Undamped simple Pendulum written in Hamiltonian form












1














Show that the equation of motion for an undamped simple pendulum



$y'' + frac{g}{l}sin(y)=0$



can be written in Hamiltonian form.



So this is the Euler-Lagrange form of the simple pendulum equation, which I know can be written in Hamiltonian form, and from there I can extract the canonical/Hamilton equations, nonetheless, if I attempt to write this as a system of first order differential equations and attempt to define the potential and kinetic energy I become stuck. So far I have the first order equations



$y_1'=y_2$



$y_2'=-frac{g}{l}sin(y_1)$



nonetheless, if I define the potential as



$y_2'=-frac{g}{lm}sin(y_1) = -frac{1}{m}frac{dU(y)}{dy}$



taking the kinetic energy as T=$frac{my_2^2}{2}$,



then the total energy would read



$E(y_1,y_2) = U(y_1) + frac{my_2^2}{2}$,



$E(y_1,y_2) = frac{g}{l}cos(y_1) + frac{my_2^2}{2}$,



hence, the Hamiltonian form would read



$y_1' = my_2$ and $y_2' = -frac{g}{ml}sin(y_1)$



Is the Total energy equation correct? As well as the Hamilton equations? Any help would be greatly appreciated.






differential-equations proof-writing hamilton-equations







share|cite|improve this question





























    1














    Show that the equation of motion for an undamped simple pendulum



    $y'' + frac{g}{l}sin(y)=0$



    can be written in Hamiltonian form.



    So this is the Euler-Lagrange form of the simple pendulum equation, which I know can be written in Hamiltonian form, and from there I can extract the canonical/Hamilton equations, nonetheless, if I attempt to write this as a system of first order differential equations and attempt to define the potential and kinetic energy I become stuck. So far I have the first order equations



    $y_1'=y_2$



    $y_2'=-frac{g}{l}sin(y_1)$



    nonetheless, if I define the potential as



    $y_2'=-frac{g}{lm}sin(y_1) = -frac{1}{m}frac{dU(y)}{dy}$



    taking the kinetic energy as T=$frac{my_2^2}{2}$,



    then the total energy would read



    $E(y_1,y_2) = U(y_1) + frac{my_2^2}{2}$,



    $E(y_1,y_2) = frac{g}{l}cos(y_1) + frac{my_2^2}{2}$,



    hence, the Hamiltonian form would read



    $y_1' = my_2$ and $y_2' = -frac{g}{ml}sin(y_1)$



    Is the Total energy equation correct? As well as the Hamilton equations? Any help would be greatly appreciated.






    differential-equations proof-writing hamilton-equations







    share|cite|improve this question



























      1












      1








      1







      Show that the equation of motion for an undamped simple pendulum



      $y'' + frac{g}{l}sin(y)=0$



      can be written in Hamiltonian form.



      So this is the Euler-Lagrange form of the simple pendulum equation, which I know can be written in Hamiltonian form, and from there I can extract the canonical/Hamilton equations, nonetheless, if I attempt to write this as a system of first order differential equations and attempt to define the potential and kinetic energy I become stuck. So far I have the first order equations



      $y_1'=y_2$



      $y_2'=-frac{g}{l}sin(y_1)$



      nonetheless, if I define the potential as



      $y_2'=-frac{g}{lm}sin(y_1) = -frac{1}{m}frac{dU(y)}{dy}$



      taking the kinetic energy as T=$frac{my_2^2}{2}$,



      then the total energy would read



      $E(y_1,y_2) = U(y_1) + frac{my_2^2}{2}$,



      $E(y_1,y_2) = frac{g}{l}cos(y_1) + frac{my_2^2}{2}$,



      hence, the Hamiltonian form would read



      $y_1' = my_2$ and $y_2' = -frac{g}{ml}sin(y_1)$



      Is the Total energy equation correct? As well as the Hamilton equations? Any help would be greatly appreciated.






      differential-equations proof-writing hamilton-equations







      share|cite|improve this question















      Show that the equation of motion for an undamped simple pendulum



      $y'' + frac{g}{l}sin(y)=0$



      can be written in Hamiltonian form.



