Black Body radiation and Planck's radiation distribution

[Cocoleia]

Homework Statement

For blackbody radiation, rewrite the Planck radiation distribution u(ω)dω as u(λ)dλ
and sketch a graph of u(λ), which is the energy per unit volume, per unit wavelength.
Also find the limiting behaviour of u(λ) as λ → 0 and as λ → ∞.

Relevant Equations

Planck's Law

I am quite confused, as I start this question. I can easily find the following when searching up Planck's law:

However, this is not u. My prof is quite unclear and sometimes chooses his own variables as he sees fit, so i am not sure if this would be equivalent to what he is looking for u(λ)dλ

However, I also see this lovely table

yet again, not sure if any of these things are related to what I am supposed to be looking for.

Finally, I see an expression as a function of u

So would I have to take the expression of B(λ) and multiply by 4pi/c to get the expression I am looking for.

For the other parts of the question, I need this part to do them.

Thanks

 

[Charles Link]

See https://www.physicsforums.com/threa...n-plancks-law-derivation.961396/#post-6097855
I believe this thread and the "links" it contains should answer your question. I'd be happy to try to answer any additional questions you might have after reading through this thread and its "links".

 

[Charles Link]

One additional relation you need is $|B_{\lambda}(\lambda, T) \, d \lambda|=|B_{\nu}(\nu,T) \, d \nu |$ where $\nu \lambda=c$ so that $d \lambda=-\frac{c}{\nu^2} \, d \nu$, and similarly $d \nu=-\frac{c}{\lambda^2} \, d \lambda$. (The minus sign of course disappears with the absolute value signs $| \, |$ ).

 

[Cocoleia]

One additional relation you need is $|B_{\lambda}(\lambda, T) \, d \lambda|=|B_{\nu}(\nu,T) \, d \nu |$ where $\nu \lambda=c$ so that $d \lambda=-\frac{c}{\nu^2} \, d \nu$, and similarly $d \nu=-\frac{c}{\lambda^2} \, d \lambda$. (The minus sign of course disappears with the absolute value signs $| \, |$ ).

I have been trying to use what you mentioned in your previous two comments. Thank you, I think it was very useful. Do you mind looking over what I did to see if it makes sense?

Thank you for your input.

 

[Charles Link]

The only correction I can see is the minus sign (a minus 1 factor) that shows up in $\frac{d \nu}{d \lambda }$ is superfluous.
Here we are dealing with positive $u_{\lambda}(\lambda,T)$ and $u_{\nu}(\nu, T)$. That is why I put the absolute values around the $B$ relations above.
The picture is too small to readily see it all even with a magnifying glass that I have, but I checked the first couple of lines you have=especially the $\frac{c}{4}$ factor, and yes, you got that part correct. The rest is just algebra=I trust you did the algebra correctly.

 

[Charles Link]

With the bigger picture I see a couple of errors. You already put forth the necessary effort, and this is really a somewhat advanced assignment, so I'm simply going to write out what is the necessary parts of the solution.
The quantity $B_{\lambda}(\lambda, T)$ is usually designated by the letter $L$ and written as $L_{\lambda}(\lambda,T)$. There is a somewhat tricky part here that you got incorrect. I need to check my previous posts to make sure I got them right...Give me a minute, and then I will jump to the next reply box...

 

[Charles Link]

And yes, my previous posts are correct...In this "link" https://www.physicsforums.com/threa...n-plancks-law-derivation.961396/#post-6097855 post 2, about halfway down "Thereby there is a division by $\pi$ along with a factor of $\frac{c}{4}$ in going from the 2nd expression to the 3rd expression. This $\frac{c}{4 \pi}$ factor is correct in your original post=it is completely incorrect on your handwritten paper. In simple terms: $\pi B=\frac{c}{4} u$. The factor of $\pi$ as well as the $\frac{c}{4}$ are explained in the "links".

