Does time speed up or slow down near a black hole?












6














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





























    6














    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



























      6












      6








      6







      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 black-holes differential-geometry metric-tensor time






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 at 11:56









      knzhou

      40.8k11114196




      40.8k11114196










      asked Dec 17 at 0:08









      Luke Polson

      5216




      5216






















          3 Answers
          3






          active

          oldest

          votes


















          12














          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
            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



















          9














          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
            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



















          1














          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.






          share|cite|improve this answer





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "151"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448763%2fdoes-time-speed-up-or-slow-down-near-a-black-hole%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            12














            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
              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
















            12














            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
              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














            12












            12








            12






            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 Dec 17 at 1:27









            Thorondor

            1,07720




            1,07720








            • 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




              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











            9














            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
              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
















            9














            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
              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














            9












            9








            9






            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 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


















            • 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











            1














            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.






            share|cite|improve this answer


























              1














              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.






              share|cite|improve this answer
























                1












                1








                1






                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Elio Fabri

                2,0691111




                2,0691111






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Physics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f448763%2fdoes-time-speed-up-or-slow-down-near-a-black-hole%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    數位音樂下載

                    When can things happen in Etherscan, such as the picture below?

                    格利澤436b