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I've been looking at all this for the last 2 days and I can't sleep anymore unless I understand it. I think I come to the below conclusion. Much appreciated.

Planck says that the below formula returns the following: Energy density of radiation per unit wavelength(I missed other units, not needed for now) is given by:

$$B_饾渾(T)=\frac{2hc^2}{位^5}\,\frac{1}{\mathrm{e}^{hc/位k_BT}-1}.$$

From this moment of, whatever I write is my observation and need to confirm if I'm right. Hence, the question basically is whatever I write below, whether it's correct or not.

Let's imagine, 位 is 3m. So, directly putting it inside the formula, let's say it gives us 500. In my opinion, 500 is the radiation energy emitted by range of [3, 3 + 螖z] where 螖z is very very small. The reason of this is figuring out density at an exact point is wrong, as point has no length. So the above formula in my opinion gives radiation emitted by a very, very, small chunk line near 3. hence [3, 3 + 螖z]. and I presume this is the notion of any energy density function ever. They don't give a density at a point(the point you input), they give the density of very small area near that point(at least how we should interpret it).

Now, if we want to figure out the energy emitted between our ranges, let's say between 3 and 3.3, we got a couple of choices:

Choice 1: I found this formula on the internet which looks and states the following:

Energy density of radiation with wavelengths in the range of [位, 螖位] is given by.

$$B_饾渾(T)螖位=\frac{2hc^2}{位^5}\,\frac{螖位}{\mathrm{e}^{hc/位k_BT}-1}.$$

Note that the formula doesn't include integrals. For sure, I can use integrals to get a better answer, but let's just say I use this since the formula is shown without integral. Now, in our case, 螖位 = 0.3 . So 500 * 0.3 = 150 (蟻 * x = E). (Remember, we imagined that putting 3m in the 位 gives us 500). The problem with this is we assume that the density of the line [3, 3.3] is the same as the density(500) of [3, 3 + 螖z]. Which ofc is not right, but we approximate it and I have seen people do it like this without integrals. not an exact answer, but still will be close to the answer even more as our range becomes shorter, because the bigger the range, we treat this bigger range to have the same density as the [3, 3 + 螖z] and this causes 500 to be multipled by a bigger range, hence becomes even wrong with big ranges.

Choice 2: That's when integral comes into play. Integrating the function in the 2nd formula, we get a correct answer.

Thoughts ? Even if one word or thought is incorrect, I'd appreciate to let me know.

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  • $\begingroup$ You have a frequency subscript, and then a wavelength in the formula. $\endgroup$
    – JEB
    Commented Apr 21, 2023 at 22:25
  • $\begingroup$ Sorry , couldn't find it . could you let me know where exactly ? $\endgroup$
    – Nika
    Commented Apr 21, 2023 at 22:27
  • $\begingroup$ It should be $B_{\lambda}(T)$. $\endgroup$
    – JEB
    Commented Apr 21, 2023 at 22:28
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    $\begingroup$ I think that most (all?) of your confusion arises because you ignore units, both in this question and in your previous questions on this topic. (Example: 鈥淟et's say the end result I got was 700.鈥) $\endgroup$
    – Ghoster
    Commented Apr 21, 2023 at 23:01
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    $\begingroup$ The formula says that unit is energy per unit wavelength. Where are you reading that $B_\lambda$ is energy per unit wavelength? It isn鈥檛. It is power per unit area per unit of solid angle per unit wavelength. $\endgroup$
    – Ghoster
    Commented Apr 22, 2023 at 0:03

2 Answers 2

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Your description in paragraph 2 of the correct equation below it is false. The formula you've quoted is for (wavelength-dependent) spectral radiance. Your misrepresentation results in an cascade of mistakes. Everything below the bolded paragraph is false. See: wikipedia's table for radiometry units.

Wavelength-dependent Spectral Radiance, the function for which you quoted the equation, is the wavelength derivative of Radiance. It tells you the ratio of the change in Radiance to the change in wavelength at a particular wavelength. It has units of power per unit solid angle per unit area per unit wavelength.

Radiance is the derivative of Irradiance with respect to solid angle. It tells you the ratio of change in power per unit area received by an aperture to change in the solid angle subtended by the aperture at a particular solid angle subtended by the aperture. It has units of power per unit solid angle per unit area.

Irradiance is the derivative of radiant flux with respect to area. It tells you the ratio of the change of power emitted by a partial surface to the change of the area of the surface under consideration at a particular area. It has units of power per unit area.

Radiant flux is the time derivative of radiant energy. It tells you the ratio of the change in the amount of energy emitted by a closed surface to the change in elapsed time by at a given time, which is to say, power.

If you want to get energy out of that expression of Planck's law, you need to integrate it with respect to wavelength over some finite change in wavelength to get Radiance over that wavelength range. Then integrate the Radiance with respect to solid angle over some finite change in solid angle (probably zero to the relevant measured solid angle) to get the Irradiance in that wavelength range received by the aperture. Then integrate the Irradiance with respect to area over some finite change in area (probably zero to the whole measured area) to get Radiant Flux in that wavelength range emitted by the area that is visible to that aperture. Then integrate the Radiant Flux with respect to time over some finite change in time to get energy emitted by the surface in that wavelength range from the area that is visible to that aperture in that time.

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  • $\begingroup$ Thank you, but in terms of our question, to me, it doesn't answer anything. What the formula is provided in wikipedia, I was following it :) I didn't want to talk about integralls/derivatives on this question at all $\endgroup$
    – Nika
    Commented Apr 22, 2023 at 11:10
  • $\begingroup$ I asked basically what per unit wavelength means. if my assumptions are wrong, then what's the correct way ? (without integrals/derivatives) $\endgroup$
    – Nika
    Commented Apr 22, 2023 at 11:51
  • $\begingroup$ @GiorgiLagidze Per unit wavelength just refers to the units of the quantity, like how velocity is distance per unit time. Any units will do, just as you can express the same velocity in feet per second or kilometers per hour. The deeper meaning is that the quantity relates a rate of change. Calculus is the mathematical tool for calculating the relationships between rates of change and the quantities that are changing. $\endgroup$
    – g s
    Commented Apr 22, 2023 at 14:46
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Most of your question is about how a density works.

Since you don't seem to understand what the physical quantity specified by $B$ is, here's how it works:

$$[B]=[action][velocity]^2/[L]^5 $$

Action is energy times time, and velocity is length per time, so:

$$[B]=[Energy][T][L/T]^2/[L]^5 $$ $$[B]=[Energy]/[T]/[L]^3 $$

So it's clearly not an energy density. Energy per time is power, so:

$$[B]=[Power]/[L]^3 $$

which looks like a power volume density, but one of the length dimensions is wavelength, so it鈥檚 a spectral density of power radiated from a (projected) surface area.

The projected parts is because there is a dimensionless per steradian implicit in the derivation. Note that this is not 鈥減er angle鈥, it鈥檚 angle squared, or steradian. So to get total energy radiated you integrate over solid angle, wavelength, area and time.

It may be informative to do the dimensional analysis with $B_{\nu}$, starting with $h$ having units of energy per frequency.

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