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Does Doppler effect happen instantly?
The Doppler effect in a medium like air (sound) versus the electromagnetic Doppler effectsome questions regarding Doppler shifting versus absorption-emissionSpaceship Doppler frequencyDoppler shift for a uniformly accelerating observerDoppler shift and special relativityDoppler shift of light from stars and galaxiesSpectral lines and the Doppler effectEffect on pulsar signals of Earth's orbitDoppler effect equation derivationDoppler Shift in Michelson-Interferometer
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$begingroup$
Assume that a far star sends light toward a receiver. If we move this antenna such that it accelerate first for a moment, and then it moves with constant speed, we can see that the frequency of the received light will be shifted instantly in antenna's frame(toward blue or red, doesn't matter). The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted? After all, the light transmitter is far like 100 light years away, so when antenna moves, in antenna's frame we would expect some kind of delay to see a movement for the the far star (transmitter) and ofcourse the Doppler shift effect (antenna is at the rest in its frame, so the only reason for receiving doppler shifted frequency would be the movement of transmitter itself in this frame). But there is no such delay in formulas at least.
If you don't get what i am trying to say note that in every frame, Doppler shift happens because of transmitter movements (every frame consider itself at the rest!). If transmitter starts its movement while it's 100 light years away, we will see transmitter movement and Doppler shift effect 100 years later. However, when antenna moves, there is no such delay which is very strange. It is as if transmitter doesn't move, but frequency has been changed out of thin air.
I understand that in accelerated systems laws of physics changes, metric is different and etc. However, even by knowing accelerated frames metrics (like Rindler and such), i can't show that there is indeed a solution for this problem. Because after all, in reality antenna will recieve Doppler shifted light even after it maintains its speed and becomes inertia
I won't accept an answer without math even though i might upvote one. Everybody can say that accelerated systems are different, i need a thorough solution.
special-relativity causality doppler-effect
asked 9 hours ago
paradoxyparadoxy
17916
$endgroup$
|
show 1 more comment
$begingroup$
Assume that a far star sends light toward a receiver. If we move this antenna such that it accelerate first for a moment, and then it moves with constant speed, we can see that the frequency of the received light will be shifted instantly in antenna's frame(toward blue or red, doesn't matter). The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted? After all, the light transmitter is far like 100 light years away, so when antenna moves, in antenna's frame we would expect some kind of delay to see a movement for the the far star (transmitter) and ofcourse the Doppler shift effect (antenna is at the rest in its frame, so the only reason for receiving doppler shifted frequency would be the movement of transmitter itself in this frame). But there is no such delay in formulas at least.
If you don't get what i am trying to say note that in every frame, Doppler shift happens because of transmitter movements (every frame consider itself at the rest!). If transmitter starts its movement while it's 100 light years away, we will see transmitter movement and Doppler shift effect 100 years later. However, when antenna moves, there is no such delay which is very strange. It is as if transmitter doesn't move, but frequency has been changed out of thin air.
I understand that in accelerated systems laws of physics changes, metric is different and etc. However, even by knowing accelerated frames metrics (like Rindler and such), i can't show that there is indeed a solution for this problem. Because after all, in reality antenna will recieve Doppler shifted light even after it maintains its speed and becomes inertia
I won't accept an answer without math even though i might upvote one. Everybody can say that accelerated systems are different, i need a thorough solution.
special-relativity causality doppler-effect
asked 9 hours ago
paradoxyparadoxy
17916
$endgroup$
1
$begingroup$
Your use of the word "dilation" doesn't make sense to me. Do you perhaps mean "delay"?
$endgroup$
– Acccumulation
9 hours ago
1
$begingroup$
@Acccumulation My first language is not english, you are completely right. Thanks.
$endgroup$
– paradoxy
8 hours ago
$begingroup$
But acceleration isn't relative, the receiver knows it accelerated from its old inertial frame to its new one.
$endgroup$
– PM 2Ring
8 hours ago
$begingroup$
Just in case please note that a receiver can also consider himself moving in a frame of the stationary source. At least it is exactly what is going on in the Einstein’s 1905 paper (§7 Theory of Doppler principle and aberration) hermes.ffn.ub.es/luisnavarro/nuevo_maletin/… Mr. Einstein clearly says that “an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $nu$,“ and „the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light“
$endgroup$
– Albert
7 hours ago
2
$begingroup$
@Albert is right. In the frame of an inertial observer that observer is at rest. But relativity does not say that the observer is ignorant of any motion relative to other frames.
