Why doesn't the escape of electromagnetic waves from a microwave depend on the reference frame, because of...
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If an observer traveling towards a microwave oven at almost the speed of light blue shifts the microwaves enough to be visible light, how can the mesh on the oven door still stop to waves from escaping the oven?
And conversely, if an observer traveling at almost the speed of light away from a microwave oven red shifts the visible light enough to be microwaves, how can the mesh on the oven door still allow the waves to escape the oven?
electromagnetism special-relativity microwaves redshift
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
asked Dec 21 '18 at 20:49
samwasamwa
714
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show 2 more comments
$begingroup$
If an observer traveling towards a microwave oven at almost the speed of light blue shifts the microwaves enough to be visible light, how can the mesh on the oven door still stop to waves from escaping the oven?
And conversely, if an observer traveling at almost the speed of light away from a microwave oven red shifts the visible light enough to be microwaves, how can the mesh on the oven door still allow the waves to escape the oven?
electromagnetism special-relativity microwaves redshift
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
asked Dec 21 '18 at 20:49
samwasamwa
714
$endgroup$
7
$begingroup$
The screen moves with the oven. From it's point of view, the radiation is in the microwave region, and it won't let any out. It would be different if you could jury rig the oven to operate with the door removed, and then you carried the door with you while you did the experiment. In that case, the screen's point of view would be the same as your point of view. It would see blue light, and it would let the blue light through the holes.
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– Solomon Slow
Dec 21 '18 at 20:56
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@SolomonSlow What you are saying is right, but there is more in this question. Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not), regardless of the frame of reference we are using. I hope that StudyStudy or someone else more well-versed in relativistic electrodynamics than me would write an answer.
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– b.Lorenz
Dec 21 '18 at 20:56
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It's going to come back to the Faraday tensor....en.m.wikipedia.org/wiki/Electromagnetic_tensor
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– user214814
Dec 21 '18 at 20:58
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That does not answer the question. It is clear that the light will not come out from the start.
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– my2cts
Dec 21 '18 at 21:08
$begingroup$
"Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not" - sure, and the answer in this case is "not" in all frames of reference. You will not see microwaves getting out when standing in front of it, when flying towards it, or with a goat, or on a boat. Where is the mystery.
$endgroup$
– Maury Markowitz
Dec 21 '18 at 21:10
|
show 2 more comments
$begingroup$
If an observer traveling towards a microwave oven at almost the speed of light blue shifts the microwaves enough to be visible light, how can the mesh on the oven door still stop to waves from escaping the oven?
And conversely, if an observer traveling at almost the speed of light away from a microwave oven red shifts the visible light enough to be microwaves, how can the mesh on the oven door still allow the waves to escape the oven?
electromagnetism special-relativity microwaves redshift
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
asked Dec 21 '18 at 20:49
samwasamwa
714
$endgroup$
If an observer traveling towards a microwave oven at almost the speed of light blue shifts the microwaves enough to be visible light, how can the mesh on the oven door still stop to waves from escaping the oven?
And conversely, if an observer traveling at almost the speed of light away from a microwave oven red shifts the visible light enough to be microwaves, how can the mesh on the oven door still allow the waves to escape the oven?
electromagnetism special-relativity microwaves redshift
electromagnetism special-relativity microwaves redshift
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
asked Dec 21 '18 at 20:49
samwasamwa
714
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
asked Dec 21 '18 at 20:49
samwasamwa
714
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
edited Dec 22 '18 at 12:47
knzhou
45.5k11122219
45.5k11122219
asked Dec 21 '18 at 20:49
samwasamwa
714
asked Dec 21 '18 at 20:49
samwasamwa
714
asked Dec 21 '18 at 20:49
samwasamwa
714
714
7
$begingroup$
The screen moves with the oven. From it's point of view, the radiation is in the microwave region, and it won't let any out. It would be different if you could jury rig the oven to operate with the door removed, and then you carried the door with you while you did the experiment. In that case, the screen's point of view would be the same as your point of view. It would see blue light, and it would let the blue light through the holes.
$endgroup$
– Solomon Slow
Dec 21 '18 at 20:56
$begingroup$
@SolomonSlow What you are saying is right, but there is more in this question. Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not), regardless of the frame of reference we are using. I hope that StudyStudy or someone else more well-versed in relativistic electrodynamics than me would write an answer.
$endgroup$
– b.Lorenz
Dec 21 '18 at 20:56
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It's going to come back to the Faraday tensor....en.m.wikipedia.org/wiki/Electromagnetic_tensor
$endgroup$
– user214814
Dec 21 '18 at 20:58
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That does not answer the question. It is clear that the light will not come out from the start.
$endgroup$
– my2cts
Dec 21 '18 at 21:08
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"Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not" - sure, and the answer in this case is "not" in all frames of reference. You will not see microwaves getting out when standing in front of it, when flying towards it, or with a goat, or on a boat. Where is the mystery.
