As I understand it, the 'radiant intensity' is already in terms of energy (rather than number of photons) as it is measured in watts per steradian. It looks like V(gamma) is designed to weight according to human perception of luminosity (and in particular peaks at 1.0 at 540*10^12 Hz, i.e. green light, and is zero outside the visible interval).

Sent: Saturday, November 17, 2018 at 7:58 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Candela

I recall that in college ('68-'72) a standard joke when someone asked "What's new?" was to answer "E/h" (since E = h nu describes Planck's constant h (I think) ... where nu is the frequency of a photon.

If so, I'd guess that for an arbitrary (constant) frequency nu (OK, gamma), V(gamma) just needs to be multiplied by the ratio gamma / (540*10^12 Hz), or in other words for any frequencies gamma_1, gamma_2:

V(gamma_1) / V(gamma_2) = gamma_1 / gamma_2.

(If this is dumb, sorry about that.)

—Dan

Adam Goucher wrote: ----- ...

"The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540*10^12 Hz and that has a radiant intensity in that direction of 1/683 watt per steradian."

where luminous intensity is defined as:

I_v = 683 V(gamma) I_e

where I_e is the radiant intensity (in watts per steradian) and V(gamma) is the 'standard luminosity function' (which must be equal to exactly 1 when gamma = 540*10^12 Hz, for these definitions to be consistent).

But this raises the question: what is V(gamma) in general? I've found a tabulation of values for integer * 10^-9 metre wavelengths in the visible interval:

http://donklipstein.com/photopic.html

but no indication as to how to compute V(gamma) for any of the 2^(aleph_null) other wavelengths beside the 401 provided. Any ideas? -----

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