Talk:Emissivity

The thermal emissivity page wasn't doing much, so I patched together material from other articles which linked directly to emissivity (which means that that article isn't doing much, either, but at least it's all in one place). As far as I know, it's all thermal emissivity. --Aranel 21:49, 15 Aug 2004 (UTC)

The Definition is Wrong !

The word 'power' should be substituted for 'energy' throughout. Another Wikki cock-up! A.J Smith

$\epsilon$ or $e$?

I wrote the original article on emissivity, on the portuguese Wikipedia. I refer to emissivity using the greek letter epsilon ($\epsilon$), instead of the mathmatical format of e ($e$). So, anyone has any idea on which one works best? --Arthur Albano 13:26, 18 July 2006 (UTC)

I've been trying to find out what is most common and my physics textbooks use $e$. They are pretty elementary though. I have found some mention of $\epsilon$ on the web though.
—Apis (talk) 19:45, 28 January 2009 (UTC)

Infrared Emissivity

A search for 'infrared emissivity' returns 0 results in Wikipedia. If somebody is aware they could add, as the term returns many results in Google - Sridhar Nov 12, 2005

It's a narrow yet vague concept. Emissivity is defined independently for each wavelength over the whole electromagnetic spectrum. On the one hand the infrared is just a tiny part of the spectrum, hence narrow, on the other it nevertheless embraces a continuum of wavelengths ranging over more than an order of magnitude, hence vague. --Vaughan Pratt (talk) 06:57, 13 July 2010 (UTC)

2 Walls

Anyone know where the information on the emissivity of 2 walls comes from? Joseph449008 (talk) 14:00, 29 January 2010 (UTC)

No idea, but it's easy to see that it's true. The formula gives the proportion of radiation (of whatever wavelength the two emissivities are for) inside one wall that enters the other wall. It takes into account Kirchhoff's law of thermal radiation, that reflectivity is complementary to emissivity. The reflectivity allows computing the amount of radiation bouncing back and forth forever in between the two walls as a result of reflection. I'll try to say something coherent about this in the article.

To find the formula, assume unit radiation inside the first wall, and write R1 = 1-E1, R2 = 1-E2 for the two reflectivities. Let A be the radiation from the first wall to the second, and B the radiation the other way. Then A = E1 + R1*B (sum of emitted and reflected radiation) while B = R2*A (just the reflected part of A). Eliminating B gives A = E1/(1-R1*R2). What is emitted into the second wall is then E2*A = E1*E2/(1-R1*R2) and it's now just algebra to verify that this is the formula in the article (expand R1 and R2, simplify, then divide both top and bottom by E1*E2). The formula is symmetric so it gives the same result for radiation from inside the first wall to inside the second. Note that "inside" means "just inside" so that absorption does not become an issue. --Vaughan Pratt (talk) 05:17, 13 July 2010 (UTC)

albedo

relation to albedo..? Cesiumfrog (talk) 07:36, 27 April 2010 (UTC)

By Kirchhoff's law of thermal radiation for a given wavelength, emissivity is simply one minus reflectivity. Naively albedo is just reflectivity, but in practice it's much more complicated and there's no simple connection. While there are various notions of albedo, for example geometric albedo and Bond albedo, none of them are for a single wavelength, whence there is no notion of albedo for which Kirchhoff's law can be trusted.

For a body whose reflectivity is independent of wavelength, Kirchhoff's law does hold, but in all other situations it's at best a rough approximation. If this were not so the formula for effective temperature, which assumes unit emissivity and therefore underestimates the temperature of a planet with no atmosphere, could solve this very simply by just dropping the factor 1-A, which would then factor in the emissivity. --Vaughan Pratt (talk) 06:50, 13 July 2010 (UTC)