# Photons in 1 cubic meter at temp T

1. Dec 30, 2004

### marcus

Benzun started a thread with this question:

I have replied to that question, and want to extend the question (without diverting Benzun's original thead).

the related question is, if you have a hollow box or cavity at some temperature T, then how many photons per cubic meter are in it?

there is a definite number of photons (per unit volume) associated with each temperature.

please confirm this if you can, IIRC the ENERGY DENSITY in an empty space at temp T is equal to

$$\frac{\pi^2}{15} \frac{k^4 T^4}{\hbar^3 c^3}$$

notice that this is related to but different from the Stef-Boltz. radiation law brightness, which is energy per unit time per unit area:

$$\frac{\pi^2}{60} \frac{k^4 T^4}{\hbar^3 c^2}$$

Now to find the NUMBER OF PHOTONS (per unit volume) all we do is
divide the energy density by the average energy per photon at temp T,
which IIRC is equal to 2.701 kT. As i recall this is a fact about thermal radiation.

If this is right then the number of photons per cubic meter that is in the room with you is given by

$$\frac{1}{2.701} \frac{\pi^2}{15} \frac{k^3 T^3}{\hbar^3 c^3}$$

where T is the absolute temperature which, I am hoping, is a comfortable T = 293 kelvin or thereabouts.

2. Dec 30, 2004

### Janitor

For those of us too lazy to look for a calculator and a units table, how many photons in a c.m. would that be at 293 K?

3. Dec 30, 2004

### marcus

the first time I got that the number of thermal photons per cubic meter is about
8 x 1014

so I am guessing that in the room around me there are
800 trillion photons per cubic meter.

the visible light photons (tho individually much stronger) are far fewer
so dont even have to count them.

it would be nice to have someone check this figure of 800 trillion

-------later-----
the second time
I got 500 trillion per cubic meter
and it is too late at night to figure out my error
maybe someone else will do it, I'm hitting the sack

Last edited: Dec 31, 2004
4. Dec 31, 2004

### marcus

shucks, I made a careless mistake the first time.
the 800 trillion is wrong
it is 500 trillion per cubic meter

5. Jan 1, 2005

### dextercioby

It is more of $$6\cdot 10^{14}$$ photons/cubic meter at 293 K as i'll rigurously prove.On the continent we call that number as '600 billion'.

Okay:everybody who takes a decent course on statistical mechanics,when discussing,thermal radiation computes the average number of photons inside an enclosed box of volume V at the temperature K according to this formula/logics (see below for the formula,my 5-th post in this thread):
....................................................................................

,where the first square paranthesis means the number of uniparticle states from the interval $(\omega,\omega+d\omega)$,the one from the second paranthesis is the average occupation number of a uniparticle state of frequency $\omega$ and the symbol from the second integral means the average number of particles/photons with the frequency in the interval $(\omega,\omega+d\omega)$.

Therefore:
$$\langle N\rangle =\frac{V}{\pi^{2}c^{3}}\frac{(kT)^{3}}{\hbar^{3}}\int_{0}^{+\infty} \frac{x^{2}}{e^{x}-1} dx$$(1)
,where i made the obvious substitution
$$\frac{\hbar\omega}{kT}\rightarrow x$$ (2)

The integral in (1) needs to be computed.

FIRST METHOD:
I make use of the formula number (1) from here to get

$$\int_{0}^{+\infty} \frac{x^{2}}{e^{x}-1} dx=\Gamma(3)\zeta(3)$$ (3)

SECOND METHOD:

$$\int_{0}^{+\infty} \frac{x^{2}}{e^{x}-1} dx=\int_{0}^{+\infty} \frac{x^{2}e^{-x}}{1-e^{-x}} dx$$ (4)

For x>0,the inverse of the denominator in (4) can be expanded into series according to
$$\frac{1}{1-e^{-x}}=\sum_{k=0}^{+\infty} e^{-kx}$$ (5)

Puttin (5) into (4),one gets
$$\int_{0}^{+\infty} \frac{x^{2}}{e^{x}-1} dx=\sum_{n=1}^{+\infty}\int_{0}^{+\infty} x^{2}e^{-nx} dx$$ (6)

