Notes/vanderVeen Aug5

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Crash Course in Glacier Dynamics

Kees van der Veen, University of Kansas August 5, 2009 Portland Summer Modeling School

What's the objective of an ice sheet model?

  • Understand evolution of ice sheet given some forcing (global warming, etc.)

Fundamental equations: conservation of xxx

  • Mass
  • Energy
  • Momentum

Conservation of Mass: Continuity Equation

  • What comes in (flux, basal freezing if, accumulation if) to some control volume must go out (flux, basal melting if, ablation if).
  • Assumption: ice is incompressible, so density is constant. Mass conservation ~ volume conservation
  • Assumption: ignore firn layer (100-150m in Antarctica, less in Greenland)

 M \Delta x + H(x)U(x) - H(x + \Delta x) U(x + \Delta x) = \frac{\Delta H}{\Delta t} \Delta x

 \frac{\Delta H}{\Delta t} = -\frac{H(x + \Delta x)U(x+\Delta x) - H(x)U(x)}{\Delta x} + M

Shrink timestep & spatial step to infinitessimal to write as differential equation

 \frac{\partial H}{\partial t} = -\frac{\partial}{\partial x} HU + M

Conservation of Momentum: Newton's second law

  •  F = ma , with zero acceleration
  • so the sum of all forces must be zero.
  • stresses are easier to work with than forces: stress is force per unit area
  • Nine stress components:  \sigma_{ij}
  • i: plane perpendicular to axis (x)
  • j: direction of stress
  • Stress tensor is symmetric, so  \sigma_{ij} = \sigma_{ji} and there are really only six distinct stress components
  • 3 equations with 6 unknowns

Force balance in z

 F_z = 0

 \sigma_{zz}(z + \Delta z) \Delta x \Delta y + \sigma_{xz}(z+\Delta z) \Delta z \Delta y + \sigma_{yz}(z + \Delta z) \Delta x \Delta z = \rho g \Delta x \Delta y \Delta z

 \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} = -\rho g

Force balance in x

 \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z} = 0

Force balance in y

(we won't concern ourselves with the transverse y direction, which also means we can eliminate the y-derivative terms in the z and x equations above)

Newton's First Law: action / reaction

  • What drives glacier flow? Gravity is the "action".
  • What is the response? Resistance to flow is the "reaction".

Exercise in deriving force balance in the horizontal

Integrate the force balance equation in x over the depth of the ice column to eventually derive an expression relating driving stress, basal drag, and longitudinal stress gradients.

Mathways, start with:

 \int_b^h \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z} dz = \int_b^h 0 dz

The integral in  \sigma_{xz} is easy because it is just the difference of the stress at the surface and the bed -- the two dz's cancel. But we have to use the Leibniz rule to work with the  \sigma_{xx} term, because the limits of integration h and b are really h(z) and b(z) - they depend on z, which we're trying to integrate over. Shucks.

Using the Leibniz Rule,  \int_b^h \frac{\partial \sigma_{xx}}{\partial x} dz = \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz - \sigma_{xx}(s)\frac{\partial s}{\partial x} + \sigma_{xx}(b)\frac{\partial b}{\partial x} .

The entire balance equation is thus

 \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz - \sigma_{xx}(s)\frac{\partial s}{\partial x} + \sigma_{xx}(b)\frac{\partial b}{\partial x} + \sigma_{xz}(s) - \sigma_{xz}(b) = 0 .

We can work on the first term, by writing out what  \sigma_{xx} is.

 \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz = \frac{\partial}{\partial x} \int_b^h [\R_{xx} - \rho g (h-z)] dz

We don't know anything about how  R_{xx} varies in z, so we won't touch that. But we can integrate the second term:

  \int_b^h \rho g (h-z) dz = \frac{1}{2} \rho g H^2

So the entire force balance equation is now

 \frac{\partial}{\partial x} \int_b^h R_{xx} dz - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x} - \rho g H\frac{\partial b}{\partial x}  + R_{xs}(s) - R_{xz}(b) = 0

We can simplify the terms involving  \rho g H now. Those are:

 - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] and  - \rho g H\frac{\partial b}{\partial x} .

If we differentiate the first term, we get  - \frac{1}{2} \frac{\partial H}{\partial x}[\rho g H] . We should note that since  H = s - b , it is also true that  \frac{\partial H}{\partial x} = \frac{\partial s}{\partial x} - \frac{\partial b}{\partial x} .

This lets us compare the first and second  \rho g H terms: their sum is just  - \rho g H \frac{\partial s}{\partial x} . This is the definition of the driving stress,  \tau_d .

So now the balance equation is  \frac{\partial}{\partial x} \int_b^h R_{xx} dz - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x} - \tau_d  + R_{xs}(s) - R_{xz}(b) = 0