Mechanics:
Heat and Thermodynamics (Macroscopic Large Scale Phenomena)
Kinetic Theory ( Statistical mechanics) ( Microscopic Small Scale Phenomena)
Average over microscopic properties give rise to macroscopic properties
Eg. Temperature is a measure of the average molecular kinetic energy. Pressure is average force on wall of container due to molecular collisons.
Electromagnetic Theory: Maxwell and the Wave Equation.
Radiation: Classically due to oscillating charge dipoles. Radiation from oscillators cause heat transfer through EM waves. Radiation transfer occur at all non zero Temperature T and is in the infra red part of the EM spectrum.
At thermal equilibrium the black body radiates and also absorbs at the same rate and the spectrum of radiation is
continuous and depends on the temperature T and the nature
of the material.
The power radiated by a black body is given by the Stefan-Boltzmann Law as
P=e σ A (T4 -T04)
where T0 is the ambient temperature, e is the emissivity and σ is the Stefan constant. By Kirchoff's Law e=a at thermal equilibrium where a is the absorptivity of the body. For an ideal black body radiator e=a=1 and the spectrum is universal and dependent only on the temperature.
A graphite cavity with thick walls and a small hole may be approximated to
an ideal black body such that it is perfect absorber and absorbs all radiation incident on it. The cavity walls re radiate and at thermal equilibrium the cavity is full of radiation at te ambient temperature T. If this cavity is heated to T then the hole must radiate like a black body by Kirchoffs Law because a perfect absrober
is also a perfect emitter. So the radiation from the hole may be understood by analysing the cavity radiation.
The energy density ( energy/volume) of the cavity radiation ρT(ν) is proportional to RT(ν) where R is the spectral radiancy (energy radiated/unit time/unit area) of the black body. A graph of the spectral radiancy was
obtained by Lummer and Pringsheim at various temperatures in deg K through experiemnets.
Spectral Radiancy Curves with Frequency
Image From: thermal-survey.co.uk via Google Images
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The characteristics of these curves are
1. Very little power radiated for low frequency P=0 for ν=0
2. Increases with increase in frequency and reaches a maxima for νmax
3. Drops with further increase and asymptotes to zero for ν infinity.
4. The peak frequency νmax shifts to higher frequecies for higher temperatures.
The total radiancy RT=∫ RT(ν) dν=σ T4
Wiens Law: W=T λmax where W is the Wien's constant. This is
a theoretical fit to an experimental curve and the value of W is thus obtained.
Instead of "Degrees Kelvin", we should use only "Kelvin"
ReplyDeleteThat is how it's defined by the BIPM.
http://www.bipm.org/en/CGPM/db/10/3/
I
That is just the nomenclature. Its for convenience and to avoid confusion. It has little to do with the Physics.
ReplyDeleteSir will we study the derivation of Planck's?
ReplyDeleteDoes anyone have the eisberg e book??
ReplyDeleteIn the line "Radiation transfer occur at all non zero Temperature T and is in the infra red part of the EM spectrum." , the word maximum is probably missing. As discussed in the class, isn't the maximum radiation in the infra red part?
ReplyDeleteelaborating with regard to the doubt i had on friday, which got cleared by the end of the class:
ReplyDeletei thought that "density" cannot be an infinitesimal value. the catch was that there are two densities, spectral density and spatial density.
while talking about spatial density, if we include all frequencies it becomes a finite value but if we talk about contribution to spatial density by a particular frequency it is understandably an infinitesimally small value.
so the quantity written was not really any "density" but contribution to the density by a particular frequency.
http://wwwitp.physik.tu-berlin.de/brandes/public_html/qm/umist_qm/node3.html
ReplyDeletelink for the experiment
Regarding the doubt raised in the class regarding the Nature of ρ(ν)dν which is defined as the energy density within the frequency range of ν to ν + dν:
ReplyDeleteI believe that both the quantities are infinitesimal. In, ρ(ν)dν, ρ(ν) is a finite quantity and in the graphs shown above, it stands for the height of the curve from the x-axis. ρ(ν)dν would physically stand for the area of an infinitesimally thin rectangular slice under the curve.
Independently of the above, the energy density within the frequency range of ν to ν + dν is also infinitesimal. Notice the infinitesimal range 'ν to ν + dν'.
