From: mtx014@coventry.ac.uk (Robert Low) Subject: Re: Tensor-question-turned-"simple"-chain-rule-question (and bonus PS) Date: 3 Mar 1999 19:55:54 -0000 Newsgroups: sci.math Keywords: acceleration is not a contravariant tensor In article <36DD0573.53AA2C3D@mpx.com.au>, Daniel wrote: >The response I got to my original post said that the object with the >x- and y- acceleration components is not a contravariant tensor. I >disagree. Suit yourself. But you'll have a lot of fun trying to get them to obey the tensor transformation law. >I can do the question perfectly if the tensor components are velocity, >rather >than acceleration. That's because velocity *is* a contravariant vector. Leesen carefully, I shall explain zees only wance. The chain rule for differentiation that you use to express dr/dt and dtheta/dt in terms of dx/dt and dy/dt says that dx/dt = dx/dr dr/dt + dx/dtheta dtheta/dt dy/dt = dy/dr dr/dt + dy/dtheta dtheta/dt This is precisely telling you that velocity transforms as a contravariant vector. Differentiating again wrt t gives d^2x/dt^2 = dx/dr d^2r/dt^2 + dx/dtheta d^2theta/dt^2 +d/dt(dx/dr) dr/dt + d/dt (dx/dtheta) dtheta/dt d^2y/dt^2 = something similar And when you work out d/dt(dx/dr) in terms of d(r,theta)/d(x,y), dr/dt and dtheta/dt, you get (dr/dt)^2, (dtheta/dt)^2, and dr/dt dtheta/dt terms. You seem to think there is something wrong here: there isn't. That's just how it is. If you want to get a contravariant vector out of acceleration, you need to use the connection. See a book on differential geometry for more help. At any rate, Thi computation is patently telling you that (d^2x/dt^2,d^2y/dt^2) is *not* a contravariant vector. These objects only transform via the tensor transformation law if the coordinate transformation is linear. -- Rob. http://www.mis.coventry.ac.uk/~mtx014/ ============================================================================== From: phunt@interpac.net Subject: Re: Tensor-question-turned-"simple"-chain-rule-question (and bonus PS) Date: Wed, 03 Mar 1999 22:35:28 GMT Newsgroups: sci.math To: entropic@mpx.com.au In article <36DD0573.53AA2C3D@mpx.com.au>, Daniel wrote: > Hullo again! > > The response I got to my original post said that the object with the > x- and y- acceleration components is not a contravariant tensor. I > disagree. > Sorry you're wrong. See problem 1.6.1 in "Tensor Analysis on Manifolds," by Richard L. Bishop and Samuel I. Goldberg. You will find that a velocity is indeed a (tangent) contravariant vector, but its derivative is not. /ph - - - - - - - [deletia --djr]