Question Turning on the light traveling through the dark space!

Jan 25, 2020
31
4
1,555
How to apply the red-shifting-of-light explanation of the darkness of the sky to the space between the Earth and the Moon: we see this area dark though the Sun's light travels through it?
 
Last edited:
  • Like
Reactions: bearnard1616
Jan 25, 2020
31
4
1,555
True or not: if we could fill the area between Earth and the Moon with any material - air, for instance - we would get this area of the sky illuminated?
 
Jan 25, 2020
31
4
1,555
Contemplation of the various images of our universe reveals that the distinctive darkness of the universe cannot be attributed to the red-shifting-of-light explanation. In these images, we see cosmic objects—those that emit light and those reflect light—appearing amid absolute darkness. During a total solar eclipse, for instance, the gases that make up the outermost atmosphere of the Sun, the corona, are illuminated; but just beyond the corona, the space is dark, although the Sun’s light is indisputably traveling through that area.

The space that separates the Earth and the Moon offers another good example: the Sun’s light has no effect of illumination throughout this area. It is only when the Sun’s light contacts with the Moon’s surface that its illuminating effect is “turned on.” Similarly, the space beyond the Moon, which is undoubtedly filled with the Sun’s light, appears dark. The light traveling through it remains invisible until it illuminates the surface of a planet, a comet, or any other object that may be traveling through this area of space.

Wouldn’t these examples suggest that where there is material in the universe, there is illumination, rather than where there is light? Wouldn’t they further suggest that the very radiation that creates visible light is invisible; and that visible light―instead of being considered as visible radiation―may be simply just a visible effect ensuing from the contact of some invisible radiation(s) emitted by hot bodies with an object?
 
Jan 25, 2020
31
4
1,555
In the photographs taken by astronauts on the surface of the Moon, a likely affirmative answer to these two questions can be found. In these photographs, which were taken during the Moon’s two-week-long daytime, we see the astronauts against a dark background of space, while the Sun illuminates the ground on which they are standing. If the Moon had an atmosphere through which the Sun’s radiations had traveled, the astronauts would have appeared against an illuminated background. But such material—gases for an atmosphere—was not found, and so the effect—illumination—did not occur. For that very reason, if a photograph was taken for the Sun from the Moon’s surface, the space between the Moon and the Sun would not have appeared illuminated; in other words, a looker at the Sun from the Moon’s surface would see all the area between the Sun and the Moon is in absolute darkness.
 
Jan 25, 2020
31
4
1,555
Would the seven colors into which the white light we receive from the Sun is split up when passing through a prism indicate that the white light we receive from the Sun ensues from the contact between seven invisible radiations emitted by the Sun, each one of these radiations is responsible for one of the rainbow colors?
 
Jan 25, 2020
31
4
1,555
Since the Sun is a source of electromagnetic radiations, could each one of the six electromagnetic radiations other than what we call light―gamma, x-ray, ultraviolet, infrared, microwave, radio wave― be responsible for one of the rainbow colors into which the white light received from the Sun split? If it is so, the seventh color should represent one unknown type of electromagnetic radiation!
 
Jan 25, 2020
31
4
1,555
Along with what we call visible light, the electromagnetic spectrum includes six different radiations, three of which are of wavelength shorter than that of what we call light, while the three others are of a wavelength longer than that of light. Note that James Clerk Maxwell’s equation proves that all forms of electromagnetic waves travel at the speed of light, and there is a characteristic phenomenon which light shares with the other forms of electromagnetic radiations―the process of so-called diffraction, that is, light and all the electromagnetic radiations bend to the sides when they pass through a small hole, and then spread out.
 

bearnard1616

BANNED
Nov 18, 2020
74
27
80
How to apply the red-shifting-of-light explanation of the darkness of the sky to the space between the Earth and the Moon: we see this area dark though the Sun's light travels through it?
I assume it would be good to use a light class launch vehicle, intended for Sun- Synchronous Orbit or microravity spacecrafts like satellites, micrasatellites or guided or unguided light-class launch vehicles like Skylark Nano II
Thus, we can get appropriate intel about darkness of the sky to the space between the Earth and the Moon
 
  • Like
Reactions: F. M. Mossa
Jan 25, 2020
31
4
1,555
Does the device that researcher in Finland and Belarus produced, which they called “metasheet”―a material into which were embedded wire helices capable of absorbing electromagnetic radiation in a very narrow band of wavelengths, but remain transparent for the others in the spectrum; the metasheet worked for microwave radiation, and a modified design was obtained to work for visible light― support my notion that light is produced at the interference point of the electromagnetic radiations when contact with an object or travel through a medium?
 
Jan 25, 2020
31
4
1,555
Do the radiations emitted by the Sun have no effect of illumination nor of heating up until they come in touch with a substance: is it for this reason that outer space remains dark and at minus 270 centigrade?
 
Jan 25, 2020
31
4
1,555
Do the radiations emitted by the Sun have no effect of illumination nor of heating up until they come in touch with a substance: is it for this reason that outer space remains dark and at minus 270 centigrade?
The density of the medium through which the Sun's radiations travel and the angle at which this medium is hit by the radiations are important factors in determining the degree of heating this medium will reach: right angle and dense medium are hotter.
 