$endgroup$
– Maury Markowitz
Dec 21 '18 at 21:10
|
show 2 more comments
7
$begingroup$
The screen moves with the oven. From it's point of view, the radiation is in the microwave region, and it won't let any out. It would be different if you could jury rig the oven to operate with the door removed, and then you carried the door with you while you did the experiment. In that case, the screen's point of view would be the same as your point of view. It would see blue light, and it would let the blue light through the holes.
$endgroup$
– Solomon Slow
Dec 21 '18 at 20:56
$begingroup$
@SolomonSlow What you are saying is right, but there is more in this question. Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not), regardless of the frame of reference we are using. I hope that StudyStudy or someone else more well-versed in relativistic electrodynamics than me would write an answer.
$endgroup$
– b.Lorenz
Dec 21 '18 at 20:56
$begingroup$
It's going to come back to the Faraday tensor....en.m.wikipedia.org/wiki/Electromagnetic_tensor
$endgroup$
– user214814
Dec 21 '18 at 20:58
$begingroup$
That does not answer the question. It is clear that the light will not come out from the start.
$endgroup$
– my2cts
Dec 21 '18 at 21:08
$begingroup$
"Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not" - sure, and the answer in this case is "not" in all frames of reference. You will not see microwaves getting out when standing in front of it, when flying towards it, or with a goat, or on a boat. Where is the mystery.
$endgroup$
– Maury Markowitz
Dec 21 '18 at 21:10
7
7
$begingroup$
The screen moves with the oven. From it's point of view, the radiation is in the microwave region, and it won't let any out. It would be different if you could jury rig the oven to operate with the door removed, and then you carried the door with you while you did the experiment. In that case, the screen's point of view would be the same as your point of view. It would see blue light, and it would let the blue light through the holes.
$endgroup$
– Solomon Slow
Dec 21 '18 at 20:56
$begingroup$
The screen moves with the oven. From it's point of view, the radiation is in the microwave region, and it won't let any out. It would be different if you could jury rig the oven to operate with the door removed, and then you carried the door with you while you did the experiment. In that case, the screen's point of view would be the same as your point of view. It would see blue light, and it would let the blue light through the holes.
$endgroup$
– Solomon Slow
Dec 21 '18 at 20:56
$begingroup$
@SolomonSlow What you are saying is right, but there is more in this question. Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not), regardless of the frame of reference we are using. I hope that StudyStudy or someone else more well-versed in relativistic electrodynamics than me would write an answer.
$endgroup$
– b.Lorenz
Dec 21 '18 at 20:56
$begingroup$
@SolomonSlow What you are saying is right, but there is more in this question. Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not), regardless of the frame of reference we are using. I hope that StudyStudy or someone else more well-versed in relativistic electrodynamics than me would write an answer.
$endgroup$
– b.Lorenz
Dec 21 '18 at 20:56
$begingroup$
It's going to come back to the Faraday tensor....en.m.wikipedia.org/wiki/Electromagnetic_tensor
$endgroup$
– user214814
Dec 21 '18 at 20:58
$begingroup$
It's going to come back to the Faraday tensor....en.m.wikipedia.org/wiki/Electromagnetic_tensor
$endgroup$
– user214814
Dec 21 '18 at 20:58
$begingroup$
That does not answer the question. It is clear that the light will not come out from the start.
$endgroup$
– my2cts
Dec 21 '18 at 21:08
$begingroup$
That does not answer the question. It is clear that the light will not come out from the start.
$endgroup$
– my2cts
Dec 21 '18 at 21:08
$begingroup$
"Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not" - sure, and the answer in this case is "not" in all frames of reference. You will not see microwaves getting out when standing in front of it, when flying towards it, or with a goat, or on a boat. Where is the mystery.
$endgroup$
– Maury Markowitz
Dec 21 '18 at 21:10
$begingroup$
"Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not" - sure, and the answer in this case is "not" in all frames of reference. You will not see microwaves getting out when standing in front of it, when flying towards it, or with a goat, or on a boat. Where is the mystery.
$endgroup$
– Maury Markowitz
Dec 21 '18 at 21:10
|
show 2 more comments
5 Answers
5
active
oldest
votes
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First of all, let's get a better picture of why microwave doors keep waves inside the oven in the first place. Using a generous amount of hand-waving: imagine that an electromagnetic wave is incident on a circular hole in the microwave oven door. Furthermore imagine that at some moment in time, the electric field is pointed towards the right side of the hole. Then electrons will move towards the left side, creating a new wave in the electric field. However, it takes some time for the wave to travel all the way around the edge of the hole. If the distance around the hole is roughly the same size as the wavelength of the incident wave [1], then the new wave is exactly out of phase with the incident wave. Thus, outside of the box, the incident wave gets canceled out.
All right, how does this look in the fast-moving observer's frame of reference? Well, the incident and response waves are propagating in the same direction, so the Doppler effect is exactly the same for both. [2] Therefore, they still cancel each other out in the observer's frame.
