Wien's Law of Displacement

Curves of dependences of the spectral radiation density of absolutely black bodies with different temperatures on the wavelength. It is seen that with increasing temperature, the maximum spectral density shifts to the short-wavelength part of the spectrum. It is this feature that describes the law of Wines.

The Wien law of displacement establishes the dependence of the wavelength at which the flux of black body energy radiation reaches its maximum, on the temperature of the black body.

Wilhelm Wien first deduced this law in 1893 , by applying the laws of thermodynamics to electromagnetic radiation .

Content

1 General view of the Wien displacement law
2 Conclusion of the law
3 Examples
4 See also
5 notes
6 Literature
7 References

General view of the Wien bias law

\lambda _{max}=b/T,

{\ displaystyle \ lambda _ {max} = b / T,}

\lambda _{max}=b/T,

Where $\lambda _{max}$ ${\ displaystyle \ lambda _ {max}}$ $\lambda _{{max}}$ Is the radiation wavelength with maximum intensity, and $T$ ${\ displaystyle T}$ $T$ - temperature. Coefficient $b={ch \over k\alpha }$ ${\ displaystyle b = {ch \ over k \ alpha}}$ $b={ch \over k\alpha }$ (where c is the speed of light in vacuum , h is the Planck constant , k is the Boltzmann constant , α ≈ 4,965114 ... is a constant value (the root of the equation ${\frac {\alpha }{5}}=1-e^{-\alpha }$ ${\ displaystyle {\ frac {\ alpha} {5}} = 1-e ^ {- \ alpha}}$ ${\frac {\alpha }{5}}=1-e^{-\alpha }$ )), called the Wien constant , in the International System of Units (SI) has a value of 0.002898 m · K.

For the frequency of light $\nu$ ${\ displaystyle \ nu}$ $\nu$ (in hertz ) the law of Wien's displacement looks like:

\nu _{\max }={\alpha  \over h}kT\approx (5,879\times 10^{10}\ \mathrm {)} \cdot T,

{\ displaystyle \ nu _ {\ max} = {\ alpha \ over h} kT \ approx (5,879 \ times 10 ^ {10} \ \ mathrm {)} \ cdot T,}

\nu _{\max }={\alpha \over h}kT\approx (5,879\times 10^{{10}}\ {\mathrm )}\cdot T,

where α ≈ 2.821439 ... is a constant value (the root of the equation ${\frac {\alpha }{3}}=1-e^{-\alpha }$ ${\ displaystyle {\ frac {\ alpha} {3}} = 1-e ^ {- \ alpha}}$ ${\frac {\alpha }{3}}=1-e^{{-\alpha }}$ ), k is the Boltzmann constant , h is the Planck constant , T is the temperature (in kelvins ).

The difference in the numerical constants here is due to the difference between the exponents in the Planck distribution recorded for the wavelength and radiation frequency: in one case, $\lambda ^{-5}$ ${\ displaystyle \ lambda ^ {- 5}}$ $\lambda ^{-5}$ , in another - $\omega ^{3}\sim \lambda ^{-3}$ ${\ displaystyle \ omega ^ {3} \ sim \ lambda ^ {- 3}}$ $\omega ^{3}\sim \lambda ^{-3}$ . This difference, in turn, occurs due to the nonlinearity of the relationship between frequency and wavelength: $\omega ={\frac {2\pi c}{\lambda }}$ ${\ displaystyle \ omega = {\ frac {2 \ pi c} {\ lambda}}}$ $\omega ={\frac {2\pi c}{\lambda }}$ , and ${\frac {d}{d\omega }}=-{\frac {\lambda ^{2}}{2\pi c}}{\frac {d}{d\lambda }}.$ ${\ displaystyle {\ frac {d} {d \ omega}} = - {\ frac {\ lambda ^ {2}} {2 \ pi c}} {\ frac {d} {d \ lambda}}.}$ ${\frac {d}{d\omega }}=-{\frac {\lambda ^{2}}{2\pi c}}{\frac {d}{d\lambda }}.$

Conclusion of the Law

For conclusion, you can use the expression of the Planck radiation law for a completely black body recorded for wavelengths :

B(\lambda ,T)={2hc \over \lambda ^{5}}{1 \over e^{hc/\lambda kT}-1}.

{\ displaystyle B (\ lambda, T) = {2hc \ over \ lambda ^ {5}} {1 \ over e ^ {hc / \ lambda kT} -1}.}

B(\lambda ,T)={2hc \over \lambda ^{5}}{1 \over e^{{hc/\lambda kT}}-1}.

