How does light 'choose' between wave and particle behaviour? Announcing the arrival of Valued...
Denied boarding although I have proper visa and documentation. To whom should I make a complaint?
What would you call this weird metallic apparatus that allows you to lift people?
Do any jurisdictions seriously consider reclassifying social media websites as publishers?
Do wooden building fires get hotter than 600°C?
Do I really need to have a message in a novel to appeal to readers?
Is it a good idea to use CNN to classify 1D signal?
Is it fair for a professor to grade us on the possession of past papers?
What are the out-of-universe reasons for the references to Toby Maguire-era Spider-Man in Into the Spider-Verse?
What's the meaning of "fortified infraction restraint"?
What order were files/directories outputted in dir?
Generate an RGB colour grid
Why does the remaining Rebel fleet at the end of Rogue One seem dramatically larger than the one in A New Hope?
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
How come Sam didn't become Lord of Horn Hill?
Is grep documentation about ignoring case wrong, since it doesn't ignore case in filenames?
Can the Great Weapon Master feat's damage bonus and accuracy penalty apply to attacks from the Spiritual Weapon spell?
How do I use the new nonlinear finite element in Mathematica 12 for this equation?
An adverb for when you're not exaggerating
Sending unknown callers to voice mail automatically?
Should I use a zero-interest credit card for a large one-time purchase?
Is there any word for a place full of confusion?
Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?
Why is it faster to reheat something than it is to cook it?
Chinese Seal on silk painting - what does it mean?
How does light 'choose' between wave and particle behaviour?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Moderator Election Q&A - Question CollectionCan a single photon behave in a non-local way?Is light particle of wave?If matter and light have dual-nature, shouldn't we able to explain the observed phenomenons using either wave or particle?Have they really photographed light behaving both as a particle and a wave?Why does a real/virtual photon interact only with charged particle?How does wave-particle duality describe Photoelectric effect?Why wave theory cannot explain photoelectric effect and provides evidence for particle nature of light?Dual nature of matter and radiationHow to decide whether light shows particle of wave nature?Relationship Between Saturation Current and Light Intensity According to Wave Theory of LightQuantization and wave-particle dualism of light
$begingroup$
Light exhibits wave behaviour in phenomenon such as interference but particle behaviour in the photoelectric effect. How does light 'choose' where to be a wave and where to be a particle?
quantum-mechanics wave-particle-duality
New contributor
$endgroup$
add a comment |
$begingroup$
Light exhibits wave behaviour in phenomenon such as interference but particle behaviour in the photoelectric effect. How does light 'choose' where to be a wave and where to be a particle?
quantum-mechanics wave-particle-duality
New contributor
$endgroup$
2
$begingroup$
Light has no choice here. Photoelectric effect is a manifestly particle-like phenomenon by default. If you perform an experiment and observe photoelectric effect, we ascribe a 'particle-like' nature to the process. There is no wave-like photoelectric effect, because we defined it to refer to processes that involve particles knocking off each other.
$endgroup$
– Avantgarde
8 hours ago
3
$begingroup$
Not worth an answer but a simple analogy may help you understand some of the actual answers. Think of a cylinder, with one of its flat face on a table. If you look at it from the side, it's a rectangle. If you look at it from the top, it's a a circle. Both observations are true but the differences come from the fact that your way of measuring is not very precise and will heavily influence the way you perceive the observed object.
$endgroup$
– Echox
7 hours ago
$begingroup$
This question is briefly addressed in the last paragraph of this answer: physics.stackexchange.com/a/469623
$endgroup$
– Chiral Anomaly
2 hours ago
add a comment |
$begingroup$
Light exhibits wave behaviour in phenomenon such as interference but particle behaviour in the photoelectric effect. How does light 'choose' where to be a wave and where to be a particle?
quantum-mechanics wave-particle-duality
New contributor
$endgroup$
Light exhibits wave behaviour in phenomenon such as interference but particle behaviour in the photoelectric effect. How does light 'choose' where to be a wave and where to be a particle?
quantum-mechanics wave-particle-duality
quantum-mechanics wave-particle-duality
New contributor
New contributor
New contributor
asked 12 hours ago
d_gd_g
774
774
New contributor
New contributor
2
$begingroup$
Light has no choice here. Photoelectric effect is a manifestly particle-like phenomenon by default. If you perform an experiment and observe photoelectric effect, we ascribe a 'particle-like' nature to the process. There is no wave-like photoelectric effect, because we defined it to refer to processes that involve particles knocking off each other.
$endgroup$
– Avantgarde
8 hours ago
3
$begingroup$
Not worth an answer but a simple analogy may help you understand some of the actual answers. Think of a cylinder, with one of its flat face on a table. If you look at it from the side, it's a rectangle. If you look at it from the top, it's a a circle. Both observations are true but the differences come from the fact that your way of measuring is not very precise and will heavily influence the way you perceive the observed object.
$endgroup$
– Echox
7 hours ago
$begingroup$
This question is briefly addressed in the last paragraph of this answer: physics.stackexchange.com/a/469623
$endgroup$
– Chiral Anomaly
2 hours ago
add a comment |
2
$begingroup$
Light has no choice here. Photoelectric effect is a manifestly particle-like phenomenon by default. If you perform an experiment and observe photoelectric effect, we ascribe a 'particle-like' nature to the process. There is no wave-like photoelectric effect, because we defined it to refer to processes that involve particles knocking off each other.
$endgroup$
– Avantgarde
8 hours ago
3
$begingroup$
Not worth an answer but a simple analogy may help you understand some of the actual answers. Think of a cylinder, with one of its flat face on a table. If you look at it from the side, it's a rectangle. If you look at it from the top, it's a a circle. Both observations are true but the differences come from the fact that your way of measuring is not very precise and will heavily influence the way you perceive the observed object.
$endgroup$
– Echox
7 hours ago
$begingroup$
This question is briefly addressed in the last paragraph of this answer: physics.stackexchange.com/a/469623
$endgroup$
– Chiral Anomaly
2 hours ago
2
2
$begingroup$
Light has no choice here. Photoelectric effect is a manifestly particle-like phenomenon by default. If you perform an experiment and observe photoelectric effect, we ascribe a 'particle-like' nature to the process. There is no wave-like photoelectric effect, because we defined it to refer to processes that involve particles knocking off each other.
$endgroup$
– Avantgarde
8 hours ago
$begingroup$
Light has no choice here. Photoelectric effect is a manifestly particle-like phenomenon by default. If you perform an experiment and observe photoelectric effect, we ascribe a 'particle-like' nature to the process. There is no wave-like photoelectric effect, because we defined it to refer to processes that involve particles knocking off each other.
$endgroup$
– Avantgarde
8 hours ago
3
3
$begingroup$
Not worth an answer but a simple analogy may help you understand some of the actual answers. Think of a cylinder, with one of its flat face on a table. If you look at it from the side, it's a rectangle. If you look at it from the top, it's a a circle. Both observations are true but the differences come from the fact that your way of measuring is not very precise and will heavily influence the way you perceive the observed object.
$endgroup$
– Echox
7 hours ago
$begingroup$
Not worth an answer but a simple analogy may help you understand some of the actual answers. Think of a cylinder, with one of its flat face on a table. If you look at it from the side, it's a rectangle. If you look at it from the top, it's a a circle. Both observations are true but the differences come from the fact that your way of measuring is not very precise and will heavily influence the way you perceive the observed object.
$endgroup$
– Echox
7 hours ago
$begingroup$
This question is briefly addressed in the last paragraph of this answer: physics.stackexchange.com/a/469623
$endgroup$
– Chiral Anomaly
2 hours ago
$begingroup$
This question is briefly addressed in the last paragraph of this answer: physics.stackexchange.com/a/469623
$endgroup$
– Chiral Anomaly
2 hours ago
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
Light always behaves as a wave. Particles can be thought of as a combination of waves, a wave packet. What determines which behavior you'll get is the length scale of the system the light is interacting with and the wavelength of the impinging light.
Suppose you have a series of Bowling Balls suspended in the air making a wall, but with spaces between. Shoot a bunch of 2mm diameter ball bearings (BBs) at the bowling balls. Some will pass through the wall, hitting no bowling balls. Some will hit a ball and bounce in some direction.
Pay close attention to where the BBs come from and their initial speed and where they end up, and you will be able to tell the shape, size, and position of the bowling balls.
Reverse the problem. Have a bunch of suspended BBs in the air and shoot bowling balls at them. Each bowling ball will hit multiple BBs. The outgoing trajectory won't tell you much about the BBs.
Strike a tuning fork, it vibrates, making a characteristic sound. Play that note at high volume, you can set the tuning fork to vibrating. It doesn't start to vibrate at just any wavelength.
