Why don't electromagnetic waves interact with each other?
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My exact question is that what refers to this phenomenon? I saw also Richards Feynman's video in that he talks about light and says that if we look at something those light waves that come from that thing are not disturbed from any other electromagnetic waves and explains this kind of way that if I can see things clearly, in front of me, although if someone stand in the right of me, can also clearly see any thing in the left of me, our light waves cross each other but the are not disturbed by each other. This is a kinda cool explanation but I don't understand that exactly, because I am not convinced that if those two electromagnetic waves would interact then I couldn't see the thing in front of me clearly
electromagnetic-radiation
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add a comment |
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My exact question is that what refers to this phenomenon? I saw also Richards Feynman's video in that he talks about light and says that if we look at something those light waves that come from that thing are not disturbed from any other electromagnetic waves and explains this kind of way that if I can see things clearly, in front of me, although if someone stand in the right of me, can also clearly see any thing in the left of me, our light waves cross each other but the are not disturbed by each other. This is a kinda cool explanation but I don't understand that exactly, because I am not convinced that if those two electromagnetic waves would interact then I couldn't see the thing in front of me clearly
electromagnetic-radiation
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Imagine light beams of flashlights were water jets. When you cross two of them, they scatter on each other (interact), so you don't see anything clearly.
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– safesphere
Apr 7 at 7:44
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Please provide a link of the video.
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– SRS
Apr 7 at 8:42
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Link: youtu.be/P1ww1IXRfTA?t=2372
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– isarandi
Apr 7 at 15:58
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I suppose one could say that they do interact in the sense that the electromagnetic field at each point in time and space reachable by both waves is the result of both waves. But due to the underlying principles of field propagation it just so happens that after all the messy interference both waves emerge "unperturbed" as if the interference never had happened ;-). (These principles are discussed in G. Smith's answer.)
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– Peter A. Schneider
Apr 7 at 17:02
add a comment |
$begingroup$
My exact question is that what refers to this phenomenon? I saw also Richards Feynman's video in that he talks about light and says that if we look at something those light waves that come from that thing are not disturbed from any other electromagnetic waves and explains this kind of way that if I can see things clearly, in front of me, although if someone stand in the right of me, can also clearly see any thing in the left of me, our light waves cross each other but the are not disturbed by each other. This is a kinda cool explanation but I don't understand that exactly, because I am not convinced that if those two electromagnetic waves would interact then I couldn't see the thing in front of me clearly
electromagnetic-radiation
$endgroup$
My exact question is that what refers to this phenomenon? I saw also Richards Feynman's video in that he talks about light and says that if we look at something those light waves that come from that thing are not disturbed from any other electromagnetic waves and explains this kind of way that if I can see things clearly, in front of me, although if someone stand in the right of me, can also clearly see any thing in the left of me, our light waves cross each other but the are not disturbed by each other. This is a kinda cool explanation but I don't understand that exactly, because I am not convinced that if those two electromagnetic waves would interact then I couldn't see the thing in front of me clearly
electromagnetic-radiation
electromagnetic-radiation
edited Apr 7 at 8:41
SRS
6,862435126
6,862435126
asked Apr 6 at 23:53
Bálint TataiBálint Tatai
38728
38728
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Imagine light beams of flashlights were water jets. When you cross two of them, they scatter on each other (interact), so you don't see anything clearly.
$endgroup$
– safesphere
Apr 7 at 7:44
$begingroup$
Please provide a link of the video.
$endgroup$
– SRS
Apr 7 at 8:42
$begingroup$
Link: youtu.be/P1ww1IXRfTA?t=2372
$endgroup$
– isarandi
Apr 7 at 15:58
$begingroup$
I suppose one could say that they do interact in the sense that the electromagnetic field at each point in time and space reachable by both waves is the result of both waves. But due to the underlying principles of field propagation it just so happens that after all the messy interference both waves emerge "unperturbed" as if the interference never had happened ;-). (These principles are discussed in G. Smith's answer.)
$endgroup$
– Peter A. Schneider
Apr 7 at 17:02
add a comment |
$begingroup$
Imagine light beams of flashlights were water jets. When you cross two of them, they scatter on each other (interact), so you don't see anything clearly.
$endgroup$
– safesphere
Apr 7 at 7:44
$begingroup$
Please provide a link of the video.
$endgroup$
– SRS
Apr 7 at 8:42
$begingroup$
Link: youtu.be/P1ww1IXRfTA?t=2372
$endgroup$
– isarandi
Apr 7 at 15:58
$begingroup$
I suppose one could say that they do interact in the sense that the electromagnetic field at each point in time and space reachable by both waves is the result of both waves. But due to the underlying principles of field propagation it just so happens that after all the messy interference both waves emerge "unperturbed" as if the interference never had happened ;-). (These principles are discussed in G. Smith's answer.)
$endgroup$
– Peter A. Schneider
Apr 7 at 17:02
$begingroup$
Imagine light beams of flashlights were water jets. When you cross two of them, they scatter on each other (interact), so you don't see anything clearly.
