Energy Density Notation Confusion

[laser1]

Homework Statement

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Relevant Equations

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$$u(\nu, T) = \frac{8 \pi \nu^2}{c^3} \cdot \frac{h \nu}{e^{h \nu / k T} - 1}$$ Now using the relation $c=\nu \lambda$, $$u(\lambda, T)=8 \pi h \frac{1}{\lambda^3} \cdot \frac{1}{e^{hc / (\lambda k T)} - 1}$$ Now this is still per unit frequency. To get it to per unit wavelength, then $$ u(\lambda, T) = \frac{8 \pi h c}{\lambda^5 \left(e^{\frac{hc}{\lambda k T}} - 1\right)}$$ My question is for notation: how to distinguish between the 2nd and 3rd equation?

 

[Frabjous]

Doesn’t going from 2 to 3 wrongly imply c/λ²=1?

 

[laser1]

Doesn’t going from 2 to 3 wrongly imply c/λ²=1?

You are correct, sorry, I should have mentioned that the u's are not the same. The 2nd equation u is not the same as 3rd equation u. 2nd equation is per unit frequency, 3rd equation is per unit wavelength. My question is how to distinguish that the u's are different.

 

[Frabjous]

Do you have a reference for this notation? If it is from multiple sources, you are just observing author’s discretion.

 

[kuruman]

My question is for notation: how to distinguish between the 2nd and 3rd equation?

There is nothing to distinguish. The second equation is incorrect and the third is correct. You are forgetting that $u(\text{whatever,T})$ is a distribution function. It represents energy $d\epsilon$ within an interval $d(\text{whatever,T})$. So, if whatever = frequency, the starting equation is $$d\epsilon=u(\nu,T)d\nu.$$ Then, $$d\epsilon= \frac{8 \pi h \nu^3}{c^3} \cdot \frac{1}{e^{h \nu / k T} - 1}d\nu.$$ Now you replace the frequency symbols with the wavelength symbols using $\nu=c/\lambda$ to get $$d\epsilon= \frac{8 \pi }{\lambda^3} \cdot \frac{1}{e^{\frac{h c}{ \lambda k T}} - 1}\left( -\frac{c}{\lambda^2}\right)d\lambda$$which gives the distribution function $$u(\lambda,T)=\frac{d\epsilon}{d\lambda}=\frac{8 \pi h c }{\lambda^5} \cdot \frac{1}{e^{\frac{h c}{ \lambda k T}} - 1}.$$ The negative sign is dropped because all it says is that $d\epsilon$ is positive when the frequency is increasing and negative when the wavelength is increasing.

On edit:
I should have clarified that $d\epsilon$ is an energy density interval i.e. energy per unit volume as dimensional analysis shows.

 

[vela]

You are correct, sorry, I should have mentioned that the u's are not the same. The 2nd equation u is not the same as 3rd equation u. 2nd equation is per unit frequency, 3rd equation is per unit wavelength. My question is how to distinguish that the u's are different.

The usual way—use a different letter, add a subscript, etc.

 

[vela]

The second equation is incorrect and the third is correct.

A mathematician would disagree. The $u$ in $u(\nu,T)$ and in $u(\lambda,T)$ represent different functions, so they should, strictly speaking, be denoted differently.

 

[kuruman]

The second equation is incorrect and the third is correct.

A mathematician would disagree. The $u$ in $u(\nu,T)$ and in $u(\lambda,T)$ represent different functions, so they should, strictly speaking, be denoted differently.

I have seen $u_{\nu}$ and $u_{\lambda}$ in textbooks. To me the distinction is obvious when I look to the right hand side and see whether there is a frequency or a wavelength. Nevertheless, the OP's second equation is mathematically and dimensionally incorrect as a distribution function.

OP's first equation is energy density per frequency interval. Dimensionally, when multiplied by frequency units it gives energy density (Joules/m³). Likewise, when the third equation is multiplied by units of length, it also gives energy density units.

OP's second equation is none of the above and a physicist would say that it is dimensionally incorrect. This problem arose when OP, in trying to convert from a per frequency interval to a per wavelength interval, did not take into account that the distribution functions are basically derivatives and, therefore, the two intervals are related by the chain rule.

 

[laser1]

OP's second equation is none of the above

If you multiply the second equation by frequency units it gives units of energy density $(J/m^3)$. I would say that the second equation is energy density per frequency interval, but expressed in terms of wavelength rather than frequency.

 

[kuruman]

If you multiply the second equation by frequency units it gives units of energy density $(J/m^3)$. I would say that the second equation is energy density per frequency interval, but expressed in terms of wavelength rather than frequency.

Sure, but then if you plot this, you have energy density per frequency interval as a function of wavelength. How useful is that?

 

[laser1]

Sure, but then if you plot this, you have energy density per frequency interval as a function of wavelength. How useful is that?

Fair, very good point.