Weighty Problems (A tutorial)

Hmmm, another paper has emerged with a big role for dynamic topography as a cause for deformation in the western U.S. (Becker et al., Nature, 2015), and if you read the press releases and resulting news coverage you’d think this was the Big Answer to earthquakes away from plate boundaries.

Sorry, don’t think so.  But GG hasn’t had the time to really go through this paper in detail, and in any event feels kind of bad for picking on Thorsten earlier, who is a perfectly pleasant fellow. So let’s stand back and consider the root cause here, which is good old dynamic topography.  A simple test a lot of us like to apply is to look for free-air gravity anomalies that should be associated with dynamic topography.  While GG was involved with a paper that dealt with this is rather gory detail, let’s think of this really simply: what is a free-air anomaly, where and how can we use it, and what does it show where we can use it?

First, what it is.  Gravity at its most basic is an inverse-square law, so the gravitational acceleration is GM/r2, where G is Newton’s constant and M is the mass of Earth in this case. Because the earth rotates and is not quite a sphere, the acceleration varies with latitude. When we are talking about relatively small differences in elevation (and after correcting for tides) we can linearize this to g = g0Cz, where g0 is the value at sea level and C is a constant (basically 2GM/r03) and z is the elevation above sea level. The free air anomaly is simply to take the measured value of gravity, corrected for latitude, and add Cz to it.

Now, so what?  Consider isostasy.  If we have a point at an elevation h on rock with a density ρ1, if we are in isostatic equilibrium then there is a low density body at depth where the mass deficit equals the excess mass of the elevated area. So if we just pretend that this is an extra thickness of crust d extending into a mantle of density ρ2, then hρ1=d2 – ρ1) or d=hρ1/(ρ2 – ρ1).

Simple crustal isostasy.

Simple crustal isostasy. (“r”s are rhos (ρ) below)

What free-air gravity would we get from this? First, let’s pretend that the width of our elevated plateau is quite large compared to its thickness, which isn’t a bad approximation for scales where isostasy works. The gravitational attraction of an infinite slab of thickness h is 2πGΔρh. Note that there is no dependence on the depth of the slab (because it is infinite in extent).  So here we add the two contributions that differ from the sites to the right, which we set to be 0:

g =  2πG(ρ1h+(ρ1 – ρ2)d)
= 2πGh(ρ1+(ρ1-ρ2)ρ1/(ρ2 – ρ1))
= 2πGh(ρ1 – ρ1) = 0

So in the extreme of isostatic equilibrium over a broad area, the free air anomaly is zero.  Cool.

How about a plateau that is solely supported by flow of material in the mantle? This might push up the Moho and the surface, so it might look like this:

Dynamic

Gravity here now adds:

g =  2πG(ρ1h+(ρ2 – ρ1)h)
= 2πGhρ2

which, for a mantle density of 3250 kg/m³, works out to 0.14 mGal/m or 140 mGal/km of elevation. You can look on a free-air gravity map and see that the 1.5 km high Colorado Plateau and the more than 4 km high Tibetan Plateau have nothing like that kind of anomaly averaged over their whole area.

This is of course unfair because dynamic topography flows are driven by density anomalies in the mantle, so there is some low-density object below for this case.  The thing is, once we put our density anomaly at some depth in the mantle, our approximation of an infinite slab looks poorer and poorer. So let’s be somewhat more accurate and figure out the gravity from a finite length slab of material:

Dynamic uplift over a low density body at depth

Dynamic uplift over a low density body at depth

Now so long as the thickness of our compensating body t is a lot less than d, the shape of the resulting gravity anomaly will depend on the ratio w/d (when we plot x/d on the x-axis) and the amplitude on (ρ3 – ρ2)t. If we start with the total mass deficit in the body equalling the mass surplus in our elevated plateau+uplifted Moho, we can see how close we come to matching the gravity anomaly from the plateau and uplifted Moho (which we’ll continue to pretend is an infinite slab):

Plot of how closely the gravity anomaly of a finite width slab compares with that of an infinite slab

Plot of how closely the gravity anomaly of a finite width slab compares with that of an infinite slab

So imagine that we have a plateau 500 km wide we think is supported by a load 500 km deep. Our slab width/slab depth is 1, so the gravity anomaly from our load is under a third of that of the displaced surface material, so our gravity anomaly is (2/3)*2πGρ2h or about 93 mGal/km. Still a pretty large number. To get down into the 20 mGal/km range we need the gravitational attraction of the slab to be about 85% of the surface load, which means slab width/slab depth = 8 from the plot above, so if our plateau if 500 km wide, we need the load to be at a depth of 500/8 = 62.5 km, which is pretty much an isostatic depth.  Even a 1000 km wide plateau would need the load to be at a depth of 125 km, which is a lithospheric depth. (And if you work with a disc of material instead of a slab extending infinitely in and out of the section, this gets even worse).

Does this mean there is no way to have dynamic loads? Well, not entirely, but it makes it hard. Now you could argue that the mass deficit of the dynamic load needs to be of higher magnitude than the surface load in order to generate the total flow field, which will be wider than the body, so you could relax the depth somewhat, but you then will run into the problem that you have to tune things to a particular depth (you can tune by adjusting the viscosity structure of the earth).

So, let’s look at one example.  The High Plains in the central U.S. are a puzzle and seem like something you could imagine being made by dynamic effects.  Here is an east-west profile of the free air anomaly of surface points:

East-west profile of free air anomaly--Rockies left of -105, Plains to east (right) of that.

East-west profile of free air anomaly–Rockies left of -105, Plains to east (right) of that.

Pretty darn close to zero in the Plains (the bump near 97 is the mid-continent rift). But, you say, look at the Rockies–maybe this works there?

The problem in the Rockies (and nearly any other mountainous area) is that surface free air measurements cannot be corrected for topography.  While a point on a mountaintop will have a positive contribution to the free air gravity from the mass of the mountains below, a point in a valley will have a negative contribution from those same mountains. A traditional terrain correction doesn’t help. So you get a lot of scatter, and it turns out that to correct all this out is pretty gruesome. (An alternative is to use satellite based gravity).

So there are some thoughts on gravity and topography.  Presumably some of you will let GG know if there are mistakes, as can happen when pounding something like this out on the spur of the moment…

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