$endgroup$
– Aaron Stevens
7 hours ago
|
show 1 more comment
$begingroup$
Assume that a far star sends light toward a receiver. If we move this antenna such that it accelerate first for a moment, and then it moves with constant speed, we can see that the frequency of the received light will be shifted instantly in antenna's frame(toward blue or red, doesn't matter). The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted? After all, the light transmitter is far like 100 light years away, so when antenna moves, in antenna's frame we would expect some kind of delay to see a movement for the the far star (transmitter) and ofcourse the Doppler shift effect (antenna is at the rest in its frame, so the only reason for receiving doppler shifted frequency would be the movement of transmitter itself in this frame). But there is no such delay in formulas at least.
If you don't get what i am trying to say note that in every frame, Doppler shift happens because of transmitter movements (every frame consider itself at the rest!). If transmitter starts its movement while it's 100 light years away, we will see transmitter movement and Doppler shift effect 100 years later. However, when antenna moves, there is no such delay which is very strange. It is as if transmitter doesn't move, but frequency has been changed out of thin air.
I understand that in accelerated systems laws of physics changes, metric is different and etc. However, even by knowing accelerated frames metrics (like Rindler and such), i can't show that there is indeed a solution for this problem. Because after all, in reality antenna will recieve Doppler shifted light even after it maintains its speed and becomes inertia
I won't accept an answer without math even though i might upvote one. Everybody can say that accelerated systems are different, i need a thorough solution.
special-relativity causality doppler-effect
asked 9 hours ago
paradoxyparadoxy
17916
$endgroup$
Assume that a far star sends light toward a receiver. If we move this antenna such that it accelerate first for a moment, and then it moves with constant speed, we can see that the frequency of the received light will be shifted instantly in antenna's frame(toward blue or red, doesn't matter). The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted? After all, the light transmitter is far like 100 light years away, so when antenna moves, in antenna's frame we would expect some kind of delay to see a movement for the the far star (transmitter) and ofcourse the Doppler shift effect (antenna is at the rest in its frame, so the only reason for receiving doppler shifted frequency would be the movement of transmitter itself in this frame). But there is no such delay in formulas at least.
If you don't get what i am trying to say note that in every frame, Doppler shift happens because of transmitter movements (every frame consider itself at the rest!). If transmitter starts its movement while it's 100 light years away, we will see transmitter movement and Doppler shift effect 100 years later. However, when antenna moves, there is no such delay which is very strange. It is as if transmitter doesn't move, but frequency has been changed out of thin air.
I understand that in accelerated systems laws of physics changes, metric is different and etc. However, even by knowing accelerated frames metrics (like Rindler and such), i can't show that there is indeed a solution for this problem. Because after all, in reality antenna will recieve Doppler shifted light even after it maintains its speed and becomes inertia
I won't accept an answer without math even though i might upvote one. Everybody can say that accelerated systems are different, i need a thorough solution.
special-relativity causality doppler-effect
special-relativity causality doppler-effect
asked 9 hours ago
paradoxyparadoxy
17916
asked 9 hours ago
paradoxyparadoxy
17916
edited 8 hours ago
paradoxy
asked 9 hours ago
paradoxyparadoxy
17916
asked 9 hours ago
paradoxyparadoxy
17916
asked 9 hours ago
paradoxyparadoxy
17916
17916
1
$begingroup$
Your use of the word "dilation" doesn't make sense to me. Do you perhaps mean "delay"?
$endgroup$
– Acccumulation
9 hours ago
1
$begingroup$
@Acccumulation My first language is not english, you are completely right. Thanks.
$endgroup$
– paradoxy
8 hours ago
$begingroup$
But acceleration isn't relative, the receiver knows it accelerated from its old inertial frame to its new one.
$endgroup$
– PM 2Ring
8 hours ago
$begingroup$
Just in case please note that a receiver can also consider himself moving in a frame of the stationary source. At least it is exactly what is going on in the Einstein’s 1905 paper (§7 Theory of Doppler principle and aberration) hermes.ffn.ub.es/luisnavarro/nuevo_maletin/… Mr. Einstein clearly says that “an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $nu$,“ and „the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light“
$endgroup$
– Albert
7 hours ago
2
$begingroup$
@Albert is right. In the frame of an inertial observer that observer is at rest. But relativity does not say that the observer is ignorant of any motion relative to other frames.