Notes:
If the hole circumference is smaller than the wavelength, there is still a path inside the metal with the same length as the incident wavelength, so the same thing happens. Waves created along paths longer and shorter than the wavelength cancel each other out. A small amount of the incident wave gets through because this cancellation process doesn't work perfectly for nonzero hole sizes. Getting a more detailed picture of the interaction would require a full numerical simulation, but this model is accurate enough for the question.
We can be a bit more precise about why the Doppler effect is the same for both the incident and response waves. The relativistic Doppler effect has two components: time dilation, which shows up as $gamma$, and the ordinary Doppler effect (due to the finite speed of light and the distance between wavefronts), which appears as $1-beta$. From the observer's perspective, the ordinary Doppler component is the same for both the incident and response waves, while time dilation shows up as the electrons in the door appearing to respond much more quickly than they do in the microwave oven frame.
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
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add a comment |
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The moving glass door has a different response to the electromagnetic field. You have to transform its electromagnetic susceptibility along. This is a forth rank Lorentz tensor say $epsilon_{munurhosigma} $ containing the dielectric and the magnetic susceptibility. It has the same symmetry as the Riemann tensor of GR. The result will be that the window is opaque. I leave it to to work this out as an interesting exercise.
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
$endgroup$
$begingroup$
where does the electromag. suceptibility enter into play? it is about the size of the holes only, that make the metal act like a faraday cage, right?
$endgroup$
– Wolphram jonny
Dec 21 '18 at 21:44
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The susceptibility of the metal must be transformed. I predict that the oven acts as a Faraday cage at the frequency of the radiation in any reference frame.
$endgroup$
– my2cts
Dec 21 '18 at 22:36
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oh I see! you seem correct, although I have no idea, lol
$endgroup$
– Wolphram jonny
Dec 22 '18 at 0:36
add a comment |
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There's a very simple intuition here: the microwaves bounce off because they're wider than the holes. In the moving frame, the microwaves are only contracted lengthwise. They're still just as wide, so they still bounce off.
If you used a frame where the microwave was moving sideways, the width of the microwaves would indeed be contracted. But then the holes would be too, by the same factor, so they still don't get out.
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
$endgroup$
$begingroup$
@b.Lorenz If you think about it a bit more, you'll realize that while it sounds superficially like a wrong explanation, it's actually right. If you make a wavepacket whose width is less wide than a hole, it will sail right through. There isn't some magical nonlocal effect that makes it bounce off.
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– knzhou
Dec 22 '18 at 14:09
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Well, that makes sense.
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– b.Lorenz
Dec 22 '18 at 14:25
add a comment |
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The answer is the wave is not actually changing. Think of it this way. When a car playing loud music drives by, the pitch increases as it approaches, sounds correct for a moment, then the pitch decreases as it moves away. The person in the car hears the same pitch the whole time.
Edit:
Additionally, imagine a car is driving 40 mph toward a wall which is moving 0 mph. The collision will occur with results consistent with a 40 mph collision. Now if a car is driving 40 mph toward another car which is driving 40 mph toward it (head on), a collision will occur with results consistent with a car driving 80 mph toward a wall which is moving 0 mph. If a car which is driving 40 mph toward another car which is driving 40 mph from behind (rear end), a collision will not even occur...so let us imagine the car in the rear is driving 41 mph...a collision will occur which is consistent with a car driving 1 mph toward a wall which is moving 0 mph.
In other words:
The music which has exited the car through the window and has been heard by someone standing outside the car, is the sound which is experienced as being different.
The music is a higher pitch while the car is moving toward the person because the speed* of the wave is increased by the speed of the car. The pitch sounds normal for a moment because the car is not moving toward or away from the person which leaves the speed* of the sound as it is, and the pitch is lower as the car is moving away because the speed* of the sound wave is reduced by the speed of the car.
Application to the microwave scenario:
The microwave oven is like the car and the electromagnetic wave (visible or microwave) is like the sound wave of the music.
The electromagnetic wave being shifted is the one which has left the interior of the microwave oven.
The shift being observed in the electromagnetic wave is caused by a change in the speed* of the wave as it is observed relative to the observer.
Edit: *speed of the waves therefore increased frequency.
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
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add a comment |
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Doppler affects the observers perception. The observed does not change, so the microwaves would still be blocked and, you could not see them at all, let alone them shifted.
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
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5 Answers
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5 Answers
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First of all, let's get a better picture of why microwave doors keep waves inside the oven in the first place. Using a generous amount of hand-waving: imagine that an electromagnetic wave is incident on a circular hole in the microwave oven door. Furthermore imagine that at some moment in time, the electric field is pointed towards the right side of the hole. Then electrons will move towards the left side, creating a new wave in the electric field. However, it takes some time for the wave to travel all the way around the edge of the hole. If the distance around the hole is roughly the same size as the wavelength of the incident wave [1], then the new wave is exactly out of phase with the incident wave. Thus, outside of the box, the incident wave gets canceled out.
