Go Back   Physics Help Forum > Physics Forums > Physics

Physics Physics Forum - General Physics Discussion and Physics News

Like Tree2Likes
Reply
 
LinkBack Thread Tools Display Modes
Old Jul 20th 2019, 05:27 AM   #1
Junior Member
 
Join Date: Jul 2019
Posts: 26
Lightbulb Rayleigh-Jeans law

Good Afternoon,

As I told you in my presentation, at 70 years old, I start to learn physics.
My brain is a little bit slow and confusedůso sorry for my (most of prabable) trivial question.

I read on a website that the law of Rayleigh-Jeans is given for the radiance function of wavelenght (λ)
Radiance (function of wavelength)=2ckT/λ square
Radiance (function of frequency)=(2kT/c square). v square with λ=c/v and the partial derivative dλ/dv=c/v square
v is the frequency...don't found to type the greek letter

My questions:
1.How to reach the second equation starting from the first one (please detailed step by step explanation ?
2.In the second equation why a partial derivative has to be added ?
3. How to reach for the partial derivative dλ/dv the value c/v square...trivial, sorry

Thanks in advance and best regards
philippe is offline   Reply With Quote
Old Jul 20th 2019, 06:11 AM   #2
Senior Member
 
Join Date: Aug 2010
Posts: 434
$\lambda$ is the wavelength in, say, meters, $\nu$ is the frequency in, say, "waves per second", and c is the wave speed in meters per second. Imagine standing by a fixed point, watching the waves go by. In "T" seconds, you will have seen waves to a total length of cT meters go by. Since each wave has length $\lambda$ you will have seen $\frac{cT}{\lambda}$ waves. So $\nu T= \frac{cT}{\lambda}$ or $\nu= \frac{c}{\lambda}$.
That can also be written as $\lambda\nu= c$ or $\lambda= \frac{c}{\nu}$.

If the wave has constant wave speed (as, for example, light in vacuum) then we could easily switch between wave length and frequency using $\lambda= c\nu$: to go from
Radiance (function of wavelength)=$\frac{2ckT}{\lambda^2}$
Radiance (function of frequency)=$\frac{2kT\nu^2}{c}$
replace $\lambda^2$ with $\left(\frac{c}{\nu}\right)^2= \frac{c^2}{\nu^2}$:

$\frac{2ckT}{\lambda^2}= \frac{2ckT}{1}\frac{\nu^2}{c^2}= \frac{2ckT\nu^2}{c^2}= \frac{2kT\nu^2}{c}$.

If the wave speed, c, is NOT constant but varies with time then the relationship between wave length and frequency varies so we need that partial derivative
topsquark likes this.
HallsofIvy is offline   Reply With Quote
Old Jul 20th 2019, 02:44 PM   #3
Junior Member
 
Join Date: Jul 2019
Posts: 26
Rayleigh-Jeans law

Dear friend,
Thanks for the reply, however is something I don't understand.
For the Radiance L (function of wave length) as denominator you write lambda square.
From what I read on the net, the denominator written is lambda exponent 4
Please, can you clarify this point

Thanks in advance and best regards
philippe is offline   Reply With Quote
Old Jul 21st 2019, 11:30 AM   #4
Senior Member
 
Join Date: Aug 2010
Posts: 434
I was just going by what you wrote:
I read on a website that the law of Rayleigh-Jeans is given for the radiance function of wavelenght (λ)
Radiance (function of wavelength)=2ckT/λ square
Radiance (function of frequency)=(2kT/c square).
.
HallsofIvy is offline   Reply With Quote
Old Jul 21st 2019, 07:01 PM   #5
Junior Member
 
Join Date: Jul 2019
Posts: 26
Sorry to bother you with that... As you see I'm a zero
I must have misread the article.
So I reformulate after having reading few more articles
L fact wave length(T) = 2ckT/ lambda power 4
Knowing lambda= c/v and d lambda=(c/v square) d v
Give d lambda= (c/v square) DV
Then Lv(T) = (2kT/c square) v square
C= light speed in vacuum
K= Boltzmann contacts
T= temperature
I don't understand how to go from first equation to second.
Please and thanks in advance to write a line by line.
Im sure that my problem is coming from this partial derivative.
As well I don't understand how from lambda=c/v reaching
d lambda= (c/v square) DV

No. My case is not desperate... It just I start from scratch at 70.
Thanks in advance and regards
Philippe
philippe is offline   Reply With Quote
Old Jul 21st 2019, 11:13 PM   #6
Junior Member
 
Join Date: Jul 2019
Posts: 26
Dear All,

After spending few time... I find and I understand.

The only question remaining is to know how Rayleigh and Jeans discover L(lambda) = (2ckT) / lambda power 4
Empirical or from theory?
Thanks and regards
philippe is offline   Reply With Quote
Old Jul 22nd 2019, 12:52 AM   #7
Junior Member
 
Join Date: Jul 2019
Posts: 26
Hello,

Continuing reading what I find on the subject
The rayleigh-Jeans law will come from the equipartition theorems?
WHAT IS THAT???
If someone can give me some clues
I remember you that Wikipedia is blocked here.
I would like to understand equipartition and the way to followed to reach the rayleigh-Jeans law

Thanks and best regards

Philippe
philippe is offline   Reply With Quote
Old Jul 22nd 2019, 02:24 AM   #8
Senior Member
 
Join Date: Oct 2017
Location: Glasgow
Posts: 426
Originally Posted by philippe View Post
Hello,

Continuing reading what I find on the subject
The rayleigh-Jeans law will come from the equipartition theorems?
WHAT IS THAT???
If someone can give me some clues
I remember you that Wikipedia is blocked here.
I would like to understand equipartition and the way to followed to reach the rayleigh-Jeans law

Thanks and best regards

Philippe
The equipartition theorem states that for every degree of freedom of a gas molecule, there exists a quantity of energy $\displaystyle \frac{1}{2}kT$ that can be absorbed as heat (i.e. to heat capacity).

There is a derivation here:
Rayleigh-Jeans Law Development

Let me know if you can access it or not.
benit13 is offline   Reply With Quote
Old Jul 23rd 2019, 03:58 AM   #9
Junior Member
 
Join Date: Jul 2019
Posts: 26
Dear All,

You are lucky to have access to Wikipedia.
Now I'm in definition
I'm searching what is the emitted flux density (Wm*2) and the emitted energy (J m*-3) it's related to my learning on black body.
As well what is the irradiance?
Of course again trivial basic questions. Sorry for that.
Thanks in advance and regards
Philippe
philippe is offline   Reply With Quote
Old Jul 23rd 2019, 04:43 AM   #10
Senior Member
 
Join Date: Oct 2017
Location: Glasgow
Posts: 426
Sorry, I replied to this in the other thread:
Introducing myself...Philippe from China
benit13 is offline   Reply With Quote
Reply

  Physics Help Forum > Physics Forums > Physics

Tags
law, rayleighjeans



Thread Tools
Display Modes



Facebook Twitter Google+ RSS Feed