Show two Lagrangians are equivalentProof that total derivative is the only function that can be added to Lagrangian without changing the EOMProof that total derivative is the only function that can be added to Lagrangian without changing the EOMWhy two different Lagrangians to derive geodesic equations?Proving independence of the lagrangian on position of a free particle using the euler-lagrange equationFrom Newtonian systems to Lagrange mechanics using Euler - Lagrange equationsLagrangians not related via a total time derivative lead to same Noether symmetries?Lagrangian in a system with a specific velocity dependent potentialWhat are Lagrange Multipliers, regarding holonomic constraints in classical mechanics?Harmonic Oscillator from a second order Lagrangian: applicationsExactly how specifically can the Lagrangian be defined?How does the Hamiltonian change if $Lto L + fracdFdt$?
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Show two Lagrangians are equivalent
Proof that total derivative is the only function that can be added to Lagrangian without changing the EOMProof that total derivative is the only function that can be added to Lagrangian without changing the EOMWhy two different Lagrangians to derive geodesic equations?Proving independence of the lagrangian on position of a free particle using the euler-lagrange equationFrom Newtonian systems to Lagrange mechanics using Euler - Lagrange equationsLagrangians not related via a total time derivative lead to same Noether symmetries?Lagrangian in a system with a specific velocity dependent potentialWhat are Lagrange Multipliers, regarding holonomic constraints in classical mechanics?Harmonic Oscillator from a second order Lagrangian: applicationsExactly how specifically can the Lagrangian be defined?How does the Hamiltonian change if $Lto L + fracdFdt$?
$begingroup$
I need to show that these two Lagrangians are equivalent:
beginalign
L(dotx,doty,x,y)&=dot x^2+dot y + x^2-y ,\
tildeL(dot x, dot y, x, y)&=dot x^2+dot y -2y^3.
endalign
It is the case iff they differ for a total derivation like $fracdFdt(x,y)$.
In this case, the difference is $x^2+y^3$ and I can't imagine such an $F(x,y)$ whose total derivative is the one above. How should I move?
I tried with the following $F(x,y)=fracx^33dot x + fracy^44dot y$, but it shouldn't have the dotted terms.
Actually, I just proved they don't give rise to the same Lagrange equations, so I can conclude they're not equivalent, right?
homework-and-exercises classical-mechanics lagrangian-formalism variational-principle action
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
asked Apr 23 at 15:52
VoBVoB
1257
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I need to show that these two Lagrangians are equivalent:
beginalign
L(dotx,doty,x,y)&=dot x^2+dot y + x^2-y ,\
tildeL(dot x, dot y, x, y)&=dot x^2+dot y -2y^3.
endalign
It is the case iff they differ for a total derivation like $fracdFdt(x,y)$.
In this case, the difference is $x^2+y^3$ and I can't imagine such an $F(x,y)$ whose total derivative is the one above. How should I move?
I tried with the following $F(x,y)=fracx^33dot x + fracy^44dot y$, but it shouldn't have the dotted terms.
Actually, I just proved they don't give rise to the same Lagrange equations, so I can conclude they're not equivalent, right?
homework-and-exercises classical-mechanics lagrangian-formalism variational-principle action
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
asked Apr 23 at 15:52
VoBVoB
1257
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I need to show that these two Lagrangians are equivalent:
beginalign
L(dotx,doty,x,y)&=dot x^2+dot y + x^2-y ,\
tildeL(dot x, dot y, x, y)&=dot x^2+dot y -2y^3.
endalign
It is the case iff they differ for a total derivation like $fracdFdt(x,y)$.
In this case, the difference is $x^2+y^3$ and I can't imagine such an $F(x,y)$ whose total derivative is the one above. How should I move?
I tried with the following $F(x,y)=fracx^33dot x + fracy^44dot y$, but it shouldn't have the dotted terms.