All right, how does this look in the fast-moving observer's frame of reference? Well, the incident and response waves are propagating in the same direction, so the Doppler effect is exactly the same for both. [2] Therefore, they still cancel each other out in the observer's frame.
Notes:
If the hole circumference is smaller than the wavelength, there is still a path inside the metal with the same length as the incident wavelength, so the same thing happens. Waves created along paths longer and shorter than the wavelength cancel each other out. A small amount of the incident wave gets through because this cancellation process doesn't work perfectly for nonzero hole sizes. Getting a more detailed picture of the interaction would require a full numerical simulation, but this model is accurate enough for the question.
We can be a bit more precise about why the Doppler effect is the same for both the incident and response waves. The relativistic Doppler effect has two components: time dilation, which shows up as $gamma$, and the ordinary Doppler effect (due to the finite speed of light and the distance between wavefronts), which appears as $1-beta$. From the observer's perspective, the ordinary Doppler component is the same for both the incident and response waves, while time dilation shows up as the electrons in the door appearing to respond much more quickly than they do in the microwave oven frame.
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
$endgroup$
add a comment |
$begingroup$
First of all, let's get a better picture of why microwave doors keep waves inside the oven in the first place. Using a generous amount of hand-waving: imagine that an electromagnetic wave is incident on a circular hole in the microwave oven door. Furthermore imagine that at some moment in time, the electric field is pointed towards the right side of the hole. Then electrons will move towards the left side, creating a new wave in the electric field. However, it takes some time for the wave to travel all the way around the edge of the hole. If the distance around the hole is roughly the same size as the wavelength of the incident wave [1], then the new wave is exactly out of phase with the incident wave. Thus, outside of the box, the incident wave gets canceled out.
All right, how does this look in the fast-moving observer's frame of reference? Well, the incident and response waves are propagating in the same direction, so the Doppler effect is exactly the same for both. [2] Therefore, they still cancel each other out in the observer's frame.
Notes:
If the hole circumference is smaller than the wavelength, there is still a path inside the metal with the same length as the incident wavelength, so the same thing happens. Waves created along paths longer and shorter than the wavelength cancel each other out. A small amount of the incident wave gets through because this cancellation process doesn't work perfectly for nonzero hole sizes. Getting a more detailed picture of the interaction would require a full numerical simulation, but this model is accurate enough for the question.
We can be a bit more precise about why the Doppler effect is the same for both the incident and response waves. The relativistic Doppler effect has two components: time dilation, which shows up as $gamma$, and the ordinary Doppler effect (due to the finite speed of light and the distance between wavefronts), which appears as $1-beta$. From the observer's perspective, the ordinary Doppler component is the same for both the incident and response waves, while time dilation shows up as the electrons in the door appearing to respond much more quickly than they do in the microwave oven frame.
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
$endgroup$
add a comment |
$begingroup$
First of all, let's get a better picture of why microwave doors keep waves inside the oven in the first place. Using a generous amount of hand-waving: imagine that an electromagnetic wave is incident on a circular hole in the microwave oven door. Furthermore imagine that at some moment in time, the electric field is pointed towards the right side of the hole. Then electrons will move towards the left side, creating a new wave in the electric field. However, it takes some time for the wave to travel all the way around the edge of the hole. If the distance around the hole is roughly the same size as the wavelength of the incident wave [1], then the new wave is exactly out of phase with the incident wave. Thus, outside of the box, the incident wave gets canceled out.
All right, how does this look in the fast-moving observer's frame of reference? Well, the incident and response waves are propagating in the same direction, so the Doppler effect is exactly the same for both. [2] Therefore, they still cancel each other out in the observer's frame.
Notes:
If the hole circumference is smaller than the wavelength, there is still a path inside the metal with the same length as the incident wavelength, so the same thing happens. Waves created along paths longer and shorter than the wavelength cancel each other out. A small amount of the incident wave gets through because this cancellation process doesn't work perfectly for nonzero hole sizes. Getting a more detailed picture of the interaction would require a full numerical simulation, but this model is accurate enough for the question.
We can be a bit more precise about why the Doppler effect is the same for both the incident and response waves. The relativistic Doppler effect has two components: time dilation, which shows up as $gamma$, and the ordinary Doppler effect (due to the finite speed of light and the distance between wavefronts), which appears as $1-beta$. From the observer's perspective, the ordinary Doppler component is the same for both the incident and response waves, while time dilation shows up as the electrons in the door appearing to respond much more quickly than they do in the microwave oven frame.
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
$endgroup$
First of all, let's get a better picture of why microwave doors keep waves inside the oven in the first place. Using a generous amount of hand-waving: imagine that an electromagnetic wave is incident on a circular hole in the microwave oven door. Furthermore imagine that at some moment in time, the electric field is pointed towards the right side of the hole. Then electrons will move towards the left side, creating a new wave in the electric field. However, it takes some time for the wave to travel all the way around the edge of the hole. If the distance around the hole is roughly the same size as the wavelength of the incident wave [1], then the new wave is exactly out of phase with the incident wave. Thus, outside of the box, the incident wave gets canceled out.
