# Black Body Work

1. Oct 7, 2006

### The Bob

Hi all,

I hope everyone is well and that life is treating you all good.

I am going to be honest with my question and say that I have not tried to do it myself yet. I, unfortunately, do not have time now to try and then repost so I do hope you will all forgive me for just stating a question. I do intent to attempt it tonight incase I can do it but for safety sake I am going to ask here as well, as on inspection it does not look to simple.

I will also say that this is for a University problem set that I have to do so, in a way it is homework. If the Mentors wish to move the thread then please do so and let me know via the messaging service please.

The problem is this: I need to differentiate the formula for the Energy Density of a Black Body:

$$U(\nu) = \frac{8 \pi \nu^3 h}{c^3 (e^{\frac{h \nu}{kT}} - 1)}$$

So I need: $$\frac{dU}{d \nu} = 0$$

As I said, I have not attempted it but please assume I understand the main rules of A-leve Mathematics. Again, I apologise for not having attempted it but I am in a rush and really cannot do it before I go. I will attempt it tonight and report what I find tomorrow but it maybe too little too late.

P.S. Please note that there is a h (for Planck's Constant) missing from the numberate of the equation and that dU by dv needs to equal 0, in the end. See next post.

Last edited: Oct 7, 2006
2. Oct 7, 2006

### The Bob

Correct equations

$$U(\nu) = \frac{8 \pi \nu^3 h}{c^3 (e^{\frac{h \nu}{kT}} - 1)}$$

and

$$\frac{dU}{d \nu} = 0$$

Cheers,

3. Oct 8, 2006

### The Bob

As promised, I attempted the differentiation myself last night.

I got to $$\frac{dU}{d \nu} = \frac{8 \pi h \nu^2}{c^3} \cdot \frac{e^{\frac{h \nu}{kT}}(3 - \frac{h \nu}{kT}) - 3}{e^{\frac{h \nu}{kT}(e^{\frac{h \nu}{kT} - 2) + 1}$$

but something tells me I did the differentiation wrong in the first place. I used the equation as a quotient but did nothing with the T, because it is a constant for different graphs but can be varied.

Can anyone shine a million watt torch's light on this problem please.

Cheers,