Why is it natural to impose the condition that the metric remains unchanged under parallel transport?

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In Wald’s General Relativity, given a metric tensor $g_ab$, and two vectors $v^a,w^b$, the author said it is “natural” to impose the condition that the $g_abv^aw^b$ is invariant under parallel transport, which leads us to “naturally” choose a derivative operator.



But I can’t see why it is “natural”. Can anyone explain this to me?










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    We would like to preserve the norm (angle) of (between) vectors under parallel transport. See physics.stackexchange.com/q/47919
    – Avantgarde
    Aug 19 at 13:11











  • Adding to Avantgarde's excellent comment: We would like to preserve norms/angles because by the equivalence principle, a freely falling observer shouldn't notice any change in lengths or angles locally (in his (approximate) Lorentz frame).
    – balu
    Aug 19 at 20:10















up vote
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down vote

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In Wald’s General Relativity, given a metric tensor $g_ab$, and two vectors $v^a,w^b$, the author said it is “natural” to impose the condition that the $g_abv^aw^b$ is invariant under parallel transport, which leads us to “naturally” choose a derivative operator.



But I can’t see why it is “natural”. Can anyone explain this to me?










share|cite|improve this question



















  • 4




    We would like to preserve the norm (angle) of (between) vectors under parallel transport. See physics.stackexchange.com/q/47919
    – Avantgarde
    Aug 19 at 13:11











  • Adding to Avantgarde's excellent comment: We would like to preserve norms/angles because by the equivalence principle, a freely falling observer shouldn't notice any change in lengths or angles locally (in his (approximate) Lorentz frame).
    – balu
    Aug 19 at 20:10













up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





In Wald’s General Relativity, given a metric tensor $g_ab$, and two vectors $v^a,w^b$, the author said it is “natural” to impose the condition that the $g_abv^aw^b$ is invariant under parallel transport, which leads us to “naturally” choose a derivative operator.



But I can’t see why it is “natural”. Can anyone explain this to me?










share|cite|improve this question















In Wald’s General Relativity, given a metric tensor $g_ab$, and two vectors $v^a,w^b$, the author said it is “natural” to impose the condition that the $g_abv^aw^b$ is invariant under parallel transport, which leads us to “naturally” choose a derivative operator.



But I can’t see why it is “natural”. Can anyone explain this to me?







general-relativity differential-geometry metric-tensor differentiation






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edited Aug 19 at 16:42









Qmechanic♦

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asked Aug 19 at 12:06









Jerry

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1435







  • 4




    We would like to preserve the norm (angle) of (between) vectors under parallel transport. See physics.stackexchange.com/q/47919
    – Avantgarde
    Aug 19 at 13:11











  • Adding to Avantgarde's excellent comment: We would like to preserve norms/angles because by the equivalence principle, a freely falling observer shouldn't notice any change in lengths or angles locally (in his (approximate) Lorentz frame).
    – balu
    Aug 19 at 20:10













  • 4




    We would like to preserve the norm (angle) of (between) vectors under parallel transport. See physics.stackexchange.com/q/47919
    – Avantgarde
    Aug 19 at 13:11











  • Adding to Avantgarde's excellent comment: We would like to preserve norms/angles because by the equivalence principle, a freely falling observer shouldn't notice any change in lengths or angles locally (in his (approximate) Lorentz frame).
    – balu
    Aug 19 at 20:10








4




4




We would like to preserve the norm (angle) of (between) vectors under parallel transport. See physics.stackexchange.com/q/47919
– Avantgarde
Aug 19 at 13:11





We would like to preserve the norm (angle) of (between) vectors under parallel transport. See physics.stackexchange.com/q/47919
– Avantgarde
Aug 19 at 13:11













Adding to Avantgarde's excellent comment: We would like to preserve norms/angles because by the equivalence principle, a freely falling observer shouldn't notice any change in lengths or angles locally (in his (approximate) Lorentz frame).
– balu
Aug 19 at 20:10





Adding to Avantgarde's excellent comment: We would like to preserve norms/angles because by the equivalence principle, a freely falling observer shouldn't notice any change in lengths or angles locally (in his (approximate) Lorentz frame).
– balu
Aug 19 at 20:10











2 Answers
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First I'll amplify on the point made by Avantgarde in a comment. Let's say you're in outer space, in free-fall conditions, and you have two gyroscopes. You orient the gyroscopes so that their axes are perpendicular. The directions of their axes can be described by three-vectors, but these three-vectors can also be promoted to four-vectors in a natural way, by requiring that they be vectors of simultaneity in your frame, i.e., they represent purely spatial displacements to you. Let these two vectors be $v^a$ and $w^b$. Then the orthogonality of their axes can be expressed as $v^aw_a=0$, which is equivalent to $g_abv^aw^b=0$.