Actually, I just proved they don't give rise to the same Lagrange equations, so I can conclude they're not equivalent, right?
homework-and-exercises classical-mechanics lagrangian-formalism variational-principle action
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
asked Apr 23 at 15:52
VoBVoB
1257
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I need to show that these two Lagrangians are equivalent:
beginalign
L(dotx,doty,x,y)&=dot x^2+dot y + x^2-y ,\
tildeL(dot x, dot y, x, y)&=dot x^2+dot y -2y^3.
endalign
It is the case iff they differ for a total derivation like $fracdFdt(x,y)$.
In this case, the difference is $x^2+y^3$ and I can't imagine such an $F(x,y)$ whose total derivative is the one above. How should I move?
I tried with the following $F(x,y)=fracx^33dot x + fracy^44dot y$, but it shouldn't have the dotted terms.
Actually, I just proved they don't give rise to the same Lagrange equations, so I can conclude they're not equivalent, right?
homework-and-exercises classical-mechanics lagrangian-formalism variational-principle action
homework-and-exercises classical-mechanics lagrangian-formalism variational-principle action
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
asked Apr 23 at 15:52
VoBVoB
1257
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
asked Apr 23 at 15:52
VoBVoB
1257
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
edited Apr 23 at 17:54
Qmechanic♦
108k122021255
108k122021255
asked Apr 23 at 15:52
VoBVoB
1257
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 23 at 15:52
VoBVoB
1257
asked Apr 23 at 15:52
VoBVoB
1257
1257
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
VoB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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$begingroup$
I leave it to OP and the reader to prove that OP's two Lagrangians are indeed classically inequivalent, but let me make the following general remarks:
Two Lagrangians $L_1$ and $L_2$ are classical equivalent iff they give the same Euler-Lagrange (EL) equations.
A sufficient condition is that the difference $L_2-L_1=fracdFdt$ is a total derivative, but it should be stressed that it is not a necessary condition, cf. e.g. my Phys.SE answer here.
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
$endgroup$
$begingroup$
Thanks, but I'm a bit confused with the textiff in 1), because I know that if I take as Lagrangians $L$ and $barL=alpha L$, they give rise to same Lagrange equations, but they're not equivalent...
$endgroup$
– VoB
Apr 23 at 18:08
$begingroup$
@VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations.
$endgroup$
– probably_someone
Apr 23 at 20:49
1
$begingroup$
Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent
$endgroup$
– VoB
Apr 23 at 20:53
add a comment |
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1 Answer
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$begingroup$
I leave it to OP and the reader to prove that OP's two Lagrangians are indeed classically inequivalent, but let me make the following general remarks:
Two Lagrangians $L_1$ and $L_2$ are classical equivalent iff they give the same Euler-Lagrange (EL) equations.
A sufficient condition is that the difference $L_2-L_1=fracdFdt$ is a total derivative, but it should be stressed that it is not a necessary condition, cf. e.g. my Phys.SE answer here.
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
$endgroup$
$begingroup$
Thanks, but I'm a bit confused with the textiff in 1), because I know that if I take as Lagrangians $L$ and $barL=alpha L$, they give rise to same Lagrange equations, but they're not equivalent...
$endgroup$
– VoB
Apr 23 at 18:08
$begingroup$
@VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations.
$endgroup$
– probably_someone
Apr 23 at 20:49
1
$begingroup$
Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent
$endgroup$
– VoB
Apr 23 at 20:53
add a comment |
$begingroup$
I leave it to OP and the reader to prove that OP's two Lagrangians are indeed classically inequivalent, but let me make the following general remarks:
Two Lagrangians $L_1$ and $L_2$ are classical equivalent iff they give the same Euler-Lagrange (EL) equations.
A sufficient condition is that the difference $L_2-L_1=fracdFdt$ is a total derivative, but it should be stressed that it is not a necessary condition, cf. e.g. my Phys.SE answer here.
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
$endgroup$
$begingroup$
Thanks, but I'm a bit confused with the textiff in 1), because I know that if I take as Lagrangians $L$ and $barL=alpha L$, they give rise to same Lagrange equations, but they're not equivalent...
