Derive a formula for blackbody radiation

In summary: I just feel sad that someone with such a big potential wasted it like this. :/In summary, my friend seems to have a lot of knowledge on physics, but he is not using it effectively. He is making jokes instead of asking questions, which is not helping him learn.
  • #1
George Isaac
11
0
when we derive a formula for blackbody radiation, we say that each electromagnetic mode has energy nhf, n is an integer, h is Planck's constant f is the frequency of the radiation. We interpret n as the number of photons per mode. However, a deeper QM analysis involves replacing the electromagnetic mode by a harmonic oscillator having energy (n+1/2)hf. What does this 1/2 mean? 1/2 a photon? We also use a normalized Boltzmann probability distribution to complete the calculation, but when we calculate the number of carriers in semiconductor electronics, we use a Fermi Dirac probablity distribution that is in no way normalized. Anybody have a sound explanation?
 
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  • #2
George Isaac said:
when we derive a formula for blackbody radiation, we say that each electromagnetic mode has energy nhf, n is an integer, h is Planck's constant f is the frequency of the radiation. We interpret n as the number of photons per mode. However, a deeper QM analysis involves replacing the electromagnetic mode by a harmonic oscillator having energy (n+1/2)hf. What does this 1/2 mean? 1/2 a photon? We also use a normalized Boltzmann probability distribution to complete the calculation, but when we calculate the number of carriers in semiconductor electronics, we use a Fermi Dirac probablity distribution that is in no way normalized. Anybody have a sound explanation?

My friend,i've read your other post as well,and i'll probably post an aswer there too.
I'm afraid that u haven't read too much physics and the one u have is struggling to stay alive in deep coma.
Do u know what a photon is...?Do you know what a phonon is...??My answer is that u don't,else u could be able to tell them apart. :wink: There is no 1/2 photon,no 1/2 phonon.There's just ground state energy of a harmonic oscillator.
When wnd where are we allowed (if so) to use Boltzmann statistics when dealing with em radiation and solid state physics...??
Do u have any idea about the physical segnificance of the Fermi-Dirac distribution function?If u did,u might have made a coonection between it and the condition for normalization.

PS.How does a normalization condition for a quantum statistical ensemble look like??
 
  • #3
Don't worry about 1/2 energies in statistical mechanics. Quantum oscillators have to have them and can't lose them. So you can assume ground-state is zero.

As for normalization of the Fermi-Dirac distribution, I believe it's normalized. You divided by the grand partition function right?
 
  • #4
dextercioby said:
My friend,i've read your other post as well,and i'll probably post an aswer there too.
I'm afraid that u haven't read too much physics and the one u have is struggling to stay alive in deep coma.
Do u know what a photon is...?Do you know what a phonon is...??My answer is that u don't,else u could be able to tell them apart. :wink: There is no 1/2 photon,no 1/2 phonon.There's just ground state energy of a harmonic oscillator.
When wnd where are we allowed (if so) to use Boltzmann statistics when dealing with em radiation and solid state physics...??
Do u have any idea about the physical segnificance of the Fermi-Dirac distribution function?If u did,u might have made a coonection between it and the condition for normalization.

PS.How does a normalization condition for a quantum statistical ensemble look like??

My friend
I would till you what George Isaac knows and what he doesn't, I assisted in teaching him in the undergraduate and now he is a research assistant. But I guess tilling you will not change a thing because you are not from those who ask to learn or criticize to create.
My advice to you: Do not advice! Do not help! You are showing off ... you have nothing to say more the questions themselves or some of the courses you seem to take "nowadays". You are not helping by throwing jokes.
As for myself I would not bother replying you anymore and I advice George to do the same; because we seek the physics behind the theory not the solution of some maths with big terms and strange conventions around. And if it is necessary we will learn more and will not be hindered by some joke tiller like you to ask to learn from those who reply to teach.
 
  • #5
sifeddin said:
My friend
I would till you what George Isaac knows and what he doesn't, I assisted in teaching him in the undergraduate and now he is a research assistant. But I guess tilling you will not change a thing because you are not from those who ask to learn or criticize to create.

