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Old Aug 16th 2018, 01:18 PM   #1
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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
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Old Aug 16th 2018, 03:37 PM   #2
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Originally Posted by Mazin View Post
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
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Old Aug 16th 2018, 06:05 PM   #3
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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.
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Old Aug 17th 2018, 03:10 PM   #4
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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.
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Old Aug 18th 2018, 10:27 AM   #5
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Originally Posted by Mazin View Post
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
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Old Aug 18th 2018, 12:52 PM   #6
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Thanks. I get that aspect. r to the 0 = 1.
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Old Aug 23rd 2018, 05:27 AM   #7
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Do you have any updates on this problem? Thanks
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Old Aug 23rd 2018, 09:18 AM   #8
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Originally Posted by Mazin View Post
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
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Old Aug 23rd 2018, 11:04 AM   #9
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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.
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Old Aug 23rd 2018, 12:11 PM   #10
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Originally Posted by Mazin View Post
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
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