = is part of the proble!}) Note that soeties we want V ( r ) only inside soe finite region of space, e.g. coaxial capacitor if so, then don t have to worry about r = solutions being finite an exaple the Coaxial Capacitor: End View of a Coaxial Capacitor a b If the r = solutions (singular at x= kρ = )!!! region is an excluded region in the proble, then ust include (i.e. allow) the N ( x) If r = region is included in proble, then ALL coefficients Cn = Dn (for all, n), V ρϕ,, z is r =. if Using/iposing BC s on surfaces, orthogonality conditions on sines, cosines, Jν ( x), N ( x) etc. can find / deterine values for all A n, B n, C n, D n coefficients!! ν ν, 5-8. All rights reserved. 9

20 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede -Diensional Circular Syetry Laplace s Equation in (Circular) Cylindrical Coordinates y ρ ϕ x V ( ρϕ, ) = V ( ρϕ, ) = R( ρ) Q( ϕ) Potential, V ( ρ, ϕ ) is independent of z (e.g. infinitely long coaxial cable) Get: Then: V V, ρ ρ ρ ρ ϕ V ( ρϕ) = ρ + = ( ρ ) ( ϕ ) Again try product solution ρ d dr d Q ρ C R = = Let C = k ρ dρ dρ Q( ϕ) dϕ d dr ( ρ ) dq ρ ρ kr( ρ) = ( ϕ ) and + kq( ϕ ) = dρ dρ dϕ Require all solutions Q ( ϕ ) to be single-valued, i.e. Q( ϕ) = Q( ϕ+ π ) because ust have V ( ϕ ) = V ( ϕ+ π). Solutions for Q ( ϕ ) are of the for: Q( ϕ ) = Acos kϕ+ Bsin kϕ Q( ϕ ) Q( ϕ kπ) dq( ϕ ) + nq( ϕ ) = Q ( ϕ) = A ( nϕ) + B ( nϕ) = + requires k = integer =, ±, ±, ± 3, ± n cos sin n n n dϕ ( ρ ) ρ ρ = d dr n n ρ ρ nr( ρ) = Rn( ρ) = Cnρ + Dnρ forn (i.e. n =,, 3,...) dρ dρ R ρ = C + D ln ρ n only for = o o o 5-8. All rights reserved.

21 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede General Solution for V ( ρϕ), = in Two Diensions: Cylindrical (a.k.a. Zonal) Haronics (, ) n cos n cos n ρ ϕ = + ρ + ρ ϕ + ρ ϕ + ρ sin( ϕ) + ρ n sin( ϕ) V V Vln an n bn n cn n dn n n= Again, apply BC s on all relevant surfaces, ipose deterine all coefficients, V, V, a n, b n, c n and d n. V r =finite, etc. these will dictate / i.e. Solve for V, V, a n, b n, c n and d n by applying all boundary conditions, V ( r ) =finite, and using orthogonality conditions / properties: π ρo π ρ o π ρ o π ρ o n a " A" dϕ dρ ρ V ρ, ϕ ρ cos nϕ n = da = d d n b " B" dϕ dρ ρ V ρ, ϕ ρ cos nϕ n = = = n c " C" dϕ dρ ρ V ρ, ϕ ρ sin nϕ n n ρ ρ ϕ dn " D" dϕ dρ ρ V ρ, ϕ ρ sin nϕ A, B, C, D are appropriate noralization factors (we will discuss later). Laplace s Equation V( r ϑϕ) ẑ V = V ( r, ϑ, ϕ ),, = In Spherical Coordinates Ο θ r θ ˆϕ ˆr ŷ ϕ ˆϕ ˆx V r ϑϕ =,, r V θ V V = + sin + = r r r r sinθ θ θ r sin θ ϕ 5-8. All rights reserved.

