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Comment by bigbillheck | Hacker News Reader

> The path loss equation violates conservation of energy ie the frequency or wavelength term depending on how it's structured cannot be in the equation.

Why is that?

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It's because energy created by the transmitter must degrade as one over R squared in the far field. The frequency (or wavelength, have your pick) has nothing to do with the energy transmitted because energy must be conserved. Putting in the frequency term then violates conservation of energy between the antennas. Then, at the receiving antenna the error of conservation of energy is then patched up by assigning a bogus 'gain' at the receiver. The transmitter and receiver are asymmetric but the path loss equation pretends that they are because that's easier for most people to understand and it works out 'end to end'.

Absolutely I agree that the geometry of the problem dictates 1/R^2 dependence, regardless of frequency. The gain, which I agree is a misleading way to think about the area, is related to the area of receive through the frequency terms. If you don't like that form of the path loss equation, I understand (I don't either!), but physics is not broken.

Where the "bogus" gain really shines, though: I can take my original receive antenna, operate it as a transmitter (so gain is now relevant), receive with my original transmit antenna (where I now care about area) and get the exact same result in terms of loss!

The formula on wiki has a distance squared term in the denominator tho?

Yes it does but it's the lambda symbol that's the impostor. An analogy of how it really works is to think of the transmitter as a water hose that is spraying water into a bucket, which is the receiver.

Short answer: it doesn't, though I understand why it's misleading. Read my response above.

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