Physics Help Forum Deriving an expression for Radiation Pressure from Black body function
 User Name Remember Me? Password

 General Physics General Physics Help Forum

 Aug 16th 2018, 12:18 PM #1 Junior Member   Join Date: Aug 2018 Posts: 7 Deriving an expression for Radiation Pressure from Black body function https://photos.app.goo.gl/LWzgtJsCXs4zCCbc7 Hello, I am reading a book about Stellar Structure and Evolution and it takes an expression for the Black body function in terms of frequency and substitutes it into a 'Pressure Integral' to derive an expression for Radiation Pressure. It just gives me the Input, and then the output and says that the Integral is easy but I can't do it. I posted similar request for help on another forum but so far no replies so I was wondering if someone could help? I guess it's an Integration question rather than anything else. The problem is on the image which I linked above. Thanks
Aug 16th 2018, 02:37 PM   #2

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,554
 Originally Posted by Mazin https://photos.app.goo.gl/LWzgtJsCXs4zCCbc7 Hello, I am reading a book about Stellar Structure and Evolution and it takes an expression for the Black body function in terms of frequency and substitutes it into a 'Pressure Integral' to derive an expression for Radiation Pressure. It just gives me the Input, and then the output and says that the Integral is easy but I can't do it. I posted similar request for help on another forum but so far no replies so I was wondering if someone could help? I guess it's an Integration question rather than anything else. The problem is on the image which I linked above. Thanks
It's handy to know this kind of integral, so take a careful look at it.

There's a little trick to this one. First, recall that if we have a geometric series with an infinite number of terms then
$\displaystyle S = \sum_{n = 1}^{\infty} r^n = \frac{r}{1 - r}$

Keep that in your back pocket for the moment.

We need to do an integral of the form
$\displaystyle \int \frac{x^3}{e^{x} - 1} ~ dx$

$\displaystyle = \int \frac{x^3 e^{-x}}{1 - e^{-x}}~dx$

Remember the sum? Well $\displaystyle \sum_{n = 1}^{\infty} \left ( e^{-x} \right ) ^n = \frac{e^{-x}}{1 - e^{-x}}$

This means your integral becomes
$\displaystyle = \int x^3 \sum_{n = 1}^{\infty} e^{-nx} ~ dx$

To a Physicist switching sums and integrals are second nature. We leave the pesky problem of the details to the Math people. So...

$\displaystyle = \sum_{n = 1}^{\infty} \int x^3 e^{-nx} ~ dx$

That's the tricky part. The rest is integration by parts and the sum deals with the zeta function, $\displaystyle \zeta (4)$.

I leave the rest to you.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Aug 16th 2018, 05:05 PM #3 Junior Member   Join Date: Aug 2018 Posts: 7 Thank you very much for this. I'll come back to it tomorrow after sleep. I see how you re-arranged the equation by factoring out e^x and then taking it to the numerator and reversing the power. Took me a while to see why that works, but now its obvious. I will look at Geometric series piece tomorrow. It's a series of little tricks you've done I think. I'll see if I can follow.
 Aug 17th 2018, 02:10 PM #4 Junior Member   Join Date: Aug 2018 Posts: 7 https://photos.app.goo.gl/3SmS28CJ49gdD5Q47 https://photos.app.goo.gl/ouPh4Rf9uyVufxqp7 Hello Dan. I found for a geometric series starting with value a, and common ratio r, the Sum is a/1-r where r < 1. I assumed that this is equivalent to r/1-r in your note, I took the Indefinite Integral but I don't know how to relate it to the Sum. I know there are formula for 1 to n, but the Sum is from 1 to Infinity. How do I get from here to the answer? I think it will relate to the Sum. Thank you for helping me.
Aug 18th 2018, 09:27 AM   #5

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,554
 Originally Posted by Mazin https://photos.app.goo.gl/3SmS28CJ49gdD5Q47 https://photos.app.goo.gl/ouPh4Rf9uyVufxqp7 Hello Dan. I found for a geometric series starting with value a, and common ratio r, the Sum is a/1-r where r < 1. I assumed that this is equivalent to r/1-r in your note, I took the Indefinite Integral but I don't know how to relate it to the Sum. I know there are formula for 1 to n, but the Sum is from 1 to Infinity. How do I get from here to the answer? I think it will relate to the Sum. Thank you for helping me.
Sorry about not getting right back to you... I've been busy.

Just to make sure it's been covered I needed an expression of the form $\displaystyle \frac{r}{1 - r}$, which was in the integral. So
$\displaystyle \sum_{n = 0}^{\infty} r^n = 1 + \sum_{n = 1}^{\infty} r^n = \frac{1}{1 - r}$

So
$\displaystyle \frac{1}{1- r} - 1 = \frac{r}{1 - r} = \sum_{n = 1}^{\infty} r^n$

I'll get back to you on the rest later today.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Aug 18th 2018, 11:52 AM #6 Junior Member   Join Date: Aug 2018 Posts: 7 Thanks. I get that aspect. r to the 0 = 1.
 Aug 23rd 2018, 04:27 AM #7 Junior Member   Join Date: Aug 2018 Posts: 7 Do you have any updates on this problem? Thanks
Aug 23rd 2018, 08:18 AM   #8

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,554
 Originally Posted by Mazin Do you have any updates on this problem? Thanks
Sorry, I had one of my "bad" couple of days. I'll get to it soon. Sorry for the delay.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Aug 23rd 2018, 10:04 AM #9 Junior Member   Join Date: Aug 2018 Posts: 7 Reassuring it's on your to do list but take your time. I'd like to know how to do it, so I can do similar problems but I can wait 2-3 weeks, or however long it takes. Hope whatever's bad doesn't trouble you too much.
Aug 23rd 2018, 11:11 AM   #10

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,554
 Originally Posted by Mazin Reassuring it's on your to do list but take your time. I'd like to know how to do it, so I can do similar problems but I can wait 2-3 weeks, or however long it takes. Hope whatever's bad doesn't trouble you too much.
It comes and goes. I'll get to it by the end of today.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post alanjkim General Physics 4 Oct 10th 2017 01:26 PM davidlaing Thermodynamics and Fluid Mechanics 17 Aug 22nd 2015 12:23 PM shamreen Theoretical Physics 3 Jun 2nd 2010 12:14 AM viciado Advanced Waves and Sound 1 Dec 15th 2009 06:13 AM richmondh2o Advanced Thermodynamics 1 Apr 6th 2009 10:46 PM