Re: a' vs a and a vs a' looks different in streamplot

*To*: mathgroup at smc.vnet.net*Subject*: [mg99672] Re: a' vs a and a vs a' looks different in streamplot*From*: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>*Date*: Sun, 10 May 2009 05:20:33 -0400 (EDT)*References*: <gu3bgr$fd$1@smc.vnet.net>

The transformation {x,y} -> {y,x} is NOT a rotation. Check this out: Reduce[RotationMatrix[\[Tau]].{a, -a - kr b} == {-a - kr b, a}, \[Tau]] A vector plot of this transformation: VectorPlot[{y - x, x - y}, {x, -5, 5}, {y, -5, 5}] Cheers -- Sjoerd On May 9, 9:31 am, sean_inc... at yahoo.com wrote: > Consider the following. > > a'[t]= -a[t] -kr -b[t] > > Let us plot the vectorfields of a' vs a. (or a vs a' for that matter) > as follows. > > Manipulate[{StreamPlot[{a, -a - kr b}, {a, -10, 10}, {b, -10, 10}, > FrameLabel -> {a, a'}], > StreamPlot[{-a - kr b, a}, {a, -10, 10}, {b, -10, 10}, > FrameLabel -> {a', a}]}, {kr, 0, 10}] > > Why do the plots of " a vs. a' " and " a' vs. a " look so different? > They aren't just 90' rotation of each other. > > Shouldn't they be? > > Thanks for any insight. > > Sean