Planck function in the scale of wavelengths take the form

and approximation of wine is given by the formula

Therefore

Means, A relative Error In the approximation Of wine = . From other side,

investigating the function of wine to the maximum, it is easy to find that

This very useful result to eat a Wein's displacement law. They are usually limited to the fact that they note the constancy of work . However, is very important numerical value

, it is more precise, then that 1 noticeably exceed this number. Actually, we have , so that is small with . The intensity in the maximum, which we (now soundly) we calculate in the approximation of wine, there is

If we do not use ourselves the approximation of wine, but to work with the precise Planckian function, then it appears that it is equal not accurately to 5, but 4.965 (prover'te!).

Everything else, including the conclusion that , it remains valid.

The long-wave approximation of Rayleigh -Djinsa, opposite to the approximation of wine, ensures relative error only with (prover'te). Let us say, accuracy in 10%

approximation of Rayleigh -Djinsa gives only with , that exceed 25 times!

As a result it occurs that entire Planckian curve , as eye on the usual graph sees it, is not distinguished from the Wien curve. However, the majority of the students (yes, perhaps, and eks- students also) erroneously assumes that the approximation of wine for of the maximum, the approximation of Rayleigh -Djinsa applicably only to the left begins to work slightly more to the right it, and maximum itself is described well only by the precise Planck formula. In reality, as we were convinced, everything entirely not so.

8.2 Planckian curves, which correspond to different T, do not intersect, since . Hence it follows that with any (fixed) of value monotonically they grow with T. Further, the height of the maximum of Planckian curve, i.e., the maximum value of intensity, is

proportional dl4 and for . This is easy to show, investigating to the

maximum the appropriate functions of Planck (see task ). However, area hearth by both curves, and and , grows (Stefan-Boltzmann law). Therefore with an increase in the temperature Planck's curve in the scale of wavelengths "is sharpened", and in the frequency scale - "is dulled".

8.3 let f(x) is differentiated at point . We consider for simplicity that also

. For obtaining the exponential approximation f(x) in the environment , i.e., idea f(x) of the form

let us enter as follows. We will consider as function . Then we have usual linearization in the environment :

Hence, raising to a higher power, we obtain exponential the approximation f(x) given above, moreover it is revealed that

Similar exponential approximations are used in physics (and, in particular, in astrophysics) literally at every turn. Unfortunately, there is neither in one known to the authors course of mathematical analysis about this nor word - although teach all this one should, even studying analysis it is not too deep. Apparently, it is considered that the student all itself this will consider, when a little "he grows up". We decided to destroy tradition and not to await, when this happens.

The given under the condition of task exponential approximation of the dependence of

blackbody intensity on T in the environment is obtained by the recently described standard method. We give computation the reader.

The exponential approximation of the Planck function, which for some reason you will find not in one textbook, makes it possible to understand many qualitative special features of solar and starry spectra. See, in particular, task .

8.4 A why, strictly, it must be equal to (3/2)kT? Indeed photon - not classical particle, which moves with the nonrelativistic speed. But only to such particles is applicable classical formula (3/2)kT.

In order to find the medium energy of one blackbody photon , it is must the bulk density of energy of the field of the emission

to divide by the number of photons per unit of volume

After making in both integrals one and the same replacement , we reveal that where

For estimation A it is possible to use the approximation of wine, i.e., to disregard 1 in comparison with in two last integrals (cf. with the consideration in the task ). Then we immediately obtain that , since (we advise this to memorize)

The latter easily proves by integration in parts. Thus, ; after calculating integrals accurately, we would find that A= 2.70, so that is final

Thus, the medium energy of one blackbody photon almost is twice more than as the medium energy of the thermal agitation of nonrelativistic particle. However, the contribution of each photon to the pressure almost exactly is the same as each particle: however, radiation pressure

, the gas pressure P = n k T, where and n - concentrations of photons and particles, respectively. (as you do think, why so it is obtained? However, this is already faster physics, than astronomy. But indeed are decided we to sometimes train you to "physical mathematics", so why not to teach barely and to "astronomical physics"?) From recently aforesaid it follows that the relation of the concentrations of photons and particles exists simultaneously (with the accuracy ) and the ratio of radiation pressure to the gas (see task ).

8.5 expression for already appeared in the solution of the previous problem:

Substitution reduces it to the form

where

Thus . In order to find the precise value of constant of proportionality, it is necessary to obtain C. As we know (see solution of the previous problem), approximately it is possible to consider that C= 2. However, the precise value C is obtained as follows:

where - Riemann's zeta function:

The number is not expressed as any "standard" constants ( , e, the Euler's constant , etc.). It

is equal .

After the substitution of all constants into the obtained above expression for we find that

We somewhat stepped back here from our usual style - to obtain faster estimation, than accurate results, and to try to avoid bulky calculations. To make this at least one time, however, it is useful.

But this is how formula can be obtained entirely simply, combining other known results. The density of radiant energy of radiation equilibrium is equal

where a - constant of radiant density:

and the medium energy of one photon (see the previous task). Therefore which immediately gives coefficient of 20 with . In reality, of course, there is nothing fundamentally new in this method calculation - simply we used a finished numerical value of a constant radiant density a (it there is, for example, in Allen [ 1 ]).

8.6 frequency of the photon, emitted upon transfer of hydrogen atom from level m to level n, are