On Wed, Jan 26, 2011 at 4:04 AM, Stephen A. Lawrence <[email protected]>wrote:
>
>
> On 01/25/2011 06:43 PM, David Jonsson wrote:
> > Hi
> >
> > We know that mean root square speed in specific perpendicular
> > directions is like this
> >
> > Vrms^2 = 3 Vrmx_x^2 = 3 Vrmx_y^2 = 3 Vrmx_z^2
> >
> > What is the relation for the mean linear speed Vavg in specific
> > directions?
>
...
> and finally, distributing the (1/N^2),
>
> Vavg^2 = (Vavg_x^2 + Vavg_y^2 + Vavg_z^2)
>
> Unless I messed it up somewhere...
>
Looks credible.
>
> More simply, average velocity is the same as total velocity up to a
> scale factor, and that's the same as momentum, so it just behaves like
> like a vector.
>
>
> >
> > And assuming isotropic distributions (Vavg_x = Vavg_y = Vavg_z) leading
> to
> > Vavg^2 = 3Vavg_x^2 + Vavg_y^2 + Vavg_z^2
> > which is like the relation on top.
>
> You presumably meant
>
> Vavg^2 = 3Vavg_x^2 = Vavg_y^2 = Vavg_z^2
>
> and that certainly looks right, if Vavg_{x,y,z} are all equal.
>
I meant Vavg^2 = Vavg_x^2 + Vavg_y^2 + Vavg_z^2
which for equal distributions in all directions is Vavg^2 = 3Vavg_x^2 =
3Vavg_y^2 = 3Vavg_z^2
So Vavg_x =Vavg_y =Vavg_z = Vavg/sqrt(3)
Not that bad anyway. I explain the shear flow on Venus by a factor 13%.
David