Jan 25, 2020
31
4
1,555
True or not: if we could fill the area between Earth and the Moon with any material - air, for instance - we would get this area of the sky illuminated?
I think that the darkness of the universe is not due to its vastness or a lack of sources of light. The radiations that crate visible light dominate every spot of the universe, but the universe remains dark only because there is no enough material to “turn on” the illuminating effect of the radiations emitted by the countless billions of stars found within.
 
Jan 25, 2020
31
4
1,555
Sweeping the Universe to keep the light traveling through it “turned off”!
I think that gravity plays a major role in keeping outer space dark by attracting dust particles and gas into the domain of nearby space objects. Earth’s gravity attracts about 15,000 tons of space debris yearly, preventing their amassing across the nearby outer space. If this amount of dust particles were to let buildup around our planet, it would have turned the night illuminated.
 
Mar 4, 2020
663
86
1,980
Our sun produces a solar wind. 99% of this wind is isolated charge,(free electrons and protons) which is accelerated out beyond Neptune. When that wind passes Neptune, it is traveling much faster than when it passed Earth. We have not found where this acceleration stops. We can find no evidence of any of this charge, recombining. Recombining charge would emit EM and would be detected. Free charge is not visible, and can not reflect visible light.

And free charge, is not attracted to any gravitational object. It appears to be immune to gravity. This wind has been blowing for billions of years. Image all the wind for all the stars for all these eons. What's the likelihood of there being more mass that is not gathered together with gravity? After all this time, most of the mass in this universe might be in this free state. In other words, there might be more matter dissolved/dispersed out in space, than in all the stars and in all the galaxies.

Perhaps, the real function of a star is to disperse mass. Not collect it.

The universe remains a very mysterious place.
 
Jan 25, 2020
31
4
1,555
Our sun produces a solar wind. 99% of this wind is isolated charge,(free electrons and protons) which is accelerated out beyond Neptune. When that wind passes Neptune, it is traveling much faster than when it passed Earth. We have not found where this acceleration stops. We can find no evidence of any of this charge, recombining. Recombining charge would emit EM and would be detected. Free charge is not visible, and can not reflect visible light.

And free charge, is not attracted to any gravitational object. It appears to be immune to gravity. This wind has been blowing for billions of years. Image all the wind for all the stars for all these eons. What's the likelihood of there being more mass that is not gathered together with gravity? After all this time, most of the mass in this universe might be in this free state. In other words, there might be more matter dissolved/dispersed out in space, than in all the stars and in all the galaxies.

Perhaps, the real function of a star is to disperse mass. Not collect it.

The universe remains a very mysterious place.
“My problem” is not with the solar wind, it is with the radiations emitted by the stars: these radiations do not have any effect of heating nor of illuminating except they travel through a medium or contact with a substance; were these radiation visible or hot, the universe would have experienced unbearable brightness and heat.
 
Mar 13, 2021
5
0
30
Contemplation of the various images of our universe reveals that the distinctive darkness of the universe cannot be attributed to the red-shifting-of-light explanation. In these images, we see cosmic objects—those that emit light and those reflect light—appearing amid absolute darkness. During a total solar eclipse, for instance, the gases that make up the outermost atmosphere of the Sun, the corona, are illuminated; but just beyond the corona, the space is dark, although the Sun’s light is indisputably traveling through that area.

The space that separates the Earth and the Moon offers another good example: the Sun’s light has no effect of illumination throughout this area. It is only when the Sun’s light contacts with the Moon’s surface that its illuminating effect is “turned on.” Similarly, the space beyond the Moon, which is undoubtedly filled with the Sun’s light, appears dark. The light traveling through it remains invisible until it illuminates the surface of a planet, a comet, or any other object that may be traveling through this area of space.

Wouldn’t these examples suggest that where there is material in the universe, there is illumination, rather than where there is light? Wouldn’t they further suggest that the very radiation that creates visible light is invisible; and that visible light―instead of being considered as visible radiation―may be simply just a visible effect ensuing from the contact of some invisible radiation(s) emitted by hot bodies with an object?
There actually is matter between the earth & moon. We just can't see it because our eyes have evolved to see what's in their environment. We see the light reflecting off objects, not the object itself.
 
Jan 25, 2020
31
4
1,555
The density of the medium through which the Sun's radiations travel and the angle at which this medium is hit by the radiations are important factors in determining the degree of heating this medium will reach: right angle and dense medium are hotter.
Does this notion give a clue to the secret why the outermost layer of the Sun's atmosphere, the corona, is hotter than the inner layer of the Sun's atmosphere, the chromosphere? Would the answer be: simply because the corona is denser than the chromosphere?
 
Mar 4, 2020
663
86
1,980
Without a media, one can not see the side of light. One can see the flashlight beam from the side, because of the air media. Out in space, you can't see side light.

The velocity of the particles in the sun's liquid state, is lower than the velocity of the particles in the gaseous atmosphere.....the less dense environment gives the particles a longer acceleration time....to go faster. Not as many collisions. But the collisions are more energetic.

