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What is the relation between a symmetry and the invariance of the Lagrangian?
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While proving that homogeneity of space implies conservation of momentum, we use the fact that homogeneity of space means that the Lagrangian of the system remains invariant under translation. Why is it so? I know that homogeneity of space means that if we perform 2 experiments a few metres apart, they will provide same results but how does that imply the invariance of the Lagrangian?
classical-mechanics lagrangian-formalism symmetry
edited Aug 8 at 15:27
Qmechanic♦
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asked Aug 8 at 13:58
Abhirup Mukherjee
1675
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up vote
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While proving that homogeneity of space implies conservation of momentum, we use the fact that homogeneity of space means that the Lagrangian of the system remains invariant under translation. Why is it so? I know that homogeneity of space means that if we perform 2 experiments a few metres apart, they will provide same results but how does that imply the invariance of the Lagrangian?
classical-mechanics lagrangian-formalism symmetry
edited Aug 8 at 15:27
Qmechanic♦
96.5k121631019
asked Aug 8 at 13:58
Abhirup Mukherjee
1675
1
Related: Do an action and its Euler-Lagrange equations have the same symmetries?
– Qmechanic♦
Aug 8 at 15:26
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
While proving that homogeneity of space implies conservation of momentum, we use the fact that homogeneity of space means that the Lagrangian of the system remains invariant under translation. Why is it so? I know that homogeneity of space means that if we perform 2 experiments a few metres apart, they will provide same results but how does that imply the invariance of the Lagrangian?
classical-mechanics lagrangian-formalism symmetry
edited Aug 8 at 15:27
Qmechanic♦
96.5k121631019
asked Aug 8 at 13:58
Abhirup Mukherjee
1675
While proving that homogeneity of space implies conservation of momentum, we use the fact that homogeneity of space means that the Lagrangian of the system remains invariant under translation. Why is it so? I know that homogeneity of space means that if we perform 2 experiments a few metres apart, they will provide same results but how does that imply the invariance of the Lagrangian?
classical-mechanics lagrangian-formalism symmetry
edited Aug 8 at 15:27
Qmechanic♦
96.5k121631019
asked Aug 8 at 13:58
Abhirup Mukherjee
1675
edited Aug 8 at 15:27
Qmechanic♦
96.5k121631019
edited Aug 8 at 15:27
Qmechanic♦
96.5k121631019
edited Aug 8 at 15:27
Qmechanic♦
96.5k121631019
96.5k121631019
asked Aug 8 at 13:58
Abhirup Mukherjee
1675
asked Aug 8 at 13:58
Abhirup Mukherjee
1675
asked Aug 8 at 13:58
Abhirup Mukherjee
1675
1675
1
Related: Do an action and its Euler-Lagrange equations have the same symmetries?
– Qmechanic♦
Aug 8 at 15:26
add a comment |Â
1
Related: Do an action and its Euler-Lagrange equations have the same symmetries?
– Qmechanic♦
Aug 8 at 15:26
1
1
Related: Do an action and its Euler-Lagrange equations have the same symmetries?
– Qmechanic♦
Aug 8 at 15:26
Related: Do an action and its Euler-Lagrange equations have the same symmetries?
– Qmechanic♦
Aug 8 at 15:26
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
10
down vote
accepted
It does not imply indeed!
The elementary and physically natural notion of symmetry you suggest can be made more precise as follows, and this is the general notion of dynamical symmetry,
S1 A symmetry of a dynamical system is a bijective transformation in the space of states such that any solution of the equation of motion is transformed into another solution of the equation of motion by it.
This notion of symmetry is independent from the Lagrangian nature (or Hamiltonian or other) of the equation of motion.
Another restricted notion is that follows, valid and very popular in the Lagrangian formulation.
S2 A symmetry is a bijective trasformation in the space of states leaving invariant the Lagrangian up to total derivatives.
The relation between these two notions is that
Given a Lagrangian, a symmetry in the sense S2 is also a symmetry in the sense S1 with respect to the solutions of Euler-Lagrange equations.
It is possible to construct counterexamples to prove that the converse implication does not hold in general.
A symmetry in the restricted Lagrangian sense S2 (more precisely a continuous symmetry i.e. a one-parameter group of symmetries) has the famous consequence that a conserved quantity exists along the motion of the system (Noether theorem).
