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 $Lto L + fracdFdt$?
Will tsunami waves travel forever if there was no land?
Pulling the rope with one hand is as heavy as with two hands?
What do the phrase "Reeyan's seacrest" and the word "fraggle" mean in a sketch?
Don’t seats that recline flat defeat the purpose of having seatbelts?
What route did the Hindenburg take when traveling from Germany to the U.S.?
Why isn't the definition of absolute value applied when squaring a radical containing a variable?
How to creep the reader out with what seems like a normal person?
Was there a shared-world project before "Thieves World"?
How would one muzzle a full grown polar bear in the 13th century?
What does it mean to express a gate in Dirac notation?
Phrase for the opposite of "foolproof"
Do I have to worry about players making “bad” choices on level up?
What was the first Intel x86 processor with "Base + Index * Scale + Displacement" addressing mode?
How to make a pipeline wait for end-of-file or stop after an error?
Packing rectangles: Does rotation ever help?
Will a top journal at least read my introduction?
What makes accurate emulation of old systems a difficult task?
How to have a sharp product image?
Combinable filters
Binary Numbers Magic Trick
What are the potential pitfalls when using metals as a currency?
how to find the equation of a circle given points of the circle
simple conditions equation
How to reduce LED flash rate (frequency)
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
1307
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
1307
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
1307
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
1307
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
1307
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
1307
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
1307
asked Apr 23 at 15:52
VoBVoB
1307
1307
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
active
oldest
votes
$begingroup$
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 |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
VoB is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f474567%2fshow-two-lagrangians-are-equivalent%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f474567%2fshow-two-lagrangians-are-equivalent%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