Density controls the acceleration time, which determines the velocity. The velocity is the amount of energy the particles has.

It's a density gradient that gives us a light bow(rainbow). Or can bow a flashlight beam thru a tank of sugar water. Another density gradient is used as evidence of a gravity bow.
 
Jan 25, 2020
31
4
1,555
Do the radiations emitted by the Sun have no effect of illumination nor of heating up until they come in touch with a substance: is it for this reason that outer space remains dark and at minus 270 centigrade?
Would the seven colors into which the white light we receive from the Sun is split up when passing through a prism indicate that the white light we receive from the Sun ensues from the contact between seven invisible radiations emitted by the Sun, each one of these radiations is responsible for one of the rainbow colors?
As I did not know that mirrors reflect heat as well as light, I made a simple experiment in which I used a lens to focus the Sun’s light reflected from a mirror (a looking glass) onto a paper, and, to my surprise, I got the paper burned. This, I think, supports my notion that what we call light is a result produced when some invisible radiations, emitted by a hot body, meet a material.
 
Jan 27, 2020
457
113
1,880
Let's first examine stellar winds.

Stellar winds are known to be radiatively driven. The radiative momentum is transferred from radiation to a gas by means of absorption. Not all chemical elements are equal absorbers, so some elements are accelerated more than the others. In the case that there is enough collisions that may redistribute the momentum over the entire gas (basically elastic collisions), the radiatively driven stellar wind behaves like a one- component fluid.

In a real radiatively driven wind the chemical species may be divided into three basic groups according to the ability to absorb radiation.

The first group consists of radiatively accelerated particles that have enough ability to absorb radiation and, consequently, are significantly accelerated by absorption of radiation in spectral lines.

The second group involves particles whose contribution to the radiative force is almost negligible compared to the first group. This group is taken along by friction with the first group. Since its role in the stellar wind is passive, we call this group as a passive component. In real stars it consists of hydrogen and helium.

The third group are free electrons, which are accelerated by Thomson scattering and by friction as well. Each component is described by a set of equations of continuity, motion, and energy.

Calculations of the three-component wind models show the same results as for the one component wind, which means that the one-component approximation is an adequate one for winds of these stars. On the other hand, there is a relatively large heating for lower effective stellar temperatures, namely for the main-sequence B type stars. An interesting fact is that for lower effective temperatures we obtain higher wind temperatures, a consequence of a frictional heating, which rises with decreasing effectiveness of elastic collisions.

Next, think of the luminosity—the energy emitted per second by the star—as an intrinsic property of the star. As that energy gets emitted, you can picture it passing through spherical shells centered on the star. In the above image, the entire spherical shell isn't illustrated, just a small section. Each shell should receive the same total amount of energy per second from the star, but since each successive sphere is larger, the light hitting an individual section of a more distant sphere will be diluted compared to the amount of light hitting an individual section of a nearby sphere. The amount of dilution is related to the surface area of the spheres, which is given by:
A = 4 π d2A = 4 π d2 .

How bright will the same light source appear to observers fixed to a spherical shell with a radius twice as large as the first shell? Since the radius of the first sphere is d, and the radius of the second sphere would be 2 x d2 x d , then the surface area of the larger sphere is larger by a factor of 4.

Inverse_square_law_.png
The Inverse Square Law, Credit: Wikimedia Commons

Thus, the equation for the apparent brightness of a light source is given by the luminosity divided by the surface area of a sphere with radius equal to your distance from the light source, or
F = L / 4 π d2F = L / 4 π d2 , where d is your distance from the light source.

The apparent brightness is often referred to more generally as the flux, and is abbreviated F (as I did above). In practical terms, flux is given in units of energy per unit time per unit area (e.g., Joules / second / square meter). Since luminosity is defined as the amount of energy emitted by the object, it is given in units of energy per unit time [e.g., Joules / second (1 Joule / second = 1 Watt)Joules / second (1 Joule / second = 1 Watt) ]. The distance between the observer and the light source is d, and should be in distance units, such as meters. You are probably familiar with the luminosity of light bulbs given in Watts (e.g., a 100 W bulb), and so you could, for example, refer to the Sun as having a luminosity of 3.9 x 1026 W3.9 x 1026 W . Given that value for the luminosity of the Sun and adopting the distance from the Sun to the Earth of 1 AU = 1.5 x 1011 m1 AU = 1.5 x 1011 m , you can calculate the Flux received on Earth by the Sun, which is:

F = 3.9 x 1026 W / 4 π (1.5 x 1011 m)2 = 1,379 W per square meterF = 3.9 x 1026 W / 4 π (1.5 x 1011 m)2 = 1,379 W per square meter

Lastly, we turn to the photons which are not identifiable separate particles that travel along trajectories. They are the quantum excitations of the EM field, and spread out as a spherical wave. When absorbed, say by individual atoms that are not coherent and may be a long way apart, the quantum wave-function of each photon collapses (or possibly branched) and each photon can then be separately identified as having been absorbed in one location: the atom. However, this does not mean that the photon has travelled along a path from the point of emission to the point of absorption. It has not. It propagated as a spherical wave, maybe isotropically (same in all directions) or maybe not.

