<p><strong>We use $\rho V_\mathrm{P}$ because it agrees with experiment.</strong></p>

<p>You point out that density and velocity are correlated and ask, "why not use only velocity contrast or density contrast?" instead of their product.</p>

<p>In truth, P-wave velocity $V_\mathrm{P}$ velocity is not necessarily <em>that</em> well correlated with density $\rho$. Look at this plot of density vs $V_\mathrm{P}$ for rocks in a North Sea oil field from Avseth et al. (2001):</p>

<p><a href="https://i.sstatic.net/hAQ2N.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hAQ2N.png" alt="enter image description here"></a></p>

<p>Clearly, some of those rocks have almost no contrast on one axis, and can only be separated on the other. Indeed, this sums up the great problem of seismic inversion — we would like to recover these quantities individually, but all we can usually get at is impedance. </p>

<p>Another example of the non-correlation of velocity and density is the so-called "fizz effect": introducing a very small amount of gas in the pore-water, less than 5%, can reduce the P-wave velocity by as much as 20%, while having almost no effect on the density. The result: a substantial change in impedance (e.g. Han &amp; Batzle 2002).</p>

<p>The other thing about the product $\rho V_\mathrm{P}$ is that the impedance contrast at an interface gives us — quantitatively — the reflection coefficient $R$ at that interface (assuming vertical incidence; it's <a href="https://agilescientific.com/blog/2011/6/6/what-is-avo.html" rel="nofollow noreferrer">more complicated</a> for other angles):</p>

<p>$$ R = \frac{Z_2 - Z_1}{Z_2 + Z_1} $$</p>

<p>where $Z_2$ and $Z_1$ are the acoustic impedances of the lower and upper layers respectively.</p>

<p>To go back to your more general question about intuition. I agree, the intuition is slippery. The analogy to electrical circuits seems obscure, and I daresay it does to other geologists. "High impedance" sounds to me like "lots of opposition"... so why is the velocity high in high impedance rocks? Perhaps because of this, I tend to focus on thinking of high impedance rocks as 'hard' and low impedance rocks as 'soft'. I find this more geologically useful. </p>

<p>So I guess I don't have a very good answer for that part of your question.</p>

<p><strong>References</strong></p>

<p>Avseth, P, T Mukerji, G Mavko, and JA Tyssekvam (2001). Rock physics and AVO analysis for lithofacies and pore fluid prediction in a North Sea oil field. SEG <em>The Leading Edge</em>, April 2001, p 429–434. <a href="https://library.seg.org/doi/abs/10.1190/1.1438968" rel="nofollow noreferrer">doi 10.1190/1.1438968</a></p>

<p>D-H Han and M Batzle (2002). Fizz water and low gas-saturated reservoirs. SEG The Leading Edge, 21(4), 395-398. <a href="https://library.seg.org/doi/abs/10.1190/1.1471605" rel="nofollow noreferrer">doi 10.1190/1.1471605</a></p>

CC BY-SA 3.0
2018-01-04T19:05:57.700Z
2018-01-04T19:05:57.700Z
Matt Hall
<p>This concept actually comes from the electrical. They define the impedance as $Z=\frac{V}{I}$. 
Where V is potential difference/ voltage and I is current.</p>

<p>In acoustics, a similar expression can be obtained using the plane wave equation. 
Since,<br>
                $$\partial_t V = -\rho^{-1}\partial_xP$$
where $P$ is pressure and $V$ is velocity.
Now take fourier transform of both sides and defined impedance as $$Z(\omega)= \frac{P(\omega)}{V(\omega)} = \rho c$$
Where, obviously, $P$ is analougous to potential difference and $V$ is analougous to current.</p>

<p>You may refer to any wave theory book for details.</p>

CC BY-SA 3.0
2017-12-26T10:04:35.227Z
2017-12-26T11:23:54.607Z
2017-12-26T11:23:54.607Z
Amartya
<p>What is the intuition behind acoustic impedance? Yes, I know that it is the opposition to a seismic wave travelling in the medium. But why it is defined as density &times; velocity?</p>

<p>The velocity of a seismic wave depends on density, so why not use only the velocity contrast or density contrast between layers instead of the impedance contrast?</p>

<geology><geophysics><seismology>
Acoustic impedance in geophysical context
shul
# Acoustic impedance in geophysical context

(Asked by: username_0 on 2017-12-25)

What is the intuition behind acoustic impedance? Yes, I know that it is the opposition to a seismic wave travelling in the medium. But why it is defined as density × velocity?

