# What is “power per unit frequency” in black body radiation? [duplicate]

What is the meaning of power radiated by a black body per unit frequency? If you have a black body with a frequency filter around it set to 530nm and you calculate the energy radiated in 1 second you will get a definite (not a differential) value. So where does "per unit frequency" come from? What is its physical meaning? Edit:
There is a physical anamoly in my argument as, suppose, at 530 nm we get 5 watt energy, so at 530.01 nm the power will approximately be the same 5 watt, and similarly for 530.001nm.
But I still don't get the flaw in my argument (even though I get that the result is wrong). What is it that I am missing?

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• – Shishir Maharana Aug 10 at 15:12
• The first answer to the question satisfies your question. – Shishir Maharana Aug 10 at 15:14
• I don't think it does. The duplicate question's answer doesn't satisfy me too as he doesn't explain why is it reasonable to look at f to f+df rather than just f. – Aman Aug 10 at 15:47
• Read his answer carefully. Looking at f just doesn't make any sense. – Shishir Maharana Aug 11 at 6:34
• @Aman, laser light is (ideally) coherent, monochromatic light, i.e., the power is concentrated in a single frequency. Contrast that with, for example, the light of an incandescent light bulb where the power is distributed over a continuum of frequencies. – Alfred Centauri Aug 11 at 13:59

What is its physical meaning?

It means that, to find the power within a given bandwidth, one integrates the power spectral density (PSD) over that bandwidth.

Assuming the PSD is essentially flat over a 1 Hz bandwidth, the power at the output of an ideal bandpass filter, centered at 530 nm and with 1 Hz bandwidth, would be the value of the PSD at 530 nm multiplied by 1 Hz.

If the bandwidth were 1 mHz, the power would be 1000 times less. For an arbitrarily small bandwidth, the power would be arbitrarily small, i.e., the power at a specific frequency (wavelength) is infinitesimal.

Keep in mind that no physical bandpass filter has infinitesimal bandwidth.

suppose, at 530 nm we get 5 watt energy

As I wrote above, the power at a specific frequency (wavelength) is infinitesimal so you won't find a finite amount of power at 530 nm unless there is a delta function in the PSD there (for example, due to the output of an ideal 530 nm laser source). But the ideal blackbody spectrum is continuous.

• my filter is hypothetical just like a black body. Can you explain a bit more? – Aman Aug 10 at 15:36
• In your hypothetical, how much energy do you get at $t=\frac 1 2$ seconds? You get zero watts, because the time duration is $0$, but there is finite power (energy per time). Likewise, over 1 second, you get zero energy at 565.65THz because the bandwidth is zero, but within a finite bandwidth you get finite energy. I suggest you do not mix frequency and wavelength until you resolve the misunderstanding. – JEB Aug 10 at 17:44
• @Aman, the internet has been down here for awhile. I've added some to my answer but I'm not quite clear on what you're looking for regarding 'a bit more'. – Alfred Centauri Aug 10 at 23:13
• @Alfred you hit on my point exactly in the edit. Just tell me why power at a 'specific' frequency is infinitesimal. – Aman Aug 11 at 12:58
• @Aman, I can't quite understand why you think it shouldn't be infinitesimal. Are you asking why is the black body spectrum continuous? Or are you asking why a continuous spectrum implies infinitesimal power at a specific frequency? – Alfred Centauri Aug 11 at 13:31

Planck's distribution function gives the intensity of the emitted radiation (emitted power per unit area) per unit of a physical quantity (frequancy, wavelength, wavenumber) from a Black Body at absolute temperature T.

You can calculate the same for a frequency interval. Let's say you wanted to set the filter from 530nm to an interval of 400nm-530nm.

https://www.spectralcalc.com/blackbody/planck_blackbody.html