Photons in 1 cubic meter at temp T

[marcus]

Benzun started a thread with this question:

hi,

just out of curiosity. What is the maximum number of photons that can be accommodated in 1 cubic meter?

-benzun

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.

 

[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?

 

[marcus]

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?

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 don't even have to count them.

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

anyway, janitor, you asked
-------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

 

[marcus]

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

 

[dextercioby]

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

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.

 

[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.

 

[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.

 

[marcus]

... 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.

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)

Is your final answer still 6E14 per cubic meter?

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

 

[marcus]

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'...

.

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

so i am puzzled by your getting 6 instead of 5.

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
your answer will agree with mine? this would be reassuring.

hopefully,
marcus

 

[Nereid]

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?

 

[dextercioby]

Is your final answer still 6E14 per cubic meter?

Obviously it is.I made the assumption explicitely.

I still get $5.1 \times 10^{14}$
so i am puzzled by your getting 6 instead of 5.
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
your answer will agree with mine? this would be reassuring.

hopefully,
marcus

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.

 

[dextercioby]

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?

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 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?

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.

 

[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.

 

[marcus]

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

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.

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

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.

 

[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 [itex] \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 $\Delta c=0$ and,since Leibniz and Newton,$\Delta 2=0$ as well.

Daniel.

 

[marcus]

...
,where,since 1983,u can set $\Delta c=0$ and,since Leibniz and ... [/itex] as well.
.

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.

 

[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.

 

[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.

 

[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 [itex] 2^{10} [/tex] which they take as 'Kilo'.
I meant the other posible way:
USA $10^{3}$:'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'.

 

[marcus]

I didn't mean that tera-exa-peta crap.That's awful. :yuck: ...PS.For me 10^{18} is 'trillion'.For you is 'quintillion'.

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.

 

[marcus]

...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.

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.

 

[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?

 

[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?

 

[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

 

[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.

 

[marcus]

...
For an unrotating black hole of mass M,its surface gravity is given by
$$\kappa=\frac{1}{4GM}$$...

I must say it adds greatly to find someone else willing to
be curious about (purely fun) questions such as the size of a green black hole!

I am not too concerned by the discrepancy between our answers, at this point, because I think it will all get reconciled once we agree on the correct formula for the "Hawking temperature"
of a non-rotating black hole

Everything must follow from the temp.

I will try to find an online source, that you can compare with yours.
If it is not too much trouble please give me a link, if you find a formula for the Hawking temperature online.

What I have is just an astronomy textbook on my shelf which says that
the temperature is given by

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

at least this makes the units work out properly, and the guy is a professor at UC berkeley. the textbook has been in use for a couple of decades so probably it is right

but obviously this is something i cannot derive myself and can only suggest that we take on faith.

BTW kappa the usual BH "surface gravity" is something we can derive and it comes out to

$$\kappa=\frac{c^4}{4GM}$$

this means that the author you are quoting is doing what they are so often doing which is to use special units in which c = 1

it could be a good rule that if there is ever any possibility of confusion then Show All the Constants

 

[dextercioby]

Your're right and I'm sorry. That is an awful books for cosmology and black holes and unexplicably i took the formulas for granted though they were using units with "c=1" and i didn't need that in my computations...
http://library.thinkquest.org/C007571/english/advance/core4.htm
and the pages that follow.

Daniel.

 

[marcus]

...
http://library.thinkquest.org/C007571/english/advance/core4.htm
and the pages that follow...

Daniel.

Thanks! that is a great link!
really superb page
even derives the hawking temperature, somehow.
http://library.thinkquest.org/C007571/english/advance/core4.htm?tqskip1=1

I am impressed with thinkquest library,

Wiki or PhysicsWorld could also be good but I did not try them and do not know if they even have anything about hawking temp.

----------

no problemo about numerical discrepancies. eventually we will get that sorted out now that we have an online source for the important thing.
everything else is probably trivial once one knows the temperature of the hole.

----------
I am developing a taste for slightly crazy physics problems. You seem to have a sense of humor as well as being about to get your PhD in physics. Is this normal? Or are most of your colleagues superserious.

Anyway you know that with a Josephson junction there is a strict correspondence between voltage across junction and the frequency of the signal.

One day a famous concert soprano comes into the lab.
She sings her high C. (the kind that is dangerous for the glassware)
and the lab techies adjust the voltage of the josephson so that it's signal
matches the Diva.

it will be some fraction of a nanovolt.

What fraction?

(actually it would be too small a fraction of a nanovolt to be possible to measure, I suppose. I do not know the limits of voltage measurement.
but anyway, theoretically what fraction of a nanovolt)

Perhaps this problem is too easy. Or not bizarre enough.

 

[dextercioby]

For a static potential,call it V_{0},and for a maximum frequency of 1100Hz,this formula:
$$h\nu=q_{e}V_{0}$$
yields
$$V_{0}=\frac{h\nu}{q_{e}}\sim 4.55pV$$
,which is very,very small indeed.I've heard of measuring picoampères,but picovolts...

Daniel.

PS.I don't know that many,the internet does.
source

 

[marcus]

...
,which is very,very small indeed.I've heard of measuring picoampères,but picovolts...

Daniel.

PS.I don't know that many,the internet does.
source

Dexter, I agree with your judgement that the voltage is unmeasureably small.
Your source gives the top of the soprano's range as 1100 Hz but i will take it to be 1000 Hz out of laziness.

the NIST gives the adopted standard for the Josephson coefficient (2e/h) as
483597.9 GHz per Volt.

http://physics.nist.gov/cuu/Constants/

http://physics.nist.gov/cgi-bin/cuu/Category?view=html&Adopted+values.x=106&Adopted+values.y=12

So that is 483597.9 Hz per nanovolt

484 kHz per nanovolt

I am essentially just reproducing what you said, at least order of magnitude.

It looks like the techies at the Bureau Internationale des Poids es Mesures will have to produce a stable voltage of

1/484 nanovolt!

if they wish to match the highnote of the soprano.

Even for people at BIPM this is probably too difficult.