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Unformatted text preview: 2 Differential calculus 2.1 Asymptotics of functions: blackbody radiation One of the hardest things to teach students is how to have a qualitative feel for the important aspects of different functions. The first thing to always do when you are studying a function is consider its asymptotics, i.e. its behavior at 0 and ∞ and possibly near other x where the function is singular. Let’s take the example of the blackbody radiation function u ( λ ). This is the density of radiation per unit wavelength emitted by a blackbody (i.e. a perfectly absorbent body) in thermal equilibrium at temperature T , u ( λ ) dλ = 8 πhc λ 5 1 e hc/λkT 1 dλ, (1) where h is Planck’s constant, c the speed of light, k Boltzmann’s constant, and λ the wavelength of light omitted. The easiest thing to think about is an object with a cavity inside, with radiation of all frequencies emitted by the blackbody filling the cavity. Actually, the first thing to do is to analyze the equation to make sure it’s dimensionally correct. h has the dimensions of angular momentum, i.e. [ h ] = mL 2 /t . The quantity hc therefore has dimensions mL 3 /t 2 , and the quantity hc/λ dimensions mL 2 /t 2 , which has dimensions of an energy. Therefore the exponent in the denominator of Eq. 1 is indeed dimensionless, as it should be before we can raise it to a power! The factor which has dimensions, hc ( dλ ) /λ 5 then has dimensions ( mL 3 /t 2 ) /L 4 , or m/ ( Lt 2 ), which is the same dimensions as [ energy ] / [ volume ], so that checks! The equation is therefore ok. Our analysis of dimensions gives us hints about how to approach understanding the function. It contains a lot of physical quantities, but most of these are con stants . We are interested in how u ( λ ) depends on wavelength λ . Let’s define a new dimensionless variable x ≡ λkT/ ( hc ), which is proportional to λ . In terms of x , the formula reads u ( x ) = 8 πhc kT hc ¶ 5 1 x ¶ 5 1 e (1 /x ) 1 (2) Let’s ignore the xindependent prefactor and just look at the function of x . Large wavelengths clearly correspond to large x and small wavelengths to small x . But “large” and “small” relative to what? The claim is now that the dimensionless function u ( x ) = ( e (1 /x ) 1) 1 /x 5 has no obvious “scale” for x in it; therefore the most important values of x (i.e. where the function is big) must be of order 1. This 1 may seem like an outlandish claim, so let’s check it. We can expand using a Taylor expansion in either x or 1 /x to find u = ‰ e (1 /x ) /x 5 x ¿ 1 1 x 4 x 1 (3) Figure 1: Energy density u of blackbody as function of reduced wavelength...
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 Fall '07
 HIRSHFELD
 Derivative, The Land, dx, dθ

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