This scenario is the best and simplest illustration of the so-called measurement or collapse problem in quantum theory, the origin of the various QM (quantum mechanics) interpretations.

Consider one photon setting out from the Andromeda galaxy as a spherical wave. 2.5 million years later it is absorbed by a camera or a human eye here on earth. The spherical wave, 5 million light-years in diameter, instantly collapses to a single point here on earth. That photon will never be detected anywhere but here. The collapse is non-local, seems to be instantaneous, but cannot be detected in any way or used to signal faster than light.

The whole thing is outrageous and incomprehensible. It cannot be what is actually happening, but the only alternative accounts are equally outrageous, or more so: the no-collapse neo-realist many-worlds interpretation; Qbism (wavefunctions are not real, it’s all in your head), Copenhagen (don’t ask), and so on.

If you look at light as a collection of little particles, you could say that dimmer light has its photons more spread out. But, they are not spread out in space while traveling. Rather, they are spread out in time and space as they are received. A sufficiently sensitive photon counter device can detect the reception of light one photon at a time. Shine light at such a device and it does not receive the light as a steady stream. Rather, it receives the light as a series of discrete bundles of energy separated by gaps in time. Similarly, shine light at a sufficiently sensitive array of photon counters, and it receives the light at point locations with spatial gaps between them. When viewed in this way, a light beam always has gaps between its photons, whether the light be very bright or very dim. Very dim light beams have larger gaps in time and space between the reception of each photon compared to brighter light beams. Light from a very distant star has spread out over a very large area and become very dim in the process. The gaps between photon reception from a very distant, dim star are therefore large. Again, it is only the reception time and locations that has gaps. There are no gaps in space between the photons as they travel.

If you look at light as a wave, then there are no gaps unless specifically placed there on purpose. Of course, if you repeatedly turn on and off a flashlight, the light beam coming from your flashlight will have gaps. Similarly, if you shine a continuous beam of light through a shutter that is repeatedly opening and closing, you can create gaps.

Then, if you shine a continuous beam of light into free space, the wave will start with no gaps and therefore develop no gaps as it travels. Waves are field oscillations that are spread out smoothly through space. Spreading out a wave over a larger area just causes the wave strength to weaken, but does not cause gaps to form. Therefore, if you look at photons as waves, spatial gaps never form in light as it travels through free space, no matter how dim it gets. The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.

A helpful way to look at photons is that they act like waves while traveling and act like particles when interacting with matter. In the context of starlight, the light travels through space for millions of years acting like a wave, and then acts like a collection of particles when hitting the photon detector, the telescope, or an eye.

When a photon is absorbed by an electron, it is completely destroyed. All its energy is imparted to the electron, which instantly jumps to a new energy level. The photon itself ceases to be. In the equations which govern this interaction, one side of the equation (for the initial state) has terms for both the electron and the photon, while the other side (representing the final state) has only one term: for the electron.

The opposite happens when an electron emits a photon. The photon is not selected from a "well" of photons living in the atom or from a basket of photons on the atom's front porch; it is created instantaneously out of the vacuum energy. The electron in the high energy level is instantly converted into a lower energy-level electron and a photon. There is no in-between state where the photon is being constructed. It instantly pops into existance.

So the question is: where does the photon come from?

Strangely, it doesn't seem to come from anywhere. The universe must put the extra energy somewhere, and because electrons in atoms are electromagnetic phenomena, a photon is born with the required energy. In a weak-force interaction, say the decay of a neutron, that energy goes into a neutrino particle which is also instantaneously created. Each force has its own carrier particles, and knows how to make them.

See: http://www.space.bas.bg/astro/Rogen2004/StPh-8.pdf

See: https://www.e-education.psu.edu/astro801/content/l4_p4.html

See: https://www.quora.com/As-light-from-a-star-spreads-out-and-weakens-do-gaps-form-between-the-photons?share=1

See: https://www.wtamu.edu/~cbaird/sq/2015/02/12/as-light-from-a-star-spreads-out-and-weakens-do-gaps-form-between-the-photons/

See: http://curious.astro.cornell.edu/about-us/137-physics/general-physics/particles-and-quantum-physics/805-how-are-photons-created-and-destroyed-advanced


Each photon therefore collapses from mostly wave-like to mostly particle-like upon being detected. Since the photons act mostly like waves while traveling, there are no gaps that develop between them while traveling. And since the photons act mostly like particles when being detected, there are gaps in the time when the photons are detected and in the locations where they are detected.

The act of detecting the light, anywhere after it has been emitted, causes it to collapse from wave-like to particle-like, and therefore introduces the gaps. A very dim light beam from a distant star has a very weak wave magnitude, which leads to large gaps in photon reception.
Hartmann352
 
Jan 25, 2020
31
4
1,555
Let's first examine stellar winds.

Stellar winds are known to be radiatively driven. The radiative momentum is transferred from radiation to a gas by means of absorption. Not all chemical elements are equal absorbers, so some elements are accelerated more than the others. In the case that there is enough collisions that may redistribute the momentum over the entire gas (basically elastic collisions), the radiatively driven stellar wind behaves like a one- component fluid.