Adopting a Hamiltonian viewpoint however the two notions of symmetry are instead equivalent, provided the used transformations are canonical transformations and one refers to the Hamiltonian instead of the Lagrangian.
A canonical transformation is a dynamical symmetry (type S1 with respect to Hamilton equations) if and only if it preserves the form of the Hamiltonian function.
In turn, these equivalent facts are equivalent to the existence of a conserved quantity along the motion of the system when the symmetry is a continuous one.
Within the Hamiltonian framework the three facts are equivalent.
In summary,
the lagrangian of any isolated system is translationally invariant in the rest space of inertial reference frames
is just a possible statement of the spatial homogeneity postulate. Surely not the most physically natural and general, which would rely on the general notion S1, rather than S2.
However, if assuming from scratch a Lagrangian formulation as in Landau-Lifsits' textbook, this theoretically advanced form of the postulate can be accepted.
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
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up vote
1
down vote
I'll just add a little to Valter Moretti's excellent answer to explain why in particular momentum conservation is connected to translation invariance. Given the continuous symmetry $deltamathbfq=epsilonmathbfK$ with $dotepsilon=0$, on-shell $$delta L=epsilon(mathbfKcdotfracpartial Lpartialmathbfq+dotmathbfKcdotfracpartial Lpartialdotmathbfq)=epsilon(mathbfKcdotdotmathbfp+dotmathbfKcdotmathbfp)=epsilonfracddtmathbfKcdotmathbfp.$$Thus $L$ is invariant iff $mathbfKcdotmathbfp$ is conserved. If this works for any constant $mathbfK$, each component of $mathbfp$ is conserved.
answered Aug 8 at 20:36
J.G.
8,27921124
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
It does not imply indeed!
The elementary and physically natural notion of symmetry you suggest can be made more precise as follows, and this is the general notion of dynamical symmetry,
S1 A symmetry of a dynamical system is a bijective transformation in the space of states such that any solution of the equation of motion is transformed into another solution of the equation of motion by it.
This notion of symmetry is independent from the Lagrangian nature (or Hamiltonian or other) of the equation of motion.
Another restricted notion is that follows, valid and very popular in the Lagrangian formulation.
S2 A symmetry is a bijective trasformation in the space of states leaving invariant the Lagrangian up to total derivatives.
The relation between these two notions is that
Given a Lagrangian, a symmetry in the sense S2 is also a symmetry in the sense S1 with respect to the solutions of Euler-Lagrange equations.
It is possible to construct counterexamples to prove that the converse implication does not hold in general.
A symmetry in the restricted Lagrangian sense S2 (more precisely a continuous symmetry i.e. a one-parameter group of symmetries) has the famous consequence that a conserved quantity exists along the motion of the system (Noether theorem).
Adopting a Hamiltonian viewpoint however the two notions of symmetry are instead equivalent, provided the used transformations are canonical transformations and one refers to the Hamiltonian instead of the Lagrangian.
A canonical transformation is a dynamical symmetry (type S1 with respect to Hamilton equations) if and only if it preserves the form of the Hamiltonian function.
In turn, these equivalent facts are equivalent to the existence of a conserved quantity along the motion of the system when the symmetry is a continuous one.
Within the Hamiltonian framework the three facts are equivalent.
In summary,
the lagrangian of any isolated system is translationally invariant in the rest space of inertial reference frames
is just a possible statement of the spatial homogeneity postulate. Surely not the most physically natural and general, which would rely on the general notion S1, rather than S2.
However, if assuming from scratch a Lagrangian formulation as in Landau-Lifsits' textbook, this theoretically advanced form of the postulate can be accepted.
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
add a comment |Â
up vote
10
down vote
accepted
It does not imply indeed!
The elementary and physically natural notion of symmetry you suggest can be made more precise as follows, and this is the general notion of dynamical symmetry,
S1 A symmetry of a dynamical system is a bijective transformation in the space of states such that any solution of the equation of motion is transformed into another solution of the equation of motion by it.
This notion of symmetry is independent from the Lagrangian nature (or Hamiltonian or other) of the equation of motion.
Another restricted notion is that follows, valid and very popular in the Lagrangian formulation.
S2 A symmetry is a bijective trasformation in the space of states leaving invariant the Lagrangian up to total derivatives.
The relation between these two notions is that
Given a Lagrangian, a symmetry in the sense S2 is also a symmetry in the sense S1 with respect to the solutions of Euler-Lagrange equations.