Now that u told me he's a Research Assistant ,it gives me credit for the jokes.I mean,there's a difference:i would have felt really stupid having made jokes at someone who hasn't studied.Someone basically didn't open a SM book and i would have felt stupid and definitely would have made no sense at all making jokes at someone who's "out of the picture".But now that u told me that he's graduated and he's "reserching" (i guess in physics,else it would have made no sense to reveal such a matter) deeply :tongue2:,i guess not knowing the difference between photons and phonons (except for the obvious one:a lousy letter) is unacceptable for someone who's "researching" and definitely a damn good reason to make (sometimes bad,i agree :wink: ) jokes.Jesus,it's outrageous! :yuck:

Daniel.

PS.If u come up with the explanation he ditched SM classes especially in order not to know the diffference between a photon and a phonon,then u're a hero... :tongue2: My hero...
PPS.I guess it doesn't take too much to become a "researcher" 'nowadays'.
 
  • #6
Please Think!

Daniel,

Actually I am researching in engineering not physics. I am an engineer not a physicist and I am researching in Optical communication systems. Physics is like a hobby to me, I am curious about how the universe works. So, you merely reflect a narrow vision when you think that everybody in this forum is an expert in theoretical physics. If I had been trained as a physicist in my undergraduate studies, I would not be seeking answers from someone like you. Reply please, don't just pretend that you didn't read this post.
 
  • #7
I would like to ask you again, have you published any work anywhere? Maybe you are a lousy physics researcher yourself!
 
  • #8
George Isaac said:
Daniel,

Actually I am researching in engineering not physics. I am an engineer not a physicist and I am researching in Optical communication systems. Physics is like a hobby to me, I am curious about how the universe works. So, you merely reflect a narrow vision when you think that everybody in this forum is an expert in theoretical physics. If I had been trained as a physicist in my undergraduate studies, I would not be seeking answers from someone like you. Reply please, don't just pretend that you didn't read this post.

In ANY case, the original eigenvalue you had for the energy state of a harmonic oscillator is wrong. It is

[tex]E_n = (n + 1/2)\hbar \omega[/tex]

This corresponds to the classical "normal modes" of oscillation. And yes, for n=0, there IS an energy value, the so-called zero-point energy. This is the natural outcome of the formulation and is a direct consequence of the uncertainty principle. It means that for a quantum harmonic oscillator, the lowest energy it can have is not zero, but rather [tex]1/2 \hbar \omega[/tex]. There are no "photons" here. Photons only comes in where there is an energy transition from one state to another.

Zz.

Edit: Actually, hf can be written as hbar*omega. So it isn't wrong, sorry. However, this is still not for the "photon". It is for the harmonic oscillator in question. So there are no "half photon" here.
 
Last edited:
  • #9
George Isaac said:
I would like to ask you again, have you published any work anywhere? Maybe you are a lousy physics researcher yourself!

Ups,i blew it! :blushing: The part with enegineering didn't cross my (narrow) mind. :tongue2: I can say:good for you and keep up the good wok in Optical Communication Systems.Though optics has a rather new branch these days,called "quantum optics".It basically deals with quantum (photonic) effects on normal optics.So i guess the notion of "photon" should be familiar to you from your business,not from the bit of physics studied in college.Or i may be wrong again...

Anyway,to your knowledge,i'm not a researcher,not yet,anyway,i have still a lot to learn and haven't decided which branch of teoretical phyiscs to chose.
I'll have to make it,before applying for a PhD.

Daniel.
 
  • #10
dextercioby said:
Now that u told me he's a Research Assistant ,it gives me credit for the jokes.I mean,there's a difference:i would have felt really stupid having made jokes at someone who hasn't studied.Someone basically didn't open a SM book and i would have felt stupid and definitely would have made no sense at all making jokes at someone who's "out of the picture".But now that u told me that he's graduated and he's "reserching" (i guess in physics,else it would have made no sense to reveal such a matter) deeply :tongue2:,i guess not knowing the difference between photons and phonons (except for the obvious one:a lousy letter) is unacceptable for someone who's "researching" and definitely a damn good reason to make (sometimes bad,i agree :wink: ) jokes.Jesus,it's outrageous! :yuck:

Daniel.