The last integral from (6) can be performed via part integration as follows
$$\int_{0}^{+\infty} x^{2}e^{-nx} dx = -\frac{1}{n}x^{2}e^{-nx}|_{0}^{+\infty}+\frac{2}{n}\int_{0}^{+\infty} xe^{-nx} dx$$
$$=-\frac{2}{n^{2}}xe^{-nx}|_{0}^{+\infty}+\frac{2}{n^{2}}\int_{0}^{+\infty} e^{-nx} dx=-\frac{2}{n^{3}}|_{0}^{+\infty}=\frac{2}{n^{3}}$$(7)

Going with (7) back to (6),one gets the integral
$$\int_{0}^{+\infty} \frac{x^{2}}{e^{x}-1} dx=2\sum_{n=1}^{+\infty}\frac{1}{n^{3}}$$ (8)

Comparison between (8) and (3) yield:
$$\Gamma(3)=2$$ (9)
$$\zeta(3)=\sum_{n=1}^{+\infty}\frac{1}{n^{3}}$$ (10)
,both formulas in accordance with the theory of "Gamma" and "Zeta" functions.

So,we have found that:
$$\langle N\rangle=\frac{2k^{3}\zeta(3)}{\pi^{2}c^{3}\hbar^{3}} VT^{3}$$ (11)

For a volume of 1 cubic meter the average number of photons at the temperature T,call it $\hat{N}$ is found to be
$$\hat{N}\sim 2.37\cdot 10^{7} T^{3}$$ photons
For T=293K,one finds
$$\hat{N}_{T=293K}\sim 5.96\cdot 10^{14}$$ photons.
Which can be rounded to $6\cdot 10^{14}$

Daniel.

Last edited: Jan 2, 2005
6. Jan 1, 2005

### dextercioby

Sorry,my friends,LaTex error. I'll post the missing part later.Apparently,there's a bug in the system. :grumpy:

The mean energy of a photon at temperature T is found to be
$$\langle U\rangle_{T} =\frac{\Gamma(4)\zeta(4)}{\Gamma(3)\zeta(3)}kT$$
,which is approximately $2.7 kT$.

Daniel.

7. Jan 1, 2005

### Janitor

Thanks guys for doing the legwork. My intuition refused to even make a guess ahead of time as to what the number would be.

8. Jan 1, 2005

### marcus

hi Daniel, I was delighted by having some confirmation of this part.
I have been using that the average photon energy at temp T is
2.701 kT

where the number 2.701 I get from the zeta function this way:

$$2.701... =\frac{3\zeta(4)}{\zeta(3)}$$

I see that your formula boils down to the same thing!

but we may still differ about the final answer (I did my calculation in a hurry and it was just approximate)

I got something between 5 and 6 that rounded down to 5E14.

Last edited: Jan 1, 2005
9. Jan 1, 2005

### marcus

Hi Daniel, I still get $5.1 \times 10^{14}$

You said in your last post that you were experiencing a "Latex error".
could it be that you are mistaken and that when you fix the error

hopefully,
marcus

10. Jan 2, 2005

### Nereid

Staff Emeritus
I say let's have more Benzun-type questions!

Not wishing to take this OT (if this post is, please let me know), ...

Is there a minimum number of photons? What would the state with this minimum number correspond to?

As T increases, what sort of effects might we see? First let's assume the box is empty of all baryons. I'm thinking about the maximum that marcus discussed on the other thread - creation of a black hole ... would something else happen well before then?

11. Jan 2, 2005

### dextercioby

Obviously it is.I made the assumption explicitely.

I used the formula number (11) from my first post and the numerical values (let's drop the units,they match,anyway)

$$k\sim 1.38\cdot 10^{-23}$$
$$\hbar\sim 10^{-34}$$
$$c\sim 3\cdot 10^{8}$$
$$\pi\sim 3.14$$
$$\zeta(3)\sim 1.202$$

And found exactly what i already stated
$$\langle N \rangle\sim 2.37\cdot 10^{7} T^{3}$$ photons in one cubic meter.
Putting T=293,it yields
$$\sim 5.96\cdot 10^{14}$$ photons in one cubic meter.

Daniel.

