polar coordinates

Polar coordinates PhL 6.25.09 1. Compute partial derivatives of Cartesian coordinates: 1 2. Compute partial derivatives of Polar coordinates: 1 3. Compute the Laplacian. 2 4. Relation between unit vectors: 4 5. Time derivatives of the unit vectors 4 6. The gradient in polar coordinates: 4 7. The divergence in polar coordinates: 5 Added section 5 on 12.14.16. In retrospect, all these results can be obtained from cylindrical coordinates if you ignore z. But let's go ahead anyway and do all the details as if we didn't know that fact. Here we go: x = r cosθ r2 = x2 + y2 y = r sinθ y/x = tanθ 1. Compute partial derivatives of Cartesian coordinates: dx = dr cosθ - r sinθdθ dy = dr sinθ +r cosθdθ ∂x/∂r = cosθ ∂x/∂θ = -r sinθ ∂y/∂r = sinθ ∂y/∂θ = r cosθ 2. Compute partial derivatives of Polar coordinates: rdr = xdx + ydy (xdy-ydx)/x2 = sec2θ dθ ∂r/∂x = x/r = cosθ ∂r/∂y = y/r = sinθ // web verified ∂θ/∂x = (-y/sec2θ x2) = (-rsinθ / [sec2θ r2cos2θ]) = -sinθ/r // web verified ∂θ/∂y = (1/sec2θ x) = (1/[sec2θ r cosθ]) = cosθ/r // web verified Summary: ∂r/∂x = cosθ ∂r/∂y = sinθ ∂θ/∂x = -sinθ/r ∂θ/∂y = cosθ/r 3. Compute the Laplacian. (a) Compute ∂x and then ∂x2 ∂x = ∂/∂x = ∂θ/∂x * ∂/∂θ + ∂r/∂x * ∂/∂r = (-sinθ/r)∂θ + cosθ ∂r ∂x2 = [(-sinθ/r)∂θ + cosθ ∂r]2 = [(sinθ/r)∂θ – cosθ ∂r]2 = [(sinθ/r)∂θ – cosθ ∂r] [(sinθ/r)∂θ – cosθ ∂r] Have to compute out all four terms (put in signs later) (sinθ/r)∂θ(sinθ/r)∂θ = (sinθ/r){ (cosθ/r)∂θ + (sinθ/r) ∂θ2} = sinθcosθ/r2∂θ + sin2θ/r2 ∂θ2 (sinθ/r)∂θ cosθ ∂r = (sinθ/r){ -sinθ ∂r + cosθ ∂θ∂r } = -sin2θ/r ∂r + sinθcosθ/r ∂θ∂r cosθ ∂r(sinθ/r)∂θ = cosθ { -sinθ/r2 ∂θ + sinθ ∂r∂θ} = - sinθcosθ/r2 ∂θ + sinθcosθ ∂r∂θ cosθ ∂r(cosθ ∂r) = cos2θ ∂r2 Add up the four terms to get ∂x2 = sinθcosθ/r2∂θ + sin2θ/r2 ∂θ2 +sin2θ/r ∂r - sinθcosθ/r ∂θ∂r + sinθcosθ/r2 ∂θ - sinθcosθ ∂r∂θ + cos2θ ∂r2 ∂x2 = 2 sinθcosθ /r2 ∂θ + sin2θ/r2 ∂θ2 + sin2θ/r ∂r -2 sinθcosθ /r ∂θ∂r + cos2θ ∂r2 This painful result agrees with a website calculation I found which I quote here: (a) Compute ∂y and then ∂y2 ∂y = ∂/∂y = ∂θ/∂y * ∂/∂θ + ∂r/∂y * ∂/∂r = (cosθ/r)∂θ +sinθ ∂r ∂y2 = [(cosθ/r)∂θ + sinθ ∂r]2 = [(cosθ/r)∂θ + sinθ ∂r]2 = [(cosθ/r)∂θ + sinθ ∂r] [(cosθ/r)∂θ + sinθ ∂r] Have to compute out all four terms (put in signs later) (cosθ/r)∂θ(cosθ/r)∂θ = (cosθ/r){ (-sinθ/r)∂θ + (cosθ/r) ∂θ2} = - sinθcosθ/r2∂θ + cos2θ/r2 ∂θ2 (cosθ/r)∂θ sinθ ∂r = (cosθ/r){ +cosθ ∂r +sinθ ∂θ∂r } = cos2θ/r ∂r + sinθcosθ/r ∂θ∂r sinθ ∂r(cosθ/r)∂θ = sinθ { -cosθ/r2 ∂θ + cosθ/r ∂r∂θ} = - sinθcosθ/r2 ∂θ + sinθcosθ/r ∂r∂θ sinθ ∂r(sinθ ∂r) = sin2θ ∂r2 Add up