Laplace Equations and Separation of Variables in Electromagnetism | PHY 481, Study notes of Physics

Download Laplace Equations and Separation of Variables in Electromagnetism | PHY 481 and more Study notes Physics in PDF only on Docsity! Lecture 20 Carl Bromberg - Prof. of Physics PHY481: Electromagnetism Solving Laplace’s equation via “Separation of Variables” 1) Cartesian coordinates ✓ 2) Spherical coordinates      3) Cylindrical coordinates     Examples    Lecture 20 Carl Bromberg - Prof. of Physics 1 Cartesian coordinates  Determine nature of solutions on the rectangle (unbounded z) – Grounded at x = –a/2 or +a/2 then Xk(x) = Akcos(kx) +Bksin(kx) – Grounded at y = –a/2 or +a/2 then Yk(y) = Akcos(ky) +Bksin(ky) – Other solution is then Uk(u) = Ckcosh(ku) + Dksinh(ku) (same k)  Use periodic (sine or cosine) solution and boundary conditions to determine legal values of k = nπx/a (same for both solutions)  Generate general solution V(x,y) = Σ Xk(x)Yk(y)  Use additional boundary conditions to determine coeficients – Periodic B.C.: Vb(x0,y) = Σ Xk(x0)Yk(y) or Vb(x,y0) = Σ Xk(x)Yk(y0) – Apply Fourier Integral to both sides: Be sure to split Fourier integral (-a,-a/2,a/2,a) – Use orthogonality on right side: – Determine coef. Ak,Bk,Ck,Dk  Substitute into V(x,y) = Σ Xk(x)kY(y) 1 a Vb(x, y0 )cos mπ x a −a a ∫ (or y version) cos nπ x a( )cos mπ x a( ) −a a ∫ = δmn (or y version) Lecture 20 Carl Bromberg - Prof. of Physics 4 Familiar problem in spherical coordinates Orthogonality of Legendre polynomials: Pm cosθ( )Pn cosθ( )d cosθ( ) −1 1 ∫ = 2δmn 2n+1 Consider a grounded conducting sphere radius, a, in a constant external field, E0 in z direction V (a,θ) = 0 = Aa  + Ba − +1( )( ) =0 ∞ ∑ P cosθ( ) V (a,θ)Pn cosθ( ) −1 1 ∫ d cosθ( ) = 0 = Aa + B a +1 ⎛ ⎝⎜ ⎞ ⎠⎟=0 ∞ ∑ P cosθ( )Pn cosθ( ) −1 1 ∫ d cosθ( ) Multiply by Pn cosθ( ) and integrate 0 = Ana n + Bna − n+1( )( )2 2n +1( ) Bn = −Ana 2n+1 Boundary condition V = 0 on sphere Boundary condition V (r →∞,θ) = −E0z = −E0r cosθ = −E0r cosθ + E0a 3 cosθ r 2 cosθ = P1(cosθ) ⇒  = 1 A1 = −E0 V (r,θ) = Ar  + B r +1 ⎛ ⎝⎜ ⎞ ⎠⎟=0 ∞ ∑ P cosθ( ) uniform field dipole E0a 3 = p 4πε0 dipole moment E field Lecture 20 Carl Bromberg - Prof. of Physics 5 Cylindrical coordinates ∇2V (r,θ) = 1 r ∂ ∂r r ∂V ∂r ⎛ ⎝⎜ ⎞ ⎠⎟ + 1 r 2 ∂2V ∂φ2 = 0 V (r,φ) = R(r)Φ(φ) r 2 V ∇2V (r,φ) = r R ∂ ∂r r ∂R ∂r ⎛ ⎝⎜ ⎞ ⎠⎟ + 1 Φ ∂2Φ ∂φ2 = 0 Rn(r) = Anr n + Bnr −n n 2 −n 2 Φn(φ) = Cn cos nφ + Dn sin nφ V (r,φ) = Aln r + B + Anr n + Bnr −n( ) Cn cos nφ + Dn sin nφ( ) n=1 ∞ ∑ R0(r) = Aln r + B Laplace’s equation in cylindrical coordinates Separation of variables r dependence φ dependence Constants sum to zero Separate solutions Solution with series to insure boundary conditions can be satisfied Lecture 20 Carl Bromberg - Prof. of Physics 6 Cylinder problem Half cylinders, radius R. V = +V 0 on right half, and V = -V 0 on left half Find potential inside and out. V (r,φ) = Aln r + B + Anr n + Bnr −n( ) Cn cos nφ + Dn sin nφ( ) n=1 ∞ ∑ VInt (r,φ) = Anr n cos nφ n=1 ∞ ∑ VExt (r,φ) = Bnr −n cos nφ n=1 ∞ ∑ cos nφ cos mφ = πδnm 0 2π ∫ Boundary conditions at r = 0 and r = ∞ Orthogonality VInt (R,φ) =VExt (R,φ) AnR n = BnR −n = cn An = cn R n ; Bn = cnR n VInt (r,φ) = cn r n R n cos nφ n=1 ∞ ∑ VExt (r,φ) = cn R n r n cos nφ n=1 ∞ ∑ Use Orthogonality c j = 4V0(−1) j π (2 j +1) sin 2 j +1( )π 2 VInt (r,φ) = 4V0 π −1( ) j 2 j +1 r 2 j+1 R 2 j+1 cos(2 j +1)φ j=0 ∞ ∑ VExt (r,φ) = 4V0 π −1( ) j 2 j +1 R 2 j+1 r 2 j+1 cos(2 j +1)φ j=0 ∞ ∑ Complete solutionBe careful to break Fourier integral into pieces !! f (φ) = V0 1 st & 4 rd quadrants −V0 2 nd & 3 th quadrants ⎧ ⎨ ⎪ ⎩⎪