What happens to "light" between the Sun and the Earth? Is space a "sea of photons"?

[Alias Moniker]

We can't see light in space because there is nothing for light to illuminate. But is there still an electromagnetic wave between the Earth and Sun or does this disturbance happen because of atmosphere?

If we looked at a cross cut of the space between the Sun and the Earth with an infrared or other light spectrum camera, would we see the electromagnetic waves of any light spectrum?

Is space empty, or is space a "sea of photons"?

Seemingly infinite stars are emitting seemingly infinite photons into space, seemingly infinitely. Space is dark because there is not a lot in space for photons to interact with, but whenever something is in space for photons to interact with, we find that photons are there. Does this mean that space is an "ocean of photons"? Does this hold with the theory of mass energy equivalence that all matter is just "slowed" or "condensed" or "less energetic" energy??

Edited July 24, 2014 by Alias Moniker

[Sensei]

Average visible photon has wavelength ~532 nm
so energy of such photon is h*c/wavelength = 6.62607e-34 * 299792458 / 532e-9 = 3.73e-019 J
Sun is emitting at Zenith 1360 J/s of energy per m^2 of Earth (including absorbed by atmosphere). (I am simplifying calcs to not take care of UV/IR/MW - let's hypothetically assume the all is from visible spectrum)

1360 J / 3.73e-019 J = ~ 3.65*10^21 photons per m^2 per second.

Knowing above, you can calculate how many photons is in volume of sphere with radius equal to distance between Sun and Earth (or any further if you want)..

 

You should learn about inverse square law, to calculate energy per area unit.

In other thread I posted mine animation showing inverse square law for a few particles spreading in the all directions from one origin.

Edited July 24, 2014 by Sensei

[Alias Moniker]

Average visible photon has wavelength ~532 nm

so energy of such photon is h*c/wavelength = 6.62607e-34 * 299792458 / 532e-9 = 3.73e-019 J

Sun is emitting at Zenith 1050 J/s of energy per m^2 of Earth (at least such amount is reaching ground at max). (I am simplifying calcs to not take care of UV/IR/MW - let's hypothetically assume the all is from visible spectrum)

 

1050 J / 3.73e-019 J = ~ 2.8e+21 photons per m^2 per second.

That's average 2812 photons per 1 nm^2 per second of your face.

 

Knowing above, you can calculate how many photons is in volume of sphere with radius equal to distance between Sun and Earth (or any further if you want)..

 

You should learn about inverse square law, to calculate energy per area unit.

In other thread I posted mine animation showing inverse square law for a few particles spreading in the all directions from one origin.

 

From suns the "photons" spread in all directions from different points of origin in space. The sun is a sphere and it's entire surface emits the photons so they are not all emitting from the same point. How does this effect things?

 

Do we agree that at all times, between the Sun and the Earth, there are photons?

[Mordred]

excellent question

The average energy density of space is extremely low, This value includes the density of photons, (radiation), dark matter and matter.

 

as the universe is extremely close to flat we can use the critical dnsity formula for a good approximation.

 

[latex]\rho_{crit} = \frac{3H^2}{8\pi G}[/latex] this will give you and average energy-density of 10^-26 kg/ m³ this corresponds to roughly 5 protons per m³

 

calculating the number of photons can be done via the temperature, using the Bose Einstein distribution distribution, I won't bore you with the formula however it will work out to roughly 410.5 photons per cm³

 

equation 32 page 76 however as this is a section of a larger article page 16

http://www.helsinki.fi/~hkurkisu/cosmology/Cosmo6.pdf

 

different answers of similar orders of magnitude can result depending on what the average wavelength you use for the photon

 

energy-density of photons overall average is roughly 4.641 *10^-31 kg/m³ this is a decent average as it also correlates to the value given in numerous textbooks and the cosmic inventory

http://arxiv.org/pdf/astro-ph/0406095v2.pdf "The Cosmic energy inventory

 

within the same orders of magnitude

Edited July 24, 2014 by Mordred

[Alias Moniker]

excellent question

The average energy density of space is extremely low, This value includes the density of photons, (radiation), dark matter and matter.