To find the extrema of this function depending on the wavelength, it should be differentiated by $\lambda$ ${\ displaystyle \ lambda}$ and equate the derivative to zero :

{\partial B \over \partial \lambda }={\frac {2hc}{\lambda ^{6}}}{1 \over e^{hc/\lambda kT}-1}\left({hc \over kT\lambda }{e^{hc/\lambda kT} \over \left(e^{hc/\lambda kT}-1\right)}-5\right)=0

{\ displaystyle {\ partial B \ over \ partial \ lambda} = {\ frac {2hc} {\ lambda ^ {6}}} {1 \ over e ^ {hc / \ lambda kT} -1} \ left ({ hc \ over kT \ lambda} {e ^ {hc / \ lambda kT} \ over \ left (e ^ {hc / \ lambda kT} -1 \ right)} - ​​5 \ right) = 0}

From this formula, we can immediately determine that the derivative approaches zero when $\lambda \rightarrow \infty$ ${\ displaystyle \ lambda \ rightarrow \ infty}$ or when $e^{hc/\lambda kT}\rightarrow \infty$ ${\ displaystyle e ^ {hc / \ lambda kT} \ rightarrow \ infty}$ what is done when $\lambda \rightarrow 0$ ${\ displaystyle \ lambda \ rightarrow 0}$ . However, both of these cases give minimum Planck functions. $B(\lambda )$ ${\ displaystyle B (\ lambda)}$ which reaches its zero for the indicated wavelengths (see the figure above). Therefore, the analysis should be continued only with the third possible case, when

{hc \over kT\lambda }{e^{hc/\lambda kT} \over \left(e^{hc/\lambda kT}-1\right)}-5=0

{\ displaystyle {hc \ over kT \ lambda} {e ^ {hc / \ lambda kT} \ over \ left (e ^ {hc / \ lambda kT} -1 \ right)} - ​​5 = 0}

Using variable substitution $x={hc \over kT\lambda }$ ${\ displaystyle x = {hc \ over kT \ lambda}}$ , this equation can be converted to

{xe^{x} \over e^{x}-1}-5=0.

{\ displaystyle {xe ^ {x} \ over e ^ {x} -1} -5 = 0.}

A numerical solution of this equation gives: ^[1]

x=4.965114231744276\ldots

{\ displaystyle x = 4.965114231744276 \ ldots}

Thus, using the change of variables and the values of the Planck , Boltzmann constants and the speed of light , we can determine the wavelength at which the radiation intensity of a black body reaches its maximum, as

\lambda _{\max }={hc \over x}{1 \over kT}={2.89776829\ldots \times 10^{-3} \over T},

{\ displaystyle \ lambda _ {\ max} = {hc \ over x} {1 \ over kT} = {2.89776829 \ ldots \ times 10 ^ {- 3} \ over T},}

where the temperature is given in kelvins , and $\lambda _{\max }$ ${\ displaystyle \ lambda _ {\ max}}$ - in meters .

Examples

According to Wien's displacement law, a black body with a human body temperature (~ 310 K ) has a maximum of thermal radiation at a wavelength of about 10 μm , which corresponds to the infrared range of the spectrum.

The CMB has an effective temperature of 2.7 K and reaches its maximum at a wavelength of 1 mm . Accordingly, this wavelength already belongs to the radio range .

Notes

↑ Solving the equation ${xe^{x} \over e^{x}-1}=n$ ${\ displaystyle {xe ^ {x} \ over e ^ {x} -1} = n}$ impossible to express with elementary functions. Its exact solution can be found using the Lambert W-function , however, in this case, it is sufficient to use the approximate solution.

Literature

BH Soffer and DK Lynch , "Some paradoxes, errors, and resolutions regarding the spectral optimization of human vision," Am. J. Phys. 67 (11), 946-953 1999.
MA Heald , “Where is the 'Wien peak'?", Am. J. Phys. 71 (12), 1322-1323 2003.

Links

The Physics World of Eric Weinstein

[1] Solving the equation ${xe^{x} \over e^{x}-1}=n$ ${\ displaystyle {xe ^ {x} \ over e ^ {x} -1} = n}$ impossible to express with elementary functions. Its exact solution can be found using the Lambert W-function , however, in this case, it is sufficient to use the approximate solution.