22 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Again, try separation of variables / try product solution: U( r) V ( r, ϑϕ, ) = P( θ) Q( ϕ) of this for!! r Q ( ϕ) d ( θ) ( θ) ( ϕ) du r U r Q dp U r P dq P θ ϕ + sin θ + = dr r sinθ dθ dθ r sin θ dϕ Multiply by r sin / U( r) P Q θ θ ϕ : ( θ) ( ϕ) du r d dp dq r sin θ + sin θ + = U( r) dr r sinθ P( θ) dθ dθ Q( ϕ) dϕ function of r+ θ only function of ϕ only Now: ( ϕ) dq( ϕ) dq = + Q ( ϕ ) = Q ϕ dϕ dϕ Solutions are of the for: Q( ϕ ) ± iϕ e Since V ( r, ϑ, ϕ) = V ( r, ϑ, ϕ± π) i.e. Q( ϕ ) = Q( ϕ± π) Then: Q ( ϕ ) ust be single-valued! Thus: du( r) dp( θ ) r U( r) dr r sinθ P( θ) dθ dθ du ( r) d dp( θ ) = sin θ + U( r) dr r sinθ P( θ) dθ dθ r sin θ = where = integer =,,, 3, d sin θ + sinθ =+ ultiply above equation by r : r du( r) d dp( θ ) = sin θ + = α U( r) dr sinθ P( θ) dθ dθ sin θ function only of r function only of θ ust hold for any/all r and θ!! ( α ) du r α U ( r) = and dr r sin ( θ ) d dp sinθ + α P ( θ) = d d sin θ θ θ θ let / define α ( +) where = integer =,,, 3, (Trust e, I know the answer...) ( + ) du r U r = and dr r ( θ ) d dp sinθ + ( + ) P ( θ) = sinθ dθ dθ sin θ 5-8. All rights reserved.

23 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Now let x = cosθ dx = d cosθ = sinθdθ Then: ( θ ) x = cos θ = sin θ sin θ = cos θ = x sinθ = x d sin θ dp + ( + ) P ( θ ) = sinθ dθ sinθ dθ sin θ d dp ( x) x + ( + ) P( x) = dx dx ( x ) Becoes: Generalized Legendre' Equation ( + ) du r General Solutions of the radial equation, U r = are of the for: dr r U r = Ar + Br + (l + A + B) are deterined by boundary conditions ( ) For = (aziuthally-syetric probles no φ-dependence) the general solution for V r, θ is of the for: aziuthally-syetric potential ( +, θ = + ) ( cosθ) V r Ar Br P = " ordinary" Legendre' Polynoial of order The coefficients A and B are deterined by the boundary conditions n.b. If no charges at r =, then B =!! Rodrigues Forula is useful for ordinary Legendre' Polynoials: d P ( x) ( x )! dx The coefficients A and B can be found / deterined by evaluating V ( r, θ ) on the conducting surfaces in the proble, e.g. suppose we want to deterine V ( r ) inside a conducting sphere of radius r = a. Then on the surface of the conducting sphere at radius r = a (an equipotential!): ( θ) ( θ) V r = a, = Aa P cos = constant Legendre' Series n.b. inside conducting sphere, e.g. there are no charges at r = B = In order to deterine coefficients, take inner product: ( + ) π A = V ( r = a, θ ) P ( cosθ) sinθdθ a noralization factor 5-8. All rights reserved. 3

24 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Orthogonality condition on P ( x) s: P x P x dx δ ( ) = + Kroenecker δ -function δ δ ' ll ' ll = for l ' l = for l ' = l n.b. The P ( cosθ ) functions for a coplete orthonoral basis set on the unit circle (r = ) for cosθ or: θ π Ordinary Legendre Polynoials P ( x) ( x cosθ ) P ( x ) = P ( x) = defined on the interval x : = x P ( x) = ( 3 x ) 3 P3 ( x) = ( 5 x 3 x) 4 P4 ( x) = ( 35 x 3 x + 3 ) P5 x = 63 x + 7 x + 5 x 8 Note: All P = even ( x) functions are even functions of x: P= even ( x) =+ P = even ( x) All P = odd ( x) functions are odd functions of x: P= odd ( x) = P = odd ( x) under x x reflection. Generally speaking, P ( x) = ( ) P ( x). If 3-D spherical coordinate proble DOES have aziuthal / ϕ -dependence, then in Associated Legendre' Equation (A.L.E.): ( θ ) d dp sinθ + ( + ) P ( θ) = x = cosθ sinθ dθ dθ sin θ Solutions to A.L.E. are Associated Legendre' Polynoials (A.L.P. s) Associated Legendre Polynoials: P ( x) ( ) ( x ) P ( x) dx " ordinary" Legendre' Polynoial =± integer i.e. =±, ±, ± 3,... but have a constraint on!!! + i.e. =, +, +,...,,, +, +,,, d All rights reserved.