The velocity of a seismic wave depends on density, so why not use only the velocity contrast or density contrast between layers instead of the impedance contrast?

**Tags:** geology, geophysics, seismology


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## 2 answers

### Answer by username_1 on 2018-01-04. Score: 1 | Accepted answer

**We use $\rho V\_\mathrm{P}$ because it agrees with experiment.**

You point out that density and velocity are correlated and ask, "why not use only velocity contrast or density contrast?" instead of their product.

In truth, P-wave velocity $V\_\mathrm{P}$ velocity is not necessarily *that* well correlated with density $\rho$. Look at this plot of density vs $V\_\mathrm{P}$ for rocks in a North Sea oil field from Avseth et al. (2001):

[![enter image description here](https://i.sstatic.net/hAQ2N.png)](https://i.sstatic.net/hAQ2N.png)

Clearly, some of those rocks have almost no contrast on one axis, and can only be separated on the other. Indeed, this sums up the great problem of seismic inversion — we would like to recover these quantities individually, but all we can usually get at is impedance.

Another example of the non-correlation of velocity and density is the so-called "fizz effect": introducing a very small amount of gas in the pore-water, less than 5%, can reduce the P-wave velocity by as much as 20%, while having almost no effect on the density. The result: a substantial change in impedance (e.g. Han & Batzle 2002).

The other thing about the product $\rho V\_\mathrm{P}$ is that the impedance contrast at an interface gives us — quantitatively — the reflection coefficient $R$ at that interface (assuming vertical incidence; it's [more complicated](https://agilescientific.com/blog/2011/6/6/what-is-avo.html) for other angles):

$$ R = \frac{Z\_2 - Z\_1}{Z\_2 + Z\_1} $$

where $Z\_2$ and $Z\_1$ are the acoustic impedances of the lower and upper layers respectively.

To go back to your more general question about intuition. I agree, the intuition is slippery. The analogy to electrical circuits seems obscure, and I daresay it does to other geologists. "High impedance" sounds to me like "lots of opposition"... so why is the velocity high in high impedance rocks? Perhaps because of this, I tend to focus on thinking of high impedance rocks as 'hard' and low impedance rocks as 'soft'. I find this more geologically useful.

So I guess I don't have a very good answer for that part of your question.

**References**

Avseth, P, T Mukerji, G Mavko, and JA Tyssekvam (2001). Rock physics and AVO analysis for lithofacies and pore fluid prediction in a North Sea oil field. SEG *The Leading Edge*, April 2001, p 429–434. [doi 10.1190/1.1438968](https://library.seg.org/doi/abs/10.1190/1.1438968)

D-H Han and M Batzle (2002). Fizz water and low gas-saturated reservoirs. SEG The Leading Edge, 21(4), 395-398. [doi 10.1190/1.1471605](https://library.seg.org/doi/abs/10.1190/1.1471605)


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### Answer by username_2 on 2017-12-26. Score: 3

This concept actually comes from the electrical. They define the impedance as $Z=\frac{V}{I}$.
Where V is potential difference/ voltage and I is current.

In acoustics, a similar expression can be obtained using the plane wave equation.
Since,  
$$\partial\_t V = -\rho^{-1}\partial\_xP$$
where $P$ is pressure and $V$ is velocity.
Now take fourier transform of both sides and defined impedance as $$Z(\omega)= \frac{P(\omega)}{V(\omega)} = \rho c$$
Where, obviously, $P$ is analougous to potential difference and $V$ is analougous to current.

You may refer to any wave theory book for details.


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