Can light be compressed?
up vote
33
down vote
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What if we take a cylindrical vessel with an inside surface completely reflecting and attach a piston such that it is also reflecting. What will happen to light if we compress it like this?
visible-light electromagnetic-radiation
add a comment |
up vote
33
down vote
favorite
What if we take a cylindrical vessel with an inside surface completely reflecting and attach a piston such that it is also reflecting. What will happen to light if we compress it like this?
visible-light electromagnetic-radiation
3
Related: What longest time ever was achieved at holding light in a closed volume?
– sumelic
Dec 1 at 4:01
add a comment |
up vote
33
down vote
favorite
up vote
33
down vote
favorite
What if we take a cylindrical vessel with an inside surface completely reflecting and attach a piston such that it is also reflecting. What will happen to light if we compress it like this?
visible-light electromagnetic-radiation
What if we take a cylindrical vessel with an inside surface completely reflecting and attach a piston such that it is also reflecting. What will happen to light if we compress it like this?
visible-light electromagnetic-radiation
visible-light electromagnetic-radiation
edited Dec 2 at 20:19
Peter Mortensen
1,91611323
1,91611323
asked Dec 1 at 3:20
user204283
17714
17714
3
Related: What longest time ever was achieved at holding light in a closed volume?
– sumelic
Dec 1 at 4:01
add a comment |
3
Related: What longest time ever was achieved at holding light in a closed volume?
– sumelic
Dec 1 at 4:01
3
3
Related: What longest time ever was achieved at holding light in a closed volume?
– sumelic
Dec 1 at 4:01
Related: What longest time ever was achieved at holding light in a closed volume?
– sumelic
Dec 1 at 4:01
add a comment |
2 Answers
2
active
oldest
votes
up vote
35
down vote
Suppose there is an amount of light (electromagnetic radiation) inside the cylinder. Note that electromagnetic radiation is composed of particles called photons, and if we consider that there is a very large number of photons inside the cylinder, we may use statistical mechanics to create a model of a photon gas. Yes, the system you describe will act like a gas, and its properties may be derived from statistics and from the properties of photons.
If a photon's frequency is $f$, its energy is $E_γ = hf$, where $h$ is Planck's constant. It is also important to remember that photons have linear momentum $$p = frac{E_γ}{c} = frac{hf}{c}$$
But the fact that photons have nonzero linear momentum implies that they will exert pressure against the cylinder's walls. Once the photon reflects on the wall, its momentum will have changed direction, and this imples that the wall has exerted a force on the photon to make it change directions. Therefore, the photon gas exerts pressure against the walls.
It can be shown that if the total energy of the photon gas is $U$, then the relationship between the pressure $P$ and the volume $V$ of the gas is $U = 3PV$.
If you push the piston, you'll do positive work and therefore give energy to the system. It can also be shown that if you push the piston very slowly (reversible process) while keeping the system isolated (adiabatic transformation), the relationship between pressure and volume will be:
$$PV^{4/3} = text{constant}$$
In other words, yes, light can be compressed and will act just like any other gas inside of a cylinder. Once you push the piston, you will feel an increase in pressure (the pressure of the photon gas increases)!
This photon gas can be used to make a simple model of stars, as is discussed in The Feynman Lectures on Physics, Vol. 1. The derivation of the other results presented before can also be found in this same book.
As pointed out in Yly's answer, the increase in energy as you push the piston will cause an increase in the frequency of the radiation, essentially causing a blueshift.
6
As a note: much of this was derived from classical electrodynamics, radiation pressure, and that kind of concepts before the concept of photons.
– Pieter
Dec 1 at 10:47
1
Indeed, you are right. All of this can be derived from classical electrodynamics. But it is easier to understand the gas-like behaviour in terms of photons nonetheless.
– João Vítor G. Lima
Dec 1 at 15:17
If E sub p represents the energy of a photon, it should be E sub gamma, since gamma is conventionally used to represent photons. Also, there is a typo. You say "fell an increase" where you mean "feel an increase."
– David Conrad
Dec 1 at 17:21
Is it possible to create a space shaped by perfect mirrors into which you can shine light that then never escapes and thus accumulates - what would cause this physical system to break down?