All right, how does this look in the fast-moving observer's frame of reference? Well, the incident and response waves are propagating in the same direction, so the Doppler effect is exactly the same for both. [2] Therefore, they still cancel each other out in the observer's frame.
Notes:
If the hole circumference is smaller than the wavelength, there is still a path inside the metal with the same length as the incident wavelength, so the same thing happens. Waves created along paths longer and shorter than the wavelength cancel each other out. A small amount of the incident wave gets through because this cancellation process doesn't work perfectly for nonzero hole sizes. Getting a more detailed picture of the interaction would require a full numerical simulation, but this model is accurate enough for the question.
We can be a bit more precise about why the Doppler effect is the same for both the incident and response waves. The relativistic Doppler effect has two components: time dilation, which shows up as $gamma$, and the ordinary Doppler effect (due to the finite speed of light and the distance between wavefronts), which appears as $1-beta$. From the observer's perspective, the ordinary Doppler component is the same for both the incident and response waves, while time dilation shows up as the electrons in the door appearing to respond much more quickly than they do in the microwave oven frame.
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
answered Dec 21 '18 at 22:21
ThorondorThorondor
1,746628
1,746628
add a comment |
add a comment |
$begingroup$
The moving glass door has a different response to the electromagnetic field. You have to transform its electromagnetic susceptibility along. This is a forth rank Lorentz tensor say $epsilon_{munurhosigma} $ containing the dielectric and the magnetic susceptibility. It has the same symmetry as the Riemann tensor of GR. The result will be that the window is opaque. I leave it to to work this out as an interesting exercise.
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
$endgroup$
$begingroup$
where does the electromag. suceptibility enter into play? it is about the size of the holes only, that make the metal act like a faraday cage, right?
$endgroup$
– Wolphram jonny
Dec 21 '18 at 21:44
$begingroup$
The susceptibility of the metal must be transformed. I predict that the oven acts as a Faraday cage at the frequency of the radiation in any reference frame.
$endgroup$
– my2cts
Dec 21 '18 at 22:36
$begingroup$
oh I see! you seem correct, although I have no idea, lol
$endgroup$
– Wolphram jonny
Dec 22 '18 at 0:36
add a comment |
$begingroup$
The moving glass door has a different response to the electromagnetic field. You have to transform its electromagnetic susceptibility along. This is a forth rank Lorentz tensor say $epsilon_{munurhosigma} $ containing the dielectric and the magnetic susceptibility. It has the same symmetry as the Riemann tensor of GR. The result will be that the window is opaque. I leave it to to work this out as an interesting exercise.
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
$endgroup$
$begingroup$
where does the electromag. suceptibility enter into play? it is about the size of the holes only, that make the metal act like a faraday cage, right?
$endgroup$
– Wolphram jonny
Dec 21 '18 at 21:44
$begingroup$
The susceptibility of the metal must be transformed. I predict that the oven acts as a Faraday cage at the frequency of the radiation in any reference frame.
$endgroup$
– my2cts
Dec 21 '18 at 22:36
$begingroup$
oh I see! you seem correct, although I have no idea, lol
$endgroup$
– Wolphram jonny
Dec 22 '18 at 0:36
add a comment |
$begingroup$
The moving glass door has a different response to the electromagnetic field. You have to transform its electromagnetic susceptibility along. This is a forth rank Lorentz tensor say $epsilon_{munurhosigma} $ containing the dielectric and the magnetic susceptibility. It has the same symmetry as the Riemann tensor of GR. The result will be that the window is opaque. I leave it to to work this out as an interesting exercise.
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
$endgroup$
The moving glass door has a different response to the electromagnetic field. You have to transform its electromagnetic susceptibility along. This is a forth rank Lorentz tensor say $epsilon_{munurhosigma} $ containing the dielectric and the magnetic susceptibility. It has the same symmetry as the Riemann tensor of GR. The result will be that the window is opaque. I leave it to to work this out as an interesting exercise.
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
edited Dec 22 '18 at 0:36
Wolphram jonny
11.1k22654
11.1k22654
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
answered Dec 21 '18 at 21:36
my2ctsmy2cts
5,7422719
5,7422719
$begingroup$
where does the electromag. suceptibility enter into play? it is about the size of the holes only, that make the metal act like a faraday cage, right?
$endgroup$
– Wolphram jonny
Dec 21 '18 at 21:44
$begingroup$
The susceptibility of the metal must be transformed. I predict that the oven acts as a Faraday cage at the frequency of the radiation in any reference frame.
$endgroup$
– my2cts
Dec 21 '18 at 22:36
$begingroup$
oh I see! you seem correct, although I have no idea, lol
$endgroup$
– Wolphram jonny
Dec 22 '18 at 0:36
add a comment |
$begingroup$
where does the electromag. suceptibility enter into play? it is about the size of the holes only, that make the metal act like a faraday cage, right?