$endgroup$
– Aaron Stevens
7 hours ago
|
show 1 more comment
1
$begingroup$
Your use of the word "dilation" doesn't make sense to me. Do you perhaps mean "delay"?
$endgroup$
– Acccumulation
9 hours ago
1
$begingroup$
@Acccumulation My first language is not english, you are completely right. Thanks.
$endgroup$
– paradoxy
8 hours ago
$begingroup$
But acceleration isn't relative, the receiver knows it accelerated from its old inertial frame to its new one.
$endgroup$
– PM 2Ring
8 hours ago
$begingroup$
Just in case please note that a receiver can also consider himself moving in a frame of the stationary source. At least it is exactly what is going on in the Einstein’s 1905 paper (§7 Theory of Doppler principle and aberration) hermes.ffn.ub.es/luisnavarro/nuevo_maletin/… Mr. Einstein clearly says that “an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $nu$,“ and „the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light“
$endgroup$
– Albert
7 hours ago
2
$begingroup$
@Albert is right. In the frame of an inertial observer that observer is at rest. But relativity does not say that the observer is ignorant of any motion relative to other frames.
$endgroup$
– Aaron Stevens
7 hours ago
1
1
$begingroup$
Your use of the word "dilation" doesn't make sense to me. Do you perhaps mean "delay"?
$endgroup$
– Acccumulation
9 hours ago
$begingroup$
Your use of the word "dilation" doesn't make sense to me. Do you perhaps mean "delay"?
$endgroup$
– Acccumulation
9 hours ago
1
1
$begingroup$
@Acccumulation My first language is not english, you are completely right. Thanks.
$endgroup$
– paradoxy
8 hours ago
$begingroup$
@Acccumulation My first language is not english, you are completely right. Thanks.
$endgroup$
– paradoxy
8 hours ago
$begingroup$
But acceleration isn't relative, the receiver knows it accelerated from its old inertial frame to its new one.
$endgroup$
– PM 2Ring
8 hours ago
$begingroup$
But acceleration isn't relative, the receiver knows it accelerated from its old inertial frame to its new one.
$endgroup$
– PM 2Ring
8 hours ago
$begingroup$
Just in case please note that a receiver can also consider himself moving in a frame of the stationary source. At least it is exactly what is going on in the Einstein’s 1905 paper (§7 Theory of Doppler principle and aberration) hermes.ffn.ub.es/luisnavarro/nuevo_maletin/… Mr. Einstein clearly says that “an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $nu$,“ and „the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light“
$endgroup$
– Albert
7 hours ago
$begingroup$
Just in case please note that a receiver can also consider himself moving in a frame of the stationary source. At least it is exactly what is going on in the Einstein’s 1905 paper (§7 Theory of Doppler principle and aberration) hermes.ffn.ub.es/luisnavarro/nuevo_maletin/… Mr. Einstein clearly says that “an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $nu$,“ and „the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light“
$endgroup$
– Albert
7 hours ago
2
2
$begingroup$
@Albert is right. In the frame of an inertial observer that observer is at rest. But relativity does not say that the observer is ignorant of any motion relative to other frames.
$endgroup$
– Aaron Stevens
7 hours ago
$begingroup$
@Albert is right. In the frame of an inertial observer that observer is at rest. But relativity does not say that the observer is ignorant of any motion relative to other frames.
$endgroup$
– Aaron Stevens
7 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted?
The light frequency was always shifted in this frame. (i.e. the frequency was always different in this frame than in the frame where the receiver is intially not moving)
You have two inertial frames, we can call them "A" and "B". Initially the receiver is at rest in frame A, and then it accelerates until it is at rest in frame B. But frame B wasn't created by this action. Frame B always existed. The only difference is that initially the receiver is at rest in frame A, so it measures the source frequency in frame A. Later the receiver is at rest in frame B, so it measures the source frequency in frame B.
