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?
>
> Could it be like this, that the components can be vector summarized to
> get the total?
> Vavg^2 = Vavg_x^2 + Vavg_y^2 + Vavg_z^2
I believe that must be the relationship, since
Vtot = sum(V) over all molecules = P/M
where 'M' is the mass of the gas and P is the momentum vector of the gas
considered as a whole.
Total momentum of the gas is
P = (X*Vtot_x + Y*Vtot_y + Z*Vtot_z) * M
where X, Y, and Z are the basis vectors. Then the magnitude of the
momentum is
|P| = sqrt(Vt_x^2 + Vt_y^2 + Vt_z^2) * M
and squared momentum is
P^2 = (Vt_x^2 + Vt_y^2 + Vt_z^2) * M^2
We also have,
Vtot^2 = P^2/M^2
And we know the average velocity is the summed velocities of the
molecules divided by N, which is the number of molecules. So,
Vavg = Vtot/N
So,
Vavg^2 = P^2/(MN)^2
Vavg^2 = (Vt_x^2 + Vt_y^2 + Vt_z^2) / N^2
and finally, distributing the (1/N^2),
Vavg^2 = (Vavg_x^2 + Vavg_y^2 + Vavg_z^2)
Unless I messed it up somewhere...
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.
>
> If so then I have to recalculate my gas shear stress derivation on
> arXiv. But since I am not sure I need some confirmation first. Please
> help.
>
> David
>
> David Jonsson, Sweden, phone callto:+46703000370
>
