# Difference between revisions of "Notes/vanderVeen Aug5"

## Crash Course in Glacier Dynamics

Kees van der Veen, University of Kansas

August 5, 2009

Portland Summer Modeling School

notes by Kristin Poinar

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 $F_z = 0$ $\sigma_{zz}(z + \Delta z) \Delta x \Delta y + \sigma_{xz}(x+\Delta x) \Delta z \Delta y + \sigma_{yz}(y + \Delta y) \Delta x \Delta z = \rho g \Delta x \Delta y \Delta z$ $\frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} = -\rho g$

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

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 $\frac{\partial}{\partial x} \int_b^h R_{xx} dz - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x} + R_{xs}(s) - R_{xz}(b) - \tau_d = 0$

We will apply the zero-stress surface boundary condition, which says that $R_{xz}(s) - R_{xx}(s) \frac{\partial s}{\partial x} = 0$.

So we can lose those terms, and the balance equation becomes $\frac{\partial}{\partial x} \int_b^h R_{xx} dz + R_{xx}(b)\frac{\partial b}{\partial x} - R_{xz}(b) + \tau_d = 0$

And, finally, we'll substitute the basal drag, $\tau_b$, for its definition: $\tau_b = R_{xz}(b) - R_{xx}(b)\frac{\partial b}{\partial x}$

to arrive at the final answer, $\frac{\partial}{\partial x} \int_b^h R_{xx} dz - \tau_b + \tau_d = 0$