# [aspect-devel] compressibility in 2d

Wolfgang Bangerth bangerth at tamu.edu
Fri Feb 10 14:04:11 PST 2017

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.

Best
W.

--
------------------------------------------------------------------------
Wolfgang Bangerth          email:                 bangerth at colostate.edu
www: http://www.math.colostate.edu/~bangerth/