Schwarzschild Geometry - Shouldn't Time Speed up near a Black Hole











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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?










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  • What makes you think there must be a flaw?
    – Buzz
    2 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
    2 hours ago

















up vote
1
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?










share|cite|improve this question






















  • What makes you think there must be a flaw?
    – Buzz
    2 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
    2 hours ago















up vote
1
down vote

favorite









up vote
1
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?










share|cite|improve this question













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 differential-geometry topology






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asked 2 hours ago









Luke Polson

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275












  • What makes you think there must be a flaw?
    – Buzz
    2 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
    2 hours ago




















  • What makes you think there must be a flaw?
    – Buzz
    2 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
    2 hours ago


















What makes you think there must be a flaw?
– Buzz
2 hours ago




What makes you think there must be a flaw?
– Buzz
2 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
2 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
2 hours ago












2 Answers
2






active

oldest

votes

















up vote
3
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).






share|cite|improve this answer

















  • 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
    1 hour 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
    1 hour 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
    55 mins 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
    31 mins 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
    30 mins ago


















up vote
5
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.






share|cite|improve this answer





















  • 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
    1 hour 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
    1 hour 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
    1 hour ago











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
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).






share|cite|improve this answer

















  • 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
    1 hour 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
    1 hour 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
    55 mins 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
    31 mins 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
    30 mins ago















up vote
3
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).






share|cite|improve this answer

















  • 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
    1 hour 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
    1 hour 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
    55 mins 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
    31 mins 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
    30 mins ago













up vote
3
down vote



accepted







up vote
3
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).






share|cite|improve this answer












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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









Thorondor

56515




56515








  • 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
    1 hour 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
    1 hour 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
    55 mins 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
    31 mins 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
    30 mins ago














  • 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
    1 hour 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
    1 hour 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
    55 mins 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
    31 mins 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
    30 mins 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
55 mins 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
55 mins 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
31 mins 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
31 mins 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
30 mins 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
30 mins ago










up vote
5
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.






share|cite|improve this answer





















  • 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
    1 hour 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
    1 hour 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
    1 hour ago















up vote
5
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.






share|cite|improve this answer





















  • 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
    1 hour 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
    1 hour 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
    1 hour ago













up vote
5
down vote










up vote
5
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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









Javier

13.9k74480




13.9k74480












  • 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
    1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago


















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