      So this is the Euler-Lagrange form of the simple pendulum equation, which I know can be written in Hamiltonian form, and from there I can extract the canonical/Hamilton equations, nonetheless, if I attempt to write this as a system of first order differential equations and attempt to define the potential and kinetic energy I become stuck. So far I have the first order equations



      $y_1'=y_2$



      $y_2'=-frac{g}{l}sin(y_1)$



      nonetheless, if I define the potential as



      $y_2'=-frac{g}{lm}sin(y_1) = -frac{1}{m}frac{dU(y)}{dy}$



      taking the kinetic energy as T=$frac{my_2^2}{2}$,



      then the total energy would read



      $E(y_1,y_2) = U(y_1) + frac{my_2^2}{2}$,



      $E(y_1,y_2) = frac{g}{l}cos(y_1) + frac{my_2^2}{2}$,



      hence, the Hamiltonian form would read



      $y_1' = my_2$ and $y_2' = -frac{g}{ml}sin(y_1)$



      Is the Total energy equation correct? As well as the Hamilton equations? Any help would be greatly appreciated.






      differential-equations proof-writing hamilton-equations




      differential-equations proof-writing hamilton-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 at 2:24

























      asked Dec 2 at 11:19









      lastgunslinger

      628




      628






















          2 Answers
          2






          active

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          0














          Define the potential a bit different



          $$
          -frac{gm}{l}sin y_1 = - frac{partial}{partial y_1}U(y_1)
          $$



          and the kinetic energy as



          $$
          T = frac{1}{2}my_2^2
          $$



          The Hamiltonian is then



          $$
          H(y_1, y_2) = frac{1}{2}my_2^2 + U(y_1)
          $$



          and the equations of motion are



          begin{eqnarray}
          frac{{rm d}y_1}{{rm d}t} &=& frac{partial H}{partial y_2} = my_2 \
          frac{{rm d}y_2}{{rm d}t} &=& -frac{partial H}{partial y_1} = -frac{partial U}{partial y_1} = -frac{gm}{l}sin y_1
          end{eqnarray}



          That being said, I don't see why you need to introduce the factor $m$. If you truly need it in your equations, remember that $H$ and $H/m$ will lead to the same equations of motion






          share|cite|improve this answer





















          • The reason I had introduced the m factor is to satisfy the equation of motion f=ma, so we have that after the 2nd DE if I define f, then a = f/m, then finding the potential from here would transfer that constant to the d/dy U, thus, we would have d/du U * 1/m.
            – lastgunslinger
            Dec 2 at 11:52










          • @lastgunslinger So I'm afraid I didn't understand what your question is then, do you mind expanding on your comment "this can't be right?" what can't be?
            – caverac
            Dec 2 at 11:54












          • What I mean by it can't be right is that the answer is too simple, that's why I have doubts concerning the answer, is it really that simple?
            – lastgunslinger
            Dec 2 at 22:29










          • @lastgunslinger :) It is that simple
            – caverac
            Dec 2 at 22:31










          • Wow, then I guess I was right? Should I not introduce the mass factor m, then?
            – lastgunslinger
            Dec 2 at 22:33



















          0














          Multiply the original equation with $y'$ and integrate to get
          $$
          C=frac{y'^2}2+frac{g}{l}(1-cos y).
          $$
          If you want to introduce further physical context, notice that $y$ is the angle, thus $y'$ the angular velocity, so that the tangential velocity is $ly'$ and thus scaling to have a kinetic energy term gives
          $$
          E = mfrac{(ly')^2}2+mgl(1-cos(y)),
          $$
          where $l(1-cos y)$ is the height over the rest position of the pendulum of length $l$ with center in $(0,l)$, so that $mgl(1-cos(y))$ is indeed the potential energy.



          The impulse variable for $y$ is then $p=ml^2y'$, so that
          begin{align}
          H(y,p)&=frac{p^2}{2ml^2}+mgl(1-cos(y))\
          y'&=H_p=frac{p}{ml^2}\
          p'&=-H_y=-mglsin(y)\
          y''=frac{p'}{ml^2}&=-frac glsin y
          end{align}
          and indeed the Hamiltonian system returns the original equation.






          share|cite|improve this answer























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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            0














            Define the potential a bit different



            $$
            -frac{gm}{l}sin y_1 = - frac{partial}{partial y_1}U(y_1)
            $$



            and the kinetic energy as



            $$
            T = frac{1}{2}my_2^2
            $$



            The Hamiltonian is then



            $$
            H(y_1, y_2) = frac{1}{2}my_2^2 + U(y_1)
            $$



            and the equations of motion are



            begin{eqnarray}
            frac{{rm d}y_1}{{rm d}t} &=& frac{partial H}{partial y_2} = my_2 \
            frac{{rm d}y_2}{{rm d}t} &=& -frac{partial H}{partial y_1} = -frac{partial U}{partial y_1} = -frac{gm}{l}sin y_1
            end{eqnarray}