To a particle, you can associate the De Broglie wavelength, $h/p$, where $p$ is the momentum. The higher the momentum, the lower the wavelength, the more particle like. Whereas macroscopic but small openings can be used with light in the double slit experiment, you need electron crystallography to demonstrate the same effects with electrons: Electron diffraction
If you have a large wavelength of light, it will interact with multiple particles of a system depending on that system. If the wavelength is sufficiently small, and so the energy sufficiently high, it can interact with a single electron instead of multiple electrons, imparting all its energy to an electron and resulting in the photoelectric effect, a particle-particle scattering effect. Change the wave length and the interaction takes on a more classical form.
In addition to the wavelengths involved you want to pay attention to the number of photons available. The fewer the photons interacting with the system, the more quantum-like it is. The higher the density of photons, the more classical the light will behave. For a more detailed explanation on the barrier between quantum vs. classical behavior, see the intro and first chapter of Griffith's text on electromagnetism.
In short, the behavior you get depends on the De Broglie wavelength of the light/particle, how many particles are inbound and how the lengths scales of the target compare with the De Broglie wavelength.
$endgroup$
$begingroup$
Any mathematical account of the correspondence principle will also give insight into how classical effects emerge in a limit from quantum ones.
$endgroup$
– J.G.
8 hours ago
1
$begingroup$
@R. Romero did you mean baseballs or ball bearings when you said "BBs" in your original answer?
$endgroup$
– binaryfunt
7 hours ago
$begingroup$
Ball bearings. sorry, I should have been clear. Tiny ball bearings a few cm in diameter.
$endgroup$
– R. Romero
7 hours ago
$begingroup$
A few centimetres in diameter is rather large for ball bearings. The BBs used in a BB gun are a few millimetres in diameter.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
Meant millimeters. Missed the edit chance there
$endgroup$
– R. Romero
4 hours ago
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't, YOU do. That's really the entire "weirdness" of quantum right there.
It's not entirely crazy. If you measure a car with a scale it will tell you it's 1200 kg, and if you measure it with a spectrometer it will tell you its red. This is perfectly natural.
The thing that makes quantum weird is that you can measure the same thing twice and get two different answers. More weird, some of those measurements are linked to each other so if you measure one the other changes.
Its as if you measured the weight of your car and the length changed. And then you measured the length and the weight changed. This precise thing happens in quantum, for instance, the position and momentum of a particle are linked in this way
In any event, the wave-or-particle nature is entirely up to you. Which nature you see is based on the experiment you use, not the photon itself.
The idea that something you do has this "radical" effect on the outcome is what drives everyone mad in QM.
$endgroup$
1
$begingroup$
This could be a nice answer if you could elaborate a bit.
$endgroup$
– Gert
12 hours ago
$begingroup$
If you can choose the state of light (for eg), would that mean that theoretically we could influence light to behave as a wave and not as photons in phenomenon such as the photoelectric effect?
$endgroup$
– d_g
12 hours ago
4
$begingroup$
Ahhh but there you go... how do you measure the photoelectric effect? By measuring its energy. Presto, it looks like a particle. Take that exact same photon and put it in a double-slit? Presto, looks like a wave. But these terms, "particle" and "wave" are the real problem, you're using classical terms that simply don't describe what's really happening. You would have a similar difficulty trying to describe a car to someone who only saw horses and tried to say the wheels were sort of like legs... how often do you need new horseshoes?
$endgroup$
– Maury Markowitz
12 hours ago
$begingroup$
If you measure the weight of car with a beam scale first and then with a spring scale, you'll get the same weight twice but in the second case your car may stretch itself under its own weight. So even in "macro" the way of measuring can have an impact on the properties of the object observed.
$endgroup$
– Echox
7 hours ago
add a comment |
$begingroup$
In fact, light is not really a wave or a particle. It is what it is; it's this strange thing that we model as a wave or a particle in order to make sense of its behaviour, depending on the scenario of interest.
At the end of the day, it's the same story with all theories in physics. Planets don't "choose" to follow Newtonian mechanics or general relativity. Instead, we can model their motion as Newtonian if we want to calculate something like where Mars will be in 2 weeks, but need to use general relativity if we want to explain why the atomic clock on a satellite runs slow compared to one on the ground.
Light doesn't "choose" to be a wave or a particle. Instead, we model it as a wave when we want to explain (or calculate) interference, but need to model it as a particle when we want to explain (or calculate) the photoelectric effect.
$endgroup$
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't. It always behaves as a wave (obeying the principle of superposition), and it always behaves as a particle (particle number being quantized).
It sounds like you may have been influenced by someone who told you that light behaves like a particle in some experiments, and like a wave in others. That's false.
$endgroup$
2
$begingroup$
How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? I don't think it is fine to say it always behaves like a wave, unless you mean to say the only consistent wave behavior is superposition
$endgroup$
– Aaron Stevens
12 hours ago
$begingroup$
@AaronStevens The photoelectric effect doesn't "disprove" a wave description. If I place a guitar on a table and dust settles on one string, which I then pluck, whether I displace no specks of dust, one of them or several depends on how hard I pluck. But what caused any such displacement? A wave transmitting energy down the string. It's just that in QM, this is also explicable in terms of individual photons' absorption. This is why I recommend learning QFT, which makes better sense of how the wave/particle aspects mesh together.
$endgroup$
– J.G.
8 hours ago
$begingroup$
@AaronStevens: How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? Could you explain why you think this?
$endgroup$
– Ben Crowell
6 hours ago
$begingroup$
@BenCrowell I agree that QFT resolves this. I was talking more along the lines of classic EM waves where you would expect the intensity of the light to influence the energy of the ejected electrons and the frequency to influence the rate at which electrons are ejected, yet we see the reverse of this.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
I'm guessing this is where the OP's confusion lies. We have some phenomena where the wave picture of light is sufficient to explain what's going on, but other times it doesn't work at all. When you are first learning it all it seems rather arbitrary without any deeper structure, where you just have to remember which picture is right to use depending on the system.
$endgroup$
– Aaron Stevens
5 hours ago
add a comment |
$begingroup$
The fundamental experiment showing the apparent contradiction is Young's double slit: How can particle characteristics be transmitted when there is only an interfering wave between the point A of emission and point B of absorption?
However, for photons in vacuum (moving at c) there is a simple answer: The spacetime interval between A and B is empty, it is zero! That means that both points A and B are adjacent. A and B may be represented by mass particles (electrons etc.) which are exchanging a momentum. The transmission is direct, without need of any intermediate particle.
In contrast, a spacetime interval cannot be observed by observers. If a light ray is transmitted from Sun to Earth, nobody will see that A (Sun) and B (Earth) are adjacent. Instead, they will observe a space distance of eight light minutes and a time interval of eight minutes, even if the spacetime interval is zero. In this situation, the light wave takes the role of a sort of "placeholder": Light waves are observed to propagate at c (according to the second postulate of special relativity), but this is mere observation.
In short, the particle characteristics may be transmitted without any photon because the spacetime interval is zero. The wave characteristics (including the propagation at c) are observation only.
By the way, for light propagating at a lower speed than c (e.g. light moving through a medium), we need quantum mechanics for the answer.
$endgroup$
add a comment |
$begingroup$
Light always behaves as a particle and waves .So there is no particular time when it can behaves like a particle but not wave and vice versa .Thus light carries both of these two nature(particle and wave) along with it (light)for all time.
Actually why is my thinking is like light has two of these nature for all time?
Reasons for considering it as *wave *
- James Clerk Maxwell proved that speed of the electromagnetic waves in free space is the same as the speed of light c=2.998 ×10^8m/s. Thereby he concluded that light consists of electromagnetic waves .So it has wave nature for all time.
- Light also shows *interference *,*diffraction *,*polarization *.
And all of these property show that light has wave nature and it is for all time.
Reasons for considering light as particle
- Max plank in 1900 introduced concept of quantum of energy .The energy exchange between radiation and surroundings is in *discrete or *quantised * form i.e energy exchange between electromagnetic waves is integer multiple of (plank constant *frequency of wave ).As light consists of electromagnetic waves then light is made of discrete bits of energy i.e photons with energy (plank constant *frequency of light).
Thereby photoelectric effect ,Compton effect all show the particle nature of light.
So radiation (electromagnetic waves) exhibit particle nature conversely material particles display wave like nature introduced by de Broglie(a moving particle has wave properties associated with it).This is confirmed experimentally by Davisson and Germer they proved that interference which is a property of waves can be obtained by with material particles such as electron.
So as radiation exhibit particle like nature but particles also exhibit
wave- like behavior .Thereby light always shows two type of natures both as waves and particles .