Last edited: Jan 2, 2005
12. Jan 2, 2005

### dextercioby

According to QM,since the photons are bosons,their quantum state occupation number varies between 0 and +infinity.So in one quantum state,there can be any number of photons whether it's zero or +infinity.So,the first of your questions is answered:0.The quantum state with 0 occupation number. The vacuum state of QFT:$|0\rangle$.

As the formula denoted in my first post with number (11) shows,the mean/average number of photons exapnds at the power 3 wrt to temperature and their mean energy increases "only" linearly,as shown by the formula including 2 zeta functions.In principle,this could go on forever and the energy in the box could be infinite.
Including gravity is not that simple.Em.radiation inside the box is treated at quantum level.That is,we assume it is quantized.However,in the Einstein equations,which are classical equations,we cannot simply put the quantized $\Theta^{\mu\nu}$,as it would really make no sense.At the right of the equal sign we would have a bunch of creation & anihilation operators and at the left a very classical and geometrical Einstein tensor.I guess we might consider em.radiation classically,and an infinite density of radiation/matter would definitely mean a black hole.But treating radiation clasically would mean other trouble.It would not obey Bose-Einstein statistics anymore,but the classical Bolzmann one.Unfortunately,when we apply Boltzmann statistics to em.radiation we end up with Rayleigh-Jeans distribution which would yield infinity considering not an infinite number of photons/stationary waves,but a finite one and the whole spectrum of frequencies.So that's why it's tricky.

Daniel.

13. Jan 2, 2005

### dextercioby

This is the missing part:
$$\langle N\rangle= \int_{0}^{+\infty}[\frac{V}{\pi^{2}c^{3}}\omega^{2}d\omega][\frac{1}{\exp(\frac{\hbar\omega}{kT})-1}]=\int_{0}^{+\infty} dN_{\omega,\omega+d\omega}$$

Daniel.

PS.The explanations are in my first post.

Last edited: Jan 2, 2005
14. Jan 2, 2005

### marcus

I see the source of the discrepancy. If you will use
$$\hbar = 1.05457 \times 10^{-34}$$
$$\hbar = 10^{-34}$$

then you will get the same answer I did, namely 5.1 x 1014

We were both using equivalent formulas, up to the point where we substituted in the values of the constants (in particular hbar).
Whew! I am relieved.

Naturally since hbar is going to be cubed, being off by over 5 percent in hbar will make a noticeable difference (over 15 percent) in the result.

15. Jan 2, 2005

### dextercioby

Yes,Marcus,you're right.I didn't take into account the number of significant digits.You may have noticed that at every constant and every numerical calculus made with those constants i made approximations (i used the $\sim [/tex] symbol),so it was obvious that the final result would have an "interval of uncertainty".I didn't compute it.As for my approx.of 'hbar',well,it's easier for me to remember it.Since most calculations from Q theories involve 'hbar' instead of 'h',it was natural to search for an approx. for this number. Since u like working with 'exact' numbers,u might have liked to compute the error in N: $$\Delta \langle N\rangle =\frac{\partial N}{\partial \hbar}\Delta\hbar +\frac{\partial N}{\partial\pi}\Delta\pi+\frac{\partial N}{\partial k}\Delta k+\frac{\partial N}{\partial\zeta(3)}\Delta \zeta(3) +\frac{\partial N}{\partial T}\Delta T+\frac{\partial N}{\partial c}\Delta c+\frac{\partial T}{\partial 2}\Delta 2$$ ,where,since 1983,u can set [itex] \Delta c=0$ and,since Leibniz and Newton,$\Delta 2=0$ as well.

Daniel.

Last edited: Jan 2, 2005
16. Jan 2, 2005

### marcus

thank god for the redefinition of the meter in 1983!
I only wish that they would hurry up and make both hbar and e exact
(unmeasurable) numbers as well.

several countries have "electric kilo" projects which would, in effect, establish the kilo on the electric standards and make exact adopted values for the Josephson (2e/hbar) and the von Klitzing (2pihbar/e^2)
(I think you know this but want to make explicit)

I am looking forward to when hbar and e go the same way as c is now.

17. Jan 2, 2005

### marcus

Daniel dextercioby!
just for fun let's both calculate the number of photons coming out of
a person's (it could be Nereid's or anybody's) mouth when she opens her mouth a little bit, like a square centimeter.

