Download Laplace's Equation in Cylindrical and Spherical Polar Coordinates: Separation of Variables and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 16 Chapter 5.1-5.3 of PS - We will not cover 5.4, 5.5 A. Separation of variables in circular co-ordinates Circular co-ordinates are the (r, φ) part of cylindrical polar co-ordinates. The Laplacian in cylindrical polar co-ordinates is given by (see Table 2.3, page 34 of PS), ∇2V = 1 r ∂ ∂r r ∂V ∂r + 1 r2 ∂2V ∂φ2 + ∂2V ∂z2 = 0 (1) If there is no dependence on z, the third term is zero, so r ∂ ∂r (r ∂V ∂r ) + ∂2V ∂φ2 = 0 (2) Assuming that V = R(r)Φ(φ), we find, r R ∂ ∂r (r ∂R ∂r ) + 1 Φ ∂2Φ ∂φ2 = 0 (3) We now separate the equation using, r R ∂ ∂r (r ∂R ∂r ) = −1 Φ ∂2Φ ∂φ2 = k2 (4) The φ equation has solutions cos(kφ), sin(kφ) when k is finite. When k = 0, it has solution C + Dφ. Noting that Φ(φ) = Φ(φ + 2π) is required to ensure that Φ is single valued, we set k = n, where n = 0, 1.... We could also use n negative, but that would just change the sign of the constants Dn in a general solution of the form Cncos(nφ) + Dnsin(nφ). The r equation has the form, r2 d2R dr2 + r dR dr − n2R = 0 (5) which for n 6= 0 has solutions R = Anr n+Bnr −n, while for n = 0, R = A+Blnr. With these observations, we find that the general solution to Laplace’s equation in circular co-ordinates (ie. cylindrical co-ordinates with no dependence on z) is, The general solution is, V (r, φ) = (A + Blnr)(C + Dφ) + ∞∑ n=1 (Anr n + Bn rn )(Cncos(nφ) + Dnsin(nφ)) (6) A simple example. Consider an uncharged conducting cylinder of radius R has its axis along the z-axis. A constant electric field, E0, is applied to the cylinder, with the field oriented along the x-axis. Assume that the cylinder is at potential V = 0 and has radius R. Find the electrostatic potential for r > R. 1 Solution. The potential as r → ∞ is −E0x, which in cylindrical co-ordinates is −E0rcosφ. The first thing to do is try the simplest solution which only contains this term at infinity, that is, V (r, φ) = (A′r + B′ r )cosφ (7) To satisfy the boundary condition at infinity, we can take A′ = −E0, giving, V (r, φ) = (−E0r + B′ r )cosφ (8) The boundary condition V = 0 at r = R implies that, V (R, φ) = 0 = (−E0R + B′ R )cosφ (9) This is satisfied provided B′ = E0R 2, so we have, V (r, φ) = (−r + R2 r )E0cosφ (10) The electric field should be normal to the surface of the conductor, so we also need to check that Eφ(R, φ) = 0. We have, Eφ = − 1 r ∂V ∂φ = (−1 + R2 r2 )E0sinφ (11) This is clearly zero at the surface of the cylinder where r = R, as required. So that’s it for this example - again a simple case where only one term in the superposition is required. B. Separation of variables in spherical polar co-ordinates The Laplacian in spherical polar co-ordinates is the most important case and is an es- sential part of quantum mechanics as well as EM. Laplace’s equation in spherical polar co-ordinates is, ∇2V = 1 r2 ∂ ∂r (r2 ∂V ∂r ) + 1 r2sinθ ∂ ∂θ (sinθ ∂V ∂θ ) + 1 r2sin2θ ∂2V ∂φ2 = 0 (12) Again lets start with the case where there is no dependence on φ, so V (r, θ, φ) → V (r, θ) = R(r)Θ(θ). Making this substitution into Laplace’s equation and dividing through by RΘ gives, ∇2V = 1 R ∂ ∂r (r2 ∂R ∂r ) = − 1 Θsinθ ∂ ∂θ (sinθ ∂Θ ∂θ ) = C = l(l + 1) (13) 2