the four terms to get ∂y2 = -sinθcosθ/r2∂θ + cos2θ/r2 ∂θ2 +cos2θ/r ∂r + sinθcosθ/r ∂θ∂r - sinθcosθ/r2 ∂θ + sinθcosθ/r ∂r∂θ + sin2θ ∂r2 ∂y2 = cos2θ/r2 ∂θ2 + cos2θ/r ∂r + 2 sinθcosθ/r ∂r∂θ + sin2θ ∂r2 - 2sinθcosθ/r2 ∂θ Summarize to this point: ∂x2 = 2 sinθcosθ /r2 ∂θ + sin2θ/r2 ∂θ2 + sin2θ/r ∂r -2 sinθcosθ /r ∂θ∂r + cos2θ ∂r2 ∂y2 = - 2sinθcosθ/r2 ∂θ + cos2θ/r2 ∂θ2 + cos2θ/r ∂r + 2 sinθcosθ/r ∂r∂θ + sin2θ ∂r2 All four sinθcosθ terms cancel, and we are left with ∂x2 + ∂y2 = ∂r2 + (1/r) ∂r + (1/r2) ∂θ2 = (1/r)∂r(r∂r) + (1/r2) ∂θ2 2 = ∂r2 + (1/r) ∂r + (1/r2) ∂θ2 = (1/r)∂r(r∂r) + (1/r2) ∂θ2 Summary: ∂x = (-sinθ/r)∂θ + cosθ ∂r ∂y = (cosθ/r)∂θ + sinθ ∂r ∂x2 = 2 sinθcosθ /r2 ∂θ + sin2θ/r2 ∂θ2 + sin2θ/r ∂r -2 sinθcosθ /r ∂θ∂r + cos2θ ∂r2 ∂y2 = - 2sinθcosθ/r2 ∂θ + cos2θ/r2 ∂θ2 + cos2θ/r ∂r + 2 sinθcosθ/r ∂r∂θ + sin2θ ∂r2 ∂x2 + ∂y2 = ∂r2 + (1/r) ∂r + (1/r2) ∂θ2 = (1/r)∂r(r∂r) + (1/r2) ∂θ2 4. Relation between unit vectors: = cosθ + sinθ = Rz(θ) = -sinθ + cosθ = Rz(θ) = cosθ – sinθ = Rz(-θ) = sinθ + cosθ = Rz(-θ) = cosθ = sinθ = -sinθ = cosθ 5. Time derivatives of the unit vectors d/dt = d/dt(cosθ + sinθ ) = -sinθ + cosθ = -sinθ [cosθ – sinθ ] + cosθ [ sinθ + cosθ ] = d/dt = d/dt( -sinθ + cosθ ) = -cosθ - sinθ = -cosθ [cosθ – sinθ ] - sinθ [ sinθ + cosθ ] - 6. The gradient in polar coordinates: = ∂x + ∂y = (cosθ – sinθ ) [(-sinθ/r)∂θ + cosθ ∂r ] + (sinθ + cosθ ) [(cosθ/r)∂θ + sinθ ∂r] = { (-sinθ cosθ /r)∂θ + cos2θ ∂r + sinθ(cosθ/r)∂θ + sin2θ ∂r } + { sin2θ/r ∂θ - sinθcosθ ∂r + cos2θ/r ∂θ + sinθcosθ ∂r} = ∂r + (1/r) ∂θ 7. The divergence in polar coordinates: f = ∂x (f) + ∂y(f) = = [(-sinθ/r)∂θ + cosθ ∂r ] f (cosθ – sinθ ) + [(cosθ/r)∂θ + sinθ ∂r] f (sinθ + cosθ ) = [(-sinθ/r)∂θ + cosθ ∂r ] (cosθ fr – sinθ fθ) + [(cosθ/r)∂θ + sinθ ∂r] (sinθ fr + cosθ fθ) = -sinθ/r ∂θ(cosθ fr) + sinθ/r ∂θ(sinθ fθ) + cos2θ ∂r fr - sinθcosθ ∂r fθ + cosθ/r ∂θ(sinθ fr) + cosθ/r ∂θ(cosθ fθ) + sin2θ ∂r fr + sinθcosθ ∂r fθ = -sinθ/r ∂θ(cosθ fr) + sinθ/r ∂θ(sinθ fθ) + ∂r fr + cosθ/r ∂θ(sinθ fr) + cosθ/r ∂θ(cosθ fθ) = -sinθ/r [ cosθ ∂θ fr - sinθ fr] + sinθ/r [ sinθ ∂θfθ + cosθ fθ] + ∂r fr + cosθ/r [ sinθ ∂θfr + cosθ fr] + cosθ/r [ cosθ ∂θ fθ - sinθ fθ] f = (1/r)fr + ∂r fr + (1/r) ∂θfθ 8. Particle Motion in Polar Coordinates r = r v = + r ∂t = + r // agrees with Marion p 31 (1.97) a = + ∂t + + r + r ∂t = + ( ) + + r + r (- ) = ( - r2) + (2 + r ) // agrees with Marion p 31 (1.98) Rewrite as v = vr + vθ vr = vθ = r a = ar + aθ ar = - r2 aθ = 2 + r If motion is constrained to a circle of radius r, set = = 0 in the above. Then v = vθ vθ = r a = ar + aθ ar = - r2 aθ = r