Timeline for What truly is spectral blackbody emissive power?
Current License: CC BY-SA 4.0
10 events
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| Oct 11 at 16:51 | comment | added | Philip Wood | Because if the emission is uniform across a band of wavelengths ($\lambda_2-\lambda_1$, say), the proportion of that power that will be emitted over a narrow band of width $\Delta \lambda$ somewhere between $\lambda_1$ and $\lambda_2$, will be $\frac {\Delta \lambda}{\lambda_2-\lambda_1}$, and if $\Delta \lambda$ is zero, that is for one discrete wavelength, the proportion of that power will be zero. | |
| Oct 11 at 10:30 | comment | added | Peter swift | @PhilipWood "Because at that particular wavelength, no power will be emitted." But why is that, sir? | |
| Mar 29, 2021 at 12:57 | vote | accept | Harshit Rajput | ||
| Mar 29, 2021 at 12:53 | comment | added | Philip Wood | I've expanded the first bit of my answer. | |
| Mar 29, 2021 at 12:52 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Mar 29, 2021 at 12:45 | comment | added | Harshit Rajput | Actually I was looking for an answer to why we can't define power for a single wavelength. | |
| Mar 29, 2021 at 12:43 | comment | added | Philip Wood | I don't think that any more detail is possible! I've looked through a few statistical mechanics textbooks, and they do seem to gloss over it. But it's implicit in the treatments of black body radiation given by Reif, Hill, Mandl, Huang... | |
| Mar 29, 2021 at 12:21 | comment | added | Harshit Rajput | Found this really interesting. This wasn't mentioned in any of my graduation level books. Could you provide me any source in which this is discussed in detail? | |
| Mar 29, 2021 at 12:02 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Mar 29, 2021 at 11:54 | history | answered | Philip Wood | CC BY-SA 4.0 |