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Source: http://www.engineeringtoolbox.com/sound-power-intensity-pressure-d_57.html

Both sound intesity and pressure level are measured in dB. Given a specific sound, are these two dB values the same?

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The first thing to understand is that dB is a logarithm of a ratio. More specifically, a logarithm of a ratio of power/power (which is why dB has no units). A Bel is the base-10 logarithm of a power ratio. A decibel slices each Bel into ten parts.

Power is power. It does not matter whether it is sound or electricity or whatever. And dB always refers to a ratio of power.

In most sound references, "dB" means "dB in reference to (as a ratio over) 'X'." Where 'X' is a reference level. I think the reference used is the sound pressure which is the threshold of human hearing. So a sound with 10 times the POWER of that reference would be 10 dB. A sound with POWER 100 times the reference would be 20 dB.

The answer to your question is: It can be. Sound intensity (if expressed as a pressure) and Sound Pressure Level (SPL) are the same. But neither is power. Sound power is proportional to sound pressure squared. Thus, if you increase sound PRESSURE by a factor of 10, you have increased POWER by a factor of 10^2 = 100. These two actions are synonomous, and result in a 20dB increase. Note that I did not explicitly state what increased by 20dB in that last sentence... because it is ALWAYS power.

So, you need to watch your units, and make sure the other guy is watching his. If he doubles his POWER level, that is a 3 dB increase. If he doubles his sound PRESSURE, that is a 6dB increase. If he doubles his sound INTENSITY, you need to nail down whether that is in units of pressure or power.

I hope that helped.

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  • $\begingroup$ I would add that sound intensity is measured in power/area units, not dB, and really shouldn't be force/area either. The related quantity measured in dB specifically should contain the word "level", e.g., sound intensity level or sound level. $\endgroup$
    – Bill N
    Commented Jul 31, 2015 at 20:34
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I think so, yes. They are both measures of the loudness of what we hear. If $L_p$ is the sound pressure level and $L_i$ is the sound intensity, then: \begin{equation} L_p =20 \log_{10}\left(\frac{p_{\mathrm{rms}}}{p_{\mathrm{ref}}}\right)\mbox{ dB} = 10 \log_{10}\left(\frac{{p_{\mathrm{{rms}}}}^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB} \end{equation} The intensity of the sound wave, however, is proportional to the square of the sound pressure, or $I \propto p_{\mathrm{rms}}^2$. This gives: \begin{equation} L_p = 10 \log_{10}\left(\frac{{p_{\mathrm{{rms}}}}^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB} = 10 \log_{10}\left(\frac{{I_{\mathrm{{rms}}}}}{{I_{\mathrm{ref}}}}\right)\mbox{ dB} = L_i \end{equation}

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  • $\begingroup$ They are related, but they are not the same. In fact, intensity is never measured in dB; it is power/area. Intensity level is measured in dB. See @Vintage answer. Secondly, they are not measures of loudness. Loudness is measured in phons, and is equal to the sound intensity level (SIL) at a frequency of 1000 Hz. Loudness is an psycho-physical-acoustic response of the ear to a particular external SIL; loudness is a human perception averaged over many people and strongly depends on frequency, as well as external SIL. 50 dB at 400 Hz is vastly different loudness from 50 dB at 1000 Hz. $\endgroup$
    – Bill N
    Commented Jul 31, 2015 at 20:44
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First of all, there is a mistake in the question. Sound intensity $I$ is not measured in dB but in W/m$^2$. Looking at the link you specified I think that you really want to compare sound intensity level $L_I$ and sound pressure level $L_p$.

Let me repeat their definitions

$$L_I = 10 \lg \frac{I}{I_0},$$

$$L_p = 10 \lg \frac{p^2}{p^2_0},$$

where $I$ is sound intensity, $I_0$ reference sound intensity, $p$ is sound pressure (RMS) and $p_0$ reference sound pressure.

In principle they could be different, depending on the defition of reference values, which are completely arbitrary! But for study of air sound, they are the same because $I_0 = 10^{-12}$ W/m$^2$ and $p_0 = 2 \times 10^{-5}$ Pa. This can be proven by taking the well-known relation for sound intensity

$$I = \frac{p^2}{\rho c},$$

where $\rho$ is density of medium and $c$ is speed of sound. Taking for air $\rho = 1.2$ kg/m$^3$ and $c = 340$ m/s, you readily obtain

$$I_0 = \frac{p_0^2}{\rho c}.$$

It is interesting to note that this is not neccessarily true for underwater acoustic, where they use different reference sound pressure and possibly different reference sound intensity too.

Now there is a big theoretical question, why do we have two definitions for the same quantity, is this really the same thing, or just made the same, or just coincidence?

Intensity tells you how sound energy propagates through space, but the only way you can measure intensity is by measuring the pressure. This is why people almost exclusively use expression sound pressure level, despite they are actually interested in sound intensity level. I thus personally believe that they are the same.

However, there are theories that say that we have power levels, that is levels in which power-related quantites appear (like sound intensity, electric power) and we have root-power levels where measurable quantities appear (like sound pressure, electric current) [1]. Note that sound pressure is square root of sound intensity and electric current is square root of electric power. So some people believe this two are completely different stuff and just conviniently made to be the same in case of the air sound.

[1] ISO 80000-1 Quantities and units - General

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