# How many photons/cm3?

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How many photons do exist by units of volume of "empty space"?

Of course there is no empty space if photons are going through it: there are photons.

So, when I look at the sun (don't do that) my eye catches a bunch of photons coming from the star. "catch" meaning that the photons hit my eye. When I look anywhere from any direction, I can catch some photons coming from some source. If I change my eye with a telescope, turning to Alpha Centauri, I catch some photons from there. I can theoretically catch photons coming from the whole observable universe.

So, at the same instant, in outer space, how many photons are coming from all sources in a single cm3?

If the stars are billions of billions, one photon from each one makes billions of billions photons not in a cm3, but in space small as 1 photon.

Logically, I guess.

That makes a lot in 1 cm3.

Or am I wrong somewhere?

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The number of photons depends on their energy and the power of the source. For each Watt, at a wavelength of 550 nm (middle of the visible spectrum) the number of photons you have is about 2.77 x 10^19 per second. If the light is of a longer wavelength you have more photons.

Sunlight gives us about 1 kW/m^2 or 0.1 W/cm^2, or a flux of 2.77 x 10^18 photons/cm^2-s if they are at 550 nm. To get the number per unit volume you divide by the speed (3 x 10^10 cm/s) so it's just under 10^8 photons/cm^3 if it's all green light.

The actual spectrum is skewed toward the low-wavelength side of visible, meaning more photons. And then you must add all the photons from the thermal radiation of the earth, which is centered in the IR. So the final answer is probably more than 10^9 photons/cm^3 (or parts of photons, since I'm ignoring the fact that some photons will have wavelengths longer than 1 cm)

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Saying that photons occupy volume would not be a completely correct statement.

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The number of photons depends on their energy and the power of the source. For each Watt, at a wavelength of 550 nm (middle of the visible spectrum) the number of photons you have is about 2.77 x 10^19 per second. If the light is of a longer wavelength you have more photons.

Sunlight gives us about 1 kW/m^2 or 0.1 W/cm^2, or a flux of 2.77 x 10^18 photons/cm^2-s if they are at 550 nm. To get the number per unit volume you divide by the speed (3 x 10^10 cm/s) so it's just under 10^8 photons/cm^3 if it's all green light.

The actual spectrum is skewed toward the low-wavelength side of visible, meaning more photons. And then you must add all the photons from the thermal radiation of the earth, which is centered in the IR. So the final answer is probably more than 10^9 photons/cm^3 (or parts of photons, since I'm ignoring the fact that some photons will have wavelengths longer than 1 cm)

10^9 only?

Is that compatible with a flux of 2.77 x 10^18 photons/cm^2?

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10^9 only?

Is that compatible with a flux of 2.77 x 10^18 photons/cm^2?

Photons move really fast. It only takes 33.3 picoseconds to go 1 cm.

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Photons move really fast. It only takes 33.3 picoseconds to go 1 cm.

Yes. But the result over volume should be more than the result over surface. No?

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Yes. But the result over volume should be more than the result over surface. No?

No, a flux is the number moving through a surface. So imagine you have a tube with a cross-section of 1 cm^2, and it's 1 light-second long. All the photons are moving in the same direction, through the end. You have 2.77 x 10^18 photons in the tube. That accounts for the flux. How many do you have in a 1 cm section? You divide by the length, which is c * 1 second.

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O.K. thanks.

So it is 2.77 x 10^18 photons/cm^2 sec

10^9 looks too few.

That is 10^3 on each edge.

(correction below)

see below

Edited by michel123456
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O.K. thanks.

So it is 2.77 x 10^18 photons/cm^2 sec

10^9 looks too few.

That is 10^3 on each edge.

on each edge? It's 2d plane..

Edited by Przemyslaw.Gruchala
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The number of photons depends on their energy and the power of the source. For each Watt, at a wavelength of 550 nm (middle of the visible spectrum) the number of photons you have is about 2.77 x 10^19 per second. If the light is of a longer wavelength you have more photons.

Sunlight gives us about 1 kW/m^2 or 0.1 W/cm^2, or a flux of 2.77 x 10^18 photons/cm^2-s if they are at 550 nm. To get the number per unit volume you divide by the speed (3 x 10^10 cm/s) so it's just under 10^8 photons/cm^3 if it's all green light.

The actual spectrum is skewed toward the low-wavelength side of visible, meaning more photons. And then you must add all the photons from the thermal radiation of the earth, which is centered in the IR. So the final answer is probably more than 10^9 photons/cm^3 (or parts of photons, since I'm ignoring the fact that some photons will have wavelengths longer than 1 cm)

Shouldn't that be the shorter the wave length the more photons you have?