New contributor
$endgroup$
$begingroup$
Ok if we say light is a particle and wave at all times, then why does it exhibit wave properties in certain cases and particle behaviour in certain phenomenon? Is it because it acts more like a wave and more like a particle in certain conditions? If so what are those conditions? Appreciate your answer btw
$endgroup$
– d_g
9 hours ago
$begingroup$
@d_g light as a whole shows two natures as i said. you can consider double silt experiment when light passes through the silt it behaves like a wave but when it strikes the screen it behaves like a particle .
$endgroup$
– Sneha Banerjee
8 hours ago
add a comment |
$begingroup$
The light doesn't chose. You as an experimenter chose which observable you want to measure, and thus which operator you use. Such measurement will result in a wavefunction collapsing into one of the eigenstates of this operator.
E.g. a positiin operator will give you a position, i.e. particle.
A momentum operator will give you a momentum, i.e. wavelike object.
New contributor
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
d_g is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f473566%2fhow-does-light-choose-between-wave-and-particle-behaviour%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Light always behaves as a wave. Particles can be thought of as a combination of waves, a wave packet. What determines which behavior you'll get is the length scale of the system the light is interacting with and the wavelength of the impinging light.
Suppose you have a series of Bowling Balls suspended in the air making a wall, but with spaces between. Shoot a bunch of 2mm diameter ball bearings (BBs) at the bowling balls. Some will pass through the wall, hitting no bowling balls. Some will hit a ball and bounce in some direction.
Pay close attention to where the BBs come from and their initial speed and where they end up, and you will be able to tell the shape, size, and position of the bowling balls.
Reverse the problem. Have a bunch of suspended BBs in the air and shoot bowling balls at them. Each bowling ball will hit multiple BBs. The outgoing trajectory won't tell you much about the BBs.
Strike a tuning fork, it vibrates, making a characteristic sound. Play that note at high volume, you can set the tuning fork to vibrating. It doesn't start to vibrate at just any wavelength.
To a particle, you can associate the De Broglie wavelength, $h/p$, where $p$ is the momentum. The higher the momentum, the lower the wavelength, the more particle like. Whereas macroscopic but small openings can be used with light in the double slit experiment, you need electron crystallography to demonstrate the same effects with electrons: Electron diffraction
If you have a large wavelength of light, it will interact with multiple particles of a system depending on that system. If the wavelength is sufficiently small, and so the energy sufficiently high, it can interact with a single electron instead of multiple electrons, imparting all its energy to an electron and resulting in the photoelectric effect, a particle-particle scattering effect. Change the wave length and the interaction takes on a more classical form.
In addition to the wavelengths involved you want to pay attention to the number of photons available. The fewer the photons interacting with the system, the more quantum-like it is. The higher the density of photons, the more classical the light will behave. For a more detailed explanation on the barrier between quantum vs. classical behavior, see the intro and first chapter of Griffith's text on electromagnetism.
In short, the behavior you get depends on the De Broglie wavelength of the light/particle, how many particles are inbound and how the lengths scales of the target compare with the De Broglie wavelength.
$endgroup$
$begingroup$
Any mathematical account of the correspondence principle will also give insight into how classical effects emerge in a limit from quantum ones.
$endgroup$
– J.G.
8 hours ago
1
$begingroup$
@R. Romero did you mean baseballs or ball bearings when you said "BBs" in your original answer?
$endgroup$
– binaryfunt
7 hours ago
$begingroup$
Ball bearings. sorry, I should have been clear. Tiny ball bearings a few cm in diameter.
$endgroup$
– R. Romero
7 hours ago
$begingroup$
A few centimetres in diameter is rather large for ball bearings. The BBs used in a BB gun are a few millimetres in diameter.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
Meant millimeters. Missed the edit chance there
$endgroup$
– R. Romero
4 hours ago
add a comment |
$begingroup$
Light always behaves as a wave. Particles can be thought of as a combination of waves, a wave packet. What determines which behavior you'll get is the length scale of the system the light is interacting with and the wavelength of the impinging light.
Suppose you have a series of Bowling Balls suspended in the air making a wall, but with spaces between. Shoot a bunch of 2mm diameter ball bearings (BBs) at the bowling balls. Some will pass through the wall, hitting no bowling balls. Some will hit a ball and bounce in some direction.
Pay close attention to where the BBs come from and their initial speed and where they end up, and you will be able to tell the shape, size, and position of the bowling balls.
Reverse the problem. Have a bunch of suspended BBs in the air and shoot bowling balls at them. Each bowling ball will hit multiple BBs. The outgoing trajectory won't tell you much about the BBs.
Strike a tuning fork, it vibrates, making a characteristic sound. Play that note at high volume, you can set the tuning fork to vibrating. It doesn't start to vibrate at just any wavelength.
To a particle, you can associate the De Broglie wavelength, $h/p$, where $p$ is the momentum. The higher the momentum, the lower the wavelength, the more particle like. Whereas macroscopic but small openings can be used with light in the double slit experiment, you need electron crystallography to demonstrate the same effects with electrons: Electron diffraction
If you have a large wavelength of light, it will interact with multiple particles of a system depending on that system. If the wavelength is sufficiently small, and so the energy sufficiently high, it can interact with a single electron instead of multiple electrons, imparting all its energy to an electron and resulting in the photoelectric effect, a particle-particle scattering effect. Change the wave length and the interaction takes on a more classical form.
In addition to the wavelengths involved you want to pay attention to the number of photons available. The fewer the photons interacting with the system, the more quantum-like it is. The higher the density of photons, the more classical the light will behave. For a more detailed explanation on the barrier between quantum vs. classical behavior, see the intro and first chapter of Griffith's text on electromagnetism.
In short, the behavior you get depends on the De Broglie wavelength of the light/particle, how many particles are inbound and how the lengths scales of the target compare with the De Broglie wavelength.
$endgroup$
$begingroup$
Any mathematical account of the correspondence principle will also give insight into how classical effects emerge in a limit from quantum ones.
$endgroup$
– J.G.
8 hours ago
1
$begingroup$
@R. Romero did you mean baseballs or ball bearings when you said "BBs" in your original answer?
$endgroup$
– binaryfunt
7 hours ago
$begingroup$
Ball bearings. sorry, I should have been clear. Tiny ball bearings a few cm in diameter.
$endgroup$
– R. Romero
7 hours ago
$begingroup$
A few centimetres in diameter is rather large for ball bearings. The BBs used in a BB gun are a few millimetres in diameter.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
Meant millimeters. Missed the edit chance there
$endgroup$
– R. Romero
4 hours ago
add a comment |
$begingroup$
Light always behaves as a wave. Particles can be thought of as a combination of waves, a wave packet. What determines which behavior you'll get is the length scale of the system the light is interacting with and the wavelength of the impinging light.
Suppose you have a series of Bowling Balls suspended in the air making a wall, but with spaces between. Shoot a bunch of 2mm diameter ball bearings (BBs) at the bowling balls. Some will pass through the wall, hitting no bowling balls. Some will hit a ball and bounce in some direction.
Pay close attention to where the BBs come from and their initial speed and where they end up, and you will be able to tell the shape, size, and position of the bowling balls.
Reverse the problem. Have a bunch of suspended BBs in the air and shoot bowling balls at them. Each bowling ball will hit multiple BBs. The outgoing trajectory won't tell you much about the BBs.
Strike a tuning fork, it vibrates, making a characteristic sound. Play that note at high volume, you can set the tuning fork to vibrating. It doesn't start to vibrate at just any wavelength.
To a particle, you can associate the De Broglie wavelength, $h/p$, where $p$ is the momentum. The higher the momentum, the lower the wavelength, the more particle like. Whereas macroscopic but small openings can be used with light in the double slit experiment, you need electron crystallography to demonstrate the same effects with electrons: Electron diffraction
If you have a large wavelength of light, it will interact with multiple particles of a system depending on that system. If the wavelength is sufficiently small, and so the energy sufficiently high, it can interact with a single electron instead of multiple electrons, imparting all its energy to an electron and resulting in the photoelectric effect, a particle-particle scattering effect. Change the wave length and the interaction takes on a more classical form.
In addition to the wavelengths involved you want to pay attention to the number of photons available. The fewer the photons interacting with the system, the more quantum-like it is. The higher the density of photons, the more classical the light will behave. For a more detailed explanation on the barrier between quantum vs. classical behavior, see the intro and first chapter of Griffith's text on electromagnetism.
In short, the behavior you get depends on the De Broglie wavelength of the light/particle, how many particles are inbound and how the lengths scales of the target compare with the De Broglie wavelength.
$endgroup$
Light always behaves as a wave. Particles can be thought of as a combination of waves, a wave packet. What determines which behavior you'll get is the length scale of the system the light is interacting with and the wavelength of the impinging light.