Now as time goes on, and you free-fall along with the gyroscopes, we expect the axes of these two gyroscopes to remain orthogonal. This is really just the equivalence principle at work. The properties of spacetime are always locally equivalent to the properties of Minkowski space. Mathematically, this means that we expect $v^aw_a$ to remain zero. This can be generalized to cases where the vectors are not necessarily spacelike or where the inner product is nonzero.



At a more philosophical/conceptual level, if this kind of inner product were to change under parallel transport, then we would probably attribute the observed change to a change in the metric. But it doesn't really make sense to talk about measuring a change in the metric, because the metric is our only tool for making measurements in the first place. So for example, if the metric were to change by a factor of 2, how would we know? That would be like saying that all objects, including rulers and clocks, changed by a factor of 2, but that would be unobservable because nothing would change relative to anything else.






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    A world line of a test particle is a physical object which, according to the equivalence principle, can be described in a local Lorentz frame as a straight line and in any coordinate frame as a geodesic.



    Local Lorentz frame

    $g_alpha beta = eta_alpha beta$ flat spacetime metric

    $partial g_alpha beta / partial x^gamma = 0$

    Note: A local Lorentz frame is the closest thing to a global Lorentz frame.

    $d^2x^alpha / dtau^2 = 0$ straight line

    $tau$ proper time



    Any coordinate frame

    $d^2x^alpha / dtau^2 + Gamma^alpha_beta gamma (dx^beta / dtau) (dx^gamma / dtau) = 0$ geodesic

    $tau$ affine parameter



    The consistency of the two representations demands $Gamma^alpha_beta gamma = 0$ in any local Lorentz frame, i.e. requires any local Lorentz frame to be a local inertial frame.



    That means that in a local Lorentz frame the covariant derivative $nabla_gamma$ shows:

    $nabla_gamma g_alpha beta = partial g_alpha beta / partial x^gamma - Gamma^mu_alpha gamma g_mu beta - Gamma^mu_beta gamma g_alpha mu = 0$

    The covariant derivative of the metric vanishes because the partial derivative and the $Gamma's$ vanish separately. However the equation is tensorial, hence it is valid in any arbitrary coordinate system.



    That is the physical basis why the metric remains unchanged under parallel transport.






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes








      up vote
      6
      down vote



      accepted










      First I'll amplify on the point made by Avantgarde in a comment. Let's say you're in outer space, in free-fall conditions, and you have two gyroscopes. You orient the gyroscopes so that their axes are perpendicular. The directions of their axes can be described by three-vectors, but these three-vectors can also be promoted to four-vectors in a natural way, by requiring that they be vectors of simultaneity in your frame, i.e., they represent purely spatial displacements to you. Let these two vectors be $v^a$ and $w^b$. Then the orthogonality of their axes can be expressed as $v^aw_a=0$, which is equivalent to $g_abv^aw^b=0$.



      Now as time goes on, and you free-fall along with the gyroscopes, we expect the axes of these two gyroscopes to remain orthogonal. This is really just the equivalence principle at work. The properties of spacetime are always locally equivalent to the properties of Minkowski space. Mathematically, this means that we expect $v^aw_a$ to remain zero. This can be generalized to cases where the vectors are not necessarily spacelike or where the inner product is nonzero.



      At a more philosophical/conceptual level, if this kind of inner product were to change under parallel transport, then we would probably attribute the observed change to a change in the metric. But it doesn't really make sense to talk about measuring a change in the metric, because the metric is our only tool for making measurements in the first place. So for example, if the metric were to change by a factor of 2, how would we know? That would be like saying that all objects, including rulers and clocks, changed by a factor of 2, but that would be unobservable because nothing would change relative to anything else.






      share|cite|improve this answer
























        up vote
        6
        down vote



        accepted










        First I'll amplify on the point made by Avantgarde in a comment. Let's say you're in outer space, in free-fall conditions, and you have two gyroscopes. You orient the gyroscopes so that their axes are perpendicular. The directions of their axes can be described by three-vectors, but these three-vectors can also be promoted to four-vectors in a natural way, by requiring that they be vectors of simultaneity in your frame, i.e., they represent purely spatial displacements to you. Let these two vectors be $v^a$ and $w^b$. Then the orthogonality of their axes can be expressed as $v^aw_a=0$, which is equivalent to $g_abv^aw^b=0$.



        Now as time goes on, and you free-fall along with the gyroscopes, we expect the axes of these two gyroscopes to remain orthogonal. This is really just the equivalence principle at work. The properties of spacetime are always locally equivalent to the properties of Minkowski space. Mathematically, this means that we expect $v^aw_a$ to remain zero. This can be generalized to cases where the vectors are not necessarily spacelike or where the inner product is nonzero.



