Lagrangian corresponding to these equations of motion [on hold]
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I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$
I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.
$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$
for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?
homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
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put on hold as off-topic by Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero Mar 5 at 1:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$
I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.
$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$
for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?
homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
$endgroup$
put on hold as off-topic by Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero Mar 5 at 1:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
Feb 28 at 21:22
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It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
Feb 28 at 21:31
$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
Feb 28 at 21:32
$begingroup$
@Ballanzor That is literally what the question is asking. I don't understand your comment.
$endgroup$
– Aaron Stevens
Mar 1 at 12:26
add a comment |
$begingroup$
I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$
I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.
$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$
for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?
homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
$endgroup$
I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$
I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.
$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$
for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?
homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators
homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
edited Mar 1 at 0:27
Qmechanic♦
106k121921212
106k121921212
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
asked Feb 28 at 21:14
TheAverageHijanoTheAverageHijano
4869
4869
put on hold as off-topic by Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero Mar 5 at 1:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero Mar 5 at 1:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Jon Custer, AccidentalFourierTransform, Kyle Kanos, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
Feb 28 at 21:22
$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
Feb 28 at 21:31
$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
Feb 28 at 21:32
$begingroup$
@Ballanzor That is literally what the question is asking. I don't understand your comment.
$endgroup$
– Aaron Stevens
Mar 1 at 12:26
add a comment |
$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
Feb 28 at 21:22
$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
Feb 28 at 21:31
$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
Feb 28 at 21:32
$begingroup$
@Ballanzor That is literally what the question is asking. I don't understand your comment.
$endgroup$
– Aaron Stevens
Mar 1 at 12:26
$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
Feb 28 at 21:22
$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
Feb 28 at 21:22
$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
Feb 28 at 21:31
$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
Feb 28 at 21:31
$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
Feb 28 at 21:32
$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
Feb 28 at 21:32
$begingroup$
@Ballanzor That is literally what the question is asking. I don't understand your comment.
$endgroup$
– Aaron Stevens
Mar 1 at 12:26
$begingroup$
@Ballanzor That is literally what the question is asking. I don't understand your comment.
$endgroup$
– Aaron Stevens
Mar 1 at 12:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I suppose the Lagrangian you found is actually
$$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
$$ddot{q}_2 - 2 q_1 q_2 = 0$$
There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
$$-ddot{q}_2 - 2 q_1 q_2 = 0$$
This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
$$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
answered Feb 28 at 21:35
GRBGRB
1,0151822
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suppose the Lagrangian you found is actually
$$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
$$ddot{q}_2 - 2 q_1 q_2 = 0$$
There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
$$-ddot{q}_2 - 2 q_1 q_2 = 0$$
This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
$$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
answered Feb 28 at 21:35
GRBGRB
1,0151822
$endgroup$
add a comment |
$begingroup$
I suppose the Lagrangian you found is actually
$$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
$$ddot{q}_2 - 2 q_1 q_2 = 0$$
There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
$$-ddot{q}_2 - 2 q_1 q_2 = 0$$
This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
$$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
answered Feb 28 at 21:35
GRBGRB
1,0151822
$endgroup$
add a comment |
$begingroup$
I suppose the Lagrangian you found is actually
$$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
$$ddot{q}_2 - 2 q_1 q_2 = 0$$
There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
$$-ddot{q}_2 - 2 q_1 q_2 = 0$$
This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
$$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
answered Feb 28 at 21:35
GRBGRB
1,0151822
$endgroup$
I suppose the Lagrangian you found is actually
$$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
$$ddot{q}_2 - 2 q_1 q_2 = 0$$
There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
$$-ddot{q}_2 - 2 q_1 q_2 = 0$$
This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
$$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
answered Feb 28 at 21:35
GRBGRB
1,0151822
answered Feb 28 at 21:35
GRBGRB
1,0151822
answered Feb 28 at 21:35
GRBGRB
1,0151822
answered Feb 28 at 21:35
GRBGRB
1,0151822
1,0151822
add a comment |
add a comment |
$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
Feb 28 at 21:22
$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
Feb 28 at 21:31
$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
Feb 28 at 21:32
$begingroup$
@Ballanzor That is literally what the question is asking. I don't understand your comment.
$endgroup$
– Aaron Stevens
Mar 1 at 12:26