Suppose you have a series of Bowling Balls suspended in the air making a wall, but with spaces between. Shoot a bunch of 2mm diameter ball bearings (BBs) at the bowling balls. Some will pass through the wall, hitting no bowling balls. Some will hit a ball and bounce in some direction.
Pay close attention to where the BBs come from and their initial speed and where they end up, and you will be able to tell the shape, size, and position of the bowling balls.
Reverse the problem. Have a bunch of suspended BBs in the air and shoot bowling balls at them. Each bowling ball will hit multiple BBs. The outgoing trajectory won't tell you much about the BBs.
Strike a tuning fork, it vibrates, making a characteristic sound. Play that note at high volume, you can set the tuning fork to vibrating. It doesn't start to vibrate at just any wavelength.
To a particle, you can associate the De Broglie wavelength, $h/p$, where $p$ is the momentum. The higher the momentum, the lower the wavelength, the more particle like. Whereas macroscopic but small openings can be used with light in the double slit experiment, you need electron crystallography to demonstrate the same effects with electrons: Electron diffraction
If you have a large wavelength of light, it will interact with multiple particles of a system depending on that system. If the wavelength is sufficiently small, and so the energy sufficiently high, it can interact with a single electron instead of multiple electrons, imparting all its energy to an electron and resulting in the photoelectric effect, a particle-particle scattering effect. Change the wave length and the interaction takes on a more classical form.
In addition to the wavelengths involved you want to pay attention to the number of photons available. The fewer the photons interacting with the system, the more quantum-like it is. The higher the density of photons, the more classical the light will behave. For a more detailed explanation on the barrier between quantum vs. classical behavior, see the intro and first chapter of Griffith's text on electromagnetism.
In short, the behavior you get depends on the De Broglie wavelength of the light/particle, how many particles are inbound and how the lengths scales of the target compare with the De Broglie wavelength.
edited 7 hours ago
answered 12 hours ago
R. RomeroR. Romero
6039
6039
$begingroup$
Any mathematical account of the correspondence principle will also give insight into how classical effects emerge in a limit from quantum ones.
$endgroup$
– J.G.
8 hours ago
1
$begingroup$
@R. Romero did you mean baseballs or ball bearings when you said "BBs" in your original answer?
$endgroup$
– binaryfunt
7 hours ago
$begingroup$
Ball bearings. sorry, I should have been clear. Tiny ball bearings a few cm in diameter.
$endgroup$
– R. Romero
7 hours ago
$begingroup$
A few centimetres in diameter is rather large for ball bearings. The BBs used in a BB gun are a few millimetres in diameter.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
Meant millimeters. Missed the edit chance there
$endgroup$
– R. Romero
4 hours ago
add a comment |
$begingroup$
Any mathematical account of the correspondence principle will also give insight into how classical effects emerge in a limit from quantum ones.
$endgroup$
– J.G.
8 hours ago
1
$begingroup$
@R. Romero did you mean baseballs or ball bearings when you said "BBs" in your original answer?
$endgroup$
– binaryfunt
7 hours ago
$begingroup$
Ball bearings. sorry, I should have been clear. Tiny ball bearings a few cm in diameter.
$endgroup$
– R. Romero
7 hours ago
$begingroup$
A few centimetres in diameter is rather large for ball bearings. The BBs used in a BB gun are a few millimetres in diameter.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
Meant millimeters. Missed the edit chance there
$endgroup$
– R. Romero
4 hours ago
$begingroup$
Any mathematical account of the correspondence principle will also give insight into how classical effects emerge in a limit from quantum ones.
$endgroup$
– J.G.
8 hours ago
$begingroup$
Any mathematical account of the correspondence principle will also give insight into how classical effects emerge in a limit from quantum ones.
$endgroup$
– J.G.
8 hours ago
1
1
$begingroup$
@R. Romero did you mean baseballs or ball bearings when you said "BBs" in your original answer?
$endgroup$
– binaryfunt
7 hours ago
$begingroup$
@R. Romero did you mean baseballs or ball bearings when you said "BBs" in your original answer?
$endgroup$
– binaryfunt
7 hours ago
$begingroup$
Ball bearings. sorry, I should have been clear. Tiny ball bearings a few cm in diameter.
$endgroup$
– R. Romero
7 hours ago
$begingroup$
Ball bearings. sorry, I should have been clear. Tiny ball bearings a few cm in diameter.
$endgroup$
– R. Romero
7 hours ago
$begingroup$
A few centimetres in diameter is rather large for ball bearings. The BBs used in a BB gun are a few millimetres in diameter.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
A few centimetres in diameter is rather large for ball bearings. The BBs used in a BB gun are a few millimetres in diameter.
$endgroup$
– PM 2Ring
4 hours ago
$begingroup$
Meant millimeters. Missed the edit chance there
$endgroup$
– R. Romero
4 hours ago
$begingroup$
Meant millimeters. Missed the edit chance there
$endgroup$
– R. Romero
4 hours ago
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't, YOU do. That's really the entire "weirdness" of quantum right there.
It's not entirely crazy. If you measure a car with a scale it will tell you it's 1200 kg, and if you measure it with a spectrometer it will tell you its red. This is perfectly natural.
The thing that makes quantum weird is that you can measure the same thing twice and get two different answers. More weird, some of those measurements are linked to each other so if you measure one the other changes.
Its as if you measured the weight of your car and the length changed. And then you measured the length and the weight changed. This precise thing happens in quantum, for instance, the position and momentum of a particle are linked in this way
In any event, the wave-or-particle nature is entirely up to you. Which nature you see is based on the experiment you use, not the photon itself.
The idea that something you do has this "radical" effect on the outcome is what drives everyone mad in QM.
$endgroup$
1
$begingroup$
This could be a nice answer if you could elaborate a bit.
$endgroup$
– Gert
12 hours ago
$begingroup$
If you can choose the state of light (for eg), would that mean that theoretically we could influence light to behave as a wave and not as photons in phenomenon such as the photoelectric effect?
$endgroup$
– d_g
12 hours ago
4
$begingroup$
Ahhh but there you go... how do you measure the photoelectric effect? By measuring its energy. Presto, it looks like a particle. Take that exact same photon and put it in a double-slit? Presto, looks like a wave. But these terms, "particle" and "wave" are the real problem, you're using classical terms that simply don't describe what's really happening. You would have a similar difficulty trying to describe a car to someone who only saw horses and tried to say the wheels were sort of like legs... how often do you need new horseshoes?
$endgroup$
– Maury Markowitz
12 hours ago
$begingroup$
If you measure the weight of car with a beam scale first and then with a spring scale, you'll get the same weight twice but in the second case your car may stretch itself under its own weight. So even in "macro" the way of measuring can have an impact on the properties of the object observed.
$endgroup$
– Echox
7 hours ago
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't, YOU do. That's really the entire "weirdness" of quantum right there.
It's not entirely crazy. If you measure a car with a scale it will tell you it's 1200 kg, and if you measure it with a spectrometer it will tell you its red. This is perfectly natural.
The thing that makes quantum weird is that you can measure the same thing twice and get two different answers. More weird, some of those measurements are linked to each other so if you measure one the other changes.
Its as if you measured the weight of your car and the length changed. And then you measured the length and the weight changed. This precise thing happens in quantum, for instance, the position and momentum of a particle are linked in this way
In any event, the wave-or-particle nature is entirely up to you. Which nature you see is based on the experiment you use, not the photon itself.
The idea that something you do has this "radical" effect on the outcome is what drives everyone mad in QM.
$endgroup$
1
$begingroup$
This could be a nice answer if you could elaborate a bit.
$endgroup$
– Gert
12 hours ago
$begingroup$
If you can choose the state of light (for eg), would that mean that theoretically we could influence light to behave as a wave and not as photons in phenomenon such as the photoelectric effect?
$endgroup$
– d_g
12 hours ago
4
$begingroup$
Ahhh but there you go... how do you measure the photoelectric effect? By measuring its energy. Presto, it looks like a particle. Take that exact same photon and put it in a double-slit? Presto, looks like a wave. But these terms, "particle" and "wave" are the real problem, you're using classical terms that simply don't describe what's really happening. You would have a similar difficulty trying to describe a car to someone who only saw horses and tried to say the wheels were sort of like legs... how often do you need new horseshoes?
$endgroup$
– Maury Markowitz
12 hours ago
$begingroup$
If you measure the weight of car with a beam scale first and then with a spring scale, you'll get the same weight twice but in the second case your car may stretch itself under its own weight. So even in "macro" the way of measuring can have an impact on the properties of the object observed.
$endgroup$
– Echox
7 hours ago
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't, YOU do. That's really the entire "weirdness" of quantum right there.
It's not entirely crazy. If you measure a car with a scale it will tell you it's 1200 kg, and if you measure it with a spectrometer it will tell you its red. This is perfectly natural.