But the frequency of the light in the frame B never changed, so there's no reason to worry about how it could have changed instantaneously.
answered 8 hours ago
The PhotonThe Photon
10.6k11935
$endgroup$
$begingroup$
I had the same answer in my mind before asking the question, however, there is a problem with this. In actual B's frame, star is really moving (because from the begining of the time, it was moving!) however, in antennta's frame who is just moved to B's frame, star would be at the rest, or are you suggesting that a very far star will start its movement in antennta's frame suddenly (which means star had moved 100y ago before antenna even starts its movement that now antenna see its movement?)Another possible answer might be discarding all of information that antenna ever had from A frame,but no!
$endgroup$
– paradoxy
7 hours ago
$begingroup$
There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been.
$endgroup$
– The Photon
7 hours ago
$begingroup$
But now that the receiver is at rest in frame B, it will start measuring the frequency of the light from the star as if it were an object at rest in frame B.
$endgroup$
– The Photon
6 hours ago
$begingroup$
RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one.
$endgroup$
– The Photon
6 hours ago
$begingroup$
I see no problem with constructing a frame for an accelerated observer (which is done by Rindler and others, as long as you keep acceleration constant though). What you are saying is like after changing of frames antenna will consider star in movement from the very begining of the time, while antenna clearly knows that it's not the case. As a matter of the fact, if it was the case, then every accelerated observers should discard every information they ever had at every moment. not very plausible, and it is not really the case (like in accelerated systems we couldn't solve equations of motion)
$endgroup$
– paradoxy
6 hours ago
|
show 1 more comment
$begingroup$
The position and velocity of a source of light are completely irrelevant once the light exists. It does not have a frequency nor a wavelength, both of those are (inertial) frame dependent quantities that transform as a 4 vector:
$$ k_mu = (omega/c, vec k) $$
Different frames see different $k_mu$, which transform according to a Lorentz transformation:
$$ k'_mu = Lambda_mu^nu k_nu $$.
answered 7 hours ago
JEBJEB
7,1111919
$endgroup$
$begingroup$
Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then?
$endgroup$
– paradoxy
7 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted?
The light frequency was always shifted in this frame. (i.e. the frequency was always different in this frame than in the frame where the receiver is intially not moving)
You have two inertial frames, we can call them "A" and "B". Initially the receiver is at rest in frame A, and then it accelerates until it is at rest in frame B. But frame B wasn't created by this action. Frame B always existed. The only difference is that initially the receiver is at rest in frame A, so it measures the source frequency in frame A. Later the receiver is at rest in frame B, so it measures the source frequency in frame B.
But the frequency of the light in the frame B never changed, so there's no reason to worry about how it could have changed instantaneously.
answered 8 hours ago
The PhotonThe Photon
10.6k11935
$endgroup$
$begingroup$
I had the same answer in my mind before asking the question, however, there is a problem with this. In actual B's frame, star is really moving (because from the begining of the time, it was moving!) however, in antennta's frame who is just moved to B's frame, star would be at the rest, or are you suggesting that a very far star will start its movement in antennta's frame suddenly (which means star had moved 100y ago before antenna even starts its movement that now antenna see its movement?)Another possible answer might be discarding all of information that antenna ever had from A frame,but no!
$endgroup$
– paradoxy
7 hours ago
$begingroup$
There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been.
$endgroup$
– The Photon
7 hours ago
$begingroup$
But now that the receiver is at rest in frame B, it will start measuring the frequency of the light from the star as if it were an object at rest in frame B.
$endgroup$
– The Photon
6 hours ago
$begingroup$
RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one.
$endgroup$
– The Photon
6 hours ago
$begingroup$
I see no problem with constructing a frame for an accelerated observer (which is done by Rindler and others, as long as you keep acceleration constant though). What you are saying is like after changing of frames antenna will consider star in movement from the very begining of the time, while antenna clearly knows that it's not the case. As a matter of the fact, if it was the case, then every accelerated observers should discard every information they ever had at every moment. not very plausible, and it is not really the case (like in accelerated systems we couldn't solve equations of motion)
$endgroup$
– paradoxy
6 hours ago
|
show 1 more comment
$begingroup$
The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted?
The light frequency was always shifted in this frame. (i.e. the frequency was always different in this frame than in the frame where the receiver is intially not moving)
You have two inertial frames, we can call them "A" and "B". Initially the receiver is at rest in frame A, and then it accelerates until it is at rest in frame B. But frame B wasn't created by this action. Frame B always existed. The only difference is that initially the receiver is at rest in frame A, so it measures the source frequency in frame A. Later the receiver is at rest in frame B, so it measures the source frequency in frame B.
