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What is the density of photons in a beam of light?

by photonkid
Tags: beam, density, light, photons
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photonkid
#1
Mar16-12, 07:30 PM
P: 29
I'm a "layman" when it comes to physics and I read in the FAQ that a photon is not a real particle but is described as one for the benefit of lay-people.

I read that 65 billion neutrinos pass through a square centimeter every second so I wondered how many photons pass through a square centimeter every second, perpendicular to the sun, when exposed to the sun. Can anyone tell me?

Wikipedia says a photon is an elementary particle and that it's a gauge boson.

So now I'm wondering, if electromagnetic radiation is an oscillating electric / magnetic field, in a beam of light, are photons "side by side" or do they follow one another? How close behind one photon is the next photon? Does the electric / magnetic field of one photon interfere with the other photons around it?

TIA
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Jolb
#2
Mar16-12, 10:34 PM
P: 419
I'll give you an approximate answer, which would probably be within a factor of 10 of the real answer.

At the earth's distance from the sun, a square meter facing directly toward the sun (not on an angle) receives a power of about 1300 watts. [One watt is one Joule per second.] To make the calculation less involved, I'll use the assumption that all of this power comes to the earth as yellow light. (Of course this is wrong; the sun radiates something closer to a blackbody spectrum, but this should give a decent approximation.) Yellow light has a wavelength of about 550nm.

The energy of a single photon is given by E=h*c/λ, where h is Planck's constant, c is the speed of light, and λ is the photon's wavelength. Thus a 550nm photon has an energy of about 3.6*10^-19 Joules. Therefore, you would need 1300/(3.6*10^-19)=3.6*10^21 photons to make 1300 Joules, and thus you would need 3.6*10^21 photons per second (all of 550nm) to match the power of the sun.

So the rough answer is that there are about 3.6*10^21 photons/second going through each square meter facing the sun at the Earth's distance. So if you stood on the equator at noon, that would be the number hitting a square meter of the ground per second. (Remember that at different times and places on the earth, these rays are striking the surface at some angle [unless it's night time when there are none], and you'd multiply 3.6*10^21 by cosine of this angle to adjust for this.)

If you want to convert to square centimeters instead of meters, it would be 3.6*10^17 photons per second per square centimeter. So the answer to your question would be, "On the order of a hundred quadrillion."


To answer your second question, you must distinguish between the classical idea of "electromagnetic waves" and the quantum mechanical idea of wave-particle photons. Photons don't exist in classical E&M--radiation is just waves. In quantum mechanics, however, we do have photons. But in quantum mechanics, the particle-wave doesn't really exist as a particle until its position is measured. So asking how the photons follow one another inside the radiation is a meaningless question. They only really exist as a particle once the radiation is measured.
photonkid
#3
Mar17-12, 06:23 AM
P: 29
Thanks for the detailed answer.

I'm trying to get a mental picture of what this radiation wave "looks" like. Is it anything like a wave on water where at the wavefront, all the water particles are in synch with each other. If you take a cross-section of a light beam exactly perpendicular to the light source, are the electric and magnetic fields exactly in synch with each other everywhere on the cross-section or are they all unsynchronized and fighting each other.

The further you go from the light source, the less "photons" per square centimeter but what is it that's actually diminishing. I can't picture how the electric / magnetic fields "die out". If it was a bunch of discrete particles then I can see how the density gets less the further away but how does the "wave" density get less?

(If the answer is complicated you don't need to try and explain.)
Thanks.

Jolb
#4
Mar18-12, 11:06 PM
P: 419
What is the density of photons in a beam of light?

Quote Quote by photonkid View Post
Thanks for the detailed answer.

I'm trying to get a mental picture of what this radiation wave "looks" like. Is it anything like a wave on water where at the wavefront, all the water particles are in synch with each other. If you take a cross-section of a light beam exactly perpendicular to the light source, are the electric and magnetic fields exactly in synch with each other everywhere on the cross-section or are they all unsynchronized and fighting each other.
If you were to take a cross section of sunlight perpendicular to the beam of sunlight, you would find that the fields are not homogenous. This is because the light radiated from the sun is not coherent: it has many different colors and polarizations (and the light comes in at different angles depending on what part of the photosphere of the sun it was generated from). Different colors and polarizations will interfere causing nonuniform patterns.

However, a beam of light from a laser is coherent: all the photons have the same color and the same polarization (though it isn't perfect: lasers are rated according to their "coherence length"). A cross section of a coherent beam from a laser would have uniform electric and magnetic fields.

The analogy to water waves is like this: dropping one tiny stone into a pond would cause a circular wave (a series of peaks and troughs) to travel outwards--a "cross section" (taken far from where the stone is dropped in) would have a uniform ripple height. [Here you're taking the cross section through the "wave front".] This is analogous to a coherent beam.

Suppose you took your cross section the same way (looking at the original stone) but instead of only dropping the first stone in, you also drop a second stone in at some distance away from the first one. The cross section perpendicular the first stone's wavefront would not be perpendicular to the second stone (unless you drop the second stone directly behind the first) and thus you would be cutting across various peaks and troughs in the second stone's wavefront. You thus woudn't see a uniform height on this cross section with two stones. Of course it would be more messy if you dropped in hundreds of billions of little stones all over the place.

This is the basic idea of interference. The latter is analogous to a noncoherent beam of light. The sun emits light of all colors from all different positions; this is analogous to dropping in stones of all different sizes at all different positions.

Don't take the analogy too far, because light also has polarization, which water waves don't. Water waves are also dispersive: a ripple with a long wavelength (from dropping in a big stone) would travel at a different speed than a ripple with a short one (from a small stone) whereas light of all wavelengths travels at the same speed through vacuum. Also, water waves have different mechanics depending on whether they're in shallow water or deep water. In physics, water waves are much more difficult to understand than EM waves!

The further you go from the light source, the less "photons" per square centimeter but what is it that's actually diminishing. I can't picture how the electric / magnetic fields "die out". If it was a bunch of discrete particles then I can see how the density gets less the further away but how does the "wave" density get less?
The wave's amplitude decreases as it moves farther from the source. Going back to the analogy of a water wave, let's again drop a stone into a pond. The wave starts out being a certain height, but as it travels outward, its height gets smaller and smaller.

For EM waves, it's the electric and magnetic field amplitudes that get smaller and smaller as they travel outwards. The energy of an EM wave (in the case of transverse electric waves in vacuum) is proportional to the electric field amplitude squared, E2. I don't want to fully explain Gauss' law to you (check out wikipedia if you're curious) but the idea is that if you enclose the sun with a sphere, all the sun's power will pass through the sphere regardless of the radius of the sphere. Hence the power per unit area must be inversely proportional to the area of the sphere=4∏(radius)2. Since the power is proportional to the E2, E must be proportional to 1/(radius).
photonkid
#5
Mar19-12, 03:08 AM
P: 29
Great. Thanks.
sophiecentaur
#6
Mar19-12, 04:00 AM
Sci Advisor
Thanks
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P: 12,238
Please try to avoid thinking of a beam of light as a shower of little bullets. Pretty well all of its behaviour can be described in wave terms. The 'particulate' nature is much more difficult to understand and the pictorial view is not helpful.


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