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marcus is offline
Jan9-05, 04:45 PM
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continuing the comparison
how many photons per unit volue are in a room at temp T. Not surprisingly it depends on the cube of the energy kT.

here's how a metric user might calculate it (in fact dextercioby just did it this way in another thread)

[tex]\text{number of photons per unit volume} = \frac{1}{2.701} \times \frac {\pi^2}{15}\frac{(kT)^3}{\hbar^3 c^3}}[/tex]

here's a version using natural units of energy and length.
the two are the same because
[tex]\hbar c =\mathbb{E}\mathbb{L}[/tex]

[tex]\text{number of photons per unit volume} = \frac{1}{2.701} \times \frac {\pi^2}{15}\frac{(kT)^3}{\mathbb{E}^3\mathbb{L}^3}[/tex]

the reason for the 2.701 is because in thermal radiation at temperature T, the average photon energy is 2.701 kT.
2.701 comes from the Riemann zeta function

[tex]2.701 =\frac{3 \zeta(4)}{\zeta(3)}[/tex]

what the E and the L in the denominator are telling is is that if you work just in terms of the natural units then you simply have to cube the kT and do the stuff with pi-square over 15 and 2.701 ( factors that nature insists be there)

So we can do an example. A common (Fahren. 49) temperature at the earth's surface is E-29, put in natural terms, cube that: E-87.
and the numerical factors come to 0.24

so the number of photons per unit volume in a space at this common temperature is
2.4 x 10-88

To interpret that it is good to remember that a PINTsize volume---the cubic handbreadth thing---is 1099 of the natural volume units. So at a more familiar scale our result is just

2.4 x 1011 photons per pint.

roughly a quarter of a trillion per pint.
the problem with using natural units is always AFTER you get the answer you have to know some familiar multiples of the natural units so that you can visualize the answer in humanscale or practical terms

getting the answer tends to be expedited by the main constants being ONE so you dont bother with them, but then afterwards you need to take a moment to assimilate or interpret the answer.

Conventional room temperature, on the natural temp scale, is a bit more than E-29, in fact it is 1.04E-29 (that is 69 or 70 Fahrenheit) so if you are curious about the photons per volume at room temp you would cube that
and the answer would be larger by a factor of 1.12----so 12 percent larger number of photons.

in rough terms tho, just to have an idea, it is still about a quarter trillion per pint.