Does time speed up or slow down near a black hole?
up vote
5
down vote
favorite
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity black-holes differential-geometry metric-tensor time
edited 19 mins ago
knzhou
40.2k11113194
asked 12 hours ago
Luke Polson
476
add a comment |
up vote
5
down vote
favorite
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity black-holes differential-geometry metric-tensor time
edited 19 mins ago
knzhou
40.2k11113194
asked 12 hours ago
Luke Polson
476
What makes you think there must be a flaw?
– Buzz
12 hours ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
11 hours ago
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity black-holes differential-geometry metric-tensor time
edited 19 mins ago
knzhou
40.2k11113194
asked 12 hours ago
Luke Polson
476
The Schwarzchild geometry is defined as
$$ds^2=-left(1-frac{2GM}{r} right)dt^2+left(1-frac{2GM}{r} right)^{-1}dr^2+r^2(dtheta^2+sin^2(theta) dphi^2)$$
Lets examine what happens close to and far away from a black hole.
For a stationary observer at $r=infty$, we get
$$dtau^2=-ds^2=left(1-frac{2GM}{infty} right)dt^2=dt^2 $$
so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $theta=pi/2$) a distance $r=r_0$ away from the black hole, we get
$$dtau^2=left(1-frac{2GM}{r_0} right)dt^2-{r_0}^2dphi^2$$
For a circular orbit, it can be shown that $r_0^2 dphi^2=frac{GM}{r_0}dt^2$ and hence
$$dtau^2=left(1-frac{3GM}{r_0} right)dt^2$$
Thus $dtau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.
Would someone be able to point out the flaw in my logic here?
general-relativity black-holes differential-geometry metric-tensor time
general-relativity black-holes differential-geometry metric-tensor time
edited 19 mins ago
knzhou
40.2k11113194
asked 12 hours ago
Luke Polson
476
edited 19 mins ago
knzhou
40.2k11113194
asked 12 hours ago
Luke Polson
476
edited 19 mins ago
knzhou
40.2k11113194
edited 19 mins ago
knzhou
40.2k11113194
edited 19 mins ago
knzhou
40.2k11113194
40.2k11113194
asked 12 hours ago
Luke Polson
476
asked 12 hours ago
Luke Polson
476
asked 12 hours ago
Luke Polson
476
476
What makes you think there must be a flaw?
– Buzz
12 hours ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
11 hours ago
add a comment |
What makes you think there must be a flaw?
– Buzz
12 hours ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
11 hours ago
What makes you think there must be a flaw?
– Buzz
12 hours ago
What makes you think there must be a flaw?
– Buzz
12 hours ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
11 hours ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
11 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
11
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 10 hours ago
Thorondor
68818
1
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
10 hours ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
10 hours ago
@LukePolson I don't think that it's different from special relativity. SR is just a special case of GR where we assume that spacetime is governed by the Minkowski metric. However, the fact that the Minkowski metric is invariant under Lorentz transformations makes it easy to confuse proper and coordinate time when learning SR, since you can often get correct results without really understanding the difference.
– Thorondor
10 hours ago
2
@LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...)
– Javier
9 hours ago
2
(...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer.
– Javier
9 hours ago
|
show 3 more comments
up vote
9
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 11 hours ago
Javier
14k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
10 hours ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
10 hours ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
10 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
up vote
11
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 10 hours ago
Thorondor
68818
1
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
10 hours ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
10 hours ago
@LukePolson I don't think that it's different from special relativity. SR is just a special case of GR where we assume that spacetime is governed by the Minkowski metric. However, the fact that the Minkowski metric is invariant under Lorentz transformations makes it easy to confuse proper and coordinate time when learning SR, since you can often get correct results without really understanding the difference.
– Thorondor
10 hours ago
2
@LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...)
– Javier
9 hours ago
2
(...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer.
– Javier
9 hours ago
|
show 3 more comments
up vote
11
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 10 hours ago
Thorondor
68818
1
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
10 hours ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
10 hours ago
@LukePolson I don't think that it's different from special relativity. SR is just a special case of GR where we assume that spacetime is governed by the Minkowski metric. However, the fact that the Minkowski metric is invariant under Lorentz transformations makes it easy to confuse proper and coordinate time when learning SR, since you can often get correct results without really understanding the difference.