PS.If u come up with the explanation he ditched SM classes especially in order not to know the diffference between a photon and a phonon,then u're a hero... :tongue2: My hero...
PPS.I guess it doesn't take too much to become a "researcher" 'nowadays'.

dextercioby said:
Quote:
Originally Posted by George Isaac
I would like to ask you again, have you published any work anywhere? Maybe you are a lousy physics researcher yourself!



Ups,i blew it! The part with enegineering didn't cross my (narrow) mind. I can say:good for you and keep up the good wok in Optical Communication Systems.Though optics has a rather new branch these days,called "quantum optics".It basically deals with quantum (photonic) effects on normal optics.So i guess the notion of "photon" should be familiar to you from your business,not from the bit of physics studied in college.Or i may be wrong again...

Anyway,to your knowledge,i'm not a researcher,not yet,anyway,i have still a lot to learn and haven't decided which branch of teoretical phyiscs to chose.
I'll have to make it,before applying for a PhD.

Daniel.

Daniel ... :tongue2: :tongue: :yuck:

Dear George
Didn't I till you. Da... isn't really worth it.
Dan... should apply for English classes 'before applying for a PhD' to learn how to read the forum pages title : "Physics Help and Math Help". The one Scientific question that he should ask himself: where should someone like the "time derivative of the acceleration" he is apply NOT 'to make (sometimes bad,i agree :grumpy: ) jokes.'?
I guess if there is such place I would be no hero to tell that he will 'ditch that class especially in order to know the difference between a photon and a phonon' in the English classes. But he will never find the answer to this in the English classes either.
 
  • #11
ZapperZ said:
In ANY case, the original eigenvalue you had for the energy state of a harmonic oscillator is wrong. It is

[tex]E_n = (n + 1/2)\hbar \omega[/tex]

This corresponds to the classical "normal modes" of oscillation. And yes, for n=0, there IS an energy value, the so-called zero-point energy. This is the natural outcome of the formulation and is a direct consequence of the uncertainty principle. It means that for a quantum harmonic oscillator, the lowest energy it can have is not zero, but rather [tex]1/2 \hbar \omega[/tex]. There are no "photons" here. Photons only comes in where there is an energy transition from one state to another.

Zz.

Edit: Actually, hf can be written as hbar*omega. So it isn't wrong, sorry. However, this is still not for the "photon". It is for the harmonic oscillator in question. So there are no "half photon" here.

Thank you for the sceintific reply
 

1. What is blackbody radiation and why is it important?

Blackbody radiation is the thermal radiation emitted by a perfect blackbody object. It is important because it helps us understand the relationship between temperature and the amount of energy emitted at different wavelengths.

2. How can a formula be derived for blackbody radiation?

A formula for blackbody radiation can be derived by using the principles of quantum mechanics and statistical mechanics, specifically the Planck distribution and the Bose-Einstein distribution. These distributions describe the energy levels and probabilities of particles in a system, which can be used to determine the energy emitted by a blackbody at different temperatures.

3. What is the Planck distribution and how is it related to blackbody radiation?

The Planck distribution is a mathematical function that describes the distribution of energy among the possible energy states of a system. In the case of blackbody radiation, the Planck distribution describes the energy levels of photons at different wavelengths and temperatures, and can be used to derive the formula for blackbody radiation.

4. Can the formula for blackbody radiation be applied to real objects?

Yes, the formula for blackbody radiation can be applied to real objects to estimate the amount of thermal radiation they emit. However, real objects are not perfect blackbodies, so the formula may need to be modified to account for factors such as surface reflectivity and shape.

5. How has the formula for blackbody radiation been used in scientific research?

The formula for blackbody radiation has been used in a variety of fields, including astrophysics, cosmology, and materials science. It has helped scientists understand the thermal properties of objects in space, such as stars and galaxies, and has been used to develop new materials with specific thermal properties for various applications.

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