            That being said, I don't see why you need to introduce the factor $m$. If you truly need it in your equations, remember that $H$ and $H/m$ will lead to the same equations of motion






            share|cite|improve this answer





















            • The reason I had introduced the m factor is to satisfy the equation of motion f=ma, so we have that after the 2nd DE if I define f, then a = f/m, then finding the potential from here would transfer that constant to the d/dy U, thus, we would have d/du U * 1/m.
              – lastgunslinger
              Dec 2 at 11:52










            • @lastgunslinger So I'm afraid I didn't understand what your question is then, do you mind expanding on your comment "this can't be right?" what can't be?
              – caverac
              Dec 2 at 11:54












            • What I mean by it can't be right is that the answer is too simple, that's why I have doubts concerning the answer, is it really that simple?
              – lastgunslinger
              Dec 2 at 22:29










            • @lastgunslinger :) It is that simple
              – caverac
              Dec 2 at 22:31










            • Wow, then I guess I was right? Should I not introduce the mass factor m, then?
              – lastgunslinger
              Dec 2 at 22:33
















            0














            Define the potential a bit different



            $$
            -frac{gm}{l}sin y_1 = - frac{partial}{partial y_1}U(y_1)
            $$



            and the kinetic energy as



            $$
            T = frac{1}{2}my_2^2
            $$



            The Hamiltonian is then



            $$
            H(y_1, y_2) = frac{1}{2}my_2^2 + U(y_1)
            $$



            and the equations of motion are



            begin{eqnarray}
            frac{{rm d}y_1}{{rm d}t} &=& frac{partial H}{partial y_2} = my_2 \
            frac{{rm d}y_2}{{rm d}t} &=& -frac{partial H}{partial y_1} = -frac{partial U}{partial y_1} = -frac{gm}{l}sin y_1
            end{eqnarray}



            That being said, I don't see why you need to introduce the factor $m$. If you truly need it in your equations, remember that $H$ and $H/m$ will lead to the same equations of motion






            share|cite|improve this answer





















            • The reason I had introduced the m factor is to satisfy the equation of motion f=ma, so we have that after the 2nd DE if I define f, then a = f/m, then finding the potential from here would transfer that constant to the d/dy U, thus, we would have d/du U * 1/m.
              – lastgunslinger
              Dec 2 at 11:52










            • @lastgunslinger So I'm afraid I didn't understand what your question is then, do you mind expanding on your comment "this can't be right?" what can't be?
              – caverac
              Dec 2 at 11:54












            • What I mean by it can't be right is that the answer is too simple, that's why I have doubts concerning the answer, is it really that simple?
              – lastgunslinger
              Dec 2 at 22:29










            • @lastgunslinger :) It is that simple
              – caverac
              Dec 2 at 22:31










            • Wow, then I guess I was right? Should I not introduce the mass factor m, then?
              – lastgunslinger
              Dec 2 at 22:33














            0












            0








            0






            Define the potential a bit different



            $$
            -frac{gm}{l}sin y_1 = - frac{partial}{partial y_1}U(y_1)
            $$



            and the kinetic energy as



            $$
            T = frac{1}{2}my_2^2
            $$



            The Hamiltonian is then



            $$
            H(y_1, y_2) = frac{1}{2}my_2^2 + U(y_1)
            $$



            and the equations of motion are



            begin{eqnarray}
            frac{{rm d}y_1}{{rm d}t} &=& frac{partial H}{partial y_2} = my_2 \
            frac{{rm d}y_2}{{rm d}t} &=& -frac{partial H}{partial y_1} = -frac{partial U}{partial y_1} = -frac{gm}{l}sin y_1
            end{eqnarray}



