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Moving objects and gravitational radiation
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I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves. In reality, the energy loss is negligible and undetectable, but in some systems, like PSR 1913+16, it is not.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force? Does the observer eventually come to a standstill?
gravitational-waves relative-motion
asked 11 hours ago
BMFBMF
1446 bronze badges
$endgroup$
add a comment
|
$begingroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves. In reality, the energy loss is negligible and undetectable, but in some systems, like PSR 1913+16, it is not.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force? Does the observer eventually come to a standstill?
gravitational-waves relative-motion
asked 11 hours ago
BMFBMF
1446 bronze badges
$endgroup$
add a comment
|
$begingroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves. In reality, the energy loss is negligible and undetectable, but in some systems, like PSR 1913+16, it is not.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force? Does the observer eventually come to a standstill?
gravitational-waves relative-motion
asked 11 hours ago
BMFBMF
1446 bronze badges
$endgroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves. In reality, the energy loss is negligible and undetectable, but in some systems, like PSR 1913+16, it is not.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force? Does the observer eventually come to a standstill?
gravitational-waves relative-motion
gravitational-waves relative-motion
asked 11 hours ago
BMFBMF
1446 bronze badges
asked 11 hours ago
BMFBMF
1446 bronze badges
asked 11 hours ago
BMFBMF
1446 bronze badges
asked 11 hours ago
BMFBMF
1446 bronze badges
asked 11 hours ago
BMFBMF
1446 bronze badges
1446 bronze badges
add a comment
|
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves.
No, GR does not predict this. If an object is moving inertially (i.e., in a straight line with constant speed), then it doesn't radiate. The easy way to see this is that we can pick a frame of reference in which the object isn't moving at all.
An accelerating object may radiate. There are technical conditions that have to be satisfied if there is to be radiation, and there are some subtleties involved in defining what qualifies as a radiation field, but basically we expect masses to radiate when they're accelerating.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force?
If the observer is accelerating, then GR predicts that there will be a back-reaction force from the observer's own gravitational radiation. This is the gravitational version of a well-established effect in electromagnetism. Yes, the observer can measure the force. The Hulse-Taylor system, which you refer to, is an example.
However, this is not a decelerating effect on an object that is simply moving. That wouldn't make sense. Just as a matter of Galilean relativity, observers in different frames of reference don't even agree on whether an object is accelerating or decelerating. (Consider the case where you're driving alongside someone on the freeway, and then they step on the brakes.)
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
$endgroup$
$begingroup$
I felt that there was something fishy about what I read. Rereading it now with your answer, I can see that the reading meant something along the lines of "a system of massive objects," and not just a massive object itself, although the wording was misleading. The only thing fishy was my interpretation. Thanks for answering!
$endgroup$
– BMF
10 hours ago
$begingroup$
Re, "observers in different frames of reference don't even agree on whether an object is accelerating or decelerating." Maybe that's because, in strictly technical terms, "acceleration" and "deceleration" both mean the same thing. Consider a baseball thrown straight up into the air. As it rises, would you say that it is "accelerating?" or "decelerating?" What would you say a second or two later as the ball falls back to Earth? OK, now, at what point during its flight would you say that the acceleration due to gravity acting on the ball changed?
$endgroup$
– Solomon Slow
10 hours ago
1
$begingroup$
@SolomonSlow: No, it's not just a linguistic thing. Observers in different frames don't agree on the sign of $d|textbf{v}|/dt$.
$endgroup$
– Ben Crowell
8 hours ago
add a comment
|
$begingroup$
General relativity predicts that accelerating massive objects emit gravitational waves, analogously to how accelerating charged particles emit electromagnetic waves. If an object is in an inertial reference frame (i.e., if it is moving with a constant velocity), it will therefore not emit gravitational waves.
PSR 1913+16 is a binary system, meaning you have two massive objects orbiting each other and therefore always undergoing acceleration.
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
$endgroup$
1
$begingroup$
Great, concise answer, thanks! My entire GR worldview has been faithfully restored!
$endgroup$
– BMF
10 hours ago
add a comment
|
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves.
No, GR does not predict this. If an object is moving inertially (i.e., in a straight line with constant speed), then it doesn't radiate. The easy way to see this is that we can pick a frame of reference in which the object isn't moving at all.