– jeromeyers
Dec 2 at 18:59
add a comment |
up vote
21
down vote
Ideally, this is essentially the same as compressing a quantum gas of any other boson. Macroscopically, there is a pressure exerted by the photon gas on the walls of the chamber, so compressing the piston will take work and thus will increase the internal energy of the photon gas. Microscopically, by compressing the chamber, we are making the wavelengths of the supported modes shorter, and thus the frequency and energy of the photons in the chamber will increase. So either way, the internal energy of the photon gas will go up.
The exact amount by which the internal energy increases depends on how the piston is compressed, e.g. adiabatically vs. diabatically.
In the specific case where the piston is compressed adiabatically, the occupation of each mode of the chamber remains unchanged. So the light in the chamber gets "blue-shifted", but the number of photons in a given mode does not change. Summarily, the light gets bluer (higher frequency).
3
This is like Wien's displacement law. It was derived before photons: en.wikipedia.org/wiki/Wien%27s_displacement_law#Discovery
– Pieter
Dec 1 at 10:43
add a comment |
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2 Answers
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2 Answers
2
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up vote
35
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Suppose there is an amount of light (electromagnetic radiation) inside the cylinder. Note that electromagnetic radiation is composed of particles called photons, and if we consider that there is a very large number of photons inside the cylinder, we may use statistical mechanics to create a model of a photon gas. Yes, the system you describe will act like a gas, and its properties may be derived from statistics and from the properties of photons.
If a photon's frequency is $f$, its energy is $E_γ = hf$, where $h$ is Planck's constant. It is also important to remember that photons have linear momentum $$p = frac{E_γ}{c} = frac{hf}{c}$$
But the fact that photons have nonzero linear momentum implies that they will exert pressure against the cylinder's walls. Once the photon reflects on the wall, its momentum will have changed direction, and this imples that the wall has exerted a force on the photon to make it change directions. Therefore, the photon gas exerts pressure against the walls.
It can be shown that if the total energy of the photon gas is $U$, then the relationship between the pressure $P$ and the volume $V$ of the gas is $U = 3PV$.
If you push the piston, you'll do positive work and therefore give energy to the system. It can also be shown that if you push the piston very slowly (reversible process) while keeping the system isolated (adiabatic transformation), the relationship between pressure and volume will be:
$$PV^{4/3} = text{constant}$$
In other words, yes, light can be compressed and will act just like any other gas inside of a cylinder. Once you push the piston, you will feel an increase in pressure (the pressure of the photon gas increases)!
This photon gas can be used to make a simple model of stars, as is discussed in The Feynman Lectures on Physics, Vol. 1. The derivation of the other results presented before can also be found in this same book.
As pointed out in Yly's answer, the increase in energy as you push the piston will cause an increase in the frequency of the radiation, essentially causing a blueshift.
6
As a note: much of this was derived from classical electrodynamics, radiation pressure, and that kind of concepts before the concept of photons.
– Pieter
Dec 1 at 10:47
1
Indeed, you are right. All of this can be derived from classical electrodynamics. But it is easier to understand the gas-like behaviour in terms of photons nonetheless.
– João Vítor G. Lima
Dec 1 at 15:17
If E sub p represents the energy of a photon, it should be E sub gamma, since gamma is conventionally used to represent photons. Also, there is a typo. You say "fell an increase" where you mean "feel an increase."
– David Conrad
Dec 1 at 17:21
Is it possible to create a space shaped by perfect mirrors into which you can shine light that then never escapes and thus accumulates - what would cause this physical system to break down?
– jeromeyers
Dec 2 at 18:59
add a comment |
up vote
35
down vote
Suppose there is an amount of light (electromagnetic radiation) inside the cylinder. Note that electromagnetic radiation is composed of particles called photons, and if we consider that there is a very large number of photons inside the cylinder, we may use statistical mechanics to create a model of a photon gas. Yes, the system you describe will act like a gas, and its properties may be derived from statistics and from the properties of photons.
If a photon's frequency is $f$, its energy is $E_γ = hf$, where $h$ is Planck's constant. It is also important to remember that photons have linear momentum $$p = frac{E_γ}{c} = frac{hf}{c}$$
But the fact that photons have nonzero linear momentum implies that they will exert pressure against the cylinder's walls. Once the photon reflects on the wall, its momentum will have changed direction, and this imples that the wall has exerted a force on the photon to make it change directions. Therefore, the photon gas exerts pressure against the walls.