            That being said, I don't see why you need to introduce the factor $m$. If you truly need it in your equations, remember that $H$ and $H/m$ will lead to the same equations of motion






            share|cite|improve this answer












            Define the potential a bit different



            $$
            -frac{gm}{l}sin y_1 = - frac{partial}{partial y_1}U(y_1)
            $$



            and the kinetic energy as



            $$
            T = frac{1}{2}my_2^2
            $$



            The Hamiltonian is then



            $$
            H(y_1, y_2) = frac{1}{2}my_2^2 + U(y_1)
            $$



            and the equations of motion are



            begin{eqnarray}
            frac{{rm d}y_1}{{rm d}t} &=& frac{partial H}{partial y_2} = my_2 \
            frac{{rm d}y_2}{{rm d}t} &=& -frac{partial H}{partial y_1} = -frac{partial U}{partial y_1} = -frac{gm}{l}sin y_1
            end{eqnarray}



            That being said, I don't see why you need to introduce the factor $m$. If you truly need it in your equations, remember that $H$ and $H/m$ will lead to the same equations of motion







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 2 at 11:45









            caverac

            13.4k21029




            13.4k21029












            • The reason I had introduced the m factor is to satisfy the equation of motion f=ma, so we have that after the 2nd DE if I define f, then a = f/m, then finding the potential from here would transfer that constant to the d/dy U, thus, we would have d/du U * 1/m.
              – lastgunslinger
              Dec 2 at 11:52










            • @lastgunslinger So I'm afraid I didn't understand what your question is then, do you mind expanding on your comment "this can't be right?" what can't be?
              – caverac
              Dec 2 at 11:54












            • What I mean by it can't be right is that the answer is too simple, that's why I have doubts concerning the answer, is it really that simple?
              – lastgunslinger
              Dec 2 at 22:29










            • @lastgunslinger :) It is that simple
              – caverac
              Dec 2 at 22:31










            • Wow, then I guess I was right? Should I not introduce the mass factor m, then?
              – lastgunslinger
              Dec 2 at 22:33


















            • The reason I had introduced the m factor is to satisfy the equation of motion f=ma, so we have that after the 2nd DE if I define f, then a = f/m, then finding the potential from here would transfer that constant to the d/dy U, thus, we would have d/du U * 1/m.
              – lastgunslinger
              Dec 2 at 11:52










            • @lastgunslinger So I'm afraid I didn't understand what your question is then, do you mind expanding on your comment "this can't be right?" what can't be?
              – caverac
              Dec 2 at 11:54












            • What I mean by it can't be right is that the answer is too simple, that's why I have doubts concerning the answer, is it really that simple?
              – lastgunslinger
              Dec 2 at 22:29










            • @lastgunslinger :) It is that simple
              – caverac
              Dec 2 at 22:31










            • Wow, then I guess I was right? Should I not introduce the mass factor m, then?
              – lastgunslinger
              Dec 2 at 22:33
















            The reason I had introduced the m factor is to satisfy the equation of motion f=ma, so we have that after the 2nd DE if I define f, then a = f/m, then finding the potential from here would transfer that constant to the d/dy U, thus, we would have d/du U * 1/m.
            – lastgunslinger
            Dec 2 at 11:52




            The reason I had introduced the m factor is to satisfy the equation of motion f=ma, so we have that after the 2nd DE if I define f, then a = f/m, then finding the potential from here would transfer that constant to the d/dy U, thus, we would have d/du U * 1/m.
            – lastgunslinger
            Dec 2 at 11:52












            @lastgunslinger So I'm afraid I didn't understand what your question is then, do you mind expanding on your comment "this can't be right?" what can't be?
            – caverac
            Dec 2 at 11:54






            @lastgunslinger So I'm afraid I didn't understand what your question is then, do you mind expanding on your comment "this can't be right?" what can't be?
            – caverac
            Dec 2 at 11:54














            What I mean by it can't be right is that the answer is too simple, that's why I have doubts concerning the answer, is it really that simple?
            – lastgunslinger
            Dec 2 at 22:29




            What I mean by it can't be right is that the answer is too simple, that's why I have doubts concerning the answer, is it really that simple?
            – lastgunslinger
            Dec 2 at 22:29












            @lastgunslinger :) It is that simple
            – caverac
            Dec 2 at 22:31




            @lastgunslinger :) It is that simple
            – caverac
            Dec 2 at 22:31












            Wow, then I guess I was right? Should I not introduce the mass factor m, then?
            – lastgunslinger
            Dec 2 at 22:33




            Wow, then I guess I was right? Should I not introduce the mass factor m, then?
            – lastgunslinger
            Dec 2 at 22:33











            0














            Multiply the original equation with $y'$ and integrate to get
            $$
            C=frac{y'^2}2+frac{g}{l}(1-cos y).
            $$
            If you want to introduce further physical context, notice that $y$ is the angle, thus $y'$ the angular velocity, so that the tangential velocity is $ly'$ and thus scaling to have a kinetic energy term gives
            $$
            E = mfrac{(ly')^2}2+mgl(1-cos(y)),
            $$
            where $l(1-cos y)$ is the height over the rest position of the pendulum of length $l$ with center in $(0,l)$, so that $mgl(1-cos(y))$ is indeed the potential energy.