        At a more philosophical/conceptual level, if this kind of inner product were to change under parallel transport, then we would probably attribute the observed change to a change in the metric. But it doesn't really make sense to talk about measuring a change in the metric, because the metric is our only tool for making measurements in the first place. So for example, if the metric were to change by a factor of 2, how would we know? That would be like saying that all objects, including rulers and clocks, changed by a factor of 2, but that would be unobservable because nothing would change relative to anything else.






        share|cite|improve this answer






















          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          First I'll amplify on the point made by Avantgarde in a comment. Let's say you're in outer space, in free-fall conditions, and you have two gyroscopes. You orient the gyroscopes so that their axes are perpendicular. The directions of their axes can be described by three-vectors, but these three-vectors can also be promoted to four-vectors in a natural way, by requiring that they be vectors of simultaneity in your frame, i.e., they represent purely spatial displacements to you. Let these two vectors be $v^a$ and $w^b$. Then the orthogonality of their axes can be expressed as $v^aw_a=0$, which is equivalent to $g_abv^aw^b=0$.



          Now as time goes on, and you free-fall along with the gyroscopes, we expect the axes of these two gyroscopes to remain orthogonal. This is really just the equivalence principle at work. The properties of spacetime are always locally equivalent to the properties of Minkowski space. Mathematically, this means that we expect $v^aw_a$ to remain zero. This can be generalized to cases where the vectors are not necessarily spacelike or where the inner product is nonzero.



          At a more philosophical/conceptual level, if this kind of inner product were to change under parallel transport, then we would probably attribute the observed change to a change in the metric. But it doesn't really make sense to talk about measuring a change in the metric, because the metric is our only tool for making measurements in the first place. So for example, if the metric were to change by a factor of 2, how would we know? That would be like saying that all objects, including rulers and clocks, changed by a factor of 2, but that would be unobservable because nothing would change relative to anything else.






          share|cite|improve this answer












          First I'll amplify on the point made by Avantgarde in a comment. Let's say you're in outer space, in free-fall conditions, and you have two gyroscopes. You orient the gyroscopes so that their axes are perpendicular. The directions of their axes can be described by three-vectors, but these three-vectors can also be promoted to four-vectors in a natural way, by requiring that they be vectors of simultaneity in your frame, i.e., they represent purely spatial displacements to you. Let these two vectors be $v^a$ and $w^b$. Then the orthogonality of their axes can be expressed as $v^aw_a=0$, which is equivalent to $g_abv^aw^b=0$.



          Now as time goes on, and you free-fall along with the gyroscopes, we expect the axes of these two gyroscopes to remain orthogonal. This is really just the equivalence principle at work. The properties of spacetime are always locally equivalent to the properties of Minkowski space. Mathematically, this means that we expect $v^aw_a$ to remain zero. This can be generalized to cases where the vectors are not necessarily spacelike or where the inner product is nonzero.



          At a more philosophical/conceptual level, if this kind of inner product were to change under parallel transport, then we would probably attribute the observed change to a change in the metric. But it doesn't really make sense to talk about measuring a change in the metric, because the metric is our only tool for making measurements in the first place. So for example, if the metric were to change by a factor of 2, how would we know? That would be like saying that all objects, including rulers and clocks, changed by a factor of 2, but that would be unobservable because nothing would change relative to anything else.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 19 at 14:38









          Ben Crowell

          44.5k3146271




          44.5k3146271




















              up vote
              0
              down vote













              A world line of a test particle is a physical object which, according to the equivalence principle, can be described in a local Lorentz frame as a straight line and in any coordinate frame as a geodesic.



              Local Lorentz frame

              $g_alpha beta = eta_alpha beta$ flat spacetime metric

              $partial g_alpha beta / partial x^gamma = 0$

              Note: A local Lorentz frame is the closest thing to a global Lorentz frame.

              $d^2x^alpha / dtau^2 = 0$ straight line

              $tau$ proper time



              Any coordinate frame

              $d^2x^alpha / dtau^2 + Gamma^alpha_beta gamma (dx^beta / dtau) (dx^gamma / dtau) = 0$ geodesic

              $tau$ affine parameter



              The consistency of the two representations demands $Gamma^alpha_beta gamma = 0$ in any local Lorentz frame, i.e. requires any local Lorentz frame to be a local inertial frame.



              That means that in a local Lorentz frame the covariant derivative $nabla_gamma$ shows:

              $nabla_gamma g_alpha beta = partial g_alpha beta / partial x^gamma - Gamma^mu_alpha gamma g_mu beta - Gamma^mu_beta gamma g_alpha mu = 0$

              The covariant derivative of the metric vanishes because the partial derivative and the $Gamma's$ vanish separately. However the equation is tensorial, hence it is valid in any arbitrary coordinate system.



