Is there a higher dimension analogue of Noether's theorem?












2












$begingroup$


So I have recently read the proof of Noether's theorem from the book variation calculus by Gelfand. Basically, what I have already seen is that for any single integral functional, if we have a transformation that keeps the functional invariant, we can derive a quantity that doesn't change along any solution of the Euler equations of the functional.



My question: Is there an analogue that work for multiple integral functional? That is, the corresponding system of Euler Lagrange equations are not ODEs, but rather PDEs. Can we define a quantity that is invariant on the whole solutions of these PDEs? The same argument used in Gelfand's proof for single integral functional clearly doesn't work. Can we have something that doesn't change not only with respect with one variable t, but is unchanged everywhere on the whole space like $R^n$ as long as we have a killing vector field for the functional?






lagrangian-formalism symmetry field-theory spacetime-dimensions noethers-theorem







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$endgroup$








  • 1




    $begingroup$
    The argument goes through exactly the same with more dimensions. Noether's theorem is routinely used in field theory, which takes place in 3+1 dimensions.
    $endgroup$
    – Javier
    2 hours ago
















2












$begingroup$


So I have recently read the proof of Noether's theorem from the book variation calculus by Gelfand. Basically, what I have already seen is that for any single integral functional, if we have a transformation that keeps the functional invariant, we can derive a quantity that doesn't change along any solution of the Euler equations of the functional.



My question: Is there an analogue that work for multiple integral functional? That is, the corresponding system of Euler Lagrange equations are not ODEs, but rather PDEs. Can we define a quantity that is invariant on the whole solutions of these PDEs? The same argument used in Gelfand's proof for single integral functional clearly doesn't work. Can we have something that doesn't change not only with respect with one variable t, but is unchanged everywhere on the whole space like $R^n$ as long as we have a killing vector field for the functional?






lagrangian-formalism symmetry field-theory spacetime-dimensions noethers-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The argument goes through exactly the same with more dimensions. Noether's theorem is routinely used in field theory, which takes place in 3+1 dimensions.
    $endgroup$
    – Javier
    2 hours ago














2












2








2





$begingroup$


So I have recently read the proof of Noether's theorem from the book variation calculus by Gelfand. Basically, what I have already seen is that for any single integral functional, if we have a transformation that keeps the functional invariant, we can derive a quantity that doesn't change along any solution of the Euler equations of the functional.



My question: Is there an analogue that work for multiple integral functional? That is, the corresponding system of Euler Lagrange equations are not ODEs, but rather PDEs. Can we define a quantity that is invariant on the whole solutions of these PDEs? The same argument used in Gelfand's proof for single integral functional clearly doesn't work. Can we have something that doesn't change not only with respect with one variable t, but is unchanged everywhere on the whole space like $R^n$ as long as we have a killing vector field for the functional?






lagrangian-formalism symmetry field-theory spacetime-dimensions noethers-theorem







share|cite|improve this question











$endgroup$




So I have recently read the proof of Noether's theorem from the book variation calculus by Gelfand. Basically, what I have already seen is that for any single integral functional, if we have a transformation that keeps the functional invariant, we can derive a quantity that doesn't change along any solution of the Euler equations of the functional.



My question: Is there an analogue that work for multiple integral functional? That is, the corresponding system of Euler Lagrange equations are not ODEs, but rather PDEs. Can we define a quantity that is invariant on the whole solutions of these PDEs? The same argument used in Gelfand's proof for single integral functional clearly doesn't work. Can we have something that doesn't change not only with respect with one variable t, but is unchanged everywhere on the whole space like $R^n$ as long as we have a killing vector field for the functional?






lagrangian-formalism symmetry field-theory spacetime-dimensions noethers-theorem




lagrangian-formalism symmetry field-theory spacetime-dimensions noethers-theorem






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share|cite|improve this question













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edited 1 hour ago









Qmechanic♦

106k121941220




106k121941220










asked 6 hours ago









zhongyuan chenzhongyuan chen

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  • 1




    $begingroup$
    The argument goes through exactly the same with more dimensions. Noether's theorem is routinely used in field theory, which takes place in 3+1 dimensions.
    $endgroup$
    – Javier
    2 hours ago














  • 1




    $begingroup$
    The argument goes through exactly the same with more dimensions. Noether's theorem is routinely used in field theory, which takes place in 3+1 dimensions.
    $endgroup$
    – Javier
    2 hours ago








1




1




$begingroup$
The argument goes through exactly the same with more dimensions. Noether's theorem is routinely used in field theory, which takes place in 3+1 dimensions.
$endgroup$
– Javier
2 hours ago




$begingroup$
The argument goes through exactly the same with more dimensions. Noether's theorem is routinely used in field theory, which takes place in 3+1 dimensions.
$endgroup$
– Javier
2 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time.



