# How many photons/cm3?

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1. I don't think it's helpful or physical to really talk about the waveform of single photons. For the discussion here it's probably better to concentrate on the particle properties.

2. The energy of the photons is inversely proportional to the wavelength as E=hd. If your source is producing 10 energy per second and your photons are 2 energy you'll create 5 photons per second, if they are 1 energy you'll produce 10 photons per second.

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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?

No. There are two current discussions — number per volume, and number per unit energy. Neither one relies on an argument involving the space between them. Photons can overlap.

$E = \frac{hc}{\lambda}$ As the wavelength decreases, the energy increases. For a given amount of energy, there are fewer photons at shorter wavelengths.

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1. I don't think it's helpful or physical to really talk about the waveform of single photons. For the discussion here it's probably better to concentrate on the particle properties.

2. The energy of the photons is inversely proportional to the wavelength as E=hd. If your source is producing 10 energy per second and your photons are 2 energy you'll create 5 photons per second, if they are 1 energy you'll produce 10 photons per second.

That i can understand: "The energy of the photons is inversely proportional to the wavelength as E=hd"

So I see an extremely long wavelength, like a flat ocean, and the result is no wave= no energy. That's O.K.

No. There are two current discussions — number per volume, and number per unit energy. Neither one relies on an argument involving the space between them. Photons can overlap.

$E = \frac{hc}{\lambda}$ As the wavelength decreases, the energy increases. For a given amount of energy, there are fewer photons at shorter wavelengths.

That i can understand too: "$E = \frac{hc}{\lambda}$ As the wavelength decreases, the energy increases."

It's the same as Klaynos reversed: when the wavelength decreases I observe an ocean with a lot of waves at short distance from each other and I see energy. O.K.

For a given amount of energy, there are fewer photons at shorter wavelengths.

that cannot fit into my mind.

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You have 1 pie. You can either split it into 5 equal parts or 10 smaller equal parts. The pie is the source. The big parts are high energy photons the small ones are low energy photons.

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You have 1 pie. You can either split it into 5 equal parts or 10 smaller equal parts. The pie is the source. The big parts are high energy photons the small ones are low energy photons.

The following is surely wrong.

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Michel.. see post #15.. I showed there what is "SM photon".

Gamma, or other high energetic photons need even more energetic sources to produce them.

There is low quantity of such high energetic particle in nature.

Collisions of them happen rarely.

So there is small number of gamma photons in nature (in comparison to lower energetic photons).

Analogy:

rat has 8+ child per year, that are each couple grams (less than mother),

elephant has one child per year or less,

whale has one child per ten years or so. whale child is like thousands of rats.

Always their mother mass is higher than quantity * mass of each child.

There is billion rats around the world,

whales is couple hundreds or thousands (depending on species) or so.

mother = source particle.

child of rat = photon with visible wavelength

child of whale = gamma photon.

Edited by Przemyslaw.Gruchala

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The following is surely wrong.

I'm not sure if the wave part is helping to confuse you. Photons are the quantized bundles of EM radiation. Wave nature is irrelevant to this part of the discussion.

If you have 10 eV of energy, you can get is as 1 photon with 10 eV, or 2 photons with 5 eV, or 5 photons with 2 eV, or 10 photons with 1 eV, etc. There is no need to read any more into this than a math problem with fractions of a whole. More parts means the parts are smaller.

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the wave part begins when stating that more energy corresponds to smaller wavelength. I try to cut the pie to obtain such a result but it doesn't come out.

There must be another graphical way to show

_many pieces of the pie corresponding to a large wavelength.

_few pieces corresponding to small wavelength.

Edited by michel123456

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the wave part begins when stating that more energy corresponds to smaller wavelength. I try to cut the pie to obtain such a result but it doesn't come out.

Your model was wrong. The pie is energy, not wavelength. Energy is inversely proportional to wavelength, meaning the sum of the wavelengths is not constant.

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Your model was wrong. The pie is energy, not wavelength. Energy is inversely proportional to wavelength, meaning the sum of the wavelengths is not constant.

This is a graph of inverse proportionality

here energy is either A axis or B, and wavelength the opposite.