It will be some large number of photons per second. I will calculate it first and then you, to see if you get the same answer.

Here I go. I will first find the answer in terms of some convenient natural units----N per unit area per unit time--- and then I will have to divide by 17.74 to put it in metric terms----N' per sq. centimeter per second.

the main formula i use, if you put in the k, hbar, and c, would be:

$$\frac{1}{2.701}\times \frac{\pi^2}{60} \times c( \frac{kT}{\hbar c})^3$$

but in these units the value of c is 109 and k/(hbar c) has the value 10, one can simply multiply the temp by 10 and cube, body temp is 1100, so multiplying by 10 gives 11000

$$\frac{1}{2.701}\times \frac{\pi^2}{60} \times 10^9 \times 11000^3$$

this calculates out to 8.1 x 1019 (per unit area per unit time)

but then to get "per sq. centimeter per second" I must divide by 17.74
and that makes it

4.6 x 1018 per sq cm per second.

So I would say that when a person opens their lips just a small amount, a sq. cm, as if to say Oooooo!
then photons come out of the mouth at the rate of
4.6 quintillion per second

HOWEVER ON THE CONTINENT I understand quintillion means something else besides 1018, and so I will just have to follow the metric continental practice and say "exa"

4.6 exa photons come out per second.

Of course they are also coming off the skin of the face, but not quite so fast because the exterior skin is cooler.

I hope you are able to confirm what I have calculated at least order of magnitude.

------footnotes-------

dextercioby points out that the number 2.701 is interesting: it comes from the zeta function

$$2.701 = \frac{3\zeta(4)}{\zeta(3)}$$

in thermal glow of temp T the average photon energy is 2.701 kT.

18. Jan 2, 2005

### dextercioby

The energetic emittance of the thermal radation is given by the Stefan (1879)-Boltzmann (1884) law

$$\epsilon =\sigma T^{4}=\frac{\pi^{2}k^{4}}{60c^{2}\hbar^{3}} T^{4}$$(1)

The mean energy of a photon is:
$$\langle E\rangle_{V}=\frac{\Gamma(4)\zeta(4)}{\Gamma(3)\zeta(3)}kT\sim \frac{3\pi^{4}}{90\cdot 1.202}kT$$(2)

The mean number of photons which are emitted by a surface of 1 square meter of a blackbody in one second is:
$$\langle N_{emitted}\rangle =\frac{\epsilon}{\langle E\rangle _{V}}\sim \frac{1.5\cdot 1.202 k^{3}}{\pi^{2}c^{2}\hbar^{3}} T^{3}$$(3)

In the case in which (again i won't put any units,it's SI-mKgs as above).
$$k\sim 1.38\cdot 10^{-23};\hbar\sim 10^{-34};c\sim 3\cdot 10^{8};T\sim 310;\pi\sim 3.14$$ (4)

One gets:
$$\langle N_{emitted}\rangle\sim 5.07\cdot 10^{22} photons\cdot m^{-2}s^{-1}$$(5)
For one centimeter squared
$$\langle N'_{emitted}\rangle\sim 5.07\cdot 10^{18} photons\cdot (cm)^{-2}s^{-1}$$(6)

Working with a more precise value for 'hbar' would yield
$$\langle N''_{emitted}\rangle\sim 4.32\cdot 10^{18} photons\cdot (cm)^{-2}s^{-1}$$(7)

Daniel.

Last edited: Jan 2, 2005
19. Jan 2, 2005

### dextercioby

I didn't mean that tera-exa-peta crap.That's awful. :yuck: It's used efficiently only in quantizing information where they use bits and powers of $2^{10} [/tex] which they take as 'Kilo'. I meant the other posible way: USA [itex] 10^{3}$:'thousand';$10^{6}$:'million';$10^{9}$:'billion';$10^{12}$:'trillion';$10^{15}$:'quadrillion';$10^{18}^$:'quintillion';$10^{21}$:'sextillion';...
EUROPE: $10^{3}$:'thousand';$10^{6}$:'million';$10^{6n}$:"n"-illion.,where "n"-illion is:"billion" (n=2) (apud "bi"->2),"trilion" (n=3) (apud "tri"->3),...

