Does time speed up or slow down near a black hole?
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
add a comment |
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
add a comment |
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
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 Dec 17 at 11:56
knzhou
41.6k11117199
41.6k11117199
asked Dec 17 at 0:08
Luke Polson
5216
5216
add a comment |
add a comment |
3 Answers
3
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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).
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
Dec 17 at 1:49
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
Dec 17 at 1:58
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
Dec 17 at 2:33
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
Dec 17 at 2:34
2
@LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$.
– Javier
Dec 17 at 11:54
|
show 3 more comments
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.
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
Dec 17 at 1:20
@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
Dec 17 at 1:26
@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
Dec 17 at 1:42
add a comment |
As far as I can see, in all this discussion a point is lacking (unless
I missed it). Nobody has clearly stated that in order to compare times
an operational way of doing the comparison is required. Usually,
when we're comparing times of clocks occupying different space
locations, this is accomplished via light signals.
Only when the comparison procedure is specified it becomes meaningful to
say "time ... is less than time ...". Or else "The quantity that does
not change between frames is $t$". In latter case the obvious question
is: "how do you know of coordinate time?" (Not on paper, but in the
lab.)
In some cases there are obvious procedures which experts tend to let
understood, but this should be avoided when less experts are involved.
It's well known that here the most frequent cause of errors is lurking.
In present problem OP did simply interchange the meanings of $t$ and
$tau$. His final formula is right but its interpretation is upside
down.
add a comment |
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3 Answers
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3 Answers
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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).
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
Dec 17 at 1:49
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
Dec 17 at 1:58
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
Dec 17 at 2:33
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
Dec 17 at 2:34
2
@LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$.
– Javier
Dec 17 at 11:54
|
show 3 more comments
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).
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
Dec 17 at 1:49
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
Dec 17 at 1:58
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
Dec 17 at 2:33
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
Dec 17 at 2:34
2
@LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$.
– Javier
Dec 17 at 11:54
|
show 3 more comments
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).
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 Dec 17 at 1:27
Thorondor
1,08720
1,08720
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
Dec 17 at 1:49
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
Dec 17 at 1:58
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
Dec 17 at 2:33
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
Dec 17 at 2:34
2
@LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$.
– Javier
Dec 17 at 11:54
|
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
Dec 17 at 1:49
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
Dec 17 at 1:58
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
Dec 17 at 2:33
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
Dec 17 at 2:34
2
@LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$.
– Javier
Dec 17 at 11:54
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
Dec 17 at 1:49
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
Dec 17 at 1:49
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
Dec 17 at 1:58
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
Dec 17 at 1:58
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
Dec 17 at 2:33
@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
Dec 17 at 2:33
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
Dec 17 at 2:34
(...) 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
Dec 17 at 2:34
2
2
@LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$.
– Javier
Dec 17 at 11:54
@LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$.
– Javier
Dec 17 at 11:54
|
show 3 more comments
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.
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
Dec 17 at 1:20
@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
Dec 17 at 1:26
@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
Dec 17 at 1:42
add a comment |
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.
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
Dec 17 at 1:20
@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
Dec 17 at 1:26
@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
Dec 17 at 1:42
add a comment |
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.
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 Dec 17 at 1:11
Javier
14.1k74481
14.1k74481
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
Dec 17 at 1:20
@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
Dec 17 at 1:26
@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
Dec 17 at 1:42
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
Dec 17 at 1:20
@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
Dec 17 at 1:26
@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
Dec 17 at 1:42
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
Dec 17 at 1:20
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
Dec 17 at 1:20
@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
Dec 17 at 1:26
@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
Dec 17 at 1:26
@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
Dec 17 at 1:42
@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
Dec 17 at 1:42
add a comment |
As far as I can see, in all this discussion a point is lacking (unless
I missed it). Nobody has clearly stated that in order to compare times
an operational way of doing the comparison is required. Usually,
when we're comparing times of clocks occupying different space
locations, this is accomplished via light signals.
Only when the comparison procedure is specified it becomes meaningful to
say "time ... is less than time ...". Or else "The quantity that does
not change between frames is $t$". In latter case the obvious question
is: "how do you know of coordinate time?" (Not on paper, but in the
lab.)
In some cases there are obvious procedures which experts tend to let
understood, but this should be avoided when less experts are involved.
It's well known that here the most frequent cause of errors is lurking.
In present problem OP did simply interchange the meanings of $t$ and
$tau$. His final formula is right but its interpretation is upside
down.
add a comment |
As far as I can see, in all this discussion a point is lacking (unless
I missed it). Nobody has clearly stated that in order to compare times
an operational way of doing the comparison is required. Usually,
when we're comparing times of clocks occupying different space
locations, this is accomplished via light signals.
Only when the comparison procedure is specified it becomes meaningful to
say "time ... is less than time ...". Or else "The quantity that does
not change between frames is $t$". In latter case the obvious question
is: "how do you know of coordinate time?" (Not on paper, but in the
lab.)
In some cases there are obvious procedures which experts tend to let
understood, but this should be avoided when less experts are involved.
It's well known that here the most frequent cause of errors is lurking.
In present problem OP did simply interchange the meanings of $t$ and
$tau$. His final formula is right but its interpretation is upside
down.
add a comment |
As far as I can see, in all this discussion a point is lacking (unless
I missed it). Nobody has clearly stated that in order to compare times
an operational way of doing the comparison is required. Usually,
when we're comparing times of clocks occupying different space
locations, this is accomplished via light signals.
Only when the comparison procedure is specified it becomes meaningful to
say "time ... is less than time ...". Or else "The quantity that does
not change between frames is $t$". In latter case the obvious question
is: "how do you know of coordinate time?" (Not on paper, but in the
lab.)
In some cases there are obvious procedures which experts tend to let
understood, but this should be avoided when less experts are involved.
It's well known that here the most frequent cause of errors is lurking.
In present problem OP did simply interchange the meanings of $t$ and
$tau$. His final formula is right but its interpretation is upside
down.
As far as I can see, in all this discussion a point is lacking (unless
I missed it). Nobody has clearly stated that in order to compare times
an operational way of doing the comparison is required. Usually,
when we're comparing times of clocks occupying different space
locations, this is accomplished via light signals.
Only when the comparison procedure is specified it becomes meaningful to
say "time ... is less than time ...". Or else "The quantity that does
not change between frames is $t$". In latter case the obvious question
is: "how do you know of coordinate time?" (Not on paper, but in the
lab.)
In some cases there are obvious procedures which experts tend to let
understood, but this should be avoided when less experts are involved.
It's well known that here the most frequent cause of errors is lurking.
In present problem OP did simply interchange the meanings of $t$ and
$tau$. His final formula is right but its interpretation is upside
down.
answered Dec 20 at 16:35
Elio Fabri
2,2971112
2,2971112
add a comment |
add a comment |
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