In a real radiatively driven wind the chemical species may be divided into three basic groups according to the ability to absorb radiation.

The first group consists of radiatively accelerated particles that have enough ability to absorb radiation and, consequently, are significantly accelerated by absorption of radiation in spectral lines.

The second group involves particles whose contribution to the radiative force is almost negligible compared to the first group. This group is taken along by friction with the first group. Since its role in the stellar wind is passive, we call this group as a passive component. In real stars it consists of hydrogen and helium.

The third group are free electrons, which are accelerated by Thomson scattering and by friction as well. Each component is described by a set of equations of continuity, motion, and energy.

Calculations of the three-component wind models show the same results as for the one component wind, which means that the one-component approximation is an adequate one for winds of these stars. On the other hand, there is a relatively large heating for lower effective stellar temperatures, namely for the main-sequence B type stars. An interesting fact is that for lower effective temperatures we obtain higher wind temperatures, a consequence of a frictional heating, which rises with decreasing effectiveness of elastic collisions.

Next, think of the luminosity—the energy emitted per second by the star—as an intrinsic property of the star. As that energy gets emitted, you can picture it passing through spherical shells centered on the star. In the above image, the entire spherical shell isn't illustrated, just a small section. Each shell should receive the same total amount of energy per second from the star, but since each successive sphere is larger, the light hitting an individual section of a more distant sphere will be diluted compared to the amount of light hitting an individual section of a nearby sphere. The amount of dilution is related to the surface area of the spheres, which is given by:
A = 4 π d2A = 4 π d2 .

How bright will the same light source appear to observers fixed to a spherical shell with a radius twice as large as the first shell? Since the radius of the first sphere is d, and the radius of the second sphere would be 2 x d2 x d , then the surface area of the larger sphere is larger by a factor of 4.

View attachment 1969
The Inverse Square Law, Credit: Wikimedia Commons

Thus, the equation for the apparent brightness of a light source is given by the luminosity divided by the surface area of a sphere with radius equal to your distance from the light source, or
F = L / 4 π d2F = L / 4 π d2 , where d is your distance from the light source.

The apparent brightness is often referred to more generally as the flux, and is abbreviated F (as I did above). In practical terms, flux is given in units of energy per unit time per unit area (e.g., Joules / second / square meter). Since luminosity is defined as the amount of energy emitted by the object, it is given in units of energy per unit time [e.g., Joules / second (1 Joule / second = 1 Watt)Joules / second (1 Joule / second = 1 Watt) ]. The distance between the observer and the light source is d, and should be in distance units, such as meters. You are probably familiar with the luminosity of light bulbs given in Watts (e.g., a 100 W bulb), and so you could, for example, refer to the Sun as having a luminosity of 3.9 x 1026 W3.9 x 1026 W . Given that value for the luminosity of the Sun and adopting the distance from the Sun to the Earth of 1 AU = 1.5 x 1011 m1 AU = 1.5 x 1011 m , you can calculate the Flux received on Earth by the Sun, which is:

F = 3.9 x 1026 W / 4 π (1.5 x 1011 m)2 = 1,379 W per square meterF = 3.9 x 1026 W / 4 π (1.5 x 1011 m)2 = 1,379 W per square meter

Lastly, we turn to the photons which are not identifiable separate particles that travel along trajectories. They are the quantum excitations of the EM field, and spread out as a spherical wave. When absorbed, say by individual atoms that are not coherent and may be a long way apart, the quantum wave-function of each photon collapses (or possibly branched) and each photon can then be separately identified as having been absorbed in one location: the atom. However, this does not mean that the photon has travelled along a path from the point of emission to the point of absorption. It has not. It propagated as a spherical wave, maybe isotropically (same in all directions) or maybe not.

This scenario is the best and simplest illustration of the so-called measurement or collapse problem in quantum theory, the origin of the various QM (quantum mechanics) interpretations.

Consider one photon setting out from the Andromeda galaxy as a spherical wave. 2.5 million years later it is absorbed by a camera or a human eye here on earth. The spherical wave, 5 million light-years in diameter, instantly collapses to a single point here on earth. That photon will never be detected anywhere but here. The collapse is non-local, seems to be instantaneous, but cannot be detected in any way or used to signal faster than light.

The whole thing is outrageous and incomprehensible. It cannot be what is actually happening, but the only alternative accounts are equally outrageous, or more so: the no-collapse neo-realist many-worlds interpretation; Qbism (wavefunctions are not real, it’s all in your head), Copenhagen (don’t ask), and so on.

If you look at light as a collection of little particles, you could say that dimmer light has its photons more spread out. But, they are not spread out in space while traveling. Rather, they are spread out in time and space as they are received. A sufficiently sensitive photon counter device can detect the reception of light one photon at a time. Shine light at such a device and it does not receive the light as a steady stream. Rather, it receives the light as a series of discrete bundles of energy separated by gaps in time. Similarly, shine light at a sufficiently sensitive array of photon counters, and it receives the light at point locations with spatial gaps between them. When viewed in this way, a light beam always has gaps between its photons, whether the light be very bright or very dim. Very dim light beams have larger gaps in time and space between the reception of each photon compared to brighter light beams. Light from a very distant star has spread out over a very large area and become very dim in the process. The gaps between photon reception from a very distant, dim star are therefore large. Again, it is only the reception time and locations that has gaps. There are no gaps in space between the photons as they travel.

