Which observer measures proper time in the twins paradox?










9












$begingroup$


I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.










share|cite|improve this question











$endgroup$
















    9












    $begingroup$


    I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.










    share|cite|improve this question











    $endgroup$














      9












      9








      9


      1



      $begingroup$


      I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.










      share|cite|improve this question











      $endgroup$




      I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.







      special-relativity reference-frames coordinate-systems inertial-frames observers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 12 '18 at 3:01









      Qmechanic

      105k121881202




      105k121881202










      asked Nov 12 '18 at 2:18









      Max WILLIAMSMax WILLIAMS

      4612




      4612




















          3 Answers
          3






          active

          oldest

          votes


















          22












          $begingroup$

          Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
          object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
          object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
          The objects may age different amounts between meetings.



          There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.



          One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.



          Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
            $endgroup$
            – WillO
            Nov 12 '18 at 15:31










          • $begingroup$
            (PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
            $endgroup$
            – WillO
            Nov 12 '18 at 17:03


















          16












          $begingroup$


          I know that proper time is defined as the time which the clock moving
          relative to that observe shows.




          This isn't quite correct so let's start here.



          First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.



          Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:




          In relativity, proper time along a timelike world line is defined as
          the time as measured by a clock following that line.




          In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.



          The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:




          An accelerated clock will measure a smaller elapsed time between two
          events than that measured by a non-accelerated (inertial) clock
          between the same two events. The twin paradox is an example of this
          effect.







          share|cite|improve this answer











          $endgroup$




















            4












            $begingroup$

            In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.



            There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation



            $$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$



            The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              P.S. the metric equation is in units of $c=1$.
              $endgroup$
              – Zack Hutchens
              Nov 12 '18 at 2:37










            • $begingroup$
              Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
              $endgroup$
              – Alfred Centauri
              Nov 12 '18 at 11:46










            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%2f440410%2fwhich-observer-measures-proper-time-in-the-twins-paradox%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









            22












            $begingroup$

            Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
            object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
            object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
            The objects may age different amounts between meetings.



            There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.



            One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.



            Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
              $endgroup$
              – WillO
              Nov 12 '18 at 15:31










            • $begingroup$
              (PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
              $endgroup$
              – WillO
              Nov 12 '18 at 17:03















            22












            $begingroup$

            Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
            object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
            object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
            The objects may age different amounts between meetings.



            There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.



            One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.



            Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
              $endgroup$
              – WillO
              Nov 12 '18 at 15:31










            • $begingroup$
              (PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
              $endgroup$
              – WillO
              Nov 12 '18 at 17:03













            22












            22








            22





            $begingroup$

            Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
            object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
            object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
            The objects may age different amounts between meetings.



            There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.



            One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.



            Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.






            share|cite|improve this answer











            $endgroup$



            Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
            object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
            object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
            The objects may age different amounts between meetings.



            There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.



            One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.



            Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 12 '18 at 17:04









            WillO

            6,88322132




            6,88322132










            answered Nov 12 '18 at 2:36









            Dan YandDan Yand

            11.1k21540




            11.1k21540











            • $begingroup$
              Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
              $endgroup$
              – WillO
              Nov 12 '18 at 15:31










            • $begingroup$
              (PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
              $endgroup$
              – WillO
              Nov 12 '18 at 17:03
















            • $begingroup$
              Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
              $endgroup$
              – WillO
              Nov 12 '18 at 15:31










            • $begingroup$
              (PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
              $endgroup$
              – WillO
              Nov 12 '18 at 17:03















            $begingroup$
            Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
            $endgroup$
            – WillO
            Nov 12 '18 at 15:31




            $begingroup$
            Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
            $endgroup$
            – WillO
            Nov 12 '18 at 15:31












            $begingroup$
            (PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
            $endgroup$
            – WillO
            Nov 12 '18 at 17:03




            $begingroup$
            (PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
            $endgroup$
            – WillO
            Nov 12 '18 at 17:03











            16












            $begingroup$


            I know that proper time is defined as the time which the clock moving
            relative to that observe shows.




            This isn't quite correct so let's start here.



            First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.



            Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:




            In relativity, proper time along a timelike world line is defined as
            the time as measured by a clock following that line.




            In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.



            The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:




            An accelerated clock will measure a smaller elapsed time between two
            events than that measured by a non-accelerated (inertial) clock
            between the same two events. The twin paradox is an example of this
            effect.







            share|cite|improve this answer











            $endgroup$

















              16












              $begingroup$


              I know that proper time is defined as the time which the clock moving
              relative to that observe shows.




              This isn't quite correct so let's start here.



              First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.



              Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:




              In relativity, proper time along a timelike world line is defined as
              the time as measured by a clock following that line.




