Kernel values are not the same as Dahlen Kernels

[sstaehler]

I have taken this part from #22, since it is a new question.

I have calculated the Kernels with (presumably) the same physical parameters as @kasra-hosseini did with the Dahlen method, and they look very similar, alas, they differ by 7 orders of magnitude (left: MC Kernel, right: MC Tony Dahlen).

Also, there are differences in the source region and with a second Fresnel zone above the kernel. On the one hand, this is something we expected, on the other hand, it might affect tomography quite significantly. Exciting times, people!

[kasra-hosseini]

Cool!
Have you done this comparison for 60degrees? That should be interesting since Dahlen kernels should be quite reasonable for that epicentral distance. However for 90 degrees, the sensitivity kernel already starts to sense the CMB...deviating from ray theory approximation.

[kasra-hosseini]

Why the colors are flipped by the way?

[sstaehler]

What do you mean by flipped?

[kasra-hosseini]

That the blue part of the kernel is positive in MC and negative in Dahlen.

[sstaehler]

That is a very good question, I'll look into how that happened.
Another question: Are the Dahlen Kernels absolute or relative?

[kasra-hosseini]

by relative you mean K.dV/V? Yes, and I am saying it because the sum over all elements will be the travel-time. Does it make sense?

[sstaehler]

I am not sure. Never got to think that through

[tnissen]

Did you honor the minus sign in the basic kernel definitions (eq. 5.13) ?

On 17/03/2015 19:09, Kasra Hosseini wrote:

That the blue part of the kernel is positive in MC and negative in Dahlen.

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#23 (comment).

Tarje

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[sstaehler]

@tnissen Right, I forgot the minus sign, thanks. Fixed in 970a532

[sstaehler]

The different amplitudes might be related to an apparent error in normalization (see #24)

[sstaehler]

After fixing #24, the difference is 10^19 instead of 10^7.

[sstaehler]

It seems that the division by amplitude of the AxiSEM run is pretty inconsistent. Fans of the code do remember this discussion.

Might be related. Stay tuned for more

[sstaehler]

Almost solved by 6afd224. Now only a factor of roughly 5 is remaining. Also interesting is how the relative amplitudes in the middle and at the ends is different for MC Kerner (left) and MC Tony (right)

[kasra-hosseini]

Awesome!

About the relative amplitude in the middle and at the ends, and in general detailed comparison in the shapes of the kernels, I would suggest to do it in a very fine inversion grid, something like the following picture. I can prepare the required inversion grids as needed.

[sstaehler]

Yes, that would be great!

[auerl]

Cool! In terms of how relative kernels in [s/m^3] compare to published plots of kernels, they are now also relatively similar 5x10^-17 (our kernels) vs 5x0^-15 (published kernels).

(for comparison, from Zhao & Chevron, 2012, in s/km^3)

[kasra-hosseini]

@sstaehler I thought that it is better to put the information for fine inversion grid here for future references.

Here is the address where you can find the fine inversion grid (ETH machine):
/home/khosseini/inversion_mesh/mesh_example/fine_20_96

The most interesting part is:
-45<=lat<=45 and lon=0 ---> edge-length = 20km
the rest: 600km

WARNING: I changed the source code of the Dahlen kernels on ETH machine to have exact latitudes (in the original version, they will be translated into geocentric..., so please do not use the version in the ETH machine for tomography :) )

An example for 60degrees epicentral distance (source latitude = 30, receiver latitude = -30, longitudes = 0, depths = 0):

The sum over all elements were:-602.57562715
Theoretical arrival time: 608.28
Error: 0.9% (which should be acceptable with the criteria defined in Tian et al, 2007)