But the frequency of the light in the frame B never changed, so there's no reason to worry about how it could have changed instantaneously.
answered 8 hours ago
The PhotonThe Photon
10.6k11935
$endgroup$
$begingroup$
I had the same answer in my mind before asking the question, however, there is a problem with this. In actual B's frame, star is really moving (because from the begining of the time, it was moving!) however, in antennta's frame who is just moved to B's frame, star would be at the rest, or are you suggesting that a very far star will start its movement in antennta's frame suddenly (which means star had moved 100y ago before antenna even starts its movement that now antenna see its movement?)Another possible answer might be discarding all of information that antenna ever had from A frame,but no!
$endgroup$
– paradoxy
7 hours ago
$begingroup$
There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been.
$endgroup$
– The Photon
7 hours ago
$begingroup$
But now that the receiver is at rest in frame B, it will start measuring the frequency of the light from the star as if it were an object at rest in frame B.
$endgroup$
– The Photon
6 hours ago
$begingroup$
RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one.
$endgroup$
– The Photon
6 hours ago
$begingroup$
I see no problem with constructing a frame for an accelerated observer (which is done by Rindler and others, as long as you keep acceleration constant though). What you are saying is like after changing of frames antenna will consider star in movement from the very begining of the time, while antenna clearly knows that it's not the case. As a matter of the fact, if it was the case, then every accelerated observers should discard every information they ever had at every moment. not very plausible, and it is not really the case (like in accelerated systems we couldn't solve equations of motion)
$endgroup$
– paradoxy
6 hours ago
|
show 1 more comment
$begingroup$
The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted?
The light frequency was always shifted in this frame. (i.e. the frequency was always different in this frame than in the frame where the receiver is intially not moving)
You have two inertial frames, we can call them "A" and "B". Initially the receiver is at rest in frame A, and then it accelerates until it is at rest in frame B. But frame B wasn't created by this action. Frame B always existed. The only difference is that initially the receiver is at rest in frame A, so it measures the source frequency in frame A. Later the receiver is at rest in frame B, so it measures the source frequency in frame B.
But the frequency of the light in the frame B never changed, so there's no reason to worry about how it could have changed instantaneously.
answered 8 hours ago
The PhotonThe Photon
10.6k11935
$endgroup$
The question is, in antenna's frame after acceleration, (which is indeed inertia) how can the frequency of the light still be shifted?
The light frequency was always shifted in this frame. (i.e. the frequency was always different in this frame than in the frame where the receiver is intially not moving)
You have two inertial frames, we can call them "A" and "B". Initially the receiver is at rest in frame A, and then it accelerates until it is at rest in frame B. But frame B wasn't created by this action. Frame B always existed. The only difference is that initially the receiver is at rest in frame A, so it measures the source frequency in frame A. Later the receiver is at rest in frame B, so it measures the source frequency in frame B.
But the frequency of the light in the frame B never changed, so there's no reason to worry about how it could have changed instantaneously.
answered 8 hours ago
The PhotonThe Photon
10.6k11935
edited 8 hours ago
answered 8 hours ago
The PhotonThe Photon
10.6k11935
answered 8 hours ago
The PhotonThe Photon
10.6k11935
answered 8 hours ago
The PhotonThe Photon
10.6k11935
10.6k11935
$begingroup$
I had the same answer in my mind before asking the question, however, there is a problem with this. In actual B's frame, star is really moving (because from the begining of the time, it was moving!) however, in antennta's frame who is just moved to B's frame, star would be at the rest, or are you suggesting that a very far star will start its movement in antennta's frame suddenly (which means star had moved 100y ago before antenna even starts its movement that now antenna see its movement?)Another possible answer might be discarding all of information that antenna ever had from A frame,but no!
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– paradoxy
7 hours ago
$begingroup$
There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been.
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– The Photon
7 hours ago
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But now that the receiver is at rest in frame B, it will start measuring the frequency of the light from the star as if it were an object at rest in frame B.
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– The Photon
6 hours ago
$begingroup$
RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one.
