[aspect-devel] compressibility in 2d

Menno Fraters menno.fraters at outlook.com
Mon Feb 13 11:28:35 PST 2017

Hey all,

Thanks for the quick reply and answers. We have been discussing this (with Wim and Cedric), and at first we still found it strange that the 2D case of the Stokes equation yields a different deviator in the matrix (1/2 vs 1/3). But then we realised that in the truly 2D case all the units should also be different anyway (e.g. density kg/m2, pressure N/m, etc) and one should actually derive the whole thing from the beginning in 2D. In Aspect and other codes we always work in 2D with true physical units, e.g. density in kg/m^3, pressure N/m2 etc, implicitly assuming a finite width of the 2D slice, i.e. we implicitly  work with 3D physics and therefore the “⅓” is okay.

I guess something like this should be added somewhere in the manual, but for now I think my question is answered.

Thanks for the help!


Menno, Wim and Cedric

From: Aspect-devel <aspect-devel-bounces at geodynamics.org> on behalf of Wolfgang Bangerth <bangerth at tamu.edu>
Sent: Friday, February 10, 2017 11:04:11 PM
To: aspect-devel at geodynamics.org
Subject: Re: [aspect-devel] compressibility in 2d

On 02/10/2017 02:18 PM, cedric thieulot wrote:
> So we’ve been talking about it with Menno today and it looks indeed
> that the literature is unanimous on this. However, if \dot{\bm
> \epsilon} is a 2x2 tensor/matrix the formula with 1/3 does not make
> it traceless. I/we suppose that there is a somewhat subtle argument
> here, but so far it has eluded us.

The argument is that you really have a 3-dimensional velocity vector,
but that the assumption that flow is 2-dimensional means that (i) the
z-component of the velocity is zero, and (ii) all derivatives of the
first two components with respect to z are also zero. So the gradient
(and symmetric gradient) of the real velocity has the form

   [ s_xx s_xy 0 ]
   [ s_yx s_yy 0 ]
   [ 0    0    0 ]

If you use this and subtract 1/3 of the trace times the identity, then
the resulting 3x3 tensor is trace-free.

> Furthermore, if one considers an infinite domain in, say, the z
> direction , dv_z/dz will be for sure nul, so that \dot{epsilon}_{zz}
> is nul but if we now look at the deviatoric strainrate,
> \dot{epsilon}_{zz}^{dev} will receive the value -1/3 (e_xx+e_yy).
> What does this mean ?

Yes, it is nonzero. But if you then again take the divergence of it
(where all z-derivatives are zero), then it doesn't matter that there
was something in the zz-spot.


Wolfgang Bangerth          email:                 bangerth at colostate.edu
                            www: http://www.math.colostate.edu/~bangerth/
Wolfgang Bangerth's Homepage - math.colostate.edu<http://www.math.colostate.edu/~bangerth/>
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