Geophysics: Calculating Inner Core Radius

[s7b]

I have a question about some geophysics...


When we want to calculate the radius of an inner core how do you do it. If I were to be given the radius of the whole planet as well as density and the velocity of the Pwave, where do I begin to solve. Also let's say the angular distances where P-waves are observed are also given.

 

[billiards]

If you have the P-wave velocity profile, you will see the velocity gradually increases with a few little jumps in it, then it suddenly drops down dramatically and then steadily begins to increase, and then there is another giant leap up in the velocity. That last giant leap up is the velocity going from the outer core to the inner core -- the radius of the inner core is radius of the Earth minus the depth of this giant leap.

Incidentally, if you didnt have the velocity profile, but you had observations of P-waves from all over the world, you could invert the traveltime data to find the velocity profile which would then show you the inner core depth in the way I described above.

You might also want to look at the moment of inertia of Earth, as that constrains density distribution which might be useful in finding inner core radius.

 

[K^2]

The actual wave solution is incredibly complex, but I think you can get a descent estimate based on the following assumptions.

1) There is only one sharp change in density, and that defines the core. (The crust is not going to make much of a difference for this approach, as it is sufficiently thin.)

2) Geometrical optics approach is sufficient. So we don't have to talk about diffractions of the wave, just refractions at the interfaces.

I'm going to use terms from geometrical optics, so I'll call the path a P-wave took to get from where it originated to where it was detected a "beam".

The problem is cylindrically symmetric, so we can consider a 2D problem only. So picture Earth as two concentric circles with radii R₁<R₂. R₂ is Earth's radius, and R₁ is what you are trying to find. Inner circle is the core with some density giving speed of P-wave propagation c₁, and the outside of that circle the P-wave propagates at c₂. I will denote the center of the two circles as point O.

Consider a beam that leaves point A on the outer circle and hits the core at point B. Call angle OAB θ₂ and AOB θ₁. Distance from A to B is d. Then:

R_1 sin(\theta_1) = d sin(\theta_2)

R_1 cos(\theta_1) + d cos(\theta_2) = R1 + R2

After entering the core, the beam is refracted and exits the core at point C. Refraction will depend on wave speed propagation. If we call angle of incidence α and angle of refraction β, we can define the following two relations.

\alpha = \theta_1 + \theta_2

\frac{sin(\beta)}{c_1} = \frac{sin(\alpha)}{c_2}

First is purely geometric, second is due to Snell's Law of refraction, which holds for P-waves in this approximation.

After exiting core at C, the beam continues to point D on outer circle, where it is detected. Note that angle OBC is the angle of refraction β, and it is equal to angle OCB. Angle BOC is then 180°-2β. That gives us angle AOD = AOB + BOC + COD = BOC + 2*AOD. And that's the angle from epicenter to the point where the wave was detected. Let me denote it as γ.

\gamma = 180° - 2 (\beta - \theta_1)

So you can find point of exit as function of θ₁. Though, not a trivial one. Now you need the time. You already know that the beam will travel distance d, defined by an earlier equation, before it hits the core. It will travel the same distance after leaving the core. What's left is to compute distance within the core. Call that distance b. Again, using triangle BOC for reference, we can see that the following holds.

b = 2 R_1 cos(\beta)

And this finally gives us time.

t = 2 \frac{d}{c_2} + \frac{b}{c_1}

Now you can take your data and fit it to the unknown R₁ and θ₁ on known γ and t. It's going to be an ugly fit, but presumably you'd have a computer program do it, so there shouldn't be any problems. If you have good data from opposite side of the planet with precise times, you should be able to get a very good estimate for the radius.

 

[billiards]

True that it is a complex problem. In fact it is more complex than you make out.

1) There is only one sharp change in density, and that defines the core. (The crust is not going to make much of a difference for this approach, as it is sufficiently thin.)

There are two profound interfaces. The first you correctly identified is the core/mantle boundary; the second is the outer/inner core boundary, which I believe is the boundary the questioner is interested in.

2) Geometrical optics approach is sufficient. So we don't have to talk about diffractions of the wave, just refractions at the interfaces.

Yes, geometrical ray theory gives a pretty good approximation to the true finite-frequency wavefield, which is useful in determining a useful background velocity model of the Earth.

Inner circle is the core with some density giving speed of P-wave propagation c₁, and the outside of that circle the P-wave propagates at c₂.

This is much too simple. In fact the velocity increases smoothly with depth in the Earth, this will have profound implications to the paths that rays take, and hence the observations you make at the surface.

 

[K^2]

Well, generalizing to two interfaces is simple enough, but if you introduce variable density, you are pretty much stuck doing finite element analysis on the thing. You can still get away with a 2D case, and if we assume that density varies with radius only, this can be solved easily enough, but it's a purely numerical problem now. You make a guess for the density, construct the response, and then start adjusting the density function until you hit the correct response.