$endgroup$
– Wolphram jonny
Dec 21 '18 at 21:44
$begingroup$
The susceptibility of the metal must be transformed. I predict that the oven acts as a Faraday cage at the frequency of the radiation in any reference frame.
$endgroup$
– my2cts
Dec 21 '18 at 22:36
$begingroup$
oh I see! you seem correct, although I have no idea, lol
$endgroup$
– Wolphram jonny
Dec 22 '18 at 0:36
$begingroup$
where does the electromag. suceptibility enter into play? it is about the size of the holes only, that make the metal act like a faraday cage, right?
$endgroup$
– Wolphram jonny
Dec 21 '18 at 21:44
$begingroup$
where does the electromag. suceptibility enter into play? it is about the size of the holes only, that make the metal act like a faraday cage, right?
$endgroup$
– Wolphram jonny
Dec 21 '18 at 21:44
$begingroup$
The susceptibility of the metal must be transformed. I predict that the oven acts as a Faraday cage at the frequency of the radiation in any reference frame.
$endgroup$
– my2cts
Dec 21 '18 at 22:36
$begingroup$
The susceptibility of the metal must be transformed. I predict that the oven acts as a Faraday cage at the frequency of the radiation in any reference frame.
$endgroup$
– my2cts
Dec 21 '18 at 22:36
$begingroup$
oh I see! you seem correct, although I have no idea, lol
$endgroup$
– Wolphram jonny
Dec 22 '18 at 0:36
$begingroup$
oh I see! you seem correct, although I have no idea, lol
$endgroup$
– Wolphram jonny
Dec 22 '18 at 0:36
add a comment |
$begingroup$
There's a very simple intuition here: the microwaves bounce off because they're wider than the holes. In the moving frame, the microwaves are only contracted lengthwise. They're still just as wide, so they still bounce off.
If you used a frame where the microwave was moving sideways, the width of the microwaves would indeed be contracted. But then the holes would be too, by the same factor, so they still don't get out.
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
$endgroup$
$begingroup$
@b.Lorenz If you think about it a bit more, you'll realize that while it sounds superficially like a wrong explanation, it's actually right. If you make a wavepacket whose width is less wide than a hole, it will sail right through. There isn't some magical nonlocal effect that makes it bounce off.
$endgroup$
– knzhou
Dec 22 '18 at 14:09
$begingroup$
Well, that makes sense.
$endgroup$
– b.Lorenz
Dec 22 '18 at 14:25
add a comment |
$begingroup$
There's a very simple intuition here: the microwaves bounce off because they're wider than the holes. In the moving frame, the microwaves are only contracted lengthwise. They're still just as wide, so they still bounce off.
If you used a frame where the microwave was moving sideways, the width of the microwaves would indeed be contracted. But then the holes would be too, by the same factor, so they still don't get out.
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
$endgroup$
$begingroup$
@b.Lorenz If you think about it a bit more, you'll realize that while it sounds superficially like a wrong explanation, it's actually right. If you make a wavepacket whose width is less wide than a hole, it will sail right through. There isn't some magical nonlocal effect that makes it bounce off.
$endgroup$
– knzhou
Dec 22 '18 at 14:09
$begingroup$
Well, that makes sense.
$endgroup$
– b.Lorenz
Dec 22 '18 at 14:25
add a comment |
$begingroup$
There's a very simple intuition here: the microwaves bounce off because they're wider than the holes. In the moving frame, the microwaves are only contracted lengthwise. They're still just as wide, so they still bounce off.
If you used a frame where the microwave was moving sideways, the width of the microwaves would indeed be contracted. But then the holes would be too, by the same factor, so they still don't get out.
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
$endgroup$
There's a very simple intuition here: the microwaves bounce off because they're wider than the holes. In the moving frame, the microwaves are only contracted lengthwise. They're still just as wide, so they still bounce off.
If you used a frame where the microwave was moving sideways, the width of the microwaves would indeed be contracted. But then the holes would be too, by the same factor, so they still don't get out.
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
answered Dec 22 '18 at 12:46
knzhouknzhou
45.5k11122219
45.5k11122219
$begingroup$
@b.Lorenz If you think about it a bit more, you'll realize that while it sounds superficially like a wrong explanation, it's actually right. If you make a wavepacket whose width is less wide than a hole, it will sail right through. There isn't some magical nonlocal effect that makes it bounce off.
$endgroup$
– knzhou
Dec 22 '18 at 14:09
$begingroup$
Well, that makes sense.
$endgroup$
– b.Lorenz
Dec 22 '18 at 14:25
add a comment |
$begingroup$
@b.Lorenz If you think about it a bit more, you'll realize that while it sounds superficially like a wrong explanation, it's actually right. If you make a wavepacket whose width is less wide than a hole, it will sail right through. There isn't some magical nonlocal effect that makes it bounce off.
$endgroup$
– knzhou
Dec 22 '18 at 14:09
$begingroup$
Well, that makes sense.