$endgroup$
– safesphere
Apr 7 at 7:44
$begingroup$
Imagine light beams of flashlights were water jets. When you cross two of them, they scatter on each other (interact), so you don't see anything clearly.
$endgroup$
– safesphere
Apr 7 at 7:44
$begingroup$
Please provide a link of the video.
$endgroup$
– SRS
Apr 7 at 8:42
$begingroup$
Please provide a link of the video.
$endgroup$
– SRS
Apr 7 at 8:42
$begingroup$
Link: youtu.be/P1ww1IXRfTA?t=2372
$endgroup$
– isarandi
Apr 7 at 15:58
$begingroup$
Link: youtu.be/P1ww1IXRfTA?t=2372
$endgroup$
– isarandi
Apr 7 at 15:58
$begingroup$
I suppose one could say that they do interact in the sense that the electromagnetic field at each point in time and space reachable by both waves is the result of both waves. But due to the underlying principles of field propagation it just so happens that after all the messy interference both waves emerge "unperturbed" as if the interference never had happened ;-). (These principles are discussed in G. Smith's answer.)
$endgroup$
– Peter A. Schneider
Apr 7 at 17:02
$begingroup$
I suppose one could say that they do interact in the sense that the electromagnetic field at each point in time and space reachable by both waves is the result of both waves. But due to the underlying principles of field propagation it just so happens that after all the messy interference both waves emerge "unperturbed" as if the interference never had happened ;-). (These principles are discussed in G. Smith's answer.)
$endgroup$
– Peter A. Schneider
Apr 7 at 17:02
add a comment |
1 Answer
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active
oldest
votes
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Here are three explanations of how to understand “why” electromagnetic waves don’t directly interact electromagnetically with each other, which are all equivalent to each other:
Maxwell’s equations are linear in the electric and magnetic fields, and in their sources, so the superposition of two solutions is also a solution. (For example, in Coulomb’s Law you can just add up the fields of multiple charges.)
Photons do not carry any electric charge and do not have their own electromagnetic field. (Note: By contrast, gluons do carry color charge and do interact with each other.)
The gauge group for electromagnetism is an abelian (i.e., commutative) group. (Gauge groups are something you learn about in more advanced physics courses.)
Notice that I said photons don’t directly interact with each other. They do indirectly interact via virtual electrons and positrons (or other charged particle-antiparticle pairs). Until you get to extremely intense electric and magnetic fields, this is a very tiny effect and was only recently measured.
An even tinier effect, which we will probably never be able to detect, is the gravitational interaction of electromagnetic waves or photons. Physicists believe there would be a gravitational interaction because electromagnetic waves and photons carry energy and momentum, even though photons are massless.
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Isn't that "photons are massless at rest"?
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– CJ Dennis
Apr 7 at 6:51
6
$begingroup$
@CJDennis Photons can't be at rest.
$endgroup$
– SRS
Apr 7 at 8:43
add a comment |
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$begingroup$
Here are three explanations of how to understand “why” electromagnetic waves don’t directly interact electromagnetically with each other, which are all equivalent to each other:
Maxwell’s equations are linear in the electric and magnetic fields, and in their sources, so the superposition of two solutions is also a solution. (For example, in Coulomb’s Law you can just add up the fields of multiple charges.)
Photons do not carry any electric charge and do not have their own electromagnetic field. (Note: By contrast, gluons do carry color charge and do interact with each other.)
The gauge group for electromagnetism is an abelian (i.e., commutative) group. (Gauge groups are something you learn about in more advanced physics courses.)
Notice that I said photons don’t directly interact with each other. They do indirectly interact via virtual electrons and positrons (or other charged particle-antiparticle pairs). Until you get to extremely intense electric and magnetic fields, this is a very tiny effect and was only recently measured.
An even tinier effect, which we will probably never be able to detect, is the gravitational interaction of electromagnetic waves or photons. Physicists believe there would be a gravitational interaction because electromagnetic waves and photons carry energy and momentum, even though photons are massless.
$endgroup$
$begingroup$
Isn't that "photons are massless at rest"?
$endgroup$
– CJ Dennis
Apr 7 at 6:51
6
$begingroup$
@CJDennis Photons can't be at rest.
$endgroup$
– SRS
Apr 7 at 8:43
add a comment |
$begingroup$
Here are three explanations of how to understand “why” electromagnetic waves don’t directly interact electromagnetically with each other, which are all equivalent to each other:
Maxwell’s equations are linear in the electric and magnetic fields, and in their sources, so the superposition of two solutions is also a solution. (For example, in Coulomb’s Law you can just add up the fields of multiple charges.)
Photons do not carry any electric charge and do not have their own electromagnetic field. (Note: By contrast, gluons do carry color charge and do interact with each other.)
The gauge group for electromagnetism is an abelian (i.e., commutative) group. (Gauge groups are something you learn about in more advanced physics courses.)