The "pie" is not energy but the product of energy with wavelength and gives a constant.

So the 'pie' is hc.

Edited by michel123456

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Michel, do you think that photon particle has "length" c?

Edited by Przemyslaw.Gruchala

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This is a graph of inverse proportionality

here energy is either A axis or B, and wavelength the opposite.The "pie" is not energy but the product of energy with wavelength and gives a constant.So the 'pie' is hc.

The source energy, the pie is not on that figure. We have a source for which we set the source energy and the photon frequency. If we keep the sourece energy the same as we increase the photon frequency the number of photons produced reduces.

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The source energy, the pie is not on that figure. We have a source for which we set the source energy and the photon frequency. If we keep the sourece energy the same as we increase the photon frequency the number of photons produced reduces.

Yes I see that. The graph is only for one photon.

Michel, do you think that photon particle has "length" c?

where does that come from?

No, in this kind of diagrams units are different on each axis. if you have meters on the one (wavelength) then you have some other unit on the other and the result gives something elselika a "grandmother clock"

So for each photon corresponds a constant value equal to hc.

what is hc?

Edited by michel123456

pie A: E= h*5

pie B: E=h*10

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Yes I see that. The graph is only for one photon.

But we're not talking about one photon. We're talking about one unit of energy. So if the Y axis is energy, and you have 3 units of energy, you can see that one photon is sufficient for a wavelength of 1/3. But if the wavelength is 1, you need 3 of them.

where does that come from?

No, in this kind of diagrams units are different on each axis. if you have meters on the one (wavelength) then you have some other unit on the other and the result gives something elselika a "grandmother clock"

So for each photon corresponds a constant value equal to hc.

what is hc?

hc is the constant of proportionality between energy and wavelength.

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No. There are two current discussions — number per volume, and number per unit energy. Neither one relies on an argument involving the space between them. Photons can overlap.

$E = \frac{hc}{\lambda}$ As the wavelength decreases, the energy increases. For a given amount of energy, there are fewer photons at shorter wavelengths.

"Photons can overlap."

so can energy overlap too?

Edited by michel123456

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"Photons can overlap."

so can energy overlap too?

Energy is a property not a thing. Photons which have energy can overlap.

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Energy is a property not a thing. Photons which have energy can overlap.

In an experiment with 2 laser beams from 2 different sources intersecting.

A detector perpendicular to beam A will measure the energy of beam A only and a detector perpendicular to beam B will measure the energy of beam B only.

At the intersection point there is the energy of the 2 beams, i suppose, since there are overlapping photons coming from both sources. No?

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In an experiment with 2 laser beams from 2 different sources intersecting.

A detector perpendicular to beam A will measure the energy of beam A only and a detector perpendicular to beam B will measure the energy of beam B only.

At the intersection point there is the energy of the 2 beams, i suppose, since there are overlapping photons coming from both sources. No?

That sounds correct. In this situation talking about the energy density is probably sensible, but the statement would still be correct.

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Or, put another way, if two people shine their flashlights on a surface and the beams overlap, the spot gets brighter.

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If that is correct, then back to the OP question:

in each minuscule point of outer space there are photons coming from billions and billions of stars & galaxies all around, that overlap each other. All those photons coming from all directions carry a little bit of energy, very little but multiplied by billions of sources.

How much of this concentrates into a cm3 of outer space?

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If that is correct, then back to the OP question:

in each minuscule point of outer space there are photons coming from billions and billions of stars & galaxies all around, that overlap each other. All those photons coming from all directions carry a little bit of energy, very little but multiplied by billions of sources.

How much of this concentrates into a cm3 of outer space?

Hasn't swansont anawered that in the first few posts?

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Hasn't swansont anawered that in the first few posts?

He answered about one source (the sun for example) but there are overlapping photons coming from everywhere.

Edited by michel123456

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Scale the number I gave by how much power you have.

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All the photons from everywhere is pretty much the same as starlight at night.

According to this

http://en.wikipedia.org/wiki/Lux

it's about 8 orders of magnitude less than sunlight so you can divide Swansont's number by 100,000,000 to get a reasonable estimate.

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