It can be shown that if the total energy of the photon gas is $U$, then the relationship between the pressure $P$ and the volume $V$ of the gas is $U = 3PV$.
If you push the piston, you'll do positive work and therefore give energy to the system. It can also be shown that if you push the piston very slowly (reversible process) while keeping the system isolated (adiabatic transformation), the relationship between pressure and volume will be:
$$PV^{4/3} = text{constant}$$
In other words, yes, light can be compressed and will act just like any other gas inside of a cylinder. Once you push the piston, you will feel an increase in pressure (the pressure of the photon gas increases)!
This photon gas can be used to make a simple model of stars, as is discussed in The Feynman Lectures on Physics, Vol. 1. The derivation of the other results presented before can also be found in this same book.
As pointed out in Yly's answer, the increase in energy as you push the piston will cause an increase in the frequency of the radiation, essentially causing a blueshift.
6
As a note: much of this was derived from classical electrodynamics, radiation pressure, and that kind of concepts before the concept of photons.
– Pieter
Dec 1 at 10:47
1
Indeed, you are right. All of this can be derived from classical electrodynamics. But it is easier to understand the gas-like behaviour in terms of photons nonetheless.
– João Vítor G. Lima
Dec 1 at 15:17
If E sub p represents the energy of a photon, it should be E sub gamma, since gamma is conventionally used to represent photons. Also, there is a typo. You say "fell an increase" where you mean "feel an increase."
– David Conrad
Dec 1 at 17:21
Is it possible to create a space shaped by perfect mirrors into which you can shine light that then never escapes and thus accumulates - what would cause this physical system to break down?
– jeromeyers
Dec 2 at 18:59
add a comment |
up vote
35
down vote
up vote
35
down vote
Suppose there is an amount of light (electromagnetic radiation) inside the cylinder. Note that electromagnetic radiation is composed of particles called photons, and if we consider that there is a very large number of photons inside the cylinder, we may use statistical mechanics to create a model of a photon gas. Yes, the system you describe will act like a gas, and its properties may be derived from statistics and from the properties of photons.
If a photon's frequency is $f$, its energy is $E_γ = hf$, where $h$ is Planck's constant. It is also important to remember that photons have linear momentum $$p = frac{E_γ}{c} = frac{hf}{c}$$
But the fact that photons have nonzero linear momentum implies that they will exert pressure against the cylinder's walls. Once the photon reflects on the wall, its momentum will have changed direction, and this imples that the wall has exerted a force on the photon to make it change directions. Therefore, the photon gas exerts pressure against the walls.
It can be shown that if the total energy of the photon gas is $U$, then the relationship between the pressure $P$ and the volume $V$ of the gas is $U = 3PV$.
If you push the piston, you'll do positive work and therefore give energy to the system. It can also be shown that if you push the piston very slowly (reversible process) while keeping the system isolated (adiabatic transformation), the relationship between pressure and volume will be:
$$PV^{4/3} = text{constant}$$
In other words, yes, light can be compressed and will act just like any other gas inside of a cylinder. Once you push the piston, you will feel an increase in pressure (the pressure of the photon gas increases)!
This photon gas can be used to make a simple model of stars, as is discussed in The Feynman Lectures on Physics, Vol. 1. The derivation of the other results presented before can also be found in this same book.
As pointed out in Yly's answer, the increase in energy as you push the piston will cause an increase in the frequency of the radiation, essentially causing a blueshift.
Suppose there is an amount of light (electromagnetic radiation) inside the cylinder. Note that electromagnetic radiation is composed of particles called photons, and if we consider that there is a very large number of photons inside the cylinder, we may use statistical mechanics to create a model of a photon gas. Yes, the system you describe will act like a gas, and its properties may be derived from statistics and from the properties of photons.
If a photon's frequency is $f$, its energy is $E_γ = hf$, where $h$ is Planck's constant. It is also important to remember that photons have linear momentum $$p = frac{E_γ}{c} = frac{hf}{c}$$
But the fact that photons have nonzero linear momentum implies that they will exert pressure against the cylinder's walls. Once the photon reflects on the wall, its momentum will have changed direction, and this imples that the wall has exerted a force on the photon to make it change directions. Therefore, the photon gas exerts pressure against the walls.