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– The Photon
6 hours ago
$begingroup$
I see no problem with constructing a frame for an accelerated observer (which is done by Rindler and others, as long as you keep acceleration constant though). What you are saying is like after changing of frames antenna will consider star in movement from the very begining of the time, while antenna clearly knows that it's not the case. As a matter of the fact, if it was the case, then every accelerated observers should discard every information they ever had at every moment. not very plausible, and it is not really the case (like in accelerated systems we couldn't solve equations of motion)
$endgroup$
– paradoxy
6 hours ago
|
show 1 more comment
$begingroup$
I had the same answer in my mind before asking the question, however, there is a problem with this. In actual B's frame, star is really moving (because from the begining of the time, it was moving!) however, in antennta's frame who is just moved to B's frame, star would be at the rest, or are you suggesting that a very far star will start its movement in antennta's frame suddenly (which means star had moved 100y ago before antenna even starts its movement that now antenna see its movement?)Another possible answer might be discarding all of information that antenna ever had from A frame,but no!
$endgroup$
– paradoxy
7 hours ago
$begingroup$
There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been.
$endgroup$
– The Photon
7 hours ago
$begingroup$
But now that the receiver is at rest in frame B, it will start measuring the frequency of the light from the star as if it were an object at rest in frame B.
$endgroup$
– The Photon
6 hours ago
$begingroup$
RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one.
$endgroup$
– The Photon
6 hours ago
$begingroup$
I see no problem with constructing a frame for an accelerated observer (which is done by Rindler and others, as long as you keep acceleration constant though). What you are saying is like after changing of frames antenna will consider star in movement from the very begining of the time, while antenna clearly knows that it's not the case. As a matter of the fact, if it was the case, then every accelerated observers should discard every information they ever had at every moment. not very plausible, and it is not really the case (like in accelerated systems we couldn't solve equations of motion)
$endgroup$
– paradoxy
6 hours ago
$begingroup$
I had the same answer in my mind before asking the question, however, there is a problem with this. In actual B's frame, star is really moving (because from the begining of the time, it was moving!) however, in antennta's frame who is just moved to B's frame, star would be at the rest, or are you suggesting that a very far star will start its movement in antennta's frame suddenly (which means star had moved 100y ago before antenna even starts its movement that now antenna see its movement?)Another possible answer might be discarding all of information that antenna ever had from A frame,but no!
$endgroup$
– paradoxy
7 hours ago
$begingroup$
I had the same answer in my mind before asking the question, however, there is a problem with this. In actual B's frame, star is really moving (because from the begining of the time, it was moving!) however, in antennta's frame who is just moved to B's frame, star would be at the rest, or are you suggesting that a very far star will start its movement in antennta's frame suddenly (which means star had moved 100y ago before antenna even starts its movement that now antenna see its movement?)Another possible answer might be discarding all of information that antenna ever had from A frame,but no!
$endgroup$
– paradoxy
7 hours ago
$begingroup$
There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been.
$endgroup$
– The Photon
7 hours ago
$begingroup$
There is no "antenna's frame". There is frame A and frame B. The antenna is initially at rest in frame A, then later it is at rest in frame B. When the antenna changes velocity to be at rest in frame B, it doesn't change anything about the star in frame B. The star is still moving in frame B the way it always has been.
$endgroup$
– The Photon
7 hours ago
$begingroup$
But now that the receiver is at rest in frame B, it will start measuring the frequency of the light from the star as if it were an object at rest in frame B.
$endgroup$
– The Photon
6 hours ago
$begingroup$
But now that the receiver is at rest in frame B, it will start measuring the frequency of the light from the star as if it were an object at rest in frame B.
$endgroup$
– The Photon
6 hours ago
$begingroup$
RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one.
$endgroup$
– The Photon
6 hours ago
$begingroup$
RE "there is no 'antenna's frame'", of course what I mean is the antenna's frame is not an inertial frame. Trying to analyze what happens in this non-inertial frame is beyond my knowledge. But you shouldn't expect it to behave like an inertial frame, because it isn't one.
