AFSim 仿真系统----《普朗克黑体辐射定律的区间积累》

news2024/9/23 5:31:48

参考文献

《普朗克黑体辐射函数的积分》，W. K. Widger, Jr. 和 M. P. Woodall，发表在《美国气象学会公报》，第57卷，第10期，1976年10月，页码1217-1219

上述参考文献描述了一种在不依赖于需要大量迭代的数值方法的情况下，对普朗克黑体辐射定律在一定波长范围内进行积分的机制。本文将在此重复推导，以供练习。

普朗克黑体辐射定律通俗讲解

普朗克黑体辐射定律是物理学中描述黑体辐射的重要理论。要理解这个定律，首先需要明确几个概念：

黑体：理想中，黑体是一个可以完全吸收所有入射辐射的物体。它没有反射或透过光，因此它的辐射只是与它的温度有关。
辐射：任何物体在一定温度下都会发出电磁辐射，温度越高，辐射的能量和强度通常越大。

定律内容

普朗克黑体辐射定律给出了黑体在某一温度下，单位波长的辐射强度是如何变化的。简单来说：

温度越高，辐射的能量越大：当黑体的温度升高时，它所发出的辐射会在更短的波长上有更高的强度。这意味着，热得越发的物体（例如火焰）会发出蓝光，而冷却的物体（例如冰块）则会发出红外辐射。
波长和强度关系：该定律提供了一个公式，计算在不同波长下（例如红色、蓝色等）黑体辐射的强度。普朗克的公式显示，辐射强度在某些波长上是有峰值的，通常这个峰值会随着温度的改变而改变。

实际应用

普朗克黑体辐射定律在天文学、气象学和材料科学中得到广泛应用。例如，它帮助科学家理解星体的辐射、地球的温度变化，甚至在图像处理和传感器设计中也能见到它的应用。

总结

普朗克黑体辐射定律是理解物体如何以不同方式释放热能的重要工具，尤其是在研究温度和光性能方面。这个定律不仅奠定了量子物理的基础，也帮助我们更好地认识宇宙中物体的辐射特性。

Planck’s Law of Black Body Radiation

Planck’s law of black body radiation states:

$B(T,\lambda) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{kT\lambda}} - 1}$

Or in terms of the radiation constants:

$B(T,\lambda) = \frac{c_{1L}}{\lambda^5} \frac{1}{e^{\frac{c_2}{T\lambda}}-1} \text{ (units: } \frac{W}{m^2 \cdot sr \cdot m} \text{)}$

Where the radiations constants are defined as:

$c_{1L} = 2hc^2 \text{ and } c_2 = \frac{hc}{k}$

This may be integrated to determine the radiance over a band of wavelengths:

$B_b (T) = \int_{\lambda_1}^{\lambda_2}B(T,\lambda)d\lambda \text{ (units:} \frac{W}{m^2 \cdot sr} \text{)}$

For a discrete spectral band, this may be replaced by:

$B_b (T) = \int_{\lambda_1}^{\lambda_2}B(T,\lambda)d\lambda = \int_{\lambda_1}^{\infty}B(T,\lambda)d\lambda - \int_{\lambda_2}^{\infty}B(T,\lambda)d\lambda$

Assume a substitution of variables:

$x = \frac{hc}{kT\lambda} \text{ or } \lambda = \frac{hc}{kTx}$

$dx = -\frac{hc}{kT\lambda^2} d\lambda \text{ or } d\lambda = -\frac{kT\lambda^2}{hc} dx = -\frac{kT}{hc} \frac{h^2 c^2}{k^2 T^2 x^2} dx = -\frac{hc}{kTx^2}dx$

Performing the substitution (ignoring the limits for now):

$\int \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{kT\lambda}}-1} d\lambda = \int -2hc^2 \frac{k^5 T^5 x^5}{h^5 c^5} \frac{1}{e^x-1} \frac{hc}{kTx^2}dx$

Now for some algebraic manipulation:

$&=\int -2hc^2 \frac{k^5 T^5 x^5}{h^5 c^5} \frac{hc}{kTx^2} \frac{1}{e^x-1} dx \\ &= -2hc^2 \frac{k^4 T^4}{h^4 c^4} \int x^3 \frac{1}{e^x-1}dx \\ &= -\frac{c_{1L}}{c_2^4} T^4 \int x^3 \frac{1}{e^x-1} dx$

Multiple top and bottom by

$e^{-x}$

$= -\frac{c_{1L}}{c_2^4} T^4 \int x^3 \frac{e^{-x}}{1-e^{-x}} dx$

The series expansion:

$(1 \pm z)^{-n} = 1 \mp nz + \frac{n(n+1) z^2}{2!} \mp \frac{n(n+1)(n+2) z^3}{3!} + \ldots (z^2 < 1)$

If we let

$z=e^{-x}$

and

$\frac{1}{1-e^{-x}} = (1-e^{-x})^{-1} = 1 + e^{-x} + e^{-2x} + e^{-3x} + \ldots$

Note that

is always greater than zero because all of the factors that define it are positive. As a result,

and

, as required. The integral can then be rewritten:

$&= -\frac{c_{1L}}{c_2^4} T^4 \int x^3 e^{-x} (1 + e^{-x} + e^{-2x} + e^{-3x} + e^{-4x} + \ldots )dx \\ &= -\frac{c_{1L}}{c_2^4} T^4 \sum_{n=1}^\infty \int x^3 e^{-nx} dx$

The integral can be explicitly evaluated using integration by parts:

$\int x^m e^{ax} dx = e^{ax} \sum_{r=0}^m (-1)^r \frac{m!x^{m-r}}{(m-r)!a^{r+1}}$

With

and

we have:

$\int x^3 e^{-nx} dx &= e^{-nx} [-\frac{3!x^3}{3!{(-n)}^1} + \frac{3!x^2}{2!{(-n)}^2} - \frac{3!x}{1!{(-n)}^3}+ \frac{3!}{0!{(-n)}^4}] \\ &= e^{-nx} [\frac{x^3}{n} + \frac{3x^2}{n^2} + \frac{6x}{n^3} + \frac{6}{n^4}]$

Substituting this for the integral in the summation we have:

$= -\frac{c_{1L}}{c_2^4} T^4 \sum_{n=1}^{\infty } \int x^3 e^{-nx} dx = -\frac{c_{1L}}{c_2^4} T^4 \sum_{n=1}^{\infty}e^{-nx} [\frac{x^3}{n} + \frac{3x^2}{n^2} + \frac{6x}{n^3} + \frac{6}{n^4}]$

The paper had an additional simplification that was stated to require fewer iterations for the same accuracy. We will not be using it because it varied from the direct numeric integration for low temperatures (<100 K) and short wavelengths (short than infrared), and the performance improvements were small or not observed. But I will include it here for completeness.

Note the following series:

$ln(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \ldots (-1 < z < 1)$

Let

$z=-e^{-x}$

$ln(1-e^{-x}) = -e^{-x}-\frac{(e^{-x})^2}{2}-\frac{(e^{-x})^3}{3} - \ldots - \frac{(e^{-x})^n}{n}$

The first term of the summation can then be replaced:

$\sum_{n=1}^\infty e^{-nx} \frac{x^3}{n} = x^3 \sum_{n=1}^{\infty} \frac{(e^{-x} )^n}{n} = -x^3 ln(1-e^{-x})$

And the integral can be updated:

$&= -\frac{c_{1L}}{c_2^4} T^4 [-x^3 ln(1-e^{-x})] \sum_{n=1}^{\infty} e^{-nx} [\frac{3x^2}{n^2} + \frac{6x}{n^3} + \frac{6}{n^4}] \\ &= \frac{c_{1L}}{c_2^4} T^4 [x^3 ln(1-e^{-x})] \sum_{n=1}^{\infty} e^{-nx} [\frac{3x^2}{n^2} + \frac{6x}{n^3} + \frac{6}{n^4}]$