definition of observer and time measured by different observers in general relativity












5












$begingroup$


An observer in general relativity is defined as a future directed timelike worldline
begin{align*}
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
end{align*}
together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
begin{align}
g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
end{align}
Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
begin{align}
tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
end{align}
However,
begin{align}
g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
end{align}
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.



This is all standard definition. Suppose, we have another observer $delta$:
begin{align*}
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
end{align*}
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
begin{align}
tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
end{align}
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.



However, I know that my conclusion is wrong. Can you point out where I went astray?






general-relativity observers







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    An observer in general relativity is defined as a future directed timelike worldline
    begin{align*}
    gamma:I subset mathbb R &to M \
    lambda &mapsto gamma(lambda)
    end{align*}
    together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
    begin{align}
    g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
    end{align}
    Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
    begin{align}
    tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
    end{align}
    However,
    begin{align}
    g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
    end{align}
    which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.



    This is all standard definition. Suppose, we have another observer $delta$:
    begin{align*}
    delta:I subset mathbb R &to M \
    lambda &mapsto delta(lambda)
    end{align*}
    and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
    begin{align}
    tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
    end{align}
    From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.



    However, I know that my conclusion is wrong. Can you point out where I went astray?






    general-relativity observers







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      An observer in general relativity is defined as a future directed timelike worldline
      begin{align*}
      gamma:I subset mathbb R &to M \
      lambda &mapsto gamma(lambda)
      end{align*}
      together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
      begin{align}
      g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
      end{align}
      Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
      begin{align}
      tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
      end{align}
      However,
      begin{align}
      g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
      end{align}
      which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.



      This is all standard definition. Suppose, we have another observer $delta$:
      begin{align*}
      delta:I subset mathbb R &to M \
      lambda &mapsto delta(lambda)
      end{align*}
      and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
      begin{align}
      tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
      end{align}
      From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.



      However, I know that my conclusion is wrong. Can you point out where I went astray?






      general-relativity observers







      share|cite|improve this question









      $endgroup$




      An observer in general relativity is defined as a future directed timelike worldline
      begin{align*}
      gamma:I subset mathbb R &to M \
      lambda &mapsto gamma(lambda)
      end{align*}
      together with an orthonormal basis $e_a(lambda) in T_{gamma(lambda)}M$ where $e_0(lambda)= v_{gamma, gamma(lambda)}$ and
      begin{align}
      g_{gamma(lambda)}(e_a(lambda),e_b(lambda))=eta_{ab}~. qquad (1)
      end{align}
      Here, $v_{gamma, gamma(lambda)}$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
      begin{align}
      tau_gamma = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)})}~.
      end{align}
      However,
      begin{align}
      g_{gamma(lambda)}(v_{gamma, gamma(lambda)},v_{gamma, gamma(lambda)}) = g_{gamma(lambda)}(e_0(lambda),e_0(lambda))=1 qquad (2)
      end{align}
      which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.



      This is all standard definition. Suppose, we have another observer $delta$:
      begin{align*}
      delta:I subset mathbb R &to M \
      lambda &mapsto delta(lambda)
      end{align*}
      and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
      begin{align}
      tau_delta = int_{lambda_0}^{lambda_1} dlambda sqrt{g_{delta(lambda)}(v_{delta, delta(lambda)},v_{delta, delta(lambda)})}~.
      end{align}
      From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.



      However, I know that my conclusion is wrong. Can you point out where I went astray?






      general-relativity observers




      general-relativity observers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 16 hours ago









      damaihatidamaihati

      683




      683






















          1 Answer
          1






          active

          oldest

          votes


















          6












          $begingroup$

          Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.



          Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
            $endgroup$
            – Javier
            9 hours ago












          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%2f470705%2fdefinition-of-observer-and-time-measured-by-different-observers-in-general-relat%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.



          Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
            $endgroup$
            – Javier
            9 hours ago
















          6












          $begingroup$

          Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.



          Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
            $endgroup$
            – Javier
            9 hours ago














          6












          6








          6





          $begingroup$

          Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.



          Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!






          share|cite|improve this answer









          $endgroup$



          Your conclusion is correct, because what you are doing by saying that $g(v_{gamma,gamma(lambda)},v_{gamma,gamma(lambda)}) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tilde{lambda}$ of the curve $gamma$ that have $g(v_{gamma,gamma(tilde{lambda})},v_{gamma,gamma(tilde{lambda})}) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.



          Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 16 hours ago









          VoidVoid

          10.7k1757




          10.7k1757












          • $begingroup$
            I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
            $endgroup$
            – Javier
            9 hours ago


















          • $begingroup$
            I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
            $endgroup$
            – Javier
            9 hours ago
















          $begingroup$
          I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
          $endgroup$
          – Javier
          9 hours ago




          $begingroup$
          I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
          $endgroup$
          – Javier
          9 hours ago


















          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%2f470705%2fdefinition-of-observer-and-time-measured-by-different-observers-in-general-relat%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

          Knooppunt Holsloot

          Image
          Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
          Read more

          Altaar (religie)

          Image
          Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
          Read more

          Gregoriusmis

          Image
          Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
          Read more