In electromagnetic waves, the energy formula is $E=hf$, that means frequency increasing will increase energy and so on the propagation distance. In underwater acoustic waves, low frequencies can travel larger distances comparing to higher frequencies.

  • The question is: isn't the relation $E=hf$ is applicable in underwater acoustics?
  • If it is applicable, does it mean that an increase of carrier frequency will increase the transmission distance of underwater acoustic waves?
  • $\begingroup$ AFAIK, $E=hf$ only applies to single, or at most a few, photons. Actual physical signals have so many photons that their energy is not measured like that, but rather by the amount of photons. But, I'm no expert in this field. $\endgroup$
    – MBaz
    Apr 5, 2018 at 20:53

3 Answers 3


In electromagnetic waves, the energy E=h*f,

No. This only applies to a single photon and not the entire wave, which consists of trillions times trillions of photons.

that means the frequency increase will increase the energy and so on the propagation distance.

No. Energy of the wave equals photon of photon times number of photons. In practice the quantized nature of the Energy makes no difference since the quantization is so small. At 5 GHz the energy of a single photon is about $ 10^{-24}$ Joule

In underwater acoustic waves, low frequencies can travel larger distances compating with higher frequencies.

Generally yes, but that has to do with absorption, path loss, diffusion and diffraction. Interactions with a medium and objects are indeed a function of frequency, albeit a fairly complicated one.

The question is: isn't the upper relation E=h*f is applicable in underwater acoustics?

No. Totally different physical mechanism.

If it is applicable, isn't that mean the increase of carrier frequency will increase the transmission distance for underwater acoustic waves?

No. It's not applicable.

  • $\begingroup$ Thanks @Hilmar for this clear answer, but i want to know if their is a relation between the underwater acoustic carrier frequency and the transmission power, and also its relation with propagation distance? $\endgroup$
    – user24907
    Apr 7, 2018 at 18:30

For a sine wave of amplitude $A$ and frequency $f=1/T$, its energy from $-T/2$ to $T/2$ is $A^2T/2$. Thus the power depends only on amplitude. The energy over one period does depend on frequency, but in the same observation duration, two sines having same $A$ should have the same energy (assuming observation duration is multiple of their periods).

In communications, we don't transfer energy but signal. Then power, not energy, matters.

Propagation loss is characterized grossly by distance (pathloss model) and absorption that the latter explains the choice of low frequency in underwater acoustics communications. For example, frequencies >= 1 MHz are absorbed very quickly whereas the absorption of low frequencies is much weaker. This is physics and the explaination would be complicated.

FYI, you can use this calculator for sea absorption.

And no, $E=hf$ is for photon, I have not met people who uses quantum mechanics for underwater acoustic waves.

  • $\begingroup$ thanks alot for your helpful answer, could you tell me the relation between the power and carrier frequency in underwater acoustic communications? $\endgroup$
    – user24907
    Apr 7, 2018 at 17:49

In underwater acoustics, there are two first order effects describing propagation in an isotropic freespace (ignore sound speed dependency, no near boundaries)
The first is inverse square law (for power), like in electromagetics The second is absorption. Vibration is converted to heat, which is not the ideal first order isotropic propagation situation in electromagnetics.


Nielsen, Richard O. "Sonar signal processing. Artech House." Inc., Norwood, MA (1991).

around page 207 there is a detailed derivation for absorption loss versus frequency and the optimal center frequencies for a target as a function of range.

As an aside the Friis equation is NOT used in SONAR. Actually, the homogeneous solution to the isotropic wave equation in electromagnetics doesn't say that higher frequencies need more energy to travel further.


the solution is

$$ u(r,t)=\frac{1}{r}f(r-ct) \quad \text{level not power} $$ which has no frequency dependency.

The main conceptual difference is that what Friis calls path loss , is to the output of the antenna, given a convention on how gain is stated.


The term "path loss" is not the same in acoustics and electromagnetic communications. Physics books tend to state path loss as in acoustics. Friis does not include absorption loss. Low frequencies do travel farther in terrestrial radio (which is not free space) but that is because lower frequencies tend to diffract around objects. Higher frequencies tend towards line of sight paths and often are blocked. Lower frequencies also penetrate buildings better.

Another consideration for useful range is that ambient noise has a minimum around 100KHz

Also at low frequencies, propagation resembles propagation in a wave guide. Frequencies need to be above a characteristic cut off in order to propagate.


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