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I want to know that the difference is between the following equations:

$$ \frac{\delta h_d}{\delta t} = -\nabla.(\vec{v_s} h_d) = -u_s(\frac{\delta h_d}{\delta x}) -v_s(\frac{\delta h_d}{\delta y}) - h_d(\frac{\delta u_s}{\delta x}) - h_d(\frac{\delta v_s}{\delta y})$$

$$ \frac{\delta h_d}{\delta t} = -\vec{v_s}.\nabla(h_d) = -u_s(\frac{\delta h_d}{\delta x}) -v_s(\frac{\delta h_d}{\delta y})$$

Here, $\vec{v_s}$ is the 2D velocity vector with components $u_s$ and $v_s$, and $h_d$ the thickness of a rock layer that is being advected. How does the addition of the last 2 terms in the first equation change the physical meaning of the equation? Does it make a ddifference if the velocity field is constant or non-constant in time?

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The second equation only holds if the material is incompressible; i.e. if $\nabla\cdot {\bf v}=0$. I imagine that rock is rather incompressibe under most circustances. It will change density if it gets hot though --- for example in subduction.

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  • $\begingroup$ I thought that too, yet I found a paper that uses the first equation with the following comment: "The debris layer is transported with the ice flow downstream towards the glacier terminus. This debris advection is governed by the calculated surface flow field and therefore includes compressive or extensive flow and thus dynamic thickening or thinning of the debris cover. We calculate debris advection only at the glacier surface", see Eq. 3 of frontiersin.org/articles/10.3389/feart.2021.710276/full $\endgroup$
    – Yoni Verhaegen
    Commented Jul 4, 2022 at 14:58
  • $\begingroup$ Good point. I read $h_d$ as the density, but of course the thickness can change easily. This is like the depth in the equations for shallow water waves. Water is incompressible, but the depth changes when there is vertical velocity, which is not included in the 2d velocity field. $\endgroup$
    – mike stone
    Commented Jul 4, 2022 at 15:01
  • $\begingroup$ Okay, so in the end the first equation can be used despite the rock being incompressible? $\endgroup$
    – Yoni Verhaegen
    Commented Jul 4, 2022 at 15:20
  • $\begingroup$ Yes. It is correct in tha case. And also the velocities can depend on time. $\endgroup$
    – mike stone
    Commented Jul 4, 2022 at 21:43

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