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Is there a higher dimension analogue of Noether's theorem?
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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
$endgroup$
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
$endgroup$
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
$endgroup$
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
edited 1 hour ago
Qmechanic♦
106k121941220
106k121941220
asked 6 hours ago
zhongyuan chenzhongyuan chen
232
232
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 |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
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
$$
S = int d^4 x mathcal{L}.
$$
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.
$endgroup$
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
$$
S = int d^4 x mathcal{L}.
$$
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.
$endgroup$
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.
edited 58 mins ago
answered 1 hour ago
user1379857user1379857
2,322826
2,322826
add a comment |
add a comment |
$begingroup$
Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.
$endgroup$
add a comment |
$begingroup$
Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.
$endgroup$
add a comment |
$begingroup$
Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.
$endgroup$
Yes, already Noether herself considered field theory in $n$ dimensions in her seminal 1918 paper.
answered 1 hour ago
Qmechanic♦Qmechanic
106k121941220
106k121941220
add a comment |
add a comment |
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$begingroup$
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