The thing that makes quantum weird is that you can measure the same thing twice and get two different answers. More weird, some of those measurements are linked to each other so if you measure one the other changes.
Its as if you measured the weight of your car and the length changed. And then you measured the length and the weight changed. This precise thing happens in quantum, for instance, the position and momentum of a particle are linked in this way
In any event, the wave-or-particle nature is entirely up to you. Which nature you see is based on the experiment you use, not the photon itself.
The idea that something you do has this "radical" effect on the outcome is what drives everyone mad in QM.
$endgroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't, YOU do. That's really the entire "weirdness" of quantum right there.
It's not entirely crazy. If you measure a car with a scale it will tell you it's 1200 kg, and if you measure it with a spectrometer it will tell you its red. This is perfectly natural.
The thing that makes quantum weird is that you can measure the same thing twice and get two different answers. More weird, some of those measurements are linked to each other so if you measure one the other changes.
Its as if you measured the weight of your car and the length changed. And then you measured the length and the weight changed. This precise thing happens in quantum, for instance, the position and momentum of a particle are linked in this way
In any event, the wave-or-particle nature is entirely up to you. Which nature you see is based on the experiment you use, not the photon itself.
The idea that something you do has this "radical" effect on the outcome is what drives everyone mad in QM.
edited 12 hours ago
answered 12 hours ago
Maury MarkowitzMaury Markowitz
4,240625
4,240625
1
$begingroup$
This could be a nice answer if you could elaborate a bit.
$endgroup$
– Gert
12 hours ago
$begingroup$
If you can choose the state of light (for eg), would that mean that theoretically we could influence light to behave as a wave and not as photons in phenomenon such as the photoelectric effect?
$endgroup$
– d_g
12 hours ago
4
$begingroup$
Ahhh but there you go... how do you measure the photoelectric effect? By measuring its energy. Presto, it looks like a particle. Take that exact same photon and put it in a double-slit? Presto, looks like a wave. But these terms, "particle" and "wave" are the real problem, you're using classical terms that simply don't describe what's really happening. You would have a similar difficulty trying to describe a car to someone who only saw horses and tried to say the wheels were sort of like legs... how often do you need new horseshoes?
$endgroup$
– Maury Markowitz
12 hours ago
$begingroup$
If you measure the weight of car with a beam scale first and then with a spring scale, you'll get the same weight twice but in the second case your car may stretch itself under its own weight. So even in "macro" the way of measuring can have an impact on the properties of the object observed.
$endgroup$
– Echox
7 hours ago
add a comment |
1
$begingroup$
This could be a nice answer if you could elaborate a bit.
$endgroup$
– Gert
12 hours ago
$begingroup$
If you can choose the state of light (for eg), would that mean that theoretically we could influence light to behave as a wave and not as photons in phenomenon such as the photoelectric effect?
$endgroup$
– d_g
12 hours ago
4
$begingroup$
Ahhh but there you go... how do you measure the photoelectric effect? By measuring its energy. Presto, it looks like a particle. Take that exact same photon and put it in a double-slit? Presto, looks like a wave. But these terms, "particle" and "wave" are the real problem, you're using classical terms that simply don't describe what's really happening. You would have a similar difficulty trying to describe a car to someone who only saw horses and tried to say the wheels were sort of like legs... how often do you need new horseshoes?
$endgroup$
– Maury Markowitz
12 hours ago
$begingroup$
If you measure the weight of car with a beam scale first and then with a spring scale, you'll get the same weight twice but in the second case your car may stretch itself under its own weight. So even in "macro" the way of measuring can have an impact on the properties of the object observed.
$endgroup$
– Echox
7 hours ago
1
1
$begingroup$
This could be a nice answer if you could elaborate a bit.
$endgroup$
– Gert
12 hours ago
$begingroup$
This could be a nice answer if you could elaborate a bit.
$endgroup$
– Gert
12 hours ago
$begingroup$
If you can choose the state of light (for eg), would that mean that theoretically we could influence light to behave as a wave and not as photons in phenomenon such as the photoelectric effect?
$endgroup$
– d_g
12 hours ago
$begingroup$
If you can choose the state of light (for eg), would that mean that theoretically we could influence light to behave as a wave and not as photons in phenomenon such as the photoelectric effect?
$endgroup$
– d_g
12 hours ago
4
4
$begingroup$
Ahhh but there you go... how do you measure the photoelectric effect? By measuring its energy. Presto, it looks like a particle. Take that exact same photon and put it in a double-slit? Presto, looks like a wave. But these terms, "particle" and "wave" are the real problem, you're using classical terms that simply don't describe what's really happening. You would have a similar difficulty trying to describe a car to someone who only saw horses and tried to say the wheels were sort of like legs... how often do you need new horseshoes?
$endgroup$
– Maury Markowitz
12 hours ago
$begingroup$
Ahhh but there you go... how do you measure the photoelectric effect? By measuring its energy. Presto, it looks like a particle. Take that exact same photon and put it in a double-slit? Presto, looks like a wave. But these terms, "particle" and "wave" are the real problem, you're using classical terms that simply don't describe what's really happening. You would have a similar difficulty trying to describe a car to someone who only saw horses and tried to say the wheels were sort of like legs... how often do you need new horseshoes?
$endgroup$
– Maury Markowitz
12 hours ago
$begingroup$
If you measure the weight of car with a beam scale first and then with a spring scale, you'll get the same weight twice but in the second case your car may stretch itself under its own weight. So even in "macro" the way of measuring can have an impact on the properties of the object observed.
$endgroup$
– Echox
7 hours ago
$begingroup$
If you measure the weight of car with a beam scale first and then with a spring scale, you'll get the same weight twice but in the second case your car may stretch itself under its own weight. So even in "macro" the way of measuring can have an impact on the properties of the object observed.
$endgroup$
– Echox
7 hours ago
add a comment |
$begingroup$
In fact, light is not really a wave or a particle. It is what it is; it's this strange thing that we model as a wave or a particle in order to make sense of its behaviour, depending on the scenario of interest.
At the end of the day, it's the same story with all theories in physics. Planets don't "choose" to follow Newtonian mechanics or general relativity. Instead, we can model their motion as Newtonian if we want to calculate something like where Mars will be in 2 weeks, but need to use general relativity if we want to explain why the atomic clock on a satellite runs slow compared to one on the ground.
Light doesn't "choose" to be a wave or a particle. Instead, we model it as a wave when we want to explain (or calculate) interference, but need to model it as a particle when we want to explain (or calculate) the photoelectric effect.
$endgroup$
add a comment |
$begingroup$
In fact, light is not really a wave or a particle. It is what it is; it's this strange thing that we model as a wave or a particle in order to make sense of its behaviour, depending on the scenario of interest.
At the end of the day, it's the same story with all theories in physics. Planets don't "choose" to follow Newtonian mechanics or general relativity. Instead, we can model their motion as Newtonian if we want to calculate something like where Mars will be in 2 weeks, but need to use general relativity if we want to explain why the atomic clock on a satellite runs slow compared to one on the ground.
Light doesn't "choose" to be a wave or a particle. Instead, we model it as a wave when we want to explain (or calculate) interference, but need to model it as a particle when we want to explain (or calculate) the photoelectric effect.
$endgroup$
add a comment |
$begingroup$
In fact, light is not really a wave or a particle. It is what it is; it's this strange thing that we model as a wave or a particle in order to make sense of its behaviour, depending on the scenario of interest.
At the end of the day, it's the same story with all theories in physics. Planets don't "choose" to follow Newtonian mechanics or general relativity. Instead, we can model their motion as Newtonian if we want to calculate something like where Mars will be in 2 weeks, but need to use general relativity if we want to explain why the atomic clock on a satellite runs slow compared to one on the ground.
Light doesn't "choose" to be a wave or a particle. Instead, we model it as a wave when we want to explain (or calculate) interference, but need to model it as a particle when we want to explain (or calculate) the photoelectric effect.
$endgroup$
In fact, light is not really a wave or a particle. It is what it is; it's this strange thing that we model as a wave or a particle in order to make sense of its behaviour, depending on the scenario of interest.
At the end of the day, it's the same story with all theories in physics. Planets don't "choose" to follow Newtonian mechanics or general relativity. Instead, we can model their motion as Newtonian if we want to calculate something like where Mars will be in 2 weeks, but need to use general relativity if we want to explain why the atomic clock on a satellite runs slow compared to one on the ground.
Light doesn't "choose" to be a wave or a particle. Instead, we model it as a wave when we want to explain (or calculate) interference, but need to model it as a particle when we want to explain (or calculate) the photoelectric effect.
edited 8 hours ago
answered 8 hours ago
binaryfuntbinaryfunt
5551520
5551520
add a comment |
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't. It always behaves as a wave (obeying the principle of superposition), and it always behaves as a particle (particle number being quantized).