25 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede Also: = ( ) ( ) ( + )! P x P x! Orthogonality condition for P ( x) for fixed : We now define noralized P( θ) Q The Spherical Haronics Y ( θ, ϕ ) surface of the unit sphere (r = ) * Note that Y ( θ, ϕ) = ( ) Y ( θ, ϕ) Y ( θ, ϕ ) ( + )! δ ( + ) ( )! P ( x) P ( x) dx = Kroenecker δ -function ϕ functions known as Spherical Haronics: ( + ) ( ) ( + )! iϕ Y ( θϕ, ) P ( cosθ) e 4 π! = Q for a coplete orthonoral set of basis vectors on the coplex conjugate Noralization and Orthogonality Condition: π π * d sin d Y (, ) Y (, ) Ω= 4π * i.e. dω Y (, ) Y (, ) = ϕ θ θ θϕ θϕ = δ δ ( ϕ ) i.e. i i where i θ ϕ θ ϕ δ δ Ω= dω = sin = = θdθdϕ * Copleteness Relation: Y ( θ, ϕ) Y ( θ, ϕ) = δ ( cosθ cos θ' ) δ ( ϕ ϕ' ) DIRAC δ -functions 5-8. All rights reserved. 5

26 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede = Y 4π Y ( θ, ϕ ) Spherical Haronics * = Use Y ( θ, ϕ) = ( ) Y ( θ, ϕ) in order to obtain Y, Y, Y3 3, Y3, Y etc. 3 = Y = 3 sin θ 8π i e ϕ Y Y = Y = Y 3 = cos θ Note: 4π 5 sin 4 π i = θe ϕ Y ( θ ϕ) 5 sin θ cos θ 8π = θ i e ϕ 5 3 cos 4π ( + ) ( ) π ( + )! Y ( θϕ, ) P ( cosθ) e 4! ( + ) ( θ), = P cos 4π iϕ Y = sin θ 4 4π 3 3i e ϕ = 3. Y Y = sin θ cos θ 4 π i e ϕ = sin θ 5cos 4 4π π. etc. ( θ ) 3 Y3 = cos θ cosθ i e ϕ All rights reserved.

27 UIUC Physics 435 EM Fields & Sources I Fall Seester, 7 Lecture Notes 7 Prof. Steven Errede General Solution for Laplace s Equation V( r, θϕ, ) ( θ ϕ) + = = in Spherical Polar Coordinates ( ) ( θ ϕ) + V r,, = A r B r + Y, Coefficients A and are deterined by / fro Boundary Conditions on spherical surface(s) If V V( θϕ, ) B = on surface (e.g. at r = a) (i.e. no charge at r = in proble B =, + = = = Then: V ( θ, ϕ) A Y ( θ, ϕ) 4π * And: A = dωy ( θ, ϕ) V ( θ, ϕ) Note: on surface (r = a). on surface (r = a). V( θ, ϕ) " north" pole = ( + ) = = A ( + ) 4π 4π ( cos θ ) ( θϕ, ) A = dωp V 4π ) General Coents: The ethod of separation of variables used in Laplace s equation V = in rectangular, cylindrical and spherical coordinates shows up again in Poisson s Equation ρ free V = and also in the wave equation (valid for all classical wave phenoena) ε o ψ ( rt, ) ψ ( rt, ) + = and in Schrödinger s wave equation Hψ = Eψ in Quantu c t Mechanics probles. These equations will appear again and again, in one for or another for E&M, Classical Mechanics, Quantu Mechanics courses as well as for Classical / Newtonian Gravity probles For ore detailed inforation e.g. on separation of variables and solutions to 3-D Wave ψ Equation ψ = in rectangular, cylindrical and spherical coordinates see Prof. S. c t Errede lecture notes (Lecture IV parts & ) on (sound) waves in -D, -D, 3D Physics 46 Acoustical Physics of Music website: and also see/read his Fourier Analysis Lectures on this website, if interested All rights reserved. 7