Physics question.

I was thinking about the greenhouse effect, and this question
occured to me: in the space around me, what's more common,
atoms or photons?


--
Sleepalot aa #1385

In message , Sleepalot
writes

I was thinking about the greenhouse effect, and this question
occured to me: in the space around me, what's more common,
atoms or photons?


In the universe as a whole the photon to baryon ratio is 10^9. So the
question that now needs to be answered is by how much is the baryon
density (and photon density) enhanced in the space around you, and what
is the average number of baryons per atom in that region. The answer to
this depends on how big a space you're thinking of.
--
Stewart Robert Hinsley

Stewart Robert Hinsley wrote:

In message , Sleepalot
writes

I was thinking about the greenhouse effect, and this question
occured to me: in the space around me, what's more common,
atoms or photons?


In the universe as a whole the photon to baryon ratio is 10^9.

Lol. Ok, I've looked up "baryon" - it's neutrons and protons, but
not electrons, yes? Is the universe still amost entirely hydrogen?


So the
question that now needs to be answered is by how much is the baryon
density (and photon density) enhanced in the space around you, and what
is the average number of baryons per atom in that region. The answer to
this depends on how big a space you're thinking of.

I did have a further think, bearing in mind this is a
near-the-Earth-surface situation. So, around here
1 litre of air contains about 1 Avogadro number of
atoms.

For photons, I'm working from my understanding of
cathode ray tubes: an electron hits an atom of
phosphor, causing the emmision of a photon.

So an amp of cathode current (way too high) would be
a coulomb of electons per second giving an equal
number of photons.

So (say) 10mA would be 10^16 photons per second
in 10^23 atoms .....no that's wrong because those
photons wouldn't hang around that long.

Does this make sense? Am I on the right track?


--
Sleepalot aa #1385

In message , Sleepalot
writes
Stewart Robert Hinsley wrote:

In message , Sleepalot
writes

I was thinking about the greenhouse effect, and this question
occured to me: in the space around me, what's more common,
atoms or photons?


In the universe as a whole the photon to baryon ratio is 10^9.

Lol. Ok, I've looked up "baryon" - it's neutrons and protons, but
not electrons, yes? Is the universe still amost entirely hydrogen?

Wikipedia says 75%. (I wouldn't be surprised if it gives other numbers
elsewhere.)


So the
question that now needs to be answered is by how much is the baryon
density (and photon density) enhanced in the space around you, and what
is the average number of baryons per atom in that region. The answer to
this depends on how big a space you're thinking of.

I did have a further think, bearing in mind this is a
near-the-Earth-surface situation. So, around here
1 litre of air contains about 1 Avogadro number of
atoms.

Fide Wikipedia density at sea level is 1.2g/L. A mole of atmosphere has
a mass of 28.97g. Hence a litre contains roughly 1/24th of a mole of
molecules, or 1/12th of a mole of atoms.

For photons, I'm working from my understanding of
cathode ray tubes: an electron hits an atom of
phosphor, causing the emmision of a photon.

So an amp of cathode current (way too high) would be
a coulomb of electons per second giving an equal
number of photons.

You'd want to take the efficiency of the CRT into account. I don't know
how efficient they are, but LCD displays require considerably less
power, so I infer that CRTs are not particularly efficient. It might be
more convenient to work from the energy of the photons (E=hv) and the
luminosity of the display.

So (say) 10mA would be 10^16 photons per second
in 10^23 atoms .....no that's wrong because those
photons wouldn't hang around that long.

There's all sorts of other photons zipping around - the cosmic microwave
background, radio waves from all sorts of artificial sources and heat
from all surfaces, as well as visible light. I would think that you can
neglect X-ray and gamma photons, but I'm not sure that you can neglect
the longer wavelength parts of the spectra.

You're right that if you're talking about the cubic meter in front of a
CRT you have to divide by 3x10^8 (plus some geometrical correction
factor).

Does this make sense? Am I on the right track?


