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


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



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