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I am new to GIS and just get the idea of slowness. I can understand why people need the slowness in GIS rather than just the reciprocal of velocity. However, I feel the definition of slowness is weird if we further introduce horizontal and vertical slowness. By intuition, the horizontal/vertical slowness should be the reciprocal of horizontal/vertical speed, and either horizontal or vertical wave speed should be smaller than the wave speed, due as a component. But from MIT 12.510 Introduction to Seismology, it seems that wave speed is smaller than both horizontal wave speed and vertical wave speed.

In summary, only one of the definitions for either wave speed or slowness can enjoy the regular decomposition property, is this true?

screenshot from the reference

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I don't know if I fully grasp your question; but I can confirm that slowness and velocity vectors are kind of weird things :-)

Let me describe a simple example (and not worry too much about the mathematical details). If in the figure above, if you point your wave not diagonally but just straight down, what happens? Well, $dx$ becomes infinitely large when the incidence angle becomes $i=0^\circ$, because there is no observable thing happening along the upper horizontal surface, the wave only moves down! At the same time, $dz$ becomes exactly equal to $ds$ when the incidence angle $i=0^\circ$. So, $dx=\infty$ while $dz= ds$, right? I have drawn that in the figure below, for a monochromatic (single-frequency) wave, such that for all the 3 cases, the length of the vector $\vec{p}$ is identical! I have added $c_x$ and $c_z$ as the apparent velocities.

enter image description here

So for a wave moving straight down, the wave moves in 2D for a grand total distance of $ds$, it propagated 0 in the direction of $dx$, and $ds$ in the direction of $dz$. Now your research already suggests that either of the following relations should hold -- both basically expressions of the Pythagorean rule: $$ dx^2 + dz^2 = ds^2, \tag{a} $$ or $$ \frac{1}{dx^2} + \frac{1}{dz^2} = \frac{1}{ds^2} \tag{b} $$ If we use our extreme example $dx=\infty$ and $dz=ds$, then only equation b can hold (remember, $1/\infty=0$).

If we multiply every term in equation $b$ with $dt^2$, we get: $$ \frac{dt^2}{dx^2} + \frac{dt^2}{dz^2} = \frac{dt^2}{ds^2}, \tag{b'} $$ or realizing that $dx/dt=c_x$ and $dt/dx=1/c_x=p_x$ we get: $$ p_x^2 + p_z^2 = p^2 = \frac{1}{c^2}, \tag{b"} $$

With those details out of the way, we have seen the following: $p_x=0$ but $c_x=\infty$, and $p_z=1/c_z$ and $c_z=c$. So we have found that both $\{c_x,c_z\}\geq c$, i.e., both apparent velocities were greater than or equal to the original wave speed! On the other hand, $\{p_x,p_z\}\leq p$, i.e., the apparent slowness components were less than or equal to the original slowness.

And that is the result that holds generally. The apparent velocity components are always equal to or greater than the original velocity; while the apparent slowness components are always equal to or less than the original slowness. This is why equations (b), (b') and (b") give the right decomposition rule, it's the only one where $p_x$ and $p_y$ are less than the full $p$! This is why we can think of such a thing as a slowness vector $\vec{p}=(p_x,p_y)$, because the modulus of that vector $\lvert \vec{p} \rvert^2 = p_x^2+p_y^2$ (which is eq. (b")) behaves exactly like a vector should!

I hope that maybe along the way I answered your question somehow? My apologies for the rather mathematical take on things, but it's the only way I ever understood slowness vectors!

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  • $\begingroup$ Thanks for your explanation and time on plotting those graphs! I actually have math background but limited physics background. I think the problem on the decomposition of velocity might come from, the velocity of a trace(particle) is not the same as the velocity of a wave, where the former is much easier to define. The velocity of a wave cannot track by its particle, so as you said, it is by observable receivers. $\endgroup$ – Silentmovie Jun 15 at 20:18
  • $\begingroup$ Well -- the velocity of a wave is defined only by the particle motion over time, and if the particle motion is aligned with the grid, the velocity can be measured perfectly... But yes, along an angle, these individual particle movement components are divided by the sin/cos of the angle, such that the apparent velocity appears to be much larger than they are actually! Is your question answered? $\endgroup$ – Erik Jun 17 at 8:09
  • $\begingroup$ yes, thank you. $\endgroup$ – Silentmovie Jun 17 at 20:41

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