$endgroup$
– The Photon
6 hours ago
$begingroup$
I see no problem with constructing a frame for an accelerated observer (which is done by Rindler and others, as long as you keep acceleration constant though). What you are saying is like after changing of frames antenna will consider star in movement from the very begining of the time, while antenna clearly knows that it's not the case. As a matter of the fact, if it was the case, then every accelerated observers should discard every information they ever had at every moment. not very plausible, and it is not really the case (like in accelerated systems we couldn't solve equations of motion)
$endgroup$
– paradoxy
6 hours ago
$begingroup$
I see no problem with constructing a frame for an accelerated observer (which is done by Rindler and others, as long as you keep acceleration constant though). What you are saying is like after changing of frames antenna will consider star in movement from the very begining of the time, while antenna clearly knows that it's not the case. As a matter of the fact, if it was the case, then every accelerated observers should discard every information they ever had at every moment. not very plausible, and it is not really the case (like in accelerated systems we couldn't solve equations of motion)
$endgroup$
– paradoxy
6 hours ago
|
show 1 more comment
$begingroup$
The position and velocity of a source of light are completely irrelevant once the light exists. It does not have a frequency nor a wavelength, both of those are (inertial) frame dependent quantities that transform as a 4 vector:
$$ k_mu = (omega/c, vec k) $$
Different frames see different $k_mu$, which transform according to a Lorentz transformation:
$$ k'_mu = Lambda_mu^nu k_nu $$.
answered 7 hours ago
JEBJEB
7,1111919
$endgroup$
$begingroup$
Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then?
$endgroup$
– paradoxy
7 hours ago
add a comment |
$begingroup$
The position and velocity of a source of light are completely irrelevant once the light exists. It does not have a frequency nor a wavelength, both of those are (inertial) frame dependent quantities that transform as a 4 vector:
$$ k_mu = (omega/c, vec k) $$
Different frames see different $k_mu$, which transform according to a Lorentz transformation:
$$ k'_mu = Lambda_mu^nu k_nu $$.
answered 7 hours ago
JEBJEB
7,1111919
$endgroup$
$begingroup$
Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then?
$endgroup$
– paradoxy
7 hours ago
add a comment |
$begingroup$
The position and velocity of a source of light are completely irrelevant once the light exists. It does not have a frequency nor a wavelength, both of those are (inertial) frame dependent quantities that transform as a 4 vector:
$$ k_mu = (omega/c, vec k) $$
Different frames see different $k_mu$, which transform according to a Lorentz transformation:
$$ k'_mu = Lambda_mu^nu k_nu $$.
answered 7 hours ago
JEBJEB
7,1111919
$endgroup$
The position and velocity of a source of light are completely irrelevant once the light exists. It does not have a frequency nor a wavelength, both of those are (inertial) frame dependent quantities that transform as a 4 vector:
$$ k_mu = (omega/c, vec k) $$
Different frames see different $k_mu$, which transform according to a Lorentz transformation:
$$ k'_mu = Lambda_mu^nu k_nu $$.
answered 7 hours ago
JEBJEB
7,1111919
answered 7 hours ago
JEBJEB
7,1111919
answered 7 hours ago
JEBJEB
7,1111919
answered 7 hours ago
JEBJEB
7,1111919
7,1111919
$begingroup$
Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then?
$endgroup$
– paradoxy
7 hours ago
add a comment |
$begingroup$
Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then?
$endgroup$
– paradoxy
7 hours ago
$begingroup$
Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then?
$endgroup$
– paradoxy
7 hours ago
$begingroup$
Relative speed exists only between source of light and observer. To compute doppler shift you need to know this relative speed. If you discard the source of light, then to what you assign a velocity? The light itself travels at a null path. relative speed is meaningless for it, and that lorentz boost matrix also need a relative speed between frames. Antenta itself consider itself at the rest, if v in matrix of lorentz does not belong to source of light, it is for what then?
$endgroup$
– paradoxy
7 hours ago
add a comment |
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1
$begingroup$
Your use of the word "dilation" doesn't make sense to me. Do you perhaps mean "delay"?
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– Acccumulation
9 hours ago
1
$begingroup$
@Acccumulation My first language is not english, you are completely right. Thanks.
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– paradoxy
8 hours ago
$begingroup$
But acceleration isn't relative, the receiver knows it accelerated from its old inertial frame to its new one.
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– PM 2Ring
8 hours ago
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Just in case please note that a receiver can also consider himself moving in a frame of the stationary source. At least it is exactly what is going on in the Einstein’s 1905 paper (§7 Theory of Doppler principle and aberration) hermes.ffn.ub.es/luisnavarro/nuevo_maletin/… Mr. Einstein clearly says that “an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $nu$,“ and „the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light“
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– Albert
7 hours ago
2
$begingroup$
@Albert is right. In the frame of an inertial observer that observer is at rest. But relativity does not say that the observer is ignorant of any motion relative to other frames.
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– Aaron Stevens
7 hours ago