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Undefined Hamiltonian for this particular Lagrangian


Adding a total time derivative term to the LagrangianLagrangian to HamiltonianHamiltonian mechanics really useful for numerical integration? Lagrangian can become 1st-orderFinding geodesics: Lagrangian vs HamiltonianInconsistency in Lagrangian vs Hamiltonian formalism?Gauge invariance of the HamiltonianWhat is my state in the context of Hamiltonian mechanics?Can we write a Lagrangian for the classical system with $H=kqp$?Canonical transformations preserve Hamilton's equations. Which transformation preserve the Euler-Lagrange equations?Can all canonical transformations be generated using a generating function?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?










share|cite|improve this question











$endgroup$













  • $begingroup$
    I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
    $endgroup$
    – S V
    8 hours ago

















2












$begingroup$


So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?










share|cite|improve this question











$endgroup$













  • $begingroup$
    I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
    $endgroup$
    – S V
    8 hours ago













2












2








2


1



$begingroup$


So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?










share|cite|improve this question











$endgroup$




So, this a question from a Phd qualifying examination. Given the following Lagrangan $$L=frac12dotqtextsin^2q,$$ what is the Hamiltonian for this system? So, finding the canonical momentum $p$ and then using $H=pdotq -L$, I find that it is identically zero. Now, the answer 'Zero' was a choice in the choices of answers given but the correct answer is that the Hamiltonian is undefined. I'm not sure why this should be the answer, though this is a pretty weird dynamical system. The Euler Lagrange equations just give an identity and the Hamiltonian equations also seem rather off. The Hamiltonian equations would are $$dotp=dotq=0. $$
I'm hoping someone shed some light on this, it seems that any random Lagrangian doesn't constitute a dynamical system, so what I should I be looking for?







homework-and-exercises classical-mechanics lagrangian-formalism hamiltonian-formalism






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share|cite|improve this question








edited 8 hours ago









Qmechanic

113k13 gold badges223 silver badges1341 bronze badges




113k13 gold badges223 silver badges1341 bronze badges










asked 8 hours ago









Soumil MaulickSoumil Maulick

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938 bronze badges














  • $begingroup$
    I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
    $endgroup$
    – S V
    8 hours ago
















  • $begingroup$
    I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
    $endgroup$
    – S V
    8 hours ago















$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago




$begingroup$
I think it is also important to note that the fact the EL equations are always satisfied means that any trajectory you pick is a valid solution. In particular two systems with identical initial conditions can have radically different trajectories.
$endgroup$
– S V
8 hours ago










2 Answers
2






active

oldest

votes


















6














$begingroup$

The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:



  • Finding $p$

  • Writing $H=pdot q-L$

  • Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.

For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.



However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is



$$
J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
$$



That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.






share|cite|improve this answer









$endgroup$






















    6














    $begingroup$

    1. OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.


    2. The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
      $$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      $begingroup$

      The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:



      • Finding $p$

      • Writing $H=pdot q-L$

      • Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.

      For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.



      However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is



      $$
      J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
      $$



      That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.






      share|cite|improve this answer









      $endgroup$



















        6














        $begingroup$

        The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:



        • Finding $p$

        • Writing $H=pdot q-L$

        • Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.

        For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.



        However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is



        $$
        J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
        $$



        That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.






        share|cite|improve this answer









        $endgroup$

















          6














          6










          6







          $begingroup$

          The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:



          • Finding $p$

          • Writing $H=pdot q-L$

          • Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.

          For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.



          However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is



          $$
          J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
          $$



          That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.






          share|cite|improve this answer









          $endgroup$



          The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires:



          • Finding $p$

          • Writing $H=pdot q-L$

          • Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $dot q$.

          For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,dot q)$ coordinate system and the $(q,p)$ coordinate system. This requires that $(q,p)$ is a good coordinate system, in that a state of the system can be uniquely represented by a pair of $(q,p)$ coordinates. In turn, this means that the transformation $(q,dot q)rightarrow (q,p)$ must have nonzero Jacobian.



          However, it's easy to see this is not the case. We know that $(q,p)=(q,frac12sin^2 q)$. Thus, the Jacobian is



          $$
          J=left|beginsmallmatrix1 & 0\sin(q)cos(q) & 0endsmallmatrixright| = 0
          $$



          That means we CAN'T write $p$ as a function of $q,dot q$, and there can't be a Hamiltonian $H(p,q)$ that is a function of only the coordinates $(p,q)$ because the coordinates $(p,q)$ don't specify the state of the system.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Jahan ClaesJahan Claes

          4,88012 silver badges34 bronze badges




          4,88012 silver badges34 bronze badges


























              6














              $begingroup$

              1. OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.


              2. The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
                $$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$






              share|cite|improve this answer











              $endgroup$



















                6














                $begingroup$

                1. OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.


                2. The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
                  $$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$






                share|cite|improve this answer











                $endgroup$

















                  6














                  6










                  6







                  $begingroup$

                  1. OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.


                  2. The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
                    $$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$






                  share|cite|improve this answer











                  $endgroup$



                  1. OP's Lagrangian $$L=fracddtleft(fracq4-frac18sin(2q)right)tag1$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, , cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, the Lagrangian (and Hamiltonian) formulations are ill-defined.


                  2. The Hamiltonian is defined as the Legendre transformation of the Lagrangian, and hence formally defined as
                    $$ H(q,p)~:=~sup_vinmathbbR(vp-L(q,v))~=~sup_vinmathbbRvleft(p-frac12sin^2(q)right)~=~left{beginarraylinftytext if p~neq~frac12sin^2(q),\ 0text otherwise. endarrayright. tag2$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 7 hours ago

























                  answered 8 hours ago









                  QmechanicQmechanic

                  113k13 gold badges223 silver badges1341 bronze badges




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