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on each edge? It's 2d plane..

Correct.

That's 10^2 on each edge.

It's a cube with 10^3 "points" on each edge = a cube with 10^3 by 10^3 by 10^3

=10^9

Edited by michel123456
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Shouldn't that be the shorter the wave length the more photons you have?

Shorter wavelength, higher energy, fewer photons.

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Shorter wavelength, higher energy, fewer photons.

So radio waves have more photons per " centimeter of wave" than gamma rays? Seems more than a bit counter intuitive to me but then so do many aspects of physics...

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So radio waves have more photons per " centimeter of wave" than gamma rays? Seems more than a bit counter intuitive to me but then so do many aspects of physics...

Assuming the energy flux (power of the source) is the same then yes.

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Sunlight gives us about 1 kW/m^2 or 0.1 W/cm^2, or a flux of 2.77 x 10^18 photons/cm^2-s if they are at 550 nm. To get the number per unit volume you divide by the speed (3 x 10^10 cm/s) so it's just under 10^8 photons/cm^3 if it's all green light.

See visualization below.

Imagine that highlighted polygons on the left has 1 cm width, 1 cm height.

One photon goes out, second is starting intersecting with 1 cm x 1 cm plane, after traveling 550 nm, it's also no longer intersecting with plane.

To have average 1 photon with wavelength 550 nm all the time intersecting with plane 1x1 cm we need in volume 1x1x1 cm, 1cm/550nm number of photons. 1e-2/550e-9= 18182 photons with 550 nm.

(and we still didn't count those that were reflected, and those that are parallel to plane)

Edited by Przemyslaw.Gruchala
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Assuming the energy flux (power of the source) is the same then yes.

Ok, that makes sense...

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Ok, that makes sense...

that makes no sense to me.

the less the photons the more energy?

That leads to zero photon=full energy.

I don't get it.

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that makes no sense to me.

the less the photons the more energy?

That leads to zero photon=full energy.

I don't get it.

Don't think of the wavelength as a size. I tend to work in frequency or energy it saves confusion.

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Don't think of the wavelength as a size. I tend to work in frequency or energy it saves confusion.

No photon=full energy makes no sense to me.

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that makes no sense to me.

the less the photons the more energy?

Gamma photons with high energy are created very rarely (in comparison to f.e. visible light).

After all they're created during annihilation. Electron with positron must collide to create gamma particle, etc.

You end up with single particle with energy equal to the energy of electron/positron.

That leads to zero photon=full energy.

I have no idea how you get such conclusion..

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No photon=full energy makes no sense to me.

A wavelength of 0 produces an undefined energy, it's also not physically possible.

Let me say a couple of things to try and help. Your source must have a high enough energy for the energy of your photon. Working in energies you're talking in the limit that energy tends to infinitely so your source energy must also. As we're talking about a constant energy source at some point your source would not be able to create any photons, it would not have the energy to do so.

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Possibly i didn't understand as much as i thought. my understanding is this, let's say you have a one mega watt energy source you can tune to any frequency, it would take quite a bit more radio waves to dissipate that much energy than it would gamma rays...

Am i close?

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Possibly i didn't understand as much as i thought. my understanding is this, let's say you have a one mega watt energy source you can tune to any frequency, it would take quite a bit more radio waves to dissipate that much energy than it would gamma rays...

Am i close?

It would produce a higher photon flux for radio waves than gamma rays, yes.

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that makes no sense to me.

the less the photons the more energy?

That leads to zero photon=full energy.

I don't get it.

You've dropped the important condition of having a set amount of energy (or power, for a rate of photon production). If you have fewer photons, they must each have more energy in order to fulfill that condition.

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Sorry there is wood in my head, I understand nothing.

You are all talking as if the photon was alone at distance=wavelength from the following one. As there was a gap between photons equal to the wavelength. Am I correct here?

Making the following analogy:

Say you are at sea looking at the waves. The sea is the sea (molecules of water etc.). The wave is a wave (a physical state of the sea). You seem to all talk as if the photon was a physical element called the
crest

Sure when you look at the sea shore, the crest is noticeable when the waves "die" on the rocks. You even hear the characteristic sound of the waves, so everyone could agree that the crest is "something" of different nature from all the rest of the sea. But at the end, the crest is nothing more than molecules of water, and there are plenty molecules of water in the sea, and along the wave.
So IOW I don't understand
1.why one can equalize the number of photons/ cm3 to the number of crests of a wave.
2. why there is more energy when there are less crests (when the wavelength is longer). At the end a flat surface of the sea (zero crest) equals to zero energy of the waves.

No?

Edited by michel123456

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