If you look at light as a wave, then there are no gaps unless specifically placed there on purpose. Of course, if you repeatedly turn on and off a flashlight, the light beam coming from your flashlight will have gaps. Similarly, if you shine a continuous beam of light through a shutter that is repeatedly opening and closing, you can create gaps.

Then, if you shine a continuous beam of light into free space, the wave will start with no gaps and therefore develop no gaps as it travels. Waves are field oscillations that are spread out smoothly through space. Spreading out a wave over a larger area just causes the wave strength to weaken, but does not cause gaps to form. Therefore, if you look at photons as waves, spatial gaps never form in light as it travels through free space, no matter how dim it gets. The light from a distance star indeed spreads out and weakens as it travels, but this just reduces the wave strength and does not introduce gaps.

A helpful way to look at photons is that they act like waves while traveling and act like particles when interacting with matter. In the context of starlight, the light travels through space for millions of years acting like a wave, and then acts like a collection of particles when hitting the photon detector, the telescope, or an eye.

When a photon is absorbed by an electron, it is completely destroyed. All its energy is imparted to the electron, which instantly jumps to a new energy level. The photon itself ceases to be. In the equations which govern this interaction, one side of the equation (for the initial state) has terms for both the electron and the photon, while the other side (representing the final state) has only one term: for the electron.

The opposite happens when an electron emits a photon. The photon is not selected from a "well" of photons living in the atom or from a basket of photons on the atom's front porch; it is created instantaneously out of the vacuum energy. The electron in the high energy level is instantly converted into a lower energy-level electron and a photon. There is no in-between state where the photon is being constructed. It instantly pops into existance.

So the question is: where does the photon come from?

Strangely, it doesn't seem to come from anywhere. The universe must put the extra energy somewhere, and because electrons in atoms are electromagnetic phenomena, a photon is born with the required energy. In a weak-force interaction, say the decay of a neutron, that energy goes into a neutrino particle which is also instantaneously created. Each force has its own carrier particles, and knows how to make them.

See: http://www.space.bas.bg/astro/Rogen2004/StPh-8.pdf

See: https://www.e-education.psu.edu/astro801/content/l4_p4.html

See: https://www.quora.com/As-light-from-a-star-spreads-out-and-weakens-do-gaps-form-between-the-photons?share=1

See: https://www.wtamu.edu/~cbaird/sq/2015/02/12/as-light-from-a-star-spreads-out-and-weakens-do-gaps-form-between-the-photons/

See: http://curious.astro.cornell.edu/about-us/137-physics/general-physics/particles-and-quantum-physics/805-how-are-photons-created-and-destroyed-advanced


Each photon therefore collapses from mostly wave-like to mostly particle-like upon being detected. Since the photons act mostly like waves while traveling, there are no gaps that develop between them while traveling. And since the photons act mostly like particles when being detected, there are gaps in the time when the photons are detected and in the locations where they are detected.

The act of detecting the light, anywhere after it has been emitted, causes it to collapse from wave-like to particle-like, and therefore introduces the gaps. A very dim light beam from a distant star has a very weak wave magnitude, which leads to large gaps in photon reception.
Hartmann352
Why should light dim while traveling at a speed of 300 thousand km per second through empty space? I think that the dimming of light coming from a faraway star is a problem for the human eyes: the equations you introduced refer to the distance between the observer and the light source. And what about the utilities of the instrument used in observation: if the observer uses a powerful telescope, he will see the same star, which appeared faint to his eyes, much brighter?
 
Mar 4, 2020
663
86
1,980
"Why should light dim while traveling at a speed of 300 thousand km per second through empty space?"

The propagation of light has a radial direction. Radii are not parallel. Think of spokes on a wheel, the farther from the center, the greater the spread or distance between the tips of the spokes.

That is why light dims. They call this spread divergence. Light loses it's density(intensity) with distance.

It dims.
 
Jan 27, 2020
457
113
1,880
F.M. Mossa might find the following of interest:

Some of the earliest accounts of light reflection originate from the ancient Greek mathematician Euclid, who conducted a series of experiments around 300 BC, and appears to have had a good understanding of how light is reflected. However, it wasn't until a millennium and a half later that the Arab scientist Alhazen proposed a law describing exactly what happens to a light ray when it strikes a smooth surface and then bounces off into space.

The amount of light reflected by an object, and how it is reflected, is highly dependent upon the degree of smoothness or texture of the surface. When surface imperfections are smaller than the wavelength of the incident light (as in the case of a mirror), virtually all of the light is reflected equally. However, in the real world most objects have convoluted surfaces that exhibit a diffuse reflection, with the incident light being reflected in all directions. Many of the objects that we casually view every day (people, cars, houses, animals, trees, etc.) do not themselves emit visible light but reflect incident natural sunlight and artificial light. For instance, an apple appears a shiny red color because it has a relatively smooth surface that reflects red light and absorbs other non-red (such as green, blue, and yellow) wavelengths of light. The reflection of light can be roughly categorized into two types of reflection. Specular reflection is defined as light reflected from a smooth surface at a definite angle, whereas diffuse reflection is produced by rough surfaces that tend to reflect light in all directions. There are far more occurrences of diffuse reflection than specular reflection in our everyday environment.