$endgroup$
– b.Lorenz
Dec 22 '18 at 14:25
$begingroup$
@b.Lorenz If you think about it a bit more, you'll realize that while it sounds superficially like a wrong explanation, it's actually right. If you make a wavepacket whose width is less wide than a hole, it will sail right through. There isn't some magical nonlocal effect that makes it bounce off.
$endgroup$
– knzhou
Dec 22 '18 at 14:09
$begingroup$
@b.Lorenz If you think about it a bit more, you'll realize that while it sounds superficially like a wrong explanation, it's actually right. If you make a wavepacket whose width is less wide than a hole, it will sail right through. There isn't some magical nonlocal effect that makes it bounce off.
$endgroup$
– knzhou
Dec 22 '18 at 14:09
$begingroup$
Well, that makes sense.
$endgroup$
– b.Lorenz
Dec 22 '18 at 14:25
$begingroup$
Well, that makes sense.
$endgroup$
– b.Lorenz
Dec 22 '18 at 14:25
add a comment |
$begingroup$
The answer is the wave is not actually changing. Think of it this way. When a car playing loud music drives by, the pitch increases as it approaches, sounds correct for a moment, then the pitch decreases as it moves away. The person in the car hears the same pitch the whole time.
Edit:
Additionally, imagine a car is driving 40 mph toward a wall which is moving 0 mph. The collision will occur with results consistent with a 40 mph collision. Now if a car is driving 40 mph toward another car which is driving 40 mph toward it (head on), a collision will occur with results consistent with a car driving 80 mph toward a wall which is moving 0 mph. If a car which is driving 40 mph toward another car which is driving 40 mph from behind (rear end), a collision will not even occur...so let us imagine the car in the rear is driving 41 mph...a collision will occur which is consistent with a car driving 1 mph toward a wall which is moving 0 mph.
In other words:
The music which has exited the car through the window and has been heard by someone standing outside the car, is the sound which is experienced as being different.
The music is a higher pitch while the car is moving toward the person because the speed* of the wave is increased by the speed of the car. The pitch sounds normal for a moment because the car is not moving toward or away from the person which leaves the speed* of the sound as it is, and the pitch is lower as the car is moving away because the speed* of the sound wave is reduced by the speed of the car.
Application to the microwave scenario:
The microwave oven is like the car and the electromagnetic wave (visible or microwave) is like the sound wave of the music.
The electromagnetic wave being shifted is the one which has left the interior of the microwave oven.
The shift being observed in the electromagnetic wave is caused by a change in the speed* of the wave as it is observed relative to the observer.
Edit: *speed of the waves therefore increased frequency.
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
$endgroup$
add a comment |
$begingroup$
The answer is the wave is not actually changing. Think of it this way. When a car playing loud music drives by, the pitch increases as it approaches, sounds correct for a moment, then the pitch decreases as it moves away. The person in the car hears the same pitch the whole time.
Edit:
Additionally, imagine a car is driving 40 mph toward a wall which is moving 0 mph. The collision will occur with results consistent with a 40 mph collision. Now if a car is driving 40 mph toward another car which is driving 40 mph toward it (head on), a collision will occur with results consistent with a car driving 80 mph toward a wall which is moving 0 mph. If a car which is driving 40 mph toward another car which is driving 40 mph from behind (rear end), a collision will not even occur...so let us imagine the car in the rear is driving 41 mph...a collision will occur which is consistent with a car driving 1 mph toward a wall which is moving 0 mph.
In other words:
The music which has exited the car through the window and has been heard by someone standing outside the car, is the sound which is experienced as being different.
The music is a higher pitch while the car is moving toward the person because the speed* of the wave is increased by the speed of the car. The pitch sounds normal for a moment because the car is not moving toward or away from the person which leaves the speed* of the sound as it is, and the pitch is lower as the car is moving away because the speed* of the sound wave is reduced by the speed of the car.
Application to the microwave scenario:
The microwave oven is like the car and the electromagnetic wave (visible or microwave) is like the sound wave of the music.
The electromagnetic wave being shifted is the one which has left the interior of the microwave oven.
The shift being observed in the electromagnetic wave is caused by a change in the speed* of the wave as it is observed relative to the observer.
Edit: *speed of the waves therefore increased frequency.
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
$endgroup$
add a comment |
$begingroup$
The answer is the wave is not actually changing. Think of it this way. When a car playing loud music drives by, the pitch increases as it approaches, sounds correct for a moment, then the pitch decreases as it moves away. The person in the car hears the same pitch the whole time.
Edit:
Additionally, imagine a car is driving 40 mph toward a wall which is moving 0 mph. The collision will occur with results consistent with a 40 mph collision. Now if a car is driving 40 mph toward another car which is driving 40 mph toward it (head on), a collision will occur with results consistent with a car driving 80 mph toward a wall which is moving 0 mph. If a car which is driving 40 mph toward another car which is driving 40 mph from behind (rear end), a collision will not even occur...so let us imagine the car in the rear is driving 41 mph...a collision will occur which is consistent with a car driving 1 mph toward a wall which is moving 0 mph.