              In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.



              The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:




              An accelerated clock will measure a smaller elapsed time between two
              events than that measured by a non-accelerated (inertial) clock
              between the same two events. The twin paradox is an example of this
              effect.







              share|cite|improve this answer











              $endgroup$















                16












                16








                16





                $begingroup$


                I know that proper time is defined as the time which the clock moving
                relative to that observe shows.




                This isn't quite correct so let's start here.



                First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.



                Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:




                In relativity, proper time along a timelike world line is defined as
                the time as measured by a clock following that line.




                In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.



                The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:




                An accelerated clock will measure a smaller elapsed time between two
                events than that measured by a non-accelerated (inertial) clock
                between the same two events. The twin paradox is an example of this
                effect.







                share|cite|improve this answer











                $endgroup$




                I know that proper time is defined as the time which the clock moving
                relative to that observe shows.




                This isn't quite correct so let's start here.



                First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.



                Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:




                In relativity, proper time along a timelike world line is defined as
                the time as measured by a clock following that line.




                In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.



                The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:




                An accelerated clock will measure a smaller elapsed time between two
                events than that measured by a non-accelerated (inertial) clock
                between the same two events. The twin paradox is an example of this
                effect.








                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 12 '18 at 3:24

























                answered Nov 12 '18 at 3:08









                Alfred CentauriAlfred Centauri

                48.2k350150




                48.2k350150





















                    4












                    $begingroup$

                    In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.



                    There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation



                    $$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$



                    The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.






                    share|cite|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      P.S. the metric equation is in units of $c=1$.
                      $endgroup$
                      – Zack Hutchens
                      Nov 12 '18 at 2:37










                    • $begingroup$
                      Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
                      $endgroup$
                      – Alfred Centauri
                      Nov 12 '18 at 11:46















                    4












                    $begingroup$

                    In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.



                    There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation



                    $$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$



                    The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.






                    share|cite|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      P.S. the metric equation is in units of $c=1$.
                      $endgroup$
                      – Zack Hutchens
                      Nov 12 '18 at 2:37










                    • $begingroup$
                      Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
                      $endgroup$
                      – Alfred Centauri
                      Nov 12 '18 at 11:46













                    4












                    4








                    4





                    $begingroup$

                    In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.



                    There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation



                    $$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$



                    The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.






                    share|cite|improve this answer









                    $endgroup$



                    In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.



                    There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation



                    $$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$



                    The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 12 '18 at 2:36









                    Zack HutchensZack Hutchens

                    1,8841716




                    1,8841716







                    • 1




                      $begingroup$
                      P.S. the metric equation is in units of $c=1$.
                      $endgroup$
                      – Zack Hutchens
                      Nov 12 '18 at 2:37










                    • $begingroup$
                      Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
                      $endgroup$
                      – Alfred Centauri
                      Nov 12 '18 at 11:46












                    • 1




                      $begingroup$
                      P.S. the metric equation is in units of $c=1$.
                      $endgroup$
                      – Zack Hutchens
                      Nov 12 '18 at 2:37










                    • $begingroup$
                      Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
                      $endgroup$
                      – Alfred Centauri
                      Nov 12 '18 at 11:46







                    1




                    1




                    $begingroup$
                    P.S. the metric equation is in units of $c=1$.
                    $endgroup$
                    – Zack Hutchens
                    Nov 12 '18 at 2:37




                    $begingroup$
                    P.S. the metric equation is in units of $c=1$.
                    $endgroup$
                    – Zack Hutchens
                    Nov 12 '18 at 2:37












                    $begingroup$
                    Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
                    $endgroup$
                    – Alfred Centauri
                    Nov 12 '18 at 11:46




                    $begingroup$
                    Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
                    $endgroup$
                    – Alfred Centauri
                    Nov 12 '18 at 11:46

















                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f440410%2fwhich-observer-measures-proper-time-in-the-twins-paradox%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

                    𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

                    Crossroads (UK TV series)

                    ữḛḳṊẴ ẋ,Ẩṙ,ỹḛẪẠứụỿṞṦ,Ṉẍừ,ứ Ị,Ḵ,ṏ ṇỪḎḰṰọửḊ ṾḨḮữẑỶṑỗḮṣṉẃ Ữẩụ,ṓ,ḹẕḪḫỞṿḭ ỒṱṨẁṋṜ ḅẈ ṉ ứṀḱṑỒḵ,ḏ,ḊḖỹẊ Ẻḷổ,ṥ ẔḲẪụḣể Ṱ ḭỏựẶ Ồ Ṩ,ẂḿṡḾồ ỗṗṡịṞẤḵṽẃ ṸḒẄẘ,ủẞẵṦṟầṓế