Black Body Equations (more of a math prob)

[Rob Hal]

Hi,

In one of my textbooks, I'm given these two equivalent equations for energy flux radiated from a black body, one dependent on frequency, the other on wavelength:

$$I(\lambda)d\lambda = \frac {2c^2h}{\lambda^5} \frac {1}{e^{(hc/KT\lambda)}-1}d\lambda$$

and

$$I(\nu)d\nu = \frac {2h\nu^3}{c^2} \frac {1}{e^{(hv/KT)}-1}d\nu$$

Now, I'm just trying figure out how are these equations equivalent? At first, I thought it was simple substitution of the relation $$\lambda = \frac{c}{\nu}$$, but that doesn't work... so I realize that you have to convert units by differentiating say in this case lambda, and I get the same equations with the exception of a negative sign which I can't see how it cancels.

To be quite honest, I'm just not sure I'm doing the converting right. I'm not exactly sure how to handle the left side of the equation in this case.

Any suggestions or references would be great... Thanks!

 

[Nylex]

You take the modulus when you differentiate, to get rid of the -ve sign. See http://scienceworld.wolfram.com/physics/PlanckLaw.html.

 

[Rob Hal]

Excellent...I knew it was really elementary.
Thank you very much!

 

[vincentchan]

the negative sign is due to the limit of the integral
if the lower limit of your lamda is 0 and upper limit is infinite,ie.
$$\int_{0}^\infty I(\lambda) d \lambda$$

after you do the substitution, the lower limit of v will become infinite and the upper limit will be 0! ie,
$$-\int_{\infty}^0 I(v) dv$$

in order to make the integral looks nicer, we eat the negative sign and flip the limit of the integral:
$$\int_{0}^\infty I(v) dv$$

don't worry, your calculation is completely fine...

PS. the modulus is not a must... the web page just don't want to do this argument