Yes we will certainly study Planck's distribution
ReplyDeleteYes the radiation is maximum in the infrared, in fact as high as 90 %. This is i believe discussed both in Eisberg and I also mentioned this in the
ReplyDeletelectures.
@Himanshu: Correct because its not multiplied by d ( volume) dV but d(frequency)=d (nu).
ReplyDelete@Ish: Quite correct.
Just a few thoughts --
ReplyDeleteIf e,a are constants: If a body is at a higher temperature than its surroundings, and if e>a, it will keep emitting till it reaches the surrounding temperature and will continue to radiate after that, which is not possible.
So, if e>a for T(body)>T(surroundings), then e<a for T(body)<T(surroundings) and e=a at equilibrium. So, there does not seem to be any reason for good emitters to be good absorbers away from equilibrium.
Eisberg and Resnick E-Book @ http://dl.dropbox.com/u/4342649/book.zip
ReplyDeleteFor a three dimensional wave broken into components along orthogonal axes (as we have done in the class), what shall be the frequency of the wave if the frequencies of the components are (nu)x, (nu)y and (nu)z? I think that we need to proceed by breaking the wavelength into components along the axes! Isn't it?
ReplyDeleteThe link to the Experiment shared above says that the Second Law of Thermodynamics can be used to show that the spectral energy density of black body radiation, i.e. the radiation energy per volume and per frequency interval, is only a function of the frequency and the temperature of the walls. But we have done by showing that the same holds true for the cube and then generalized the statement. Could you please elaborate on the use of Second Law?
ReplyDeleteThank You.
A wave has a single wavelength and frequency. Only the ave-vector k is a vector. Please check Griffiths for 3 d waves.
ReplyDeleteI dont really know about this line of reasoning from the second law. The standard method is the one that I have followed which chooses a simple geometry and then shows that the final result is geometry independent. I will post the reasoning if I come across it.
ReplyDeleteregarding e a discussion :
ReplyDeletethe moment it reches equilibrium e=a should be applied.
ya for vector k c= w/|k|
ReplyDeletethe equipartition theorem as I have been exposed to till now applied to the individual atomic oscillators on the walls... what is the justification for saying that the same amount of energy is contained in the standing wave that it generates? Or does the equipartition theorem hold separately and independently for radiation modes as well??
ReplyDeleteDoes the photon picture explain the slowing down of light in a medium?
ReplyDeleteThe theorem can be applied to any LARGE number of simmilar entities. Check the appendix of Resnick
ReplyDeletefor the boltzmann distribution. The fourier modes
for the waves precisely correspond to the oscillators.
Thats on Optics question and I am not an expert on that. You may ask your Optics instructors.
ReplyDeleteTrying to correlate with the previous Optics course...
ReplyDeleteEven if the walls are conducting shouldn't E(parallel) be 0 only at steady state... In this case what I am expecting is evanescence inside the metal... I could not understand why we discard possibilities like in this pic:
http://tinypic.com/r/2s12i43/7
have we assumed that the length is large compared to the skin depth lengthscale??
Sorry for the repost... this is just for documentation and the question has been answered in the comments of the 1st post where I had mistakenly posted it...
@ anonymous with few thoughts:
ReplyDeletei think its possible that you are confusing emissivity e with net emitted radiation... the LATTER would go to zero because it depends on (T-T(surroundings))^4 which goes to zero near eqm... emissivity is just the constant in that dependance...
Photon picture may explain the slowing down of light if we take into account a large number of photons. Some collides others pass through collision can be absorptive or reflective(or partially both). Statistically they slow down(or may even bend).
ReplyDeleteThe explanation for slowing down does not seem to be very clear. What advantage does it give over the wave picture to explain slowing down ?
ReplyDeleteIs it necessary that a phenomenon can be explained by both particle and wave picture?...It's not. So there is no problem at all if we are not able to find out why photon should slow down.
ReplyDeleteI don't have any idea if this gives any advantage over the wave picture, but was trying to explain the photon picture of light in refraction.
ReplyDeleteNo it is not necessary. The wave picture really involves a large number of photons. However possibly its a good idea to ask the question if a
ReplyDeletesingle photon would slow down in a medium. These are essentially quantum optics related.
Isn't it a question of this kind: what would happen to a single photon in young's double slit experiment?
ReplyDeleteWhy 1/(wavelength)^5 dependence NECESSARY for Wein's Law?
ReplyDelete