It is possible to construct counterexamples to prove that the converse implication does not hold in general.
A symmetry in the restricted Lagrangian sense S2 (more precisely a continuous symmetry i.e. a one-parameter group of symmetries) has the famous consequence that a conserved quantity exists along the motion of the system (Noether theorem).
Adopting a Hamiltonian viewpoint however the two notions of symmetry are instead equivalent, provided the used transformations are canonical transformations and one refers to the Hamiltonian instead of the Lagrangian.
A canonical transformation is a dynamical symmetry (type S1 with respect to Hamilton equations) if and only if it preserves the form of the Hamiltonian function.
In turn, these equivalent facts are equivalent to the existence of a conserved quantity along the motion of the system when the symmetry is a continuous one.
Within the Hamiltonian framework the three facts are equivalent.
In summary,
the lagrangian of any isolated system is translationally invariant in the rest space of inertial reference frames
is just a possible statement of the spatial homogeneity postulate. Surely not the most physically natural and general, which would rely on the general notion S1, rather than S2.
However, if assuming from scratch a Lagrangian formulation as in Landau-Lifsits' textbook, this theoretically advanced form of the postulate can be accepted.
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
It does not imply indeed!
The elementary and physically natural notion of symmetry you suggest can be made more precise as follows, and this is the general notion of dynamical symmetry,
S1 A symmetry of a dynamical system is a bijective transformation in the space of states such that any solution of the equation of motion is transformed into another solution of the equation of motion by it.
This notion of symmetry is independent from the Lagrangian nature (or Hamiltonian or other) of the equation of motion.
Another restricted notion is that follows, valid and very popular in the Lagrangian formulation.
S2 A symmetry is a bijective trasformation in the space of states leaving invariant the Lagrangian up to total derivatives.
The relation between these two notions is that
Given a Lagrangian, a symmetry in the sense S2 is also a symmetry in the sense S1 with respect to the solutions of Euler-Lagrange equations.
It is possible to construct counterexamples to prove that the converse implication does not hold in general.
A symmetry in the restricted Lagrangian sense S2 (more precisely a continuous symmetry i.e. a one-parameter group of symmetries) has the famous consequence that a conserved quantity exists along the motion of the system (Noether theorem).
Adopting a Hamiltonian viewpoint however the two notions of symmetry are instead equivalent, provided the used transformations are canonical transformations and one refers to the Hamiltonian instead of the Lagrangian.
A canonical transformation is a dynamical symmetry (type S1 with respect to Hamilton equations) if and only if it preserves the form of the Hamiltonian function.
In turn, these equivalent facts are equivalent to the existence of a conserved quantity along the motion of the system when the symmetry is a continuous one.
Within the Hamiltonian framework the three facts are equivalent.
In summary,
the lagrangian of any isolated system is translationally invariant in the rest space of inertial reference frames
is just a possible statement of the spatial homogeneity postulate. Surely not the most physically natural and general, which would rely on the general notion S1, rather than S2.
However, if assuming from scratch a Lagrangian formulation as in Landau-Lifsits' textbook, this theoretically advanced form of the postulate can be accepted.
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
It does not imply indeed!
The elementary and physically natural notion of symmetry you suggest can be made more precise as follows, and this is the general notion of dynamical symmetry,
S1 A symmetry of a dynamical system is a bijective transformation in the space of states such that any solution of the equation of motion is transformed into another solution of the equation of motion by it.
This notion of symmetry is independent from the Lagrangian nature (or Hamiltonian or other) of the equation of motion.
Another restricted notion is that follows, valid and very popular in the Lagrangian formulation.
S2 A symmetry is a bijective trasformation in the space of states leaving invariant the Lagrangian up to total derivatives.
The relation between these two notions is that
Given a Lagrangian, a symmetry in the sense S2 is also a symmetry in the sense S1 with respect to the solutions of Euler-Lagrange equations.
It is possible to construct counterexamples to prove that the converse implication does not hold in general.
A symmetry in the restricted Lagrangian sense S2 (more precisely a continuous symmetry i.e. a one-parameter group of symmetries) has the famous consequence that a conserved quantity exists along the motion of the system (Noether theorem).
Adopting a Hamiltonian viewpoint however the two notions of symmetry are instead equivalent, provided the used transformations are canonical transformations and one refers to the Hamiltonian instead of the Lagrangian.