It can be shown that if the total energy of the photon gas is $U$, then the relationship between the pressure $P$ and the volume $V$ of the gas is $U = 3PV$.
If you push the piston, you'll do positive work and therefore give energy to the system. It can also be shown that if you push the piston very slowly (reversible process) while keeping the system isolated (adiabatic transformation), the relationship between pressure and volume will be:
$$PV^{4/3} = text{constant}$$
In other words, yes, light can be compressed and will act just like any other gas inside of a cylinder. Once you push the piston, you will feel an increase in pressure (the pressure of the photon gas increases)!
This photon gas can be used to make a simple model of stars, as is discussed in The Feynman Lectures on Physics, Vol. 1. The derivation of the other results presented before can also be found in this same book.
As pointed out in Yly's answer, the increase in energy as you push the piston will cause an increase in the frequency of the radiation, essentially causing a blueshift.
edited Dec 1 at 17:28
answered Dec 1 at 4:13
João Vítor G. Lima
853219
853219
6
As a note: much of this was derived from classical electrodynamics, radiation pressure, and that kind of concepts before the concept of photons.
– Pieter
Dec 1 at 10:47
1
Indeed, you are right. All of this can be derived from classical electrodynamics. But it is easier to understand the gas-like behaviour in terms of photons nonetheless.
– João Vítor G. Lima
Dec 1 at 15:17
If E sub p represents the energy of a photon, it should be E sub gamma, since gamma is conventionally used to represent photons. Also, there is a typo. You say "fell an increase" where you mean "feel an increase."
– David Conrad
Dec 1 at 17:21
Is it possible to create a space shaped by perfect mirrors into which you can shine light that then never escapes and thus accumulates - what would cause this physical system to break down?
– jeromeyers
Dec 2 at 18:59
add a comment |
6
As a note: much of this was derived from classical electrodynamics, radiation pressure, and that kind of concepts before the concept of photons.
– Pieter
Dec 1 at 10:47
1
Indeed, you are right. All of this can be derived from classical electrodynamics. But it is easier to understand the gas-like behaviour in terms of photons nonetheless.
– João Vítor G. Lima
Dec 1 at 15:17
If E sub p represents the energy of a photon, it should be E sub gamma, since gamma is conventionally used to represent photons. Also, there is a typo. You say "fell an increase" where you mean "feel an increase."
– David Conrad
Dec 1 at 17:21
Is it possible to create a space shaped by perfect mirrors into which you can shine light that then never escapes and thus accumulates - what would cause this physical system to break down?
– jeromeyers
Dec 2 at 18:59
6
6
As a note: much of this was derived from classical electrodynamics, radiation pressure, and that kind of concepts before the concept of photons.
– Pieter
Dec 1 at 10:47
As a note: much of this was derived from classical electrodynamics, radiation pressure, and that kind of concepts before the concept of photons.
– Pieter
Dec 1 at 10:47
1
1
Indeed, you are right. All of this can be derived from classical electrodynamics. But it is easier to understand the gas-like behaviour in terms of photons nonetheless.
– João Vítor G. Lima
Dec 1 at 15:17
Indeed, you are right. All of this can be derived from classical electrodynamics. But it is easier to understand the gas-like behaviour in terms of photons nonetheless.
– João Vítor G. Lima
Dec 1 at 15:17
If E sub p represents the energy of a photon, it should be E sub gamma, since gamma is conventionally used to represent photons. Also, there is a typo. You say "fell an increase" where you mean "feel an increase."
– David Conrad
Dec 1 at 17:21
If E sub p represents the energy of a photon, it should be E sub gamma, since gamma is conventionally used to represent photons. Also, there is a typo. You say "fell an increase" where you mean "feel an increase."
– David Conrad
Dec 1 at 17:21
Is it possible to create a space shaped by perfect mirrors into which you can shine light that then never escapes and thus accumulates - what would cause this physical system to break down?
– jeromeyers
Dec 2 at 18:59
Is it possible to create a space shaped by perfect mirrors into which you can shine light that then never escapes and thus accumulates - what would cause this physical system to break down?