– Thorondor
10 hours ago
2
@LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...)
– Javier
9 hours ago
2
(...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer.
– Javier
9 hours ago
|
show 3 more comments
up vote
11
down vote
accepted
up vote
11
down vote
accepted
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 10 hours ago
Thorondor
68818
To expand on Javier's answer, the symbol $tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.
To avoid confusion, let's use $tau_infty$ for proper time in the frame of the stationary observer and $tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,
begin{align}
dtau_infty^2 &= dt^2 \
dtau_{orbit}^2 &= left( 1-frac{3GM}{r_0} right) dt^2
end{align}
It follows that
$$dtau_{orbit}^2 = left( 1-frac{3GM}{r_0} right) dtau_infty^2$$
which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).
answered 10 hours ago
Thorondor
68818
answered 10 hours ago
Thorondor
68818
answered 10 hours ago
Thorondor
68818
answered 10 hours ago
Thorondor
68818
68818
1
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
10 hours ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
10 hours ago
@LukePolson I don't think that it's different from special relativity. SR is just a special case of GR where we assume that spacetime is governed by the Minkowski metric. However, the fact that the Minkowski metric is invariant under Lorentz transformations makes it easy to confuse proper and coordinate time when learning SR, since you can often get correct results without really understanding the difference.
– Thorondor
10 hours ago
2
@LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...)
– Javier
9 hours ago
2
(...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer.
– Javier
9 hours ago
|
show 3 more comments
1
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
10 hours ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
10 hours ago
@LukePolson I don't think that it's different from special relativity. SR is just a special case of GR where we assume that spacetime is governed by the Minkowski metric. However, the fact that the Minkowski metric is invariant under Lorentz transformations makes it easy to confuse proper and coordinate time when learning SR, since you can often get correct results without really understanding the difference.
– Thorondor
10 hours ago
2
@LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...)
– Javier
9 hours ago
2
(...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer.
– Javier
9 hours ago
1
1
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
10 hours ago
Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames.
– Luke Polson
10 hours ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
10 hours ago
So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames.
– Luke Polson
10 hours ago
@LukePolson I don't think that it's different from special relativity. SR is just a special case of GR where we assume that spacetime is governed by the Minkowski metric. However, the fact that the Minkowski metric is invariant under Lorentz transformations makes it easy to confuse proper and coordinate time when learning SR, since you can often get correct results without really understanding the difference.
– Thorondor
10 hours ago
@LukePolson I don't think that it's different from special relativity. SR is just a special case of GR where we assume that spacetime is governed by the Minkowski metric. However, the fact that the Minkowski metric is invariant under Lorentz transformations makes it easy to confuse proper and coordinate time when learning SR, since you can often get correct results without really understanding the difference.
– Thorondor
10 hours ago
2
2
@LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...)
– Javier
9 hours ago
@LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...)
– Javier
9 hours ago
2
2
(...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer.
– Javier
9 hours ago
(...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer.
– Javier
9 hours ago
|
show 3 more comments
up vote
9
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 11 hours ago
Javier
14k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
10 hours ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
10 hours ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
10 hours ago
add a comment |
up vote
9
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 11 hours ago
Javier
14k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
10 hours ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
10 hours ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
10 hours ago
add a comment |
up vote
9
down vote
up vote
9
down vote
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 11 hours ago
Javier
14k74480
You are mixing the two times up. Proper time $tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.
answered 11 hours ago
Javier
14k74480
answered 11 hours ago
Javier
14k74480
answered 11 hours ago
Javier
14k74480
answered 11 hours ago
Javier
14k74480
14k74480
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
10 hours ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
10 hours ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
10 hours ago
add a comment |
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
10 hours ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
10 hours ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
10 hours ago
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
10 hours ago
If proper time is always measured by the observer you’re considering, how does $dS^2 =-dtau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time.
– Luke Polson
10 hours ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
10 hours ago
@LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though.
– Javier
10 hours ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
10 hours ago
@Luke Polson this answer is correct $dtau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity
– Dale
10 hours ago
add a comment |
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What makes you think there must be a flaw?
– Buzz
12 hours ago
Isn't it commonly excepted that time runs faster as one moves further away from large gravitational masses? I believe the effect is called gravitational redshift.
– Luke Polson
11 hours ago