 

as the universe is extremely close to flat we can use the critical dnsity formula for a good approximation.

 

[latex]\rho_{crit} = \frac{3H^2}{8\pi G}[/latex] this will give you and average energy-density of 10^-26 kg/ m³ this corresponds to roughly 5 protons per m³

 

calculating the number of photons can be done via the temperature, using the Bose Einstein distribution distribution, I won't bore you with the formula however it will work out to roughly 410.5 photons per cm³

 

equation 32 page 76 however as this is a section of a larger article page 16

http://www.helsinki.fi/~hkurkisu/cosmology/Cosmo6.pdf

 

different answers of similar orders of magnitude can result depending on what the average wavelength you use for the photon

 

energy-density of photons overall average is roughly 4.641 *10^-31 kg/m³ this is a decent average as it also correlates to the value given in numerous textbooks and the cosmic inventory

http://arxiv.org/pdf/astro-ph/0406095v2.pdf "The Cosmic energy inventory

 

within the same orders of magnitude

 

Cool. Does the photon have a "wavelength" in "empty space" or does it exist there as "energy particle"?

[Klaynos]

Photons in free space have several properties including wavelength. Some other properties the have are spin, momentum, energy and frequency. Some of these are related.

[Mordred]

Seemingly infinite stars are emitting seemingly infinite photons into space, seemingly infinitely. Space is dark because there is not a lot in space for photons to interact with, but whenever something is in space for photons to interact with, we find that photons are there. Does this mean that space is an "ocean of photons"? Does this hold with the theory of mass energy equivalence that all matter is just "slowed" or "condensed" or "less energetic" energy??

 

I didn't see this part previously, there is a term to define the seeming infinite photons generated by all the stars and why the universe is not all aglow. This is called Olber's paradox

 

http://en.wikipedia.org/wiki/Olbers%27_paradox

 

what it breaks down to is that the universe expands the energy-density of radiation decreases, where Olber's paradox is a static and infinite universe conjecture, this was one of the earlier indications that the universe was in fact not static and eternal.

 

http://www2.astro.psu.edu/~niel/psiwa-2006-cosmology/notes-all.pdf

Cool. Does the photon have a "wavelength" in "empty space" or does it exist there as "energy particle"?

 

 

photons have a range of wavelengths, in cosmology applications the wavelengths can vary due to gravitational redshift, cosmological redshift and Doppler redshift.

 

this article will cover the 3 types as well as the redshift to wavelength formulas of each (basic generalized formulas)

http://cosmology101.wikidot.com/redshift-and-expansion

 

side note when you see different colors your seeing different wavelengths of visible light

Edited July 24, 2014 by Mordred

[ajb]

If we looked at a cross cut of the space between the Sun and the Earth with an infrared or other light spectrum camera, would we see the electromagnetic waves of any light spectrum?

Maybe as a side remark here: the Sun is almost a black body, meaning that is does radiate at all wavelengths, however there are some dark lines seen in the specta. These are the Fraunhofer lines and correspond to absorption of the light by chemical elements in the Solar atmosphere. This gives us a fingerprint of the atoms found in the Sun's atmosphere.

Edited July 24, 2014 by ajb

[swansont]

Does this hold with the theory of mass energy equivalence that all matter is just "slowed" or "condensed" or "less energetic" energy??

 

This is a rather clumsy description of mass-energy equivalence. But no, having photons out in the universe does not violate this principle.

[radicalsymmetry]

In relation to the OP I thought it might be helpful to state that photons travel into your retinas, where they are detected by cone cells, trigger synapses, and thus your brain creates an image.

On the subject of seeing photons;

I wondered about the geometry of vision as follows.

The sun emits a photon from its far left-hand-side, and at the same time emits a photon from its far right-hand-side. Both photons travel into approximately the same location of my retina.. How can my brain distinguish which photon came from where?

Is the answer something to do with the specific (multiplicity of) wave-particles phase-position?

