<p>P and S waves are fundamentally different, when it comes to properties of the wave. An example might be that <a href="http://en.wikipedia.org/wiki/P-wave">P waves</a> can travel through fluids while S waves cannot. However, when it comes down to wave theory, these two are just different polarizations of a <a href="http://en.wikipedia.org/wiki/Mechanical_wave">mechanical wave</a>. </p>

<h3>Seismology</h3>

<p>In seismics this concept may be puzzling as we make some very fundamental assumptions. However, seismologists are very well aware that these polarizations (they call them <em>phases</em>) usually change throughout their travelpath. When you look at the image below, you can see different travelpaths and the same source will trigger P and S waves. This P wave will travel to the point P where it can be recorded as direct P wave. Upon reflection it will be partially converted to a S wave, which you can also see on the nomination PP is the P part of the reflected P wave and PS is the S part of the reflected and converted P wave. (Actually you can go on happily converting back and forth.)</p>

<p><img src="https://i.sstatic.net/G5D4p.jpg" alt="Travelpaths of waves after EQ"></p>

<h3>Seismics</h3>

<p>In seismics we get this conversion on interfaces or layer boundaries (whichever name you use). So everytime a wave is reflected it may also be partially converted into a different polarization. So essentially, your statement is wrong. The wave doesn't just convert passing through a layer, it is converted upon interaction with an interface. 
So while this wasn't a yes directly to your question let me elaborate why this conversion happens.</p>

<p><img src="https://i.sstatic.net/wBlkT.gif" alt="Reflectivity of PP and PS conversions at an interface - Source: Crewes.org"></p>

<p>This picture basically shows the rate of conversion depending on the incident angle. At 30° the rate of conversion is maximized. (The scales are terrible to read sorry.) At this interface we can see the incident wave as excitor of a new wave package. So we have to take a detour into wave excitation.</p>

<h3>Exctiting waves</h3>

<p>In marine seismics we can excite pure P waves, because we're operating in a fluid. On land you will always excite a mix of P and S waves (and variations thereupon, namely surface waves). The excitation of these waves is direction dependent. Starting a P wave in one direction will always start an S wave component perpendicular to this direction. Of course we have some energy distribution so that we get something like this:</p>

<p><img src="https://i.sstatic.net/9n4il.png" alt="Daghero, Dario, et al. &quot;Strong-coupling d-wave superconductivity in PuCoGa5 probed by point-contact spectroscopy.&quot; Nature communications 3 (2012): 786."></p>

<p>P waves in the <strong>+</strong> direction and S waves in the <strong>-</strong> direction. (This image was vor conductivity but it serves the point.)</p>

<h3>At the interface</h3>

<p>We have to look at the incident wave as an excitor at the interface, which will give us a P and a S wave response. In seismics, <a href="http://en.wikipedia.org/wiki/Focal_mechanism">wave theory</a> and other solutions you can usually see the S wave solution stands perpendicular on a P wave solution (except when we're dealing with anisotropy). So yes in the end your incident wave will start two P waves and two S waves that look like this:</p>

<p><img src="https://i.sstatic.net/Irhxt.gif" alt="Crewes.org"></p>

CC BY-SA 3.0
2014-06-16T10:22:04.107Z
2014-06-16T10:22:04.107Z
Jesper
<p>While passing through layers inside the earth some P waves get converted to S waves and then back to P waves while returning towards the surface. Is this statement true? </p>

<p>If yes, then why? (Considering that they are both fundamentally different in terms of wave theory - longitudinal vs. transverse)</p>

<geophysics><seismology><waves>
P wave to S wave conversion
shrey
# P wave to S wave conversion

(Asked by: username_0 on 2014-06-16)

While passing through layers inside the earth some P waves get converted to S waves and then back to P waves while returning towards the surface. Is this statement true?

If yes, then why? (Considering that they are both fundamentally different in terms of wave theory - longitudinal vs. transverse)

**Tags:** geophysics, seismology, waves


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## 1 answers

### Answer by username_1 on 2014-06-16. Score: 16

P and S waves are fundamentally different, when it comes to properties of the wave. An example might be that [P waves](http://en.wikipedia.org/wiki/P-wave) can travel through fluids while S waves cannot. However, when it comes down to wave theory, these two are just different polarizations of a [mechanical wave](http://en.wikipedia.org/wiki/Mechanical_wave).

### Seismology

In seismics this concept may be puzzling as we make some very fundamental assumptions. However, seismologists are very well aware that these polarizations (they call them *phases*) usually change throughout their travelpath. When you look at the image below, you can see different travelpaths and the same source will trigger P and S waves. This P wave will travel to the point P where it can be recorded as direct P wave. Upon reflection it will be partially converted to a S wave, which you can also see on the nomination PP is the P part of the reflected P wave and PS is the S part of the reflected and converted P wave. (Actually you can go on happily converting back and forth.)

![Travelpaths of waves after EQ](https://i.sstatic.net/G5D4p.jpg)

### Seismics

In seismics we get this conversion on interfaces or layer boundaries (whichever name you use). So everytime a wave is reflected it may also be partially converted into a different polarization. So essentially, your statement is wrong. The wave doesn't just convert passing through a layer, it is converted upon interaction with an interface.
So while this wasn't a yes directly to your question let me elaborate why this conversion happens.

![Reflectivity of PP and PS conversions at an interface - Source: Crewes.org](https://i.sstatic.net/wBlkT.gif)

This picture basically shows the rate of conversion depending on the incident angle. At 30° the rate of conversion is maximized. (The scales are terrible to read sorry.) At this interface we can see the incident wave as excitor of a new wave package. So we have to take a detour into wave excitation.

### Exctiting waves

In marine seismics we can excite pure P waves, because we're operating in a fluid. On land you will always excite a mix of P and S waves (and variations thereupon, namely surface waves). The excitation of these waves is direction dependent. Starting a P wave in one direction will always start an S wave component perpendicular to this direction. Of course we have some energy distribution so that we get something like this:

![Daghero, Dario, et al. "Strong-coupling d-wave superconductivity in PuCoGa5 probed by point-contact spectroscopy." Nature communications 3 (2012): 786.](https://i.sstatic.net/9n4il.png)

P waves in the **+** direction and S waves in the **-** direction. (This image was vor conductivity but it serves the point.)

### At the interface

We have to look at the incident wave as an excitor at the interface, which will give us a P and a S wave response. In seismics, [wave theory](http://en.wikipedia.org/wiki/Focal_mechanism) and other solutions you can usually see the S wave solution stands perpendicular on a P wave solution (except when we're dealing with anisotropy). So yes in the end your incident wave will start two P waves and two S waves that look like this:

![Crewes.org](https://i.sstatic.net/Irhxt.gif)


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