You could start from the solar constant. From this you can get an
estimate of the downward flux of photons from the sun. Let a be the
albedo of the Earth's surface. Let N be the downward flux. The downward
and upward energy fluxes must balance. Ignoring the greenhouse effect,
and the internal heat generation of the earth, the total flux of photons
is [(1+a) + x(1-a)]N, where x is the ratio of the energy of incoming
light and outgoing heat photons (finger in the air, x is about 10,
giving [1+a+x-ax]N = [1+x-a(x-1)]N. That comes out at about 8N.

A mobile phone generates less that 1/1000th of the solar constant. But
the energy of a photon is about 1/500000th of that of a visible light
photon. So when your mobile phone is switched the photon density in your
vicinity due to the mobile phone far exceeds that due to visible light.

To get an accurate value for the photon density you have to integrate
across the spectrum, for all sources. AM radio photons are another 1000
times less energetic that mobile phone photons, so it requires very
little power density from this for the photon density to be significant.
--
Stewart Robert Hinsley

Stewart Robert Hinsley wrote:

[snip]

Thanks for the reply. It's going to take me a while to
process all that.


--
Sleepalot aa #1385

On Thu, 17 Jul 2008 13:17:42 +0100, Sleepalot wrote:

I was thinking about the greenhouse effect, and this question
occured to me: in the space around me, what's more common,
atoms or photons?

Answer: atoms.

All we need here are rough numbers - an order of magnitude or two is
close enough.
The space around you is a box roughly 2m x 1m x1, =2m3.
As has been stated 1 mole (6e23) of air weighs about 30g, and we
know that 1m3 of air weighs about 1000g, so the space around you
contains ~ 60 moles of air, or 6e23 * 60 approx 4e25 molecules.

My astro camera (to keep it on topic) has a sensitive area of about
6mm x 6mm or 4e-6 m2. It has a well-depth of about 40,000 photons
and a QE of about 0.5 and contains 1million pixels, so each pixel
has an area of 4e-6 / 1e6 ~ 4e-12m2.
During daytime, I'd guess the sensor gets saturated in about 1mSec
So the photon flux is 4e4 photons per pixel in 1e-3 sec, or 4e7
photons per second, but double this for the 50% quantum efficiency
gives us 8e7 (call it 1e8 photons per pixel per second.)
Now multiply this by the number of pixels per square metre (1/4e-12)
gives us a flux of 2.5e19 photons per sq. metre per second.

Assume all the photons come from above (it keeps the maths simple, but
the direction doesn't matter), then the number of photons in your 2m
high box is the flux multiplied by the time they take to travel
2metres - which is 2 / 3e8 = 7e-9 seconds. Multiply this by the flux
gives a result of 7e-9 * 2.5e19 is roughly 2e11 photons in your "box"
at any one time.
Let's double that to allow for all the non-visible photons and double it
again for the hell of it. The number of photons is still a long way
short of the number of molecules.
At night it's even less.

--
.. Pete Lynch I have learned from my mistakes and
.. Marlow ... I am sure I can repeat them exactly
.. www.pete-lynch.com --- Peter Cooke.

Peter Lynch wrote:

On Thu, 17 Jul 2008 13:17:42 +0100, Sleepalot wrote:

I was thinking about the greenhouse effect, and this question
occured to me: in the space around me, what's more common,
atoms or photons?

Answer: atoms.

All we need here are rough numbers - an order of magnitude or two is
close enough.
The space around you is a box roughly 2m x 1m x1, =2m3.
As has been stated 1 mole (6e23) of air weighs about 30g, and we
know that 1m3 of air weighs about 1000g, so the space around you
contains ~ 60 moles of air, or 6e23 * 60 approx 4e25 molecules.