Light will dim based on the inverse square law, which states that a doubling of the distance will reduce the emitted phon flux by four times.

In general, we therefore multiply the distance with itself in order to calculate the enlargement of that surface area. However, a larger surface area leads to a light intensity that is inversely proportional to the square of the distance, since the same amount of light has to be distributed onto a larger surface area respectively.

Therefore, we see a decrease in light intensity.

In technical terms the inverse-square law reads as follows: The energy (in our case: light intensity) at location A (subject area) decreases inversely proportional to the square of A’s distance to the energy source.

lightfalloffsquare.jpeg

Overview-Light-Fall-Off-to-the-Square.jpeg

Per this law, light loses its brightness or luminosity as it moves away from the source. For example: when you switch on the light in one corner of the room and when you move away from the source, the light appears dim or less bright due to an increase in the distance (away from the source).

The formula of inverse-square law is given as,

Screen Shot 2022-08-14 at 9.02.24 PM.png

See: https://petapixel.com/inverse-square-law-light/

Why the sky above the Earth is bright, but space is mostly black except for stars, galaxies, planets and so on. Why isn’t space bright?

To understand that, start with why the sky s bright and blue on the Earth. It turns out most of the molecules in our atmosphere don’t really have much in the way of absorption bands in the visible spectrum. This means that light comes through the atmosphere.

Of course, if clouds and rain and volcanic ash are in the air, this changes. But consider “clear, dry air.” In that case, the molecules are smaller than wavelength of light. They don’t interact with light much because of that. But, there is an awful lot of air. Each molecule has a small probability of scattering the light a little bit. The sum total of that is that each meter of air scatters a few parts per billion of the light. But there are tens of thousands of meters of air between you and space. That means that a few percent of sunlight are scattered. Not really enough to notice the missing light at midday, but certainly enough to notice that the sky is blue. Why blue? Because the shorter the wavelength, the more likely it is that the light will be scattered by molecules. (This is not the case for larger particles. They scatter all visible light about equally. Clouds are a good example.)

But in space, unless there is a nebula or planetary atmosphere, there is not much to scatter light.

As a result, unless there is a light source at a particular position, the sky is going to be pretty dark on the Moon or outside a spacecraft.

If you were in deep space inside our galaxy, you would see stars everywhere in all directions. They would be as bright as a cloudless night on Earth, and would not twinkle. They would shine steady like planets do. Since the stars were below you as well as above, you would be able to see yourself weakly illuminated, though it wouldn't be enough to see colors.

Space is mostly just that: empty space. While it's true that space is not a perfect vacuum, but close enough that there is very little light scattering.

The term albedo comes from the Latin word albus, which means white. Albedo generally refers to visible light. However, albedo may sometimes involve infrared radiation (think heat).

Albedo is a unitless, non-dimensional value that ranges on a scale from 0 to 1. A value of 1 means that an object or surface reflects all incoming solar radiation. Surfaces that absorb all of the sunlight have a value of 0.

Surfaces like snow that reflect more light have a high albedo. Black asphalt has a low albedo because it absorbs much of the sunlight.

The overall albedo of the Moon is frequently quoted as being about 7%. This is actually the so-called Bond albedo at visible wavelengths, which refers to the fraction of the total energy impinging on a surface that is reflected in all directions. It is a concept which is useful in studies of planetary enegy balance, but has little relevance to perceived brightness, which depend entirely on the intensity reflected in a specific direction.

The NASA Moon Fact Sheet gives the Bond albedo of the Moon (presumably averaged over the entire solar spectrum, including non-visible wavelengths) as 0.11. However the CERES Earth orbiting satellite climate radiometers have measured the value to be higher and somewhere between 0.136 and 0.137, at a lunar phase angle of seven degrees.

A concept more closely related to the variations in the perceived brightness of planets due to variations in surface properties is the geometric albedo, or sometimes called the visual geometric albedo when referring specifically to the band of visible wavelengths. Geometric albedo is determined by comparing the light received from an entire spherical planet to that expected from an idealized "perfect" reflectance diffusing disk of the same same cross section, with a light source directly behind the detector. This is an indication, for example, of the total intensity of moonlight reflected back towards the Earth near Full Moon. The NASA Moon Fact Sheet gives the visual geometric albedo of the Moon as 0.12. However this number probably doesn't include the opposition surge which can increase the reflectance of the lunar surface by 50% or more when the light source and detector are precisely aligned.

