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

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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

edited 1 hour ago

Qmechanic♦

106k121941220

asked 6 hours ago

zhongyuan chenzhongyuan chen

232

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 1 hour ago

Qmechanic♦

106k121941220

asked 6 hours ago

zhongyuan chenzhongyuan chen

232

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 1 hour ago

Qmechanic♦

106k121941220

asked 6 hours ago

zhongyuan chenzhongyuan chen

232

$endgroup$

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

106k121941220

asked 6 hours ago

zhongyuan chenzhongyuan chen

232

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

106k121941220

asked 6 hours ago

zhongyuan chenzhongyuan chen

232

edited 1 hour ago

Qmechanic♦

106k121941220

edited 1 hour ago

Qmechanic♦

106k121941220

asked 6 hours ago

zhongyuan chenzhongyuan chen

232

asked 6 hours ago

zhongyuan chenzhongyuan chen

232

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
$$
S = int d^4 x mathcalL.
$$
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 $sum_mu fracdd x^mu J^mu = 0$. 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

edited 1 hour ago

answered 1 hour ago

user1379857user1379857

2,322826

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this answer

answered 1 hour ago

Qmechanic♦Qmechanic

106k121941220

$endgroup$

add a comment | 

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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
$$
S = int d^4 x mathcalL.
$$
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 $sum_mu fracdd x^mu J^mu = 0$. 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

edited 1 hour ago

answered 1 hour ago

user1379857user1379857

2,322826

$endgroup$

add a comment | 

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
$$
S = int d^4 x mathcalL.
$$
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 $sum_mu fracdd x^mu J^mu = 0$. 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

edited 1 hour ago

answered 1 hour ago

user1379857user1379857

2,322826

$endgroup$

add a comment | 

$begingroup$

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
$$
S = int d^4 x mathcalL.
$$
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 $sum_mu fracdd x^mu J^mu = 0$. 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

edited 1 hour ago

answered 1 hour ago

user1379857user1379857

2,322826

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

user1379857user1379857

2,322826

answered 1 hour ago

user1379857user1379857

2,322826

answered 1 hour ago

user1379857user1379857

2,322826

1

$begingroup$

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

share|cite|improve this answer

answered 1 hour ago

Qmechanic♦Qmechanic

106k121941220

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this answer

answered 1 hour ago

Qmechanic♦Qmechanic

106k121941220

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 1 hour ago

Qmechanic♦Qmechanic

106k121941220

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Qmechanic♦Qmechanic

106k121941220

answered 1 hour ago

Qmechanic♦Qmechanic

106k121941220

answered 1 hour ago

Qmechanic♦Qmechanic

106k121941220