For example, where as in the one dimensional case you would write



$$
S = int dt L
$$
in field theory you would write
$$
S = int d^4 x mathcal{L}.
$$
The equation of motion will be a PDE.



Noether's theorem, instead of giving you a conserved quantity $Q$ which satisfies $dot Q = 0$, would now give you a conserved current $J^mu$ (where $mu = 0, 1, 2, 3$ and $mu = 0$ is the time component and $mu = 1,2,3$ are the space components) which satisfies $sum_mu frac{d}{d x^mu} J^mu = 0$. You can still also find the a conserved quantity $Q$, which satisfies $dot Q = 0$, defined by



$$
Q = int d^3 x J^0
$$
and integrating over any fixed time.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time.



      For example, where as in the one dimensional case you would write



      $$
      S = int dt L
      $$
      in field theory you would write
      $$
      S = int d^4 x mathcal{L}.
      $$
      The equation of motion will be a PDE.



      Noether's theorem, instead of giving you a conserved quantity $Q$ which satisfies $dot Q = 0$, would now give you a conserved current $J^mu$ (where $mu = 0, 1, 2, 3$ and $mu = 0$ is the time component and $mu = 1,2,3$ are the space components) which satisfies $sum_mu frac{d}{d x^mu} J^mu = 0$. You can still also find the a conserved quantity $Q$, which satisfies $dot Q = 0$, defined by



      $$
      Q = int d^3 x J^0
      $$
      and integrating over any fixed time.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time.



        For example, where as in the one dimensional case you would write



        $$
        S = int dt L
        $$
        in field theory you would write
        $$
        S = int d^4 x mathcal{L}.
        $$
        The equation of motion will be a PDE.



        Noether's theorem, instead of giving you a conserved quantity $Q$ which satisfies $dot Q = 0$, would now give you a conserved current $J^mu$ (where $mu = 0, 1, 2, 3$ and $mu = 0$ is the time component and $mu = 1,2,3$ are the space components) which satisfies $sum_mu frac{d}{d x^mu} J^mu = 0$. You can still also find the a conserved quantity $Q$, which satisfies $dot Q = 0$, defined by



        $$
        Q = int d^3 x J^0
        $$
        and integrating over any fixed time.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time.



          For example, where as in the one dimensional case you would write



          $$
          S = int dt L
          $$
          in field theory you would write
          $$
          S = int d^4 x mathcal{L}.
          $$
          The equation of motion will be a PDE.



          Noether's theorem, instead of giving you a conserved quantity $Q$ which satisfies $dot Q = 0$, would now give you a conserved current $J^mu$ (where $mu = 0, 1, 2, 3$ and $mu = 0$ is the time component and $mu = 1,2,3$ are the space components) which satisfies $sum_mu frac{d}{d x^mu} J^mu = 0$. You can still also find the a conserved quantity $Q$, which satisfies $dot Q = 0$, defined by



          $$
          Q = int d^3 x J^0
          $$
          and integrating over any fixed time.






          share|cite|improve this answer











          $endgroup$



          In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time.



          For example, where as in the one dimensional case you would write



          $$
          S = int dt L
          $$
          in field theory you would write
          $$
          S = int d^4 x mathcal{L}.
          $$
          The equation of motion will be a PDE.



          Noether's theorem, instead of giving you a conserved quantity $Q$ which satisfies $dot Q = 0$, would now give you a conserved current $J^mu$ (where $mu = 0, 1, 2, 3$ and $mu = 0$ is the time component and $mu = 1,2,3$ are the space components) which satisfies $sum_mu frac{d}{d x^mu} J^mu = 0$. You can still also find the a conserved quantity $Q$, which satisfies $dot Q = 0$, defined by



          $$
          Q = int d^3 x J^0
          $$
          and integrating over any fixed time.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          user1379857user1379857

          2,322826




          2,322826























              1












              $begingroup$

              Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.






                  share|cite|improve this answer









                  $endgroup$



                  Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Qmechanic♦Qmechanic

                  106k121941220




                  106k121941220






























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