Notice that I said photons don’t directly interact with each other. They do indirectly interact via virtual electrons and positrons (or other charged particle-antiparticle pairs). Until you get to extremely intense electric and magnetic fields, this is a very tiny effect and was only recently measured.
An even tinier effect, which we will probably never be able to detect, is the gravitational interaction of electromagnetic waves or photons. Physicists believe there would be a gravitational interaction because electromagnetic waves and photons carry energy and momentum, even though photons are massless.
$endgroup$
$begingroup$
Isn't that "photons are massless at rest"?
$endgroup$
– CJ Dennis
Apr 7 at 6:51
6
$begingroup$
@CJDennis Photons can't be at rest.
$endgroup$
– SRS
Apr 7 at 8:43
add a comment |
$begingroup$
Here are three explanations of how to understand “why” electromagnetic waves don’t directly interact electromagnetically with each other, which are all equivalent to each other:
Maxwell’s equations are linear in the electric and magnetic fields, and in their sources, so the superposition of two solutions is also a solution. (For example, in Coulomb’s Law you can just add up the fields of multiple charges.)
Photons do not carry any electric charge and do not have their own electromagnetic field. (Note: By contrast, gluons do carry color charge and do interact with each other.)
The gauge group for electromagnetism is an abelian (i.e., commutative) group. (Gauge groups are something you learn about in more advanced physics courses.)
Notice that I said photons don’t directly interact with each other. They do indirectly interact via virtual electrons and positrons (or other charged particle-antiparticle pairs). Until you get to extremely intense electric and magnetic fields, this is a very tiny effect and was only recently measured.
An even tinier effect, which we will probably never be able to detect, is the gravitational interaction of electromagnetic waves or photons. Physicists believe there would be a gravitational interaction because electromagnetic waves and photons carry energy and momentum, even though photons are massless.
$endgroup$
Here are three explanations of how to understand “why” electromagnetic waves don’t directly interact electromagnetically with each other, which are all equivalent to each other:
Maxwell’s equations are linear in the electric and magnetic fields, and in their sources, so the superposition of two solutions is also a solution. (For example, in Coulomb’s Law you can just add up the fields of multiple charges.)
Photons do not carry any electric charge and do not have their own electromagnetic field. (Note: By contrast, gluons do carry color charge and do interact with each other.)
The gauge group for electromagnetism is an abelian (i.e., commutative) group. (Gauge groups are something you learn about in more advanced physics courses.)
Notice that I said photons don’t directly interact with each other. They do indirectly interact via virtual electrons and positrons (or other charged particle-antiparticle pairs). Until you get to extremely intense electric and magnetic fields, this is a very tiny effect and was only recently measured.
An even tinier effect, which we will probably never be able to detect, is the gravitational interaction of electromagnetic waves or photons. Physicists believe there would be a gravitational interaction because electromagnetic waves and photons carry energy and momentum, even though photons are massless.
edited Apr 7 at 0:53
answered Apr 7 at 0:26
G. SmithG. Smith
10.9k11431
10.9k11431
$begingroup$
Isn't that "photons are massless at rest"?
$endgroup$
– CJ Dennis
Apr 7 at 6:51
6
$begingroup$
@CJDennis Photons can't be at rest.
$endgroup$
– SRS
Apr 7 at 8:43
add a comment |
$begingroup$
Isn't that "photons are massless at rest"?
$endgroup$
– CJ Dennis
Apr 7 at 6:51
6
$begingroup$
@CJDennis Photons can't be at rest.
$endgroup$
– SRS
Apr 7 at 8:43
$begingroup$
Isn't that "photons are massless at rest"?
$endgroup$
– CJ Dennis
Apr 7 at 6:51
$begingroup$
Isn't that "photons are massless at rest"?
$endgroup$
– CJ Dennis
Apr 7 at 6:51
6
6
$begingroup$
@CJDennis Photons can't be at rest.
$endgroup$
– SRS
Apr 7 at 8:43
$begingroup$
@CJDennis Photons can't be at rest.
$endgroup$
– SRS
Apr 7 at 8:43
add a comment |
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$begingroup$
Imagine light beams of flashlights were water jets. When you cross two of them, they scatter on each other (interact), so you don't see anything clearly.
$endgroup$
– safesphere
Apr 7 at 7:44
$begingroup$
Please provide a link of the video.
$endgroup$
– SRS
Apr 7 at 8:42
$begingroup$
Link: youtu.be/P1ww1IXRfTA?t=2372
$endgroup$
– isarandi
Apr 7 at 15:58
$begingroup$
I suppose one could say that they do interact in the sense that the electromagnetic field at each point in time and space reachable by both waves is the result of both waves. But due to the underlying principles of field propagation it just so happens that after all the messy interference both waves emerge "unperturbed" as if the interference never had happened ;-). (These principles are discussed in G. Smith's answer.)
$endgroup$
– Peter A. Schneider
Apr 7 at 17:02