A canonical transformation is a dynamical symmetry (type S1 with respect to Hamilton equations) if and only if it preserves the form of the Hamiltonian function.
In turn, these equivalent facts are equivalent to the existence of a conserved quantity along the motion of the system when the symmetry is a continuous one.
Within the Hamiltonian framework the three facts are equivalent.
In summary,
the lagrangian of any isolated system is translationally invariant in the rest space of inertial reference frames
is just a possible statement of the spatial homogeneity postulate. Surely not the most physically natural and general, which would rely on the general notion S1, rather than S2.
However, if assuming from scratch a Lagrangian formulation as in Landau-Lifsits' textbook, this theoretically advanced form of the postulate can be accepted.
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
edited Aug 8 at 15:33
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
answered Aug 8 at 14:56
Valter Moretti
35.2k463132
35.2k463132
add a comment |Â
add a comment |Â
up vote
1
down vote
I'll just add a little to Valter Moretti's excellent answer to explain why in particular momentum conservation is connected to translation invariance. Given the continuous symmetry $deltamathbfq=epsilonmathbfK$ with $dotepsilon=0$, on-shell $$delta L=epsilon(mathbfKcdotfracpartial Lpartialmathbfq+dotmathbfKcdotfracpartial Lpartialdotmathbfq)=epsilon(mathbfKcdotdotmathbfp+dotmathbfKcdotmathbfp)=epsilonfracddtmathbfKcdotmathbfp.$$Thus $L$ is invariant iff $mathbfKcdotmathbfp$ is conserved. If this works for any constant $mathbfK$, each component of $mathbfp$ is conserved.
answered Aug 8 at 20:36
J.G.
8,27921124
add a comment |Â
up vote
1
down vote
I'll just add a little to Valter Moretti's excellent answer to explain why in particular momentum conservation is connected to translation invariance. Given the continuous symmetry $deltamathbfq=epsilonmathbfK$ with $dotepsilon=0$, on-shell $$delta L=epsilon(mathbfKcdotfracpartial Lpartialmathbfq+dotmathbfKcdotfracpartial Lpartialdotmathbfq)=epsilon(mathbfKcdotdotmathbfp+dotmathbfKcdotmathbfp)=epsilonfracddtmathbfKcdotmathbfp.$$Thus $L$ is invariant iff $mathbfKcdotmathbfp$ is conserved. If this works for any constant $mathbfK$, each component of $mathbfp$ is conserved.
answered Aug 8 at 20:36
J.G.
8,27921124
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'll just add a little to Valter Moretti's excellent answer to explain why in particular momentum conservation is connected to translation invariance. Given the continuous symmetry $deltamathbfq=epsilonmathbfK$ with $dotepsilon=0$, on-shell $$delta L=epsilon(mathbfKcdotfracpartial Lpartialmathbfq+dotmathbfKcdotfracpartial Lpartialdotmathbfq)=epsilon(mathbfKcdotdotmathbfp+dotmathbfKcdotmathbfp)=epsilonfracddtmathbfKcdotmathbfp.$$Thus $L$ is invariant iff $mathbfKcdotmathbfp$ is conserved. If this works for any constant $mathbfK$, each component of $mathbfp$ is conserved.
answered Aug 8 at 20:36
J.G.
8,27921124
I'll just add a little to Valter Moretti's excellent answer to explain why in particular momentum conservation is connected to translation invariance. Given the continuous symmetry $deltamathbfq=epsilonmathbfK$ with $dotepsilon=0$, on-shell $$delta L=epsilon(mathbfKcdotfracpartial Lpartialmathbfq+dotmathbfKcdotfracpartial Lpartialdotmathbfq)=epsilon(mathbfKcdotdotmathbfp+dotmathbfKcdotmathbfp)=epsilonfracddtmathbfKcdotmathbfp.$$Thus $L$ is invariant iff $mathbfKcdotmathbfp$ is conserved. If this works for any constant $mathbfK$, each component of $mathbfp$ is conserved.
answered Aug 8 at 20:36
J.G.
8,27921124
answered Aug 8 at 20:36
J.G.
8,27921124
answered Aug 8 at 20:36
J.G.
8,27921124
answered Aug 8 at 20:36
J.G.
8,27921124
8,27921124
add a comment |Â
add a comment |Â
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1
Related: Do an action and its Euler-Lagrange equations have the same symmetries?
– Qmechanic♦
Aug 8 at 15:26