Vector CalculusDate: 02/10/99 at 07:42:48 From: D. Denduyver Subject: Vector Calculus Can you help me and my son prove the following identity: a . (grad(a . v) - curl (v x a)) = div v, where a = constant unit vector . = scalar product x = vector product v = vector We've found that: a . curl (v x a) = a .(( a . V) v - a (V . v)) = a. (a . V)v - V . v but a . (grad(a . v)) - a . (a . V)v + V . v is not equal to V . v with V = del or nabla operator. Can you also give us a title of a textbook where we can find the proof of de (p,r) equation of an ellipse with (p, r) = pedal coordinates? Thank you very much, Dany Denduyver Date: 02/10/99 at 09:51:12 From: Doctor Mitteldorf Subject: Re: Vector Calculus My general approach for vector calculus identities is to write them out as coordinates, using the Einstein summation convention. So A.B becomes AiBi and V x F becomes EijkVjFk. For the cross product, use epsilon-ijk, which I've written Eijk. E123 = E312 = E231 = 1 and E321 = E213 = E132 = -1, and all their combinations, like E112 and E333, are zero. The particular identity you're working with is really much easier than it looks, because a is a unit vector. You can take a to be the unit vector in the 3 direction, which I'll call z, without loss of generality. a.v = v3, and (a.V) is just V3 (Notation: I'm using V for del and V3 means "del sub 3", the derivative in the z direction.) So a . grad(a . v) = V3(v3), i.e., the derivative in the z direction of the z component of v. Now for the second term: (v x a) has a component in the 1 direction equal to v2, and its component in the 2 direction is -v1. We need the 3 component of the curl of this (because we're taking the dot product of the curl with a). This is V1 (v x a)2 - V2 (vxa)1, or, using the above, -V1(v1) - V2(v2). So, collecting all terms, we have V3(v3) + V1(v1) + V2(v2), which is indeed the divergence of v. I learned vector calculus years ago from E. M. Purcell, _Electricity and Magnetism_, Vol. II of the Berkeley Physics Course. I'm not familiar with pedal coordinates, so I'm leaving this message up where another doctor might be able to direct you to an appropriate reference. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 02/10/99 at 18:29:22 From: Doctor Anthony Subject: Re: Vector Calculus I believe you have had a reply on the first question so will confine my remarks to the pedal equation of the ellipse. If phi = angle between the tangent at a point on the curve and the radius vector to that point (polar coordinates) then a simple diagram shows tan(phi) = r.d(theta)/dr sin(phi) = r.d(theta)/ds cos(phi) = dr/ds If p = perpendicular from the origin to the tangent then p = r.sin(phi) To express p in terms of r we can write 1/p^2 = 1/r^2 cosec^2(phi) = (1/r^2)[1 + cot^2(phi)] = (1/r^2)[1 + (1/r^2)(dr/d(theta))^2] 1/p^2 = 1/r^2 + (1/r^2)[dr/d(theta)]^2 .........(1) By eliminating theta between this equation and the polar equation of the curve we get the equation between p and r, the pedal equation. Example: Consider the conic with the given equation. For the curve 2a/r = 1 - cos(theta) find the (p,r) equation. Differentiating -2a/r^2 [dr/d(theta)] = sin(theta) So, r.d(theta)/dr = -2a/(r.sin(theta)) and from eqn. of curve, = -(1 - cos(theta))/sin(theta) and so, tan(phi) = -2sin^(theta/2)/[2sin(theta/2).cos(theta/2)] tan(phi) = -tan(theta/2) = tan(pi-theta/2) and phi = pi - theta/2 Then, p = r.sin(phi) = r.sin(theta/2) and from eqn. of curve = 2a.sin(theta/2)/[1-cos(theta)] = 2a.sin(theta/2)/[2.sin^2(theta/2)] = a/sin(theta/2) = a.cosec(theta/2) p^2 = a^2.cosec^2(theta/2) ..............(2) but, r = 2a/(1-cos(theta)) = a/sin^2(theta/2) therefore, cosec^2(theta/2) = r/a and from (2), p^2 = a^2.r/a p^2 = a.r which is the pedal equation of the conic. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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