An accelerating object may radiate. There are technical conditions that have to be satisfied if there is to be radiation, and there are some subtleties involved in defining what qualifies as a radiation field, but basically we expect masses to radiate when they're accelerating.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force?
If the observer is accelerating, then GR predicts that there will be a back-reaction force from the observer's own gravitational radiation. This is the gravitational version of a well-established effect in electromagnetism. Yes, the observer can measure the force. The Hulse-Taylor system, which you refer to, is an example.
However, this is not a decelerating effect on an object that is simply moving. That wouldn't make sense. Just as a matter of Galilean relativity, observers in different frames of reference don't even agree on whether an object is accelerating or decelerating. (Consider the case where you're driving alongside someone on the freeway, and then they step on the brakes.)
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
$endgroup$
$begingroup$
I felt that there was something fishy about what I read. Rereading it now with your answer, I can see that the reading meant something along the lines of "a system of massive objects," and not just a massive object itself, although the wording was misleading. The only thing fishy was my interpretation. Thanks for answering!
$endgroup$
– BMF
10 hours ago
$begingroup$
Re, "observers in different frames of reference don't even agree on whether an object is accelerating or decelerating." Maybe that's because, in strictly technical terms, "acceleration" and "deceleration" both mean the same thing. Consider a baseball thrown straight up into the air. As it rises, would you say that it is "accelerating?" or "decelerating?" What would you say a second or two later as the ball falls back to Earth? OK, now, at what point during its flight would you say that the acceleration due to gravity acting on the ball changed?
$endgroup$
– Solomon Slow
10 hours ago
1
$begingroup$
@SolomonSlow: No, it's not just a linguistic thing. Observers in different frames don't agree on the sign of $d|textbf{v}|/dt$.
$endgroup$
– Ben Crowell
8 hours ago
add a comment
|
$begingroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves.
No, GR does not predict this. If an object is moving inertially (i.e., in a straight line with constant speed), then it doesn't radiate. The easy way to see this is that we can pick a frame of reference in which the object isn't moving at all.
An accelerating object may radiate. There are technical conditions that have to be satisfied if there is to be radiation, and there are some subtleties involved in defining what qualifies as a radiation field, but basically we expect masses to radiate when they're accelerating.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force?
If the observer is accelerating, then GR predicts that there will be a back-reaction force from the observer's own gravitational radiation. This is the gravitational version of a well-established effect in electromagnetism. Yes, the observer can measure the force. The Hulse-Taylor system, which you refer to, is an example.
However, this is not a decelerating effect on an object that is simply moving. That wouldn't make sense. Just as a matter of Galilean relativity, observers in different frames of reference don't even agree on whether an object is accelerating or decelerating. (Consider the case where you're driving alongside someone on the freeway, and then they step on the brakes.)
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
$endgroup$
$begingroup$
I felt that there was something fishy about what I read. Rereading it now with your answer, I can see that the reading meant something along the lines of "a system of massive objects," and not just a massive object itself, although the wording was misleading. The only thing fishy was my interpretation. Thanks for answering!
$endgroup$
– BMF
10 hours ago
$begingroup$
Re, "observers in different frames of reference don't even agree on whether an object is accelerating or decelerating." Maybe that's because, in strictly technical terms, "acceleration" and "deceleration" both mean the same thing. Consider a baseball thrown straight up into the air. As it rises, would you say that it is "accelerating?" or "decelerating?" What would you say a second or two later as the ball falls back to Earth? OK, now, at what point during its flight would you say that the acceleration due to gravity acting on the ball changed?
$endgroup$
– Solomon Slow
10 hours ago
1
$begingroup$
@SolomonSlow: No, it's not just a linguistic thing. Observers in different frames don't agree on the sign of $d|textbf{v}|/dt$.
$endgroup$
– Ben Crowell
8 hours ago
add a comment
|
$begingroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves.
No, GR does not predict this. If an object is moving inertially (i.e., in a straight line with constant speed), then it doesn't radiate. The easy way to see this is that we can pick a frame of reference in which the object isn't moving at all.
An accelerating object may radiate. There are technical conditions that have to be satisfied if there is to be radiation, and there are some subtleties involved in defining what qualifies as a radiation field, but basically we expect masses to radiate when they're accelerating.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force?