It sounds like you may have been influenced by someone who told you that light behaves like a particle in some experiments, and like a wave in others. That's false.
$endgroup$
2
$begingroup$
How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? I don't think it is fine to say it always behaves like a wave, unless you mean to say the only consistent wave behavior is superposition
$endgroup$
– Aaron Stevens
12 hours ago
$begingroup$
@AaronStevens The photoelectric effect doesn't "disprove" a wave description. If I place a guitar on a table and dust settles on one string, which I then pluck, whether I displace no specks of dust, one of them or several depends on how hard I pluck. But what caused any such displacement? A wave transmitting energy down the string. It's just that in QM, this is also explicable in terms of individual photons' absorption. This is why I recommend learning QFT, which makes better sense of how the wave/particle aspects mesh together.
$endgroup$
– J.G.
8 hours ago
$begingroup$
@AaronStevens: How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? Could you explain why you think this?
$endgroup$
– Ben Crowell
6 hours ago
$begingroup$
@BenCrowell I agree that QFT resolves this. I was talking more along the lines of classic EM waves where you would expect the intensity of the light to influence the energy of the ejected electrons and the frequency to influence the rate at which electrons are ejected, yet we see the reverse of this.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
I'm guessing this is where the OP's confusion lies. We have some phenomena where the wave picture of light is sufficient to explain what's going on, but other times it doesn't work at all. When you are first learning it all it seems rather arbitrary without any deeper structure, where you just have to remember which picture is right to use depending on the system.
$endgroup$
– Aaron Stevens
5 hours ago
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't. It always behaves as a wave (obeying the principle of superposition), and it always behaves as a particle (particle number being quantized).
It sounds like you may have been influenced by someone who told you that light behaves like a particle in some experiments, and like a wave in others. That's false.
$endgroup$
2
$begingroup$
How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? I don't think it is fine to say it always behaves like a wave, unless you mean to say the only consistent wave behavior is superposition
$endgroup$
– Aaron Stevens
12 hours ago
$begingroup$
@AaronStevens The photoelectric effect doesn't "disprove" a wave description. If I place a guitar on a table and dust settles on one string, which I then pluck, whether I displace no specks of dust, one of them or several depends on how hard I pluck. But what caused any such displacement? A wave transmitting energy down the string. It's just that in QM, this is also explicable in terms of individual photons' absorption. This is why I recommend learning QFT, which makes better sense of how the wave/particle aspects mesh together.
$endgroup$
– J.G.
8 hours ago
$begingroup$
@AaronStevens: How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? Could you explain why you think this?
$endgroup$
– Ben Crowell
6 hours ago
$begingroup$
@BenCrowell I agree that QFT resolves this. I was talking more along the lines of classic EM waves where you would expect the intensity of the light to influence the energy of the ejected electrons and the frequency to influence the rate at which electrons are ejected, yet we see the reverse of this.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
I'm guessing this is where the OP's confusion lies. We have some phenomena where the wave picture of light is sufficient to explain what's going on, but other times it doesn't work at all. When you are first learning it all it seems rather arbitrary without any deeper structure, where you just have to remember which picture is right to use depending on the system.
$endgroup$
– Aaron Stevens
5 hours ago
add a comment |
$begingroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't. It always behaves as a wave (obeying the principle of superposition), and it always behaves as a particle (particle number being quantized).
It sounds like you may have been influenced by someone who told you that light behaves like a particle in some experiments, and like a wave in others. That's false.
$endgroup$
How does light 'choose' where to be a wave and where to be a particle?
It doesn't. It always behaves as a wave (obeying the principle of superposition), and it always behaves as a particle (particle number being quantized).
It sounds like you may have been influenced by someone who told you that light behaves like a particle in some experiments, and like a wave in others. That's false.
answered 12 hours ago
Ben CrowellBen Crowell
54.6k6165316
54.6k6165316
2
$begingroup$
How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? I don't think it is fine to say it always behaves like a wave, unless you mean to say the only consistent wave behavior is superposition
$endgroup$
– Aaron Stevens
12 hours ago
$begingroup$
@AaronStevens The photoelectric effect doesn't "disprove" a wave description. If I place a guitar on a table and dust settles on one string, which I then pluck, whether I displace no specks of dust, one of them or several depends on how hard I pluck. But what caused any such displacement? A wave transmitting energy down the string. It's just that in QM, this is also explicable in terms of individual photons' absorption. This is why I recommend learning QFT, which makes better sense of how the wave/particle aspects mesh together.
$endgroup$
– J.G.
8 hours ago
$begingroup$
@AaronStevens: How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? Could you explain why you think this?
$endgroup$
– Ben Crowell
6 hours ago
$begingroup$
@BenCrowell I agree that QFT resolves this. I was talking more along the lines of classic EM waves where you would expect the intensity of the light to influence the energy of the ejected electrons and the frequency to influence the rate at which electrons are ejected, yet we see the reverse of this.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
I'm guessing this is where the OP's confusion lies. We have some phenomena where the wave picture of light is sufficient to explain what's going on, but other times it doesn't work at all. When you are first learning it all it seems rather arbitrary without any deeper structure, where you just have to remember which picture is right to use depending on the system.
$endgroup$
– Aaron Stevens
5 hours ago
add a comment |
2
$begingroup$
How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? I don't think it is fine to say it always behaves like a wave, unless you mean to say the only consistent wave behavior is superposition
$endgroup$
– Aaron Stevens
12 hours ago
$begingroup$
@AaronStevens The photoelectric effect doesn't "disprove" a wave description. If I place a guitar on a table and dust settles on one string, which I then pluck, whether I displace no specks of dust, one of them or several depends on how hard I pluck. But what caused any such displacement? A wave transmitting energy down the string. It's just that in QM, this is also explicable in terms of individual photons' absorption. This is why I recommend learning QFT, which makes better sense of how the wave/particle aspects mesh together.
$endgroup$
– J.G.
8 hours ago
$begingroup$
@AaronStevens: How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? Could you explain why you think this?
$endgroup$
– Ben Crowell
6 hours ago
$begingroup$
@BenCrowell I agree that QFT resolves this. I was talking more along the lines of classic EM waves where you would expect the intensity of the light to influence the energy of the ejected electrons and the frequency to influence the rate at which electrons are ejected, yet we see the reverse of this.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
I'm guessing this is where the OP's confusion lies. We have some phenomena where the wave picture of light is sufficient to explain what's going on, but other times it doesn't work at all. When you are first learning it all it seems rather arbitrary without any deeper structure, where you just have to remember which picture is right to use depending on the system.
$endgroup$
– Aaron Stevens
5 hours ago
2
2
$begingroup$
How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? I don't think it is fine to say it always behaves like a wave, unless you mean to say the only consistent wave behavior is superposition
$endgroup$
– Aaron Stevens
12 hours ago
$begingroup$
How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? I don't think it is fine to say it always behaves like a wave, unless you mean to say the only consistent wave behavior is superposition
$endgroup$
– Aaron Stevens
12 hours ago
$begingroup$
@AaronStevens The photoelectric effect doesn't "disprove" a wave description. If I place a guitar on a table and dust settles on one string, which I then pluck, whether I displace no specks of dust, one of them or several depends on how hard I pluck. But what caused any such displacement? A wave transmitting energy down the string. It's just that in QM, this is also explicable in terms of individual photons' absorption. This is why I recommend learning QFT, which makes better sense of how the wave/particle aspects mesh together.
$endgroup$
– J.G.
8 hours ago
$begingroup$
@AaronStevens The photoelectric effect doesn't "disprove" a wave description. If I place a guitar on a table and dust settles on one string, which I then pluck, whether I displace no specks of dust, one of them or several depends on how hard I pluck. But what caused any such displacement? A wave transmitting energy down the string. It's just that in QM, this is also explicable in terms of individual photons' absorption. This is why I recommend learning QFT, which makes better sense of how the wave/particle aspects mesh together.
$endgroup$
– J.G.
8 hours ago
$begingroup$
@AaronStevens: How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? Could you explain why you think this?
$endgroup$
– Ben Crowell
6 hours ago
$begingroup$
@AaronStevens: How do you address the fact that the photoelectric effect doesn't make sense if you consider light as a wave though? Could you explain why you think this?