My astro camera (to keep it on topic) has a sensitive area of about
6mm x 6mm or 4e-6 m2. It has a well-depth of about 40,000 photons
and a QE of about 0.5 and contains 1million pixels, so each pixel
has an area of 4e-6 / 1e6 ~ 4e-12m2.
During daytime, I'd guess the sensor gets saturated in about 1mSec
So the photon flux is 4e4 photons per pixel in 1e-3 sec, or 4e7
photons per second, but double this for the 50% quantum efficiency
gives us 8e7 (call it 1e8 photons per pixel per second.)
Now multiply this by the number of pixels per square metre (1/4e-12)
gives us a flux of 2.5e19 photons per sq. metre per second.

Assume all the photons come from above (it keeps the maths simple, but
the direction doesn't matter), then the number of photons in your 2m
high box is the flux multiplied by the time they take to travel
2metres - which is 2 / 3e8 = 7e-9 seconds. Multiply this by the flux
gives a result of 7e-9 * 2.5e19 is roughly 2e11 photons in your "box"
at any one time.
Let's double that to allow for all the non-visible photons and double it
again for the hell of it. The number of photons is still a long way
short of the number of molecules.
At night it's even less.

Hello Peter, thanks for the reply - which I'm pleased to
say I got the jist of. I liked your method - using your camera
as a photon counter.

So, we've got e11 photons in e25 molecules, and that's
1 photon in e14 molecules - it's amazingly low, but there'll
be another one along in an instant.

Right, so if I head off towards the Sun for about 100 miles
I'm going to run out of molecules in my 2m3 while retaining
a broadly comparable photon density, yes?


--
Sleepalot aa #1385

Stewart Robert Hinsley wrote:

You could start from the solar constant. From this you can get an
estimate of the downward flux of photons from the sun. Let a be the
albedo of the Earth's surface. Let N be the downward flux. The downward
and upward energy fluxes must balance. Ignoring the greenhouse effect,
and the internal heat generation of the earth, the total flux of photons
is [(1+a) + x(1-a)]N, where x is the ratio of the energy of incoming
light and outgoing heat photons (finger in the air, x is about 10,
giving [1+a+x-ax]N = [1+x-a(x-1)]N. That comes out at about 8N.

Sorry Stewart, I'm out of my depth here. You seem to be saying
that there are (perhaps) 8 times as many photons going out (with
lower energies) as coming in. Have I got that right?



--
Sleepalot aa #1385

In message , Sleepalot
writes
Stewart Robert Hinsley wrote:

You could start from the solar constant. From this you can get an
estimate of the downward flux of photons from the sun. Let a be the
albedo of the Earth's surface. Let N be the downward flux. The downward
and upward energy fluxes must balance. Ignoring the greenhouse effect,
and the internal heat generation of the earth, the total flux of photons
is [(1+a) + x(1-a)]N, where x is the ratio of the energy of incoming
light and outgoing heat photons (finger in the air, x is about 10,
giving [1+a+x-ax]N = [1+x-a(x-1)]N. That comes out at about 8N.

Sorry Stewart, I'm out of my depth here. You seem to be saying
that there are (perhaps) 8 times as many photons going out (with
lower energies) as coming in. Have I got that right?


Yes, with caveats.

The energy reaching the surface of the Earth from the Sun is dominated
by visible and near infrared wavelengths, because

1) that's where the peak of the solar energy production is
2) radiation at many other wavelengths is absorbed by the atmosphere.

The same amount of energy, plus a small contribution from the internal
heat production of the Earth, must leave the surface of the Earth. Some
is just reflected. Some heats the surface. Some of this heat is carried
away by convection (and subsequently radiated by the atmosphere), and
some is radiated, at considerably longer wavelengths and lower energies,
with a consequent increase in the number of photons.

Because of absorption of infra-red radiation by the atmosphere there's
also an incoming flux of infra-red photons reradiated by the atmosphere.
This why I specified ignoring the greenhouse effect.

The other caveat is while the solar energy flux is dominated by visible
and near-infrared wavelengths this doesn't mean that the solar photon
flux is. AM radio wavelengths are a trillion times longer than visible
light wavelengths, so the solar energy flux has to be very small for the
photon flux at those wavelengths to be negligible.

A bit of googling suggests that it is indeed negligible.
--
Stewart Robert Hinsley