Daniel.

PS.For me 10^{18} is 'trillion'.For you is 'quintillion'.

Last edited: Jan 2, 2005
20. Jan 2, 2005

### marcus

I was just kidding when I said "exa" photons
I wish we humans did have an unambiguous convenient word for 1018 though

it seems like a good number to have a word for,
surely some other intelligent species must have an easy way to refer to it

but with us, there is the immediate confusion that you call it trillion and
I call it quintillion. I would be very happy to adopt the continental conventions, if only my fellowcountrymen would do the same.

21. Jan 2, 2005

### marcus

Daniel, i am delighted! this time we agree (in my view) quite adequately

I redid my calculation with a body temp 1097 corresponding closely to the 310 which you used and I came even closer to your answer.

In fact, with this attempt at reconciliation, I found that the photons emitted by the person---who looks slightly surprised or as if she is going to whistle---is 4.5 trillion (continental-style) per second.

22. Jan 2, 2005

### marcus

Daniel, another question has occurred to me. See if you think it worthy of us.

How massive would a black hole need to be so that the average photon in it's Hawking glow would be GREEN?

As you see I have a taste for slightly drole (but basically simple) problems.
but I do not want them to seem TOO drole to you, only a little bit whimsical. What do you think? Should we find out the mass of a "green" black hole?

23. Jan 2, 2005

### marcus

I propose to go first and ask you to see if I have made a mistake.

In the somewhat "natural" units that i find convenient it turns out that a green black hole should have a mass (unless I have made a mistake)

of 27.01 trillion (continental-style) of the roughly half-kilogram mass units

so I will convert to metric to make it easy to compare with your answer. in the system I find convenient (because the constants are exact powers of ten) the mass unit is 434 grams. So multiplying 0.434 by 27.01, I find that the green black hole mass must be

11.7 trillion kilograms

To me this seems surprisingly massive. the more massive the hole, the cooler it is. I would not expect that such a massive hole could be so hot that it glows green light. Have I made a careless error?

24. Jan 2, 2005

### marcus

IIRC the Bekenstein-Hawking temp formula (or perhaps it is solely due to Stephen Hawking?) is

$$kT = \frac{\hbar c^3}{8\pi G M}$$

now in the system I find most convenient the values of the constants are these powers of ten

|8 pi G| = 10-7
|c| = 109
|hbar| = 10-32
|e| = 10-18

$$kT = \frac{10^{-32} 10^{27}}{10^{-7} M} = \frac{100}{M}$$

the photon energy of green light is 10-17 of the system's energy unit, so I must solve this for M

$$10^{-17} = 2.701 kT = 2.701 \frac{100}{M}$$

$$M = 2.701 \times 10^{17} \times 100$$

yes, so in terms of the system's mass unit it is 2.7E19,
or 27 continental trillion

and that will come to some 11 or 12 trillion kilograms

Last edited: Jan 2, 2005
25. Jan 2, 2005

### dextercioby

The surface gravity of a rotating uncharged black hole is linked with its temperature by
$$\kappa=8\pi GT$$ (1)
For an unrotating black hole of mass M,its surface gravity is given by
$$\kappa=\frac{1}{4GM}$$(2)

Equating (1) and (2),we find
$$M=\frac{1}{32\pi G^{2}T}$$ (3)
The average energy per green photon is
$$2.7kT=h\nu_{green}=\frac{2\pi\hbar c}{\lambda_{green}}$$(4)
From (4) we find
$$T=\frac{2\pi\hbar c}{2.7k\lambda_{green}}$$(5)

Going with (5) into (3),we find
$$M=\frac{2.7k\lambda_{green}}{64\pi^{2}G^{2}\hbar c}$$
.
Putting the numbers,and taking the wavelength half of a micron,gives
$$M\sim 2.2 10^{14} Kg$$
,which is way less than your figure.I don't know,maybe i screwed up some formulas...

I used formula no.(7.155),page 223 from Carrol's course.In the text below the formula (7.156),he identifies the surface gravity the BH with the product between 8piG and T.In the formula (7.155) i took a=0 (no rotation) and wound up with formula (2).

Daniel.