If the observer is accelerating, then GR predicts that there will be a back-reaction force from the observer's own gravitational radiation. This is the gravitational version of a well-established effect in electromagnetism. Yes, the observer can measure the force. The Hulse-Taylor system, which you refer to, is an example.
However, this is not a decelerating effect on an object that is simply moving. That wouldn't make sense. Just as a matter of Galilean relativity, observers in different frames of reference don't even agree on whether an object is accelerating or decelerating. (Consider the case where you're driving alongside someone on the freeway, and then they step on the brakes.)
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
$endgroup$
I have read that GR predicts that moving, massive objects emit some of their energy as gravitational waves.
No, GR does not predict this. If an object is moving inertially (i.e., in a straight line with constant speed), then it doesn't radiate. The easy way to see this is that we can pick a frame of reference in which the object isn't moving at all.
An accelerating object may radiate. There are technical conditions that have to be satisfied if there is to be radiation, and there are some subtleties involved in defining what qualifies as a radiation field, but basically we expect masses to radiate when they're accelerating.
My question is this: if we set an observer into motion, and he emits some of his kinetic energy as GWs, can he in principle detect those GWs and feel a decelerating force?
If the observer is accelerating, then GR predicts that there will be a back-reaction force from the observer's own gravitational radiation. This is the gravitational version of a well-established effect in electromagnetism. Yes, the observer can measure the force. The Hulse-Taylor system, which you refer to, is an example.
However, this is not a decelerating effect on an object that is simply moving. That wouldn't make sense. Just as a matter of Galilean relativity, observers in different frames of reference don't even agree on whether an object is accelerating or decelerating. (Consider the case where you're driving alongside someone on the freeway, and then they step on the brakes.)
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
answered 11 hours ago
Ben CrowellBen Crowell
62k6 gold badges180 silver badges348 bronze badges
62k6 gold badges180 silver badges348 bronze badges
$begingroup$
I felt that there was something fishy about what I read. Rereading it now with your answer, I can see that the reading meant something along the lines of "a system of massive objects," and not just a massive object itself, although the wording was misleading. The only thing fishy was my interpretation. Thanks for answering!
$endgroup$
– BMF
10 hours ago
$begingroup$
Re, "observers in different frames of reference don't even agree on whether an object is accelerating or decelerating." Maybe that's because, in strictly technical terms, "acceleration" and "deceleration" both mean the same thing. Consider a baseball thrown straight up into the air. As it rises, would you say that it is "accelerating?" or "decelerating?" What would you say a second or two later as the ball falls back to Earth? OK, now, at what point during its flight would you say that the acceleration due to gravity acting on the ball changed?
$endgroup$
– Solomon Slow
10 hours ago
1
$begingroup$
@SolomonSlow: No, it's not just a linguistic thing. Observers in different frames don't agree on the sign of $d|textbf{v}|/dt$.
$endgroup$
– Ben Crowell
8 hours ago
add a comment
|
$begingroup$
I felt that there was something fishy about what I read. Rereading it now with your answer, I can see that the reading meant something along the lines of "a system of massive objects," and not just a massive object itself, although the wording was misleading. The only thing fishy was my interpretation. Thanks for answering!
$endgroup$
– BMF
10 hours ago
$begingroup$
Re, "observers in different frames of reference don't even agree on whether an object is accelerating or decelerating." Maybe that's because, in strictly technical terms, "acceleration" and "deceleration" both mean the same thing. Consider a baseball thrown straight up into the air. As it rises, would you say that it is "accelerating?" or "decelerating?" What would you say a second or two later as the ball falls back to Earth? OK, now, at what point during its flight would you say that the acceleration due to gravity acting on the ball changed?
$endgroup$
– Solomon Slow
10 hours ago
1
$begingroup$
@SolomonSlow: No, it's not just a linguistic thing. Observers in different frames don't agree on the sign of $d|textbf{v}|/dt$.
$endgroup$
– Ben Crowell
8 hours ago
$begingroup$
I felt that there was something fishy about what I read. Rereading it now with your answer, I can see that the reading meant something along the lines of "a system of massive objects," and not just a massive object itself, although the wording was misleading. The only thing fishy was my interpretation. Thanks for answering!
