# Acoustic impedance in geophysical context

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?

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): 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 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

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

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.