Can you tell me if there is any difference between these 2 units?
$\log \mathrm{quanta/cm^2/s}$ vs $\log \mathrm{photons/cm^2/s}$
Thanks
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1$\begingroup$ Photons are a type of quantua. So the latter is more vague/general. $\endgroup$– Sean E. LakeOct 4, 2022 at 15:09
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1 Answer
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"Number of photons" and "Number of quanta" are both dimensionless so
"quanta" cm$^{-2}$ s$^{-1}$ has the same units as "photons" cm$^{-2}$ s$^{-1}$.
But strictly, both expressions are wrong since:
$\log{x}= (x-1) - (1/2)(x-1)^2 + (1/3)(x-1)^3 - (1/4)(x-1)^4 + ...$
So the argument inside the logarithm should be unit-less.
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$\begingroup$ Note that $\log(a/b)\equiv\log(a)-\log(b)$, so it is possible for units to be present in the argument of logarithms, though care needs to be taken (I believe there is a PSE post on the use of units in log) $\endgroup$ Jul 3, 2019 at 14:33
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$\begingroup$ Relevant posts on why and how logarithms of dimensionful quantities can be made to work: Why is it “bad taste” to have a dimensional quantity in the argument of a logarithm or exponential function?, What is the logarithm of a kilometer? Is it a dimensionless number?, and links therein. $\endgroup$ Jan 28, 2021 at 13:16