$endgroup$
– BMF
10 hours ago
$begingroup$
I felt that there was something fishy about what I read. Rereading it now with your answer, I can see that the reading meant something along the lines of "a system of massive objects," and not just a massive object itself, although the wording was misleading. The only thing fishy was my interpretation. Thanks for answering!
$endgroup$
– BMF
10 hours ago
$begingroup$
Re, "observers in different frames of reference don't even agree on whether an object is accelerating or decelerating." Maybe that's because, in strictly technical terms, "acceleration" and "deceleration" both mean the same thing. Consider a baseball thrown straight up into the air. As it rises, would you say that it is "accelerating?" or "decelerating?" What would you say a second or two later as the ball falls back to Earth? OK, now, at what point during its flight would you say that the acceleration due to gravity acting on the ball changed?
$endgroup$
– Solomon Slow
10 hours ago
$begingroup$
Re, "observers in different frames of reference don't even agree on whether an object is accelerating or decelerating." Maybe that's because, in strictly technical terms, "acceleration" and "deceleration" both mean the same thing. Consider a baseball thrown straight up into the air. As it rises, would you say that it is "accelerating?" or "decelerating?" What would you say a second or two later as the ball falls back to Earth? OK, now, at what point during its flight would you say that the acceleration due to gravity acting on the ball changed?
$endgroup$
– Solomon Slow
10 hours ago
1
1
$begingroup$
@SolomonSlow: No, it's not just a linguistic thing. Observers in different frames don't agree on the sign of $d|textbf{v}|/dt$.
$endgroup$
– Ben Crowell
8 hours ago
$begingroup$
@SolomonSlow: No, it's not just a linguistic thing. Observers in different frames don't agree on the sign of $d|textbf{v}|/dt$.
$endgroup$
– Ben Crowell
8 hours ago
add a comment
|
$begingroup$
General relativity predicts that accelerating massive objects emit gravitational waves, analogously to how accelerating charged particles emit electromagnetic waves. If an object is in an inertial reference frame (i.e., if it is moving with a constant velocity), it will therefore not emit gravitational waves.
PSR 1913+16 is a binary system, meaning you have two massive objects orbiting each other and therefore always undergoing acceleration.
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
$endgroup$
1
$begingroup$
Great, concise answer, thanks! My entire GR worldview has been faithfully restored!
$endgroup$
– BMF
10 hours ago
add a comment
|
$begingroup$
General relativity predicts that accelerating massive objects emit gravitational waves, analogously to how accelerating charged particles emit electromagnetic waves. If an object is in an inertial reference frame (i.e., if it is moving with a constant velocity), it will therefore not emit gravitational waves.
PSR 1913+16 is a binary system, meaning you have two massive objects orbiting each other and therefore always undergoing acceleration.
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
$endgroup$
1
$begingroup$
Great, concise answer, thanks! My entire GR worldview has been faithfully restored!
$endgroup$
– BMF
10 hours ago
add a comment
|
$begingroup$
General relativity predicts that accelerating massive objects emit gravitational waves, analogously to how accelerating charged particles emit electromagnetic waves. If an object is in an inertial reference frame (i.e., if it is moving with a constant velocity), it will therefore not emit gravitational waves.
PSR 1913+16 is a binary system, meaning you have two massive objects orbiting each other and therefore always undergoing acceleration.
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
$endgroup$
General relativity predicts that accelerating massive objects emit gravitational waves, analogously to how accelerating charged particles emit electromagnetic waves. If an object is in an inertial reference frame (i.e., if it is moving with a constant velocity), it will therefore not emit gravitational waves.
PSR 1913+16 is a binary system, meaning you have two massive objects orbiting each other and therefore always undergoing acceleration.
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
answered 11 hours ago
Franz U.Franz U.
563 bronze badges
563 bronze badges
1
$begingroup$
Great, concise answer, thanks! My entire GR worldview has been faithfully restored!
$endgroup$
– BMF
10 hours ago
add a comment
|
1
$begingroup$
Great, concise answer, thanks! My entire GR worldview has been faithfully restored!
$endgroup$
– BMF
10 hours ago
1
1
$begingroup$
Great, concise answer, thanks! My entire GR worldview has been faithfully restored!
$endgroup$
– BMF
10 hours ago
$begingroup$
Great, concise answer, thanks! My entire GR worldview has been faithfully restored!
$endgroup$
– BMF
10 hours ago
add a comment
|
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