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

<p>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, <span class="math-container">$dx$</span> becomes infinitely large when the incidence angle becomes <span class="math-container">$i=0^\circ$</span>, because there is no observable thing happening along the upper horizontal surface, the wave only moves down! At the same time, <span class="math-container">$dz$</span> becomes exactly equal to <span class="math-container">$ds$</span> when the incidence angle <span class="math-container">$i=0^\circ$</span>. So, <span class="math-container">$dx=\infty$</span> while <span class="math-container">$dz= ds$</span>, 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 <span class="math-container">$\vec{p}$</span> is identical! I have added <span class="math-container">$c_x$</span> and <span class="math-container">$c_z$</span> as the <em>apparent</em> velocities.</p>

<p><a href="https://i.sstatic.net/TyQMA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TyQMA.png" alt="enter image description here"></a></p>

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

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

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

<p>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 <span class="math-container">$p_x$</span> and <span class="math-container">$p_y$</span> are less than the full <span class="math-container">$p$</span>! This is why we can think of such a thing as a slowness vector <span class="math-container">$\vec{p}=(p_x,p_y)$</span>, because the modulus of that vector <span class="math-container">$\lvert \vec{p} \rvert^2 = p_x^2+p_y^2$</span> (which is eq. (b")) behaves exactly like a vector should!</p>

<p>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!</p>

CC BY-SA 4.0
2019-06-15T20:18:49.117Z
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.
Silentmovie
CC BY-SA 4.0
2019-06-17T08:09:25.030Z
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?
Erik
CC BY-SA 4.0
2019-06-15T11:51:13.637Z
2019-06-15T13:23:51.833Z
2019-06-15T13:23:51.833Z
Erik
<p>I have no right to comment.</p>
<p>But here might be an explanation from mathematical perspective. The abnormal vector decomposition can be thought of as the wave number leakage into another dimension. Solution to a general 3D Laplace equation gives a plane wave say in <span class="math-container">$xy-$</span> plane. Since Laplace's equation is a special case of the Helmholtz equation, whereas later can have wave oscillations in all three directions. In this case, if the horizontal wavenumber <span class="math-container">$|\mathbf{k}_{xy}|=\sqrt{k_x^2+k_y^2}$</span> is less than the total wavenumber <span class="math-container">$k=\frac{\omega}{c}=\sqrt{k_x^2+k_y^2+k_z^2}$</span>. Then we can obtain a real vertical wavenumber <span class="math-container">$k_z=\sqrt{k^2-k_x^2-k_y^2}$</span> since <span class="math-container">$k^2&gt;k_x^2+k_y^2$</span>. Otherwise, <span class="math-container">$k_z$</span> will be imaginary which is why it is called slowness representing the wave wave is decaying.</p>
<p>Your correction will be appreciated!</p>

CC BY-SA 4.0
2024-07-25T13:04:40.527Z
2024-07-25T13:04:40.527Z
MathArt
<p>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 <em><a href="https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-510-introduction-to-seismology-spring-2010/lecture-notes/lec5.pdf" rel="nofollow noreferrer">MIT 12.510 Introduction to Seismology</a></em>, it seems that wave speed is smaller than both horizontal wave speed and vertical wave speed.</p>

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

<p><a href="https://i.sstatic.net/e5zfc.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/e5zfc.png" alt="screenshot from the reference"></a></p>

<seismology><waves><seismic>
wave speed is smaller than both horizontal wave speed and vertical wave speed?
Silentmovie
# wave speed is smaller than both horizontal wave speed and vertical wave speed?

(Asked by: username_0 on 2019-06-15)

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](https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-510-introduction-to-seismology-spring-2010/lecture-notes/lec5.pdf)*, 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](https://i.sstatic.net/e5zfc.png)](https://i.sstatic.net/e5zfc.png)

**Tags:** seismology, waves, seismic


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## 2 answers

### Answer by username_1 on 2019-06-15. Score: 2

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](https://i.sstatic.net/TyQMA.png)](https://i.sstatic.net/TyQMA.png)

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!

> **Comments:**
> **username_0**: 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.  
> **username_1**: 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?  

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### Answer by username_2 on 2024-07-25. Score: 0

I have no right to comment.

But here might be an explanation from mathematical perspective. The abnormal vector decomposition can be thought of as the wave number leakage into another dimension. Solution to a general 3D Laplace equation gives a plane wave say in $xy-$ plane. Since Laplace's equation is a special case of the Helmholtz equation, whereas later can have wave oscillations in all three directions. In this case, if the horizontal wavenumber $|\mathbf{k}\_{xy}|=\sqrt{k\_x^2+k\_y^2}$ is less than the total wavenumber $k=\frac{\omega}{c}=\sqrt{k\_x^2+k\_y^2+k\_z^2}$. Then we can obtain a real vertical wavenumber $k\_z=\sqrt{k^2-k\_x^2-k\_y^2}$ since $k^2>k\_x^2+k\_y^2$. Otherwise, $k\_z$ will be imaginary which is why it is called slowness representing the wave wave is decaying.

Your correction will be appreciated!


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