How do you convert watts per meter squared to decibels?

I may be comeplete wrong but aren't decibels 20Log(gain)? Also, the argument of the log must be unitless.
[tex]
=\frac{W}{m^2}=\frac{J}{m^2\cdot s}=\frac{N}{m\cdot s}
=\frac{kg\cdot m}{s^2}\frac{1}{m\cdot s}=\frac{kg}{s^3}\neq 1[/tex]

I would just look at your equation again.

 

The power density must be divided by a certain constant power density j0 to obtain dimensionless argument for log. I think they chose the minimum power density that can be heard by humans for j0 (this is supposed to be 10^-12 W/m^2). So 10*log(S)+120 is more consistently written as
10*log(S/(1W/m^2))+120=10*log(S/(10^-12 W/m^2))

20*log() is used for sound pressure amplitudes. The power density is proportional to the square of the pressure amplitude. Since log(x^2)=2*log(x), the multiplication constant must be twice smaller in the first case (so that we the same increase of decibels for the same power/pressure change).

 

Sound intensity level is defined as:
[tex]L_J = 10 \log_{10}\left( \frac{J}{J_0} \right) \text{ dB}[/tex]
where [itex]J[/itex] is the sound intensity and [itex]J_0[/itex] is the reference intensity level.

In the case that [itex]J_0 = 10^{-12} \, \frac{ \text{W}}{\text{m}^2}[/itex], the standard sound intensity reference level, the unit dB is sometimes changed to dB (SIL) where SIL stands for Sound Intensity Level.


Sound pressure level on the other hand as a slightly different formula, since sound pressure is related to sound intensity squared:
[tex]L_p = 10 \log_{10} \left( \frac{p^2}{p_0^2} \right) = 20 \log_{10} \left( \frac{p}{p_0} \right) \text{ dB}[/tex]
where [itex]p[/itex] is the sound pressure and [itex]p_0[/itex] is the reference sound pressure.

Again, in the case that [itex]p_0 = 2\times 10^{-5} \text{ Pa}[/itex] (root-mean-square pressure), the standard reference sound pressure, the unit dB is sometimes changed to dB (SPL) where SPL stands for Sound Pressure Level.


These reference soundpressure/intensity are I believe related to the human threshold of hearing at 1000 Hz. I believe it is about the soundlevel of a mosquito flying at 3m distance.


Note that the unit dB is not actually a unit, it is dimensionless.

Also note that it's perfectly possible to have a negative sound level (-5 dB for example is not uncommon). It simply means that the soundlevel is below the average threshold of human hearing.

 

when gain is a voltage ratio or a current ratio (or in the case of acoustics, the ratio of pressure deviation levels) it's 20log₁₀(gain). but if it's the ratio of power or energy (which are proportional to the square of voltage or current or acoustic pressure), then it's 10log₁₀(gain).

that was just playing fast and loose with the scalers.

actually Ed wasn't play so fast and loose:

the actual meaning of that is

[decibels] = (10dB)log(S/S₀) + 120dB

where S₀ = 1 W/m²

 

So, when measuring sound intensity in decibels using the standard for [tex]J_0[/tex], by what power does sound fall off? My intuition says it should be [tex] r^{-2}[/tex] because it spreads radially outwards but then there is air pressure and stuff like that in there.

 

J is a symbol for intensity. if energy is conserved, spherically radiating energy has to have intensity (power per unit area at a right angle to the direction of propagation) that is inverse-square. distance doubles, surface area of sphere centered at the point source increases by a factor of 4, equal power is distributed over 4 times the area, intensity is reduced by a factor of 1/4, and, since this is about power not voltage, that's -6.02 dB.

 

Sound intensity is proportional to [itex]r^{-2}[/itex] while sound pressure is propertional to [itex]r^{-1}[/itex].