<p>The slowness vector <span class="math-container">$\mathbf{s}=[\xi,\eta]$</span> (your pdf &quot;<span class="math-container">$\mathbf n=[a,b]$</span>&quot; ) consists of the horizontal component <span class="math-container">$\xi$</span> (or your <span class="math-container">$a$</span>) and vertical component <span class="math-container">$\eta$</span> (or your <span class="math-container">$b$</span>), then we have:
<span class="math-container">\begin{align}
   s=|\mathbf{s}|=\sqrt{\xi^2+\eta^2}=\frac{1}{c}
\end{align}</span>
which satisfies Pythagoras law <span class="math-container">$s^2=\xi^2+\eta^2$</span>. Explicitly, the horizontal and vertical components are related to the ray direction by
<span class="math-container">\begin{align}
   \sin(\theta)&amp;=\frac{\mathrm ds}{\mathrm dx}=c\frac{\mathrm dt}{\mathrm dx}=\frac{c}{c_x}=c\xi\\
   \cos(\theta)&amp;=\frac{\mathrm ds}{\mathrm dy}=c\frac{\mathrm dt}{\mathrm dy}=\frac{c}{c_y}=c\eta.
\end{align}</span>
Therefore
<span class="math-container">\begin{align}
   \xi&amp;=\frac{1}{c_x}=\frac{\sin(\theta)}{c}\\
   \eta&amp;=\frac{1}{c_y}=\frac{\cos(\theta)}{c}.
\end{align}</span></p>

CC BY-SA 4.0
2024-07-25T11:04:01.407Z
2024-07-25T11:04:01.407Z
MathArt
<p>I am working on a seismology assignment and it starts with the fundamentals of seismology such as the wave equation, the slowness... such as the question below:</p>
<p><em>write an equation describing the horizontal slowness (reciprocal of the velocity). This becomes a vector quantity for propagation in multiple directions.</em></p>
<p>I haven't fully grasped these concepts yet. And I looked around for lecture notes and this is what I found:
<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-introduction to seismology</a> and <a href="https://i.sstatic.net/7obCU.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7obCU.png" alt="continuum mechanics lecture notes" /></a></p>
<p>I am not sure which one is the answer to this question or what is even the difference between the two because they both seem to represent the slowness</p>

CC BY-SA 4.0
2021-03-02T12:23:57.470Z
Whoah, that seems like a lot of research! (So many tabs.)
User123
CC BY-SA 4.0
2021-03-02T13:52:34.010Z
Well... I have to x(
Rim Sleimi
<seismology><waves>
Horizontal slowness in seismology
Rim Sleimi
# Horizontal slowness in seismology

(Asked by: username_0 on 2021-02-26)

I am working on a seismology assignment and it starts with the fundamentals of seismology such as the wave equation, the slowness... such as the question below:

*write an equation describing the horizontal slowness (reciprocal of the velocity). This becomes a vector quantity for propagation in multiple directions.*

I haven't fully grasped these concepts yet. And I looked around for lecture notes and this is what I found:
[MIT-introduction to seismology](https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-510-introduction-to-seismology-spring-2010/lecture-notes/lec5.pdf) and [![continuum mechanics lecture notes](https://i.sstatic.net/7obCU.png)](https://i.sstatic.net/7obCU.png)

I am not sure which one is the answer to this question or what is even the difference between the two because they both seem to represent the slowness

**Tags:** seismology, waves

> **Comments:**
> **username_1**: Whoah, that seems like a lot of research! (So many tabs.)  
> **username_0**: Well... I have to x(  

---

## 1 answers

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

The slowness vector $\mathbf{s}=[\xi,\eta]$ (your pdf "$\mathbf n=[a,b]$" ) consists of the horizontal component $\xi$ (or your $a$) and vertical component $\eta$ (or your $b$), then we have:
\begin{align}
s=|\mathbf{s}|=\sqrt{\xi^2+\eta^2}=\frac{1}{c}
\end{align}
which satisfies Pythagoras law $s^2=\xi^2+\eta^2$. Explicitly, the horizontal and vertical components are related to the ray direction by
\begin{align}
\sin(\theta)&=\frac{\mathrm ds}{\mathrm dx}=c\frac{\mathrm dt}{\mathrm dx}=\frac{c}{c\_x}=c\xi\\
\cos(\theta)&=\frac{\mathrm ds}{\mathrm dy}=c\frac{\mathrm dt}{\mathrm dy}=\frac{c}{c\_y}=c\eta.
\end{align}
Therefore
\begin{align}
\xi&=\frac{1}{c\_x}=\frac{\sin(\theta)}{c}\\
\eta&=\frac{1}{c\_y}=\frac{\cos(\theta)}{c}.
\end{align}


---