$endgroup$
– Ben Crowell
6 hours ago
$begingroup$
@BenCrowell I agree that QFT resolves this. I was talking more along the lines of classic EM waves where you would expect the intensity of the light to influence the energy of the ejected electrons and the frequency to influence the rate at which electrons are ejected, yet we see the reverse of this.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
@BenCrowell I agree that QFT resolves this. I was talking more along the lines of classic EM waves where you would expect the intensity of the light to influence the energy of the ejected electrons and the frequency to influence the rate at which electrons are ejected, yet we see the reverse of this.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
I'm guessing this is where the OP's confusion lies. We have some phenomena where the wave picture of light is sufficient to explain what's going on, but other times it doesn't work at all. When you are first learning it all it seems rather arbitrary without any deeper structure, where you just have to remember which picture is right to use depending on the system.
$endgroup$
– Aaron Stevens
5 hours ago
$begingroup$
I'm guessing this is where the OP's confusion lies. We have some phenomena where the wave picture of light is sufficient to explain what's going on, but other times it doesn't work at all. When you are first learning it all it seems rather arbitrary without any deeper structure, where you just have to remember which picture is right to use depending on the system.
$endgroup$
– Aaron Stevens
5 hours ago
add a comment |
$begingroup$
The fundamental experiment showing the apparent contradiction is Young's double slit: How can particle characteristics be transmitted when there is only an interfering wave between the point A of emission and point B of absorption?
However, for photons in vacuum (moving at c) there is a simple answer: The spacetime interval between A and B is empty, it is zero! That means that both points A and B are adjacent. A and B may be represented by mass particles (electrons etc.) which are exchanging a momentum. The transmission is direct, without need of any intermediate particle.
In contrast, a spacetime interval cannot be observed by observers. If a light ray is transmitted from Sun to Earth, nobody will see that A (Sun) and B (Earth) are adjacent. Instead, they will observe a space distance of eight light minutes and a time interval of eight minutes, even if the spacetime interval is zero. In this situation, the light wave takes the role of a sort of "placeholder": Light waves are observed to propagate at c (according to the second postulate of special relativity), but this is mere observation.
In short, the particle characteristics may be transmitted without any photon because the spacetime interval is zero. The wave characteristics (including the propagation at c) are observation only.
By the way, for light propagating at a lower speed than c (e.g. light moving through a medium), we need quantum mechanics for the answer.
$endgroup$
add a comment |
$begingroup$
The fundamental experiment showing the apparent contradiction is Young's double slit: How can particle characteristics be transmitted when there is only an interfering wave between the point A of emission and point B of absorption?
However, for photons in vacuum (moving at c) there is a simple answer: The spacetime interval between A and B is empty, it is zero! That means that both points A and B are adjacent. A and B may be represented by mass particles (electrons etc.) which are exchanging a momentum. The transmission is direct, without need of any intermediate particle.
In contrast, a spacetime interval cannot be observed by observers. If a light ray is transmitted from Sun to Earth, nobody will see that A (Sun) and B (Earth) are adjacent. Instead, they will observe a space distance of eight light minutes and a time interval of eight minutes, even if the spacetime interval is zero. In this situation, the light wave takes the role of a sort of "placeholder": Light waves are observed to propagate at c (according to the second postulate of special relativity), but this is mere observation.
In short, the particle characteristics may be transmitted without any photon because the spacetime interval is zero. The wave characteristics (including the propagation at c) are observation only.
By the way, for light propagating at a lower speed than c (e.g. light moving through a medium), we need quantum mechanics for the answer.
$endgroup$
add a comment |
$begingroup$
The fundamental experiment showing the apparent contradiction is Young's double slit: How can particle characteristics be transmitted when there is only an interfering wave between the point A of emission and point B of absorption?
However, for photons in vacuum (moving at c) there is a simple answer: The spacetime interval between A and B is empty, it is zero! That means that both points A and B are adjacent. A and B may be represented by mass particles (electrons etc.) which are exchanging a momentum. The transmission is direct, without need of any intermediate particle.
In contrast, a spacetime interval cannot be observed by observers. If a light ray is transmitted from Sun to Earth, nobody will see that A (Sun) and B (Earth) are adjacent. Instead, they will observe a space distance of eight light minutes and a time interval of eight minutes, even if the spacetime interval is zero. In this situation, the light wave takes the role of a sort of "placeholder": Light waves are observed to propagate at c (according to the second postulate of special relativity), but this is mere observation.
In short, the particle characteristics may be transmitted without any photon because the spacetime interval is zero. The wave characteristics (including the propagation at c) are observation only.
By the way, for light propagating at a lower speed than c (e.g. light moving through a medium), we need quantum mechanics for the answer.
$endgroup$
The fundamental experiment showing the apparent contradiction is Young's double slit: How can particle characteristics be transmitted when there is only an interfering wave between the point A of emission and point B of absorption?
However, for photons in vacuum (moving at c) there is a simple answer: The spacetime interval between A and B is empty, it is zero! That means that both points A and B are adjacent. A and B may be represented by mass particles (electrons etc.) which are exchanging a momentum. The transmission is direct, without need of any intermediate particle.
In contrast, a spacetime interval cannot be observed by observers. If a light ray is transmitted from Sun to Earth, nobody will see that A (Sun) and B (Earth) are adjacent. Instead, they will observe a space distance of eight light minutes and a time interval of eight minutes, even if the spacetime interval is zero. In this situation, the light wave takes the role of a sort of "placeholder": Light waves are observed to propagate at c (according to the second postulate of special relativity), but this is mere observation.
In short, the particle characteristics may be transmitted without any photon because the spacetime interval is zero. The wave characteristics (including the propagation at c) are observation only.
By the way, for light propagating at a lower speed than c (e.g. light moving through a medium), we need quantum mechanics for the answer.
answered 10 hours ago
MoonrakerMoonraker
1,96411022
1,96411022
add a comment |
add a comment |
$begingroup$
Light always behaves as a particle and waves .So there is no particular time when it can behaves like a particle but not wave and vice versa .Thus light carries both of these two nature(particle and wave) along with it (light)for all time.
Actually why is my thinking is like light has two of these nature for all time?
Reasons for considering it as *wave *
- James Clerk Maxwell proved that speed of the electromagnetic waves in free space is the same as the speed of light c=2.998 ×10^8m/s. Thereby he concluded that light consists of electromagnetic waves .So it has wave nature for all time.
- Light also shows *interference *,*diffraction *,*polarization *.
And all of these property show that light has wave nature and it is for all time.
Reasons for considering light as particle
- Max plank in 1900 introduced concept of quantum of energy .The energy exchange between radiation and surroundings is in *discrete or *quantised * form i.e energy exchange between electromagnetic waves is integer multiple of (plank constant *frequency of wave ).As light consists of electromagnetic waves then light is made of discrete bits of energy i.e photons with energy (plank constant *frequency of light).
Thereby photoelectric effect ,Compton effect all show the particle nature of light.
So radiation (electromagnetic waves) exhibit particle nature conversely material particles display wave like nature introduced by de Broglie(a moving particle has wave properties associated with it).This is confirmed experimentally by Davisson and Germer they proved that interference which is a property of waves can be obtained by with material particles such as electron.
So as radiation exhibit particle like nature but particles also exhibit
wave- like behavior .Thereby light always shows two type of natures both as waves and particles .
New contributor
$endgroup$
$begingroup$
Ok if we say light is a particle and wave at all times, then why does it exhibit wave properties in certain cases and particle behaviour in certain phenomenon? Is it because it acts more like a wave and more like a particle in certain conditions? If so what are those conditions? Appreciate your answer btw
$endgroup$
– d_g
9 hours ago
$begingroup$
@d_g light as a whole shows two natures as i said. you can consider double silt experiment when light passes through the silt it behaves like a wave but when it strikes the screen it behaves like a particle .
$endgroup$
– Sneha Banerjee
8 hours ago
add a comment |
$begingroup$
Light always behaves as a particle and waves .So there is no particular time when it can behaves like a particle but not wave and vice versa .Thus light carries both of these two nature(particle and wave) along with it (light)for all time.
Actually why is my thinking is like light has two of these nature for all time?
Reasons for considering it as *wave *
- James Clerk Maxwell proved that speed of the electromagnetic waves in free space is the same as the speed of light c=2.998 ×10^8m/s. Thereby he concluded that light consists of electromagnetic waves .So it has wave nature for all time.
- Light also shows *interference *,*diffraction *,*polarization *.
And all of these property show that light has wave nature and it is for all time.
Reasons for considering light as particle
- Max plank in 1900 introduced concept of quantum of energy .The energy exchange between radiation and surroundings is in *discrete or *quantised * form i.e energy exchange between electromagnetic waves is integer multiple of (plank constant *frequency of wave ).As light consists of electromagnetic waves then light is made of discrete bits of energy i.e photons with energy (plank constant *frequency of light).
Thereby photoelectric effect ,Compton effect all show the particle nature of light.
So radiation (electromagnetic waves) exhibit particle nature conversely material particles display wave like nature introduced by de Broglie(a moving particle has wave properties associated with it).This is confirmed experimentally by Davisson and Germer they proved that interference which is a property of waves can be obtained by with material particles such as electron.