Edited July 24, 2014 by radicalsymmetry

[swansont]

In relation to the OP I thought it might be helpful to state that photons travel into your retinas, where they are detected by cone cells, trigger synapses, and thus your brain creates an image.

 

On the subject of seeing photons;

 

I wondered about the geometry of vision as follows.

 

The sun emits a photon from its far left-hand-side, and at the same time emits a photon from its far right-hand-side. Both photons travel into approximately the same location of my retina.. How can my brain distinguish which photon came from where?

 

Is the answer something to do with the specific (multiplicity of) wave-particles phase-position?

 

 

 

The presence of the lens means light from those two locations will not form an image at the same place on the retina. IOW if they hit the same place, you would not have an image.

[Strange]

 

The presence of the lens means light from those two locations will not form an image at the same place on the retina. IOW if they hit the same place, you would not have an image.

 

Or, at least it would be very out of focus (or just not able to resolve objects of that size).

[swansont]

 

Or, at least it would be very out of focus (or just not able to resolve objects of that size).

 

Right. I should have said you would not have a well-resolved image.

[md65536]

Is space empty, or is space a "sea of photons"?

It's not filled with long waves that occupy space. Light is quantized into packets that have a point location. Wherever you measure "where" light energy is, it will be measured as individual photons each with a single point location. A sea of photons in space would be entirely empty space with photons in it, taking up no volume.

 

Another way it's not like a sea or a volume filled with matter is that the photons don't interact. They won't spread out to fill a space, they'll just go where they're going.

 

Do we agree that at all times, between the Sun and the Earth, there are photons?

I would not agree with this (as an amateur), though I think many people would.

 

The wave-particle duality doesn't mean that light "is" a wave and it "is" a particle, only that it consistently has properties of either, and it only certainly has those properties when measured. So, when you're not measuring the particle properties of a photon, there's really no point in saying that it "is" a particle even where it's not being measured. You could easily just say that yes, if they were measured there would be photons all over between Sun and Earth at all times.

 

But a sure statement about photons in between measurements can cause problems. For example, the sun is about 8 minutes away, so you could say that there are a bunch of photons traveling at different points along an 8-minute journey between the Sun and you. But if you accelerate to near c you can shorten that journey to say 1 minute. Then there is only 1 minute worth of photons, but they're still traveling at the same speed. What happened to the other 7 minutes worth? You might imagine relative time being "adjusted", so those en-route photons are brought back into the Sun and now haven't left yet. BUT I think this is pointless. This is an example of a confusing interpretation based on the assertion that the photons must exist as particles along their entire (8-minute) journey.

 

On the other hand, if you don't measure any photons in between Sun and Earth, you only have to say that they "are" where they're measured, and the existence of photons beyond how they are measured becomes irrelevant philosophy. So I would not agree or say anything certain about the existence of photons anywhere between where they're emitted and detected (absorbed or measured).

 

 

 

Or consider a different example. Suppose two people agree that there are photons constantly traveling along straight lines from Sun to Earth. Now put a double-slit between Sun and Earth. Suppose one person says "Each photon must go through either one slit or the other" while the other disagrees and says "Each photon must go through both slits simultaneously." It depends on interpretation, and neither interpretation is meaningfully measured. Similarly, agreeing on the behavior of photons without the double-slit depends on interpretation, even if it's easier to agree due to no obvious problems with intuition.

Edited July 26, 2014 by md65536

[phyti]

We can't see light in space because there is nothing for light to illuminate. But is there still an electromagnetic wave between the Earth and Sun or does this disturbance happen because of atmosphere?

 

Is space empty, or is space a "sea of photons"?

 

 

Space contains many photons moving in random directions from random sources. Considering the earth intercepts only a small pencil of light from any source, the major portion moves past us. In this sense space is ‘full’ of light.

Questioning what portion of the sun’s radiation is intercepted by the earth, consider the earth a disc projected onto a spherical surface with a radius of 93 million mi. Assuming no errors, if the surface was scaled to the radius of the earth, the disc has the diameter of a nickel. Imagining the amount of heat received on one hemisphere on a cloudless day, the total output of the sun has a high boggle factor.