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 18 hours ago









      damaihatidamaihati

      683




      683






















          1 Answer
          1






          active

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          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
            11 hours ago












          Your Answer





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          1 Answer
          1






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          active

          oldest

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          active

          oldest

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          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
            11 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
            11 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 18 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
            11 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
            11 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
          11 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
          11 hours ago


















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