 

[Charles Link]

The quantity $M =L \pi$ is what arises in the Stefan-Boltzmann law if we integrate wavelengths from $0$ to $+\infty$. You may be familiar with $M=\sigma T^4$, so you know that $M$ is radiation (power) per unit area. The quantity $B$ or $L$ is actually more useful than $M$ to researchers who do a lot of measurements with infrared sources. The reason is that $I=L A$ and the irradiance $E$ (watts/m^2 ) at a distance $s$ on-axis from the source is given by $E=\frac{LA}{s^2} =\frac{MA}{\pi s^2}$. The extra $\pi$ in the denominator of this last equation is clumsy=researchers much prefer working with $L$. The Stefan-Boltzmann equation with $L$ reads $L=\frac{\sigma T^4}{\pi}$. This equation does thereby get the $\pi$ in the denominator, but the equation $E=\frac{LA}{s^2 }$ is important enough that the workers prefer to work with $L$ , which they are using the letter $B$. The units on $L$ are
Watts/(m^2 steradian) , but the steradian is really a dimensionless unit, because it is $\Omega=\frac{A}{r^2}$.
======================================================================================
This might be a lot of extra info, but I'm trying to give you some idea of what the $B$ they give you represents. Again the relation is $B=\frac{c}{4 \pi} u$, or what really might be easier to follow: $M=\frac{c}{4} u$. This last equation is probably the easiest to follow, but they then introduce $B$ where $B=\frac{M}{\pi}$.

 

[Cocoleia]

The quantity $M =L \pi$ is what arises in the Stefan-Boltzmann law if we integrate wavelengths from $0$ to $+\infty$. You may be familiar with $M=\sigma T^4$, so you know that $M$ is radiation per unit area. The quantity $B$ or $L$ is actually more useful than $M$ to researchers who do a lot of measurements with infrared sources. The reason is that $I=L A$ and the irradiance $E$ (watts/m^2 ) at a distance $s$ on-axis from the source is given by $E=\frac{LA}{s^2} =\frac{MA}{\pi s^2}$. The extra $\pi$ in the denominator of this last equation is clumsy=researchers much prefer working with $L$. The Stefan-Boltzmann equation with $L$ reads $L=\frac{\sigma T^4}{\pi}$. This equation does thereby get the $\pi$ in the denominator, but the equation $E=\frac{LA}{s^2 }$ is important enough that the workers prefer to work with $L$ , which they are using the letter $B$. The units on $L$ are
Watts/(m^2 steradian) , but the steradian is really a dimensionless unit, because it is $\Omega=\frac{A}{r^2}$.
======================================================================================
This might be a lot of extra info, but I'm trying to give you some idea of what the $B$ they give you represents. Again the relation is $B=\frac{c}{4 \pi} u$, or what really might be easier to follow: $M=\frac{c}{4} u$. This last equation is probably the easiest to follow, but they then introduce $B$ where $B=\frac{M}{\pi}$.

Thank you for such a detailed explanation

 

[Charles Link]

Hopefully it is helpful. If you try again, I will be happy to proof-read your results. I think your only major mistake is the equation in the rectangle in the middle of the page which should read
$B_{\nu}(\nu,T)=(\frac{c}{4 \pi}) u_{\nu}(\nu, T)$ , and the extra $d \nu$ you put on the right side with none on the left is incorrect.

 

[Charles Link]

To help you out with a couple steps: $B_{\lambda}(\lambda, T)=B_{\nu}(\nu,T) |\frac{d \nu}{d \lambda}|=(\frac{c}{4 \pi})u_{\nu}(\nu, T) (\frac{c}{\lambda^2})$.
Now just substitute $\nu=\frac{c}{\lambda}$, and you should get exactly what you need.
Be sure and see post 10 above also.
And please note: The equation with the factor of $\frac{c}{4 \pi}$ is not what you would call a standard equation. It took a little effort to show that this is indeed the correct factor. Now that we have proven that, you can use it routinely.
Hopefully, though, you read through the derivations in the "link" including the one on the particle effusion rate per unit area of $R=\frac{n \bar{v}}{4}$, so that you can see where it came from.
You should also be able to understand where the $\pi$ comes from in the denominator. This is also explained in the "links". It is important to recognize, because of the $\cos{\theta}$ fall-off, that the power radiated over a hemisphere is $P=M A=I \pi=LA \pi$, with a $\pi$ (steradians) instead of $2 \pi$ (steradians). This is important in understanding where the $\pi$ comes from.