"Even though the energy falls off as 1/r^2, the amplitude only falls off as 1/r. That's why gravitational waves fall off according to a different law than electromagnetic waves."https://twitter.com/StartsWithABang/status/1101754561260249088 …
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Replying to @Tom_Ruen
Huh??? Electromagnetic waves also fall off in amplitude as 1/r. Is Ethan forgetting his physics or am I misunderstanding this quote? https://en.wikipedia.org/wiki/Dipole_antenna … For both gravitational and electromagnetic waves, energy density goes as amplitude squared, which goes like 1/r.
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Replying to @johncarlosbaez @Tom_Ruen
Actually Ethan's blog says "The gravitational force gets weaker like 1/r^2, but gravitational waves scale as just 1/r". That's right - and again it's the same for gravitational and electric.
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Replying to @johncarlosbaez @Tom_Ruen
The difference is what shows up in your detector. For light, it's intensity or energy. For GWs, it's the amplitude of the wave that stretches or compresses space. That's why GWs scale as 1/r, and why brightness scales as 1/r^2.
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The reason is technical. There is no light detector that is fast enough to see the oscillations of an EM field in the visible (~400 THz). So the detector sees the average intensity:https://www.researchgate.net/post/Is_output_current_of_photodiode_a_DC_or_an_AC_signal …
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Nice. At about what frequency of light can we build a detector that sees the amplitude?
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Replying to @johncarlosbaez @mustgundogan and
I guess you can still comfortably measure the voltage at a radio antenna, so it has got to be somewhere between 10^9 and 10^14 Hz...
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Replying to @Quantensalat @johncarlosbaez and
But am I correct to suspect that measuring the amplitude requires a sufficient # of photons in the radiation field?
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I can tell you this.. you will at least need one photon
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