            The impulse variable for $y$ is then $p=ml^2y'$, so that
            begin{align}
            H(y,p)&=frac{p^2}{2ml^2}+mgl(1-cos(y))\
            y'&=H_p=frac{p}{ml^2}\
            p'&=-H_y=-mglsin(y)\
            y''=frac{p'}{ml^2}&=-frac glsin y
            end{align}
            and indeed the Hamiltonian system returns the original equation.






            share|cite|improve this answer




























              0














              Multiply the original equation with $y'$ and integrate to get
              $$
              C=frac{y'^2}2+frac{g}{l}(1-cos y).
              $$
              If you want to introduce further physical context, notice that $y$ is the angle, thus $y'$ the angular velocity, so that the tangential velocity is $ly'$ and thus scaling to have a kinetic energy term gives
              $$
              E = mfrac{(ly')^2}2+mgl(1-cos(y)),
              $$
              where $l(1-cos y)$ is the height over the rest position of the pendulum of length $l$ with center in $(0,l)$, so that $mgl(1-cos(y))$ is indeed the potential energy.



              The impulse variable for $y$ is then $p=ml^2y'$, so that
              begin{align}
              H(y,p)&=frac{p^2}{2ml^2}+mgl(1-cos(y))\
              y'&=H_p=frac{p}{ml^2}\
              p'&=-H_y=-mglsin(y)\
              y''=frac{p'}{ml^2}&=-frac glsin y
              end{align}
              and indeed the Hamiltonian system returns the original equation.






              share|cite|improve this answer


























                0












                0








                0






                Multiply the original equation with $y'$ and integrate to get
                $$
                C=frac{y'^2}2+frac{g}{l}(1-cos y).
                $$
                If you want to introduce further physical context, notice that $y$ is the angle, thus $y'$ the angular velocity, so that the tangential velocity is $ly'$ and thus scaling to have a kinetic energy term gives
                $$
                E = mfrac{(ly')^2}2+mgl(1-cos(y)),
                $$
                where $l(1-cos y)$ is the height over the rest position of the pendulum of length $l$ with center in $(0,l)$, so that $mgl(1-cos(y))$ is indeed the potential energy.



                The impulse variable for $y$ is then $p=ml^2y'$, so that
                begin{align}
                H(y,p)&=frac{p^2}{2ml^2}+mgl(1-cos(y))\
                y'&=H_p=frac{p}{ml^2}\
                p'&=-H_y=-mglsin(y)\
                y''=frac{p'}{ml^2}&=-frac glsin y
                end{align}
                and indeed the Hamiltonian system returns the original equation.






                share|cite|improve this answer














                Multiply the original equation with $y'$ and integrate to get
                $$
                C=frac{y'^2}2+frac{g}{l}(1-cos y).
                $$
                If you want to introduce further physical context, notice that $y$ is the angle, thus $y'$ the angular velocity, so that the tangential velocity is $ly'$ and thus scaling to have a kinetic energy term gives
                $$
                E = mfrac{(ly')^2}2+mgl(1-cos(y)),
                $$
                where $l(1-cos y)$ is the height over the rest position of the pendulum of length $l$ with center in $(0,l)$, so that $mgl(1-cos(y))$ is indeed the potential energy.



                The impulse variable for $y$ is then $p=ml^2y'$, so that
                begin{align}
                H(y,p)&=frac{p^2}{2ml^2}+mgl(1-cos(y))\
                y'&=H_p=frac{p}{ml^2}\
                p'&=-H_y=-mglsin(y)\
                y''=frac{p'}{ml^2}&=-frac glsin y
                end{align}
                and indeed the Hamiltonian system returns the original equation.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 2 at 12:37

























                answered Dec 2 at 12:16









                LutzL

                55.9k42054




                55.9k42054






























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