– jeromeyers
Dec 2 at 18:59
add a comment |
up vote
21
down vote
Ideally, this is essentially the same as compressing a quantum gas of any other boson. Macroscopically, there is a pressure exerted by the photon gas on the walls of the chamber, so compressing the piston will take work and thus will increase the internal energy of the photon gas. Microscopically, by compressing the chamber, we are making the wavelengths of the supported modes shorter, and thus the frequency and energy of the photons in the chamber will increase. So either way, the internal energy of the photon gas will go up.
The exact amount by which the internal energy increases depends on how the piston is compressed, e.g. adiabatically vs. diabatically.
In the specific case where the piston is compressed adiabatically, the occupation of each mode of the chamber remains unchanged. So the light in the chamber gets "blue-shifted", but the number of photons in a given mode does not change. Summarily, the light gets bluer (higher frequency).
3
This is like Wien's displacement law. It was derived before photons: en.wikipedia.org/wiki/Wien%27s_displacement_law#Discovery
– Pieter
Dec 1 at 10:43
add a comment |
up vote
21
down vote
Ideally, this is essentially the same as compressing a quantum gas of any other boson. Macroscopically, there is a pressure exerted by the photon gas on the walls of the chamber, so compressing the piston will take work and thus will increase the internal energy of the photon gas. Microscopically, by compressing the chamber, we are making the wavelengths of the supported modes shorter, and thus the frequency and energy of the photons in the chamber will increase. So either way, the internal energy of the photon gas will go up.
The exact amount by which the internal energy increases depends on how the piston is compressed, e.g. adiabatically vs. diabatically.
In the specific case where the piston is compressed adiabatically, the occupation of each mode of the chamber remains unchanged. So the light in the chamber gets "blue-shifted", but the number of photons in a given mode does not change. Summarily, the light gets bluer (higher frequency).
3
This is like Wien's displacement law. It was derived before photons: en.wikipedia.org/wiki/Wien%27s_displacement_law#Discovery
– Pieter
Dec 1 at 10:43
add a comment |
up vote
21
down vote
up vote
21
down vote
Ideally, this is essentially the same as compressing a quantum gas of any other boson. Macroscopically, there is a pressure exerted by the photon gas on the walls of the chamber, so compressing the piston will take work and thus will increase the internal energy of the photon gas. Microscopically, by compressing the chamber, we are making the wavelengths of the supported modes shorter, and thus the frequency and energy of the photons in the chamber will increase. So either way, the internal energy of the photon gas will go up.
The exact amount by which the internal energy increases depends on how the piston is compressed, e.g. adiabatically vs. diabatically.
In the specific case where the piston is compressed adiabatically, the occupation of each mode of the chamber remains unchanged. So the light in the chamber gets "blue-shifted", but the number of photons in a given mode does not change. Summarily, the light gets bluer (higher frequency).
Ideally, this is essentially the same as compressing a quantum gas of any other boson. Macroscopically, there is a pressure exerted by the photon gas on the walls of the chamber, so compressing the piston will take work and thus will increase the internal energy of the photon gas. Microscopically, by compressing the chamber, we are making the wavelengths of the supported modes shorter, and thus the frequency and energy of the photons in the chamber will increase. So either way, the internal energy of the photon gas will go up.
The exact amount by which the internal energy increases depends on how the piston is compressed, e.g. adiabatically vs. diabatically.
In the specific case where the piston is compressed adiabatically, the occupation of each mode of the chamber remains unchanged. So the light in the chamber gets "blue-shifted", but the number of photons in a given mode does not change. Summarily, the light gets bluer (higher frequency).
answered Dec 1 at 4:06
Yly
1,261420
1,261420
3
This is like Wien's displacement law. It was derived before photons: en.wikipedia.org/wiki/Wien%27s_displacement_law#Discovery
– Pieter
Dec 1 at 10:43
add a comment |
3
This is like Wien's displacement law. It was derived before photons: en.wikipedia.org/wiki/Wien%27s_displacement_law#Discovery
– Pieter
Dec 1 at 10:43
3
3
This is like Wien's displacement law. It was derived before photons: en.wikipedia.org/wiki/Wien%27s_displacement_law#Discovery
– Pieter
Dec 1 at 10:43
This is like Wien's displacement law. It was derived before photons: en.wikipedia.org/wiki/Wien%27s_displacement_law#Discovery
– Pieter
Dec 1 at 10:43
add a comment |
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Related: What longest time ever was achieved at holding light in a closed volume?
– sumelic
Dec 1 at 4:01