In other words:
The music which has exited the car through the window and has been heard by someone standing outside the car, is the sound which is experienced as being different.
The music is a higher pitch while the car is moving toward the person because the speed* of the wave is increased by the speed of the car. The pitch sounds normal for a moment because the car is not moving toward or away from the person which leaves the speed* of the sound as it is, and the pitch is lower as the car is moving away because the speed* of the sound wave is reduced by the speed of the car.
Application to the microwave scenario:
The microwave oven is like the car and the electromagnetic wave (visible or microwave) is like the sound wave of the music.
The electromagnetic wave being shifted is the one which has left the interior of the microwave oven.
The shift being observed in the electromagnetic wave is caused by a change in the speed* of the wave as it is observed relative to the observer.
Edit: *speed of the waves therefore increased frequency.
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
$endgroup$
The answer is the wave is not actually changing. Think of it this way. When a car playing loud music drives by, the pitch increases as it approaches, sounds correct for a moment, then the pitch decreases as it moves away. The person in the car hears the same pitch the whole time.
Edit:
Additionally, imagine a car is driving 40 mph toward a wall which is moving 0 mph. The collision will occur with results consistent with a 40 mph collision. Now if a car is driving 40 mph toward another car which is driving 40 mph toward it (head on), a collision will occur with results consistent with a car driving 80 mph toward a wall which is moving 0 mph. If a car which is driving 40 mph toward another car which is driving 40 mph from behind (rear end), a collision will not even occur...so let us imagine the car in the rear is driving 41 mph...a collision will occur which is consistent with a car driving 1 mph toward a wall which is moving 0 mph.
In other words:
The music which has exited the car through the window and has been heard by someone standing outside the car, is the sound which is experienced as being different.
The music is a higher pitch while the car is moving toward the person because the speed* of the wave is increased by the speed of the car. The pitch sounds normal for a moment because the car is not moving toward or away from the person which leaves the speed* of the sound as it is, and the pitch is lower as the car is moving away because the speed* of the sound wave is reduced by the speed of the car.
Application to the microwave scenario:
The microwave oven is like the car and the electromagnetic wave (visible or microwave) is like the sound wave of the music.
The electromagnetic wave being shifted is the one which has left the interior of the microwave oven.
The shift being observed in the electromagnetic wave is caused by a change in the speed* of the wave as it is observed relative to the observer.
Edit: *speed of the waves therefore increased frequency.
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
edited Dec 22 '18 at 18:10
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
answered Dec 22 '18 at 4:13
takintoolongtakintoolong
1194
1194
add a comment |
add a comment |
$begingroup$
Doppler affects the observers perception. The observed does not change, so the microwaves would still be blocked and, you could not see them at all, let alone them shifted.
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
$endgroup$
add a comment |
$begingroup$
Doppler affects the observers perception. The observed does not change, so the microwaves would still be blocked and, you could not see them at all, let alone them shifted.
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
$endgroup$
add a comment |
$begingroup$
Doppler affects the observers perception. The observed does not change, so the microwaves would still be blocked and, you could not see them at all, let alone them shifted.
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
$endgroup$
Doppler affects the observers perception. The observed does not change, so the microwaves would still be blocked and, you could not see them at all, let alone them shifted.
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
answered Dec 23 '18 at 12:46
Torque LeBoeufTorque LeBoeuf
1
1
add a comment |
add a comment |
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$begingroup$
The screen moves with the oven. From it's point of view, the radiation is in the microwave region, and it won't let any out. It would be different if you could jury rig the oven to operate with the door removed, and then you carried the door with you while you did the experiment. In that case, the screen's point of view would be the same as your point of view. It would see blue light, and it would let the blue light through the holes.
$endgroup$
– Solomon Slow
Dec 21 '18 at 20:56
$begingroup$
@SolomonSlow What you are saying is right, but there is more in this question. Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not), regardless of the frame of reference we are using. I hope that StudyStudy or someone else more well-versed in relativistic electrodynamics than me would write an answer.
$endgroup$
– b.Lorenz
Dec 21 '18 at 20:56
$begingroup$
It's going to come back to the Faraday tensor....en.m.wikipedia.org/wiki/Electromagnetic_tensor
$endgroup$
– user214814
Dec 21 '18 at 20:58
$begingroup$
That does not answer the question. It is clear that the light will not come out from the start.
$endgroup$
– my2cts
Dec 21 '18 at 21:08
$begingroup$
"Per relativity, the same thing has to happen (either we can detect photons by a fixed detector in front of the owen or not" - sure, and the answer in this case is "not" in all frames of reference. You will not see microwaves getting out when standing in front of it, when flying towards it, or with a goat, or on a boat. Where is the mystery.
$endgroup$
– Maury Markowitz
Dec 21 '18 at 21:10