Earth

Glaciers shape valleys

Fill a river’s V-shaped valley with a glacier six hundred metres thick and watch a hundred thousand years go by. The ice flows fastest at the middle of its surface and not at all against the bed; the shear it drags along the bed is greatest on the walls of the V and vanishes in its narrow floor, so sliding and abrasion cut the walls back and the valley widens towards a U - the old V stays drawn for comparison. Clean ice cuts far less than ice full of rock; with no sliding it cuts nothing; let the glacier advance and retreat in cycles. Every cubic metre the bed loses goes to the moraine. In a semicircular channel the solve is the closed-form Poisson solution.

By Mykhailo Makarov

Actual application image

What happens

A valley cross-section 2400 m wide on a grid of 61 by 40. The down-valley velocity of Newtonian ice (10¹³ Pa·s, slope 0.05) solves eta grad²u = −rho g sin alpha with u = 0 on the bed and du/dz = 0 at the surface; the irregular bed is met by Shortley-Weller differences and the solution by over-relaxation from the last one. The shear on the bed drives sliding in proportion - a stated simplification, the sliding does not enter the solve - and the sliding abrades the bed (Hallet), scaled by the debris in the ice and the rock’s strength. Every cubic metre goes to the moraine. Steps of 1 thousand years; one second is a thousand years at time scale 1.

Try this

Watch the V widen towards a U and read the profile exponent; compare clean and dirty ice; stop the sliding; let the glacier advance and retreat in cycles; check the semicircle against the closed form.

The model and its limits

A cross-section without along-valley flow, Newtonian ice, sliding proportional to the computed shear, a single abrasion law; plucking and rock fracture are not modelled. Profiles belong to this idealized valley.

Equations used in the module

Equations
η·∇²u = −ρ·g·sin α,  u = 0 on the bed,  ∂u/∂z = 0 at the surface
τ_b = η·|∇u| at the bed,  u_b = c_s·τ_b
dz_b/dt = −K·D/S·u_b·√(1 + (∂z_b/∂y)²)
semicircle: u = ρ·g·sin α·(R² − r²)/(4η),  τ_b = ρ·g·sin α·R/2

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