So as radiation exhibit particle like nature but particles also exhibit
wave- like behavior .Thereby light always shows two type of natures both as waves and particles .
New contributor
$endgroup$
$begingroup$
Ok if we say light is a particle and wave at all times, then why does it exhibit wave properties in certain cases and particle behaviour in certain phenomenon? Is it because it acts more like a wave and more like a particle in certain conditions? If so what are those conditions? Appreciate your answer btw
$endgroup$
– d_g
9 hours ago
$begingroup$
@d_g light as a whole shows two natures as i said. you can consider double silt experiment when light passes through the silt it behaves like a wave but when it strikes the screen it behaves like a particle .
$endgroup$
– Sneha Banerjee
8 hours ago
add a comment |
$begingroup$
Light always behaves as a particle and waves .So there is no particular time when it can behaves like a particle but not wave and vice versa .Thus light carries both of these two nature(particle and wave) along with it (light)for all time.
Actually why is my thinking is like light has two of these nature for all time?
Reasons for considering it as *wave *
- James Clerk Maxwell proved that speed of the electromagnetic waves in free space is the same as the speed of light c=2.998 ×10^8m/s. Thereby he concluded that light consists of electromagnetic waves .So it has wave nature for all time.
- Light also shows *interference *,*diffraction *,*polarization *.
And all of these property show that light has wave nature and it is for all time.
Reasons for considering light as particle
- Max plank in 1900 introduced concept of quantum of energy .The energy exchange between radiation and surroundings is in *discrete or *quantised * form i.e energy exchange between electromagnetic waves is integer multiple of (plank constant *frequency of wave ).As light consists of electromagnetic waves then light is made of discrete bits of energy i.e photons with energy (plank constant *frequency of light).
Thereby photoelectric effect ,Compton effect all show the particle nature of light.
So radiation (electromagnetic waves) exhibit particle nature conversely material particles display wave like nature introduced by de Broglie(a moving particle has wave properties associated with it).This is confirmed experimentally by Davisson and Germer they proved that interference which is a property of waves can be obtained by with material particles such as electron.
So as radiation exhibit particle like nature but particles also exhibit
wave- like behavior .Thereby light always shows two type of natures both as waves and particles .
New contributor
$endgroup$
Light always behaves as a particle and waves .So there is no particular time when it can behaves like a particle but not wave and vice versa .Thus light carries both of these two nature(particle and wave) along with it (light)for all time.
Actually why is my thinking is like light has two of these nature for all time?
Reasons for considering it as *wave *
- James Clerk Maxwell proved that speed of the electromagnetic waves in free space is the same as the speed of light c=2.998 ×10^8m/s. Thereby he concluded that light consists of electromagnetic waves .So it has wave nature for all time.
- Light also shows *interference *,*diffraction *,*polarization *.
And all of these property show that light has wave nature and it is for all time.
Reasons for considering light as particle
- Max plank in 1900 introduced concept of quantum of energy .The energy exchange between radiation and surroundings is in *discrete or *quantised * form i.e energy exchange between electromagnetic waves is integer multiple of (plank constant *frequency of wave ).As light consists of electromagnetic waves then light is made of discrete bits of energy i.e photons with energy (plank constant *frequency of light).
Thereby photoelectric effect ,Compton effect all show the particle nature of light.
So radiation (electromagnetic waves) exhibit particle nature conversely material particles display wave like nature introduced by de Broglie(a moving particle has wave properties associated with it).This is confirmed experimentally by Davisson and Germer they proved that interference which is a property of waves can be obtained by with material particles such as electron.
So as radiation exhibit particle like nature but particles also exhibit
wave- like behavior .Thereby light always shows two type of natures both as waves and particles .
New contributor
New contributor
answered 10 hours ago
Sneha BanerjeeSneha Banerjee
195
195
New contributor
New contributor
$begingroup$
Ok if we say light is a particle and wave at all times, then why does it exhibit wave properties in certain cases and particle behaviour in certain phenomenon? Is it because it acts more like a wave and more like a particle in certain conditions? If so what are those conditions? Appreciate your answer btw
$endgroup$
– d_g
9 hours ago
$begingroup$
@d_g light as a whole shows two natures as i said. you can consider double silt experiment when light passes through the silt it behaves like a wave but when it strikes the screen it behaves like a particle .
$endgroup$
– Sneha Banerjee
8 hours ago
add a comment |
$begingroup$
Ok if we say light is a particle and wave at all times, then why does it exhibit wave properties in certain cases and particle behaviour in certain phenomenon? Is it because it acts more like a wave and more like a particle in certain conditions? If so what are those conditions? Appreciate your answer btw
$endgroup$
– d_g
9 hours ago
$begingroup$
@d_g light as a whole shows two natures as i said. you can consider double silt experiment when light passes through the silt it behaves like a wave but when it strikes the screen it behaves like a particle .
$endgroup$
– Sneha Banerjee
8 hours ago
$begingroup$
Ok if we say light is a particle and wave at all times, then why does it exhibit wave properties in certain cases and particle behaviour in certain phenomenon? Is it because it acts more like a wave and more like a particle in certain conditions? If so what are those conditions? Appreciate your answer btw
$endgroup$
– d_g
9 hours ago
$begingroup$
Ok if we say light is a particle and wave at all times, then why does it exhibit wave properties in certain cases and particle behaviour in certain phenomenon? Is it because it acts more like a wave and more like a particle in certain conditions? If so what are those conditions? Appreciate your answer btw
$endgroup$
– d_g
9 hours ago
$begingroup$
@d_g light as a whole shows two natures as i said. you can consider double silt experiment when light passes through the silt it behaves like a wave but when it strikes the screen it behaves like a particle .
$endgroup$
– Sneha Banerjee
8 hours ago
$begingroup$
@d_g light as a whole shows two natures as i said. you can consider double silt experiment when light passes through the silt it behaves like a wave but when it strikes the screen it behaves like a particle .
$endgroup$
– Sneha Banerjee
8 hours ago
add a comment |
$begingroup$
The light doesn't chose. You as an experimenter chose which observable you want to measure, and thus which operator you use. Such measurement will result in a wavefunction collapsing into one of the eigenstates of this operator.
E.g. a positiin operator will give you a position, i.e. particle.
A momentum operator will give you a momentum, i.e. wavelike object.
New contributor
$endgroup$
add a comment |
$begingroup$
The light doesn't chose. You as an experimenter chose which observable you want to measure, and thus which operator you use. Such measurement will result in a wavefunction collapsing into one of the eigenstates of this operator.
E.g. a positiin operator will give you a position, i.e. particle.
A momentum operator will give you a momentum, i.e. wavelike object.
New contributor
$endgroup$
add a comment |
$begingroup$
The light doesn't chose. You as an experimenter chose which observable you want to measure, and thus which operator you use. Such measurement will result in a wavefunction collapsing into one of the eigenstates of this operator.
E.g. a positiin operator will give you a position, i.e. particle.
A momentum operator will give you a momentum, i.e. wavelike object.
New contributor
$endgroup$
The light doesn't chose. You as an experimenter chose which observable you want to measure, and thus which operator you use. Such measurement will result in a wavefunction collapsing into one of the eigenstates of this operator.
E.g. a positiin operator will give you a position, i.e. particle.
A momentum operator will give you a momentum, i.e. wavelike object.
New contributor
New contributor
answered 2 hours ago
Alex VoznyAlex Vozny
111
111
New contributor
New contributor
add a comment |
add a comment |
d_g is a new contributor. Be nice, and check out our Code of Conduct.
d_g is a new contributor. Be nice, and check out our Code of Conduct.
d_g is a new contributor. Be nice, and check out our Code of Conduct.
d_g is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f473566%2fhow-does-light-choose-between-wave-and-particle-behaviour%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Light has no choice here. Photoelectric effect is a manifestly particle-like phenomenon by default. If you perform an experiment and observe photoelectric effect, we ascribe a 'particle-like' nature to the process. There is no wave-like photoelectric effect, because we defined it to refer to processes that involve particles knocking off each other.
$endgroup$
– Avantgarde
8 hours ago
3
$begingroup$
Not worth an answer but a simple analogy may help you understand some of the actual answers. Think of a cylinder, with one of its flat face on a table. If you look at it from the side, it's a rectangle. If you look at it from the top, it's a a circle. Both observations are true but the differences come from the fact that your way of measuring is not very precise and will heavily influence the way you perceive the observed object.
$endgroup$
– Echox
7 hours ago
$begingroup$
This question is briefly addressed in the last paragraph of this answer: physics.stackexchange.com/a/469623
$endgroup$
– Chiral Anomaly
2 hours ago