              That is the physical basis why the metric remains unchanged under parallel transport.






              share|cite|improve this answer
























                up vote
                0
                down vote













                A world line of a test particle is a physical object which, according to the equivalence principle, can be described in a local Lorentz frame as a straight line and in any coordinate frame as a geodesic.



                Local Lorentz frame

                $g_alpha beta = eta_alpha beta$ flat spacetime metric

                $partial g_alpha beta / partial x^gamma = 0$

                Note: A local Lorentz frame is the closest thing to a global Lorentz frame.

                $d^2x^alpha / dtau^2 = 0$ straight line

                $tau$ proper time



                Any coordinate frame

                $d^2x^alpha / dtau^2 + Gamma^alpha_beta gamma (dx^beta / dtau) (dx^gamma / dtau) = 0$ geodesic

                $tau$ affine parameter



                The consistency of the two representations demands $Gamma^alpha_beta gamma = 0$ in any local Lorentz frame, i.e. requires any local Lorentz frame to be a local inertial frame.



                That means that in a local Lorentz frame the covariant derivative $nabla_gamma$ shows:

                $nabla_gamma g_alpha beta = partial g_alpha beta / partial x^gamma - Gamma^mu_alpha gamma g_mu beta - Gamma^mu_beta gamma g_alpha mu = 0$

                The covariant derivative of the metric vanishes because the partial derivative and the $Gamma's$ vanish separately. However the equation is tensorial, hence it is valid in any arbitrary coordinate system.



                That is the physical basis why the metric remains unchanged under parallel transport.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  A world line of a test particle is a physical object which, according to the equivalence principle, can be described in a local Lorentz frame as a straight line and in any coordinate frame as a geodesic.



                  Local Lorentz frame

                  $g_alpha beta = eta_alpha beta$ flat spacetime metric

                  $partial g_alpha beta / partial x^gamma = 0$

                  Note: A local Lorentz frame is the closest thing to a global Lorentz frame.

                  $d^2x^alpha / dtau^2 = 0$ straight line

                  $tau$ proper time



                  Any coordinate frame

                  $d^2x^alpha / dtau^2 + Gamma^alpha_beta gamma (dx^beta / dtau) (dx^gamma / dtau) = 0$ geodesic

                  $tau$ affine parameter



                  The consistency of the two representations demands $Gamma^alpha_beta gamma = 0$ in any local Lorentz frame, i.e. requires any local Lorentz frame to be a local inertial frame.



                  That means that in a local Lorentz frame the covariant derivative $nabla_gamma$ shows:

                  $nabla_gamma g_alpha beta = partial g_alpha beta / partial x^gamma - Gamma^mu_alpha gamma g_mu beta - Gamma^mu_beta gamma g_alpha mu = 0$

                  The covariant derivative of the metric vanishes because the partial derivative and the $Gamma's$ vanish separately. However the equation is tensorial, hence it is valid in any arbitrary coordinate system.



                  That is the physical basis why the metric remains unchanged under parallel transport.






                  share|cite|improve this answer












                  A world line of a test particle is a physical object which, according to the equivalence principle, can be described in a local Lorentz frame as a straight line and in any coordinate frame as a geodesic.



                  Local Lorentz frame

                  $g_alpha beta = eta_alpha beta$ flat spacetime metric

                  $partial g_alpha beta / partial x^gamma = 0$

                  Note: A local Lorentz frame is the closest thing to a global Lorentz frame.

                  $d^2x^alpha / dtau^2 = 0$ straight line

                  $tau$ proper time



                  Any coordinate frame

                  $d^2x^alpha / dtau^2 + Gamma^alpha_beta gamma (dx^beta / dtau) (dx^gamma / dtau) = 0$ geodesic

                  $tau$ affine parameter



                  The consistency of the two representations demands $Gamma^alpha_beta gamma = 0$ in any local Lorentz frame, i.e. requires any local Lorentz frame to be a local inertial frame.



                  That means that in a local Lorentz frame the covariant derivative $nabla_gamma$ shows:

                  $nabla_gamma g_alpha beta = partial g_alpha beta / partial x^gamma - Gamma^mu_alpha gamma g_mu beta - Gamma^mu_beta gamma g_alpha mu = 0$

                  The covariant derivative of the metric vanishes because the partial derivative and the $Gamma's$ vanish separately. However the equation is tensorial, hence it is valid in any arbitrary coordinate system.



                  That is the physical basis why the metric remains unchanged under parallel transport.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 19 at 16:37









                  Michele Grosso

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