A third concept is the normal albedo (or visual normal albedo) which is still more closely to the variation in in the perceived brightness of individual surface features due to variations in their reflectances. The word "normal" is used here to mean "with illumination perpendicular to the surface". Normal albedo is obtained by comparing a small sections of a planetary surface to what would be expected from a perfect diffusing reflector of the same area. Such a concept can be used to differentiate bright (high albedo) features, like the lunar highlands, from dark (low albedo) features, like the maria. However again, the retroreflectivity of many lunar materials makes it difficult to assign universally meaningful numbers. Unlike an ideal diffuser of a given reflectance, two lunar features that have equal reflectances when observed with the detector exactly aligned with the light source may be less similar when viewed at a different angle. Also the amount of brightening observed as the phase angle decreases varies with wavelength. A list of what are presumably visual normal albedo measurements can be found under Brightness of Selected Features, although it might be noted that only features near disk center are actually being evaluated at normal incidence near Full Moon.

Reflection and transmission of light waves occur because the frequencies of the light waves do not match the natural frequencies of vibration of the objects. When light waves of these frequencies strike an object, the electrons in the atoms of the object begin vibrating. But instead of vibrating in resonance at a large amplitude, the electrons vibrate for brief periods of time with small amplitudes of vibration; then the energy is reemitted as a light wave. If the object is transparent, then the vibrations of the electrons are passed on to neighboring atoms through the bulk of the material and reemitted on the opposite side of the object. Such frequencies of light waves are said to be transmitted. If the object is opaque, then the vibrations of the electrons are not passed from atom to atom through the bulk of the material. Rather the electrons of atoms on the material's surface vibrate for short periods of time and then reemit the energy as a reflected light wave. Such frequencies of light are said to be reflected.

The color of the objects that we see is largely due to the way those objects interact with light and ultimately reflect or transmit it to our eyes. The color of an object is not actually within the object itself. Rather, the color is in the light that shines upon it and is ultimately reflected or transmitted to our eyes. We know that the visible light spectrum consists of a range of frequencies, each of which corresponds to a specific color. When visible light strikes an object and a specific frequency becomes absorbed, that frequency of light will never make it to our eyes. Any visible light that strikes the object and becomes reflected or transmitted to our eyes will contribute to the color appearance of that object. So the color is not in the object itself, but in the light that strikes the object and ultimately reaches our eye. The only role that the object plays is that it might contain atoms capable of selectively absorbing one or more frequencies of the visible light that shine upon it. So if an object absorbs all of the frequencies of visible light except for the frequency associated with green light, then the object will appear green in the presence of ROYGBIV. And if an object absorbs all of the frequencies of visible light except for the frequency associated with blue light, then the object will appear blue in the presence of ROYGBIV.

reflected light.jpeg

Transparent materials are materials that allow one or more of the frequencies of visible light to be transmitted through them; whatever color(s) is/are not transmitted by such objects, are typically absorbed by them. The appearance of a transparent object is dependent upon what color(s) of light is/are incident upon the object and what color(s) of light is/are transmitted through the object.

In general, reflection, transmission and absorption depend on the wavelength of the affected radiation. Thus, these three processes can either be quantified for monochromatic radiation (in this case, the adjective "spectral" is added to the respective quantity) or for a certain kind of polychromatic radiation. For the latter, the spectral distribution of the incident radiation has to be specified. In addition, reflectance, transmittance and absorptance might also depend on polarization and geometric distribution of the incident radiation, which therefore also have to be specified.The reflectance r is defined by the ratio of reflected radiant power to incident radiant power. For a certain area element dA of the reflecting surface, the (differential) incident radiant power is given by the surface's irradiance Ee, multiplied with the size of the surface element, thus

dFe,incident = Ee dA

and the (differential) reflected radiant power is given by the exitance Me, multiplied with the size of the surface element:

dFe,reflected = Me dA

Total reflectance is further subdivided in regular reflectance rr and diffuse reflectance rd, which are given by the ratios of regularly (or specularly) reflected radiant power and diffusely reflected radiant power to incident radiant power. From this definition, it is obvious that

r = rr + rd

Being defined as ratios of radiant power values, reflectance, transmittance and absorptance are dimensionless.

Quantities such as reflectance and transmittance are used to describe the optical properties of materials. The quantities can apply to either complex radiation or to monochromatic radiation.

The optical properties of materials are not a constant since they are dependent on many parameters such as:

• thickness of the sample
• surface conditions
• angle of incidence
• temperature
• the spectral composition of the radiation (CIE standard illuminants A, B, C, D65 and other illuminants D)
• polarization effects

See: https://www.wave3.com/2021/01/22/behind-forecast-albedo-reflecting-reflecting-sunlight/

See: https://the-moon.us/wiki/Albedo

See: https://www.physicsclassroom.com/class/light/Lesson-2/Light-Absorption,-Reflection,-and-Transmission

See: https://micro.magnet.fsu.edu/primer/lightandcolor/reflectionintro.html

Light, and its corresponding photons, need a surface to be reflected from in order to be visible, and its reflectivity is determined by the surface albedo. Though flowing through space from a source, it will not be visible until reflected. A spectrum of the elected light will determine the light absorbed and the frequencies, and hence the color, of the reflected light and the color in which we view the object.
Hartmann352
 
May 8, 2022
178
3
105
This, I think, supports my notion that what we call light is a result produced when some invisible radiations, emitted by a hot body, meet a material.
I believe it is called the collapse of the wave function that releases the energy contained in the photon and becomes expressed as light and heat?
 
Last edited:

ASK THE COMMUNITY