$endgroup$
– VoB
Apr 23 at 18:08
$begingroup$
@VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations.
$endgroup$
– probably_someone
Apr 23 at 20:49
1
$begingroup$
Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent
$endgroup$
– VoB
Apr 23 at 20:53
add a comment |
$begingroup$
I leave it to OP and the reader to prove that OP's two Lagrangians are indeed classically inequivalent, but let me make the following general remarks:
Two Lagrangians $L_1$ and $L_2$ are classical equivalent iff they give the same Euler-Lagrange (EL) equations.
A sufficient condition is that the difference $L_2-L_1=fracdFdt$ is a total derivative, but it should be stressed that it is not a necessary condition, cf. e.g. my Phys.SE answer here.
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
$endgroup$
I leave it to OP and the reader to prove that OP's two Lagrangians are indeed classically inequivalent, but let me make the following general remarks:
Two Lagrangians $L_1$ and $L_2$ are classical equivalent iff they give the same Euler-Lagrange (EL) equations.
A sufficient condition is that the difference $L_2-L_1=fracdFdt$ is a total derivative, but it should be stressed that it is not a necessary condition, cf. e.g. my Phys.SE answer here.
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
answered Apr 23 at 17:17
Qmechanic♦Qmechanic
108k122021255
108k122021255
$begingroup$
Thanks, but I'm a bit confused with the textiff in 1), because I know that if I take as Lagrangians $L$ and $barL=alpha L$, they give rise to same Lagrange equations, but they're not equivalent...
$endgroup$
– VoB
Apr 23 at 18:08
$begingroup$
@VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations.
$endgroup$
– probably_someone
Apr 23 at 20:49
1
$begingroup$
Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent
$endgroup$
– VoB
Apr 23 at 20:53
add a comment |
$begingroup$
Thanks, but I'm a bit confused with the textiff in 1), because I know that if I take as Lagrangians $L$ and $barL=alpha L$, they give rise to same Lagrange equations, but they're not equivalent...
$endgroup$
– VoB
Apr 23 at 18:08
$begingroup$
@VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations.
$endgroup$
– probably_someone
Apr 23 at 20:49
1
$begingroup$
Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent
$endgroup$
– VoB
Apr 23 at 20:53
$begingroup$
Thanks, but I'm a bit confused with the textiff in 1), because I know that if I take as Lagrangians $L$ and $barL=alpha L$, they give rise to same Lagrange equations, but they're not equivalent...
$endgroup$
– VoB
Apr 23 at 18:08
$begingroup$
Thanks, but I'm a bit confused with the textiff in 1), because I know that if I take as Lagrangians $L$ and $barL=alpha L$, they give rise to same Lagrange equations, but they're not equivalent...
$endgroup$
– VoB
Apr 23 at 18:08
$begingroup$
@VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations.
$endgroup$
– probably_someone
Apr 23 at 20:49
$begingroup$
@VoB It sounds like you mean something different than Qmechanic does when you say "equivalent." According to this answer, two Lagrangians are defined to be equivalent whenever they give the same E-L equations.
$endgroup$
– probably_someone
Apr 23 at 20:49
1
1
$begingroup$
Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent
$endgroup$
– VoB
Apr 23 at 20:53
$begingroup$
Right. I say they're equivalent iff they differ for a total derivation. With this definition, two equivalent Lagrangians give rise to the same Lagrange-Equation. In my exercise, I couldn't find such a function $F$, so I checked that the don't give the same Lagrange-equations, in order to conclude the're not equivalent
$endgroup$
– VoB
Apr 23 at 20:53
add a comment |
VoB is a new contributor. Be nice, and check out our Code of Conduct.
VoB is a new contributor. Be nice, and check out our Code of Conduct.
VoB is a new contributor. Be nice, and check out our Code of Conduct.
VoB is a new contributor. Be nice, and check out our Code of Conduct.
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