Proving that a certain function is an integral of motion for a Hamiltonian












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$begingroup$



Let $H=q_1p_1-q_2p_2-aq_1^2+bq_2^2$ (with $a,b$ constant) be a Hamiltionian.



Show that $G=dfrac{p_1-aq_1}{q_2}$ is a first integral (integral of motion) of this system.




According to the proposed solution, I need to show that the Poisson Bracket $[G,H] = 0$.



I used the following method, is this valid?



Other method?



The Hamiltionian equations are:



$$dot q_1 = dfrac{partial H}{partial p_1} = q_1 qquadqquad dot q_2 = -p_2$$



$$dot p_1 = -dfrac{partial H}{partial q_1} = -p_1+2aq_1 qquadqquad dot p_2 = q_2-2bq_2$$



Then, if $G$ is a integral of motion the total time derivative must equal zero (true?)



So:



$$begin{align}
dfrac{operatorname d G}{operatorname d t} & = dfrac{1}{q_2^2}left[ q_2(dot p_1-adot q_1) - dot q_2(p_1-aq_1)right]\
&= ldots text{(substitution using the Hamiltionian equations)} \
&= 0
end{align}$$






solution-verification classical-mechanics







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$endgroup$

















    1












    $begingroup$



    Let $H=q_1p_1-q_2p_2-aq_1^2+bq_2^2$ (with $a,b$ constant) be a Hamiltionian.



    Show that $G=dfrac{p_1-aq_1}{q_2}$ is a first integral (integral of motion) of this system.




    According to the proposed solution, I need to show that the Poisson Bracket $[G,H] = 0$.



    I used the following method, is this valid?



    Other method?



    The Hamiltionian equations are:



    $$dot q_1 = dfrac{partial H}{partial p_1} = q_1 qquadqquad dot q_2 = -p_2$$



    $$dot p_1 = -dfrac{partial H}{partial q_1} = -p_1+2aq_1 qquadqquad dot p_2 = q_2-2bq_2$$



    Then, if $G$ is a integral of motion the total time derivative must equal zero (true?)



    So:



    $$begin{align}
    dfrac{operatorname d G}{operatorname d t} & = dfrac{1}{q_2^2}left[ q_2(dot p_1-adot q_1) - dot q_2(p_1-aq_1)right]\
    &= ldots text{(substitution using the Hamiltionian equations)} \
    &= 0
    end{align}$$






    solution-verification classical-mechanics







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      Let $H=q_1p_1-q_2p_2-aq_1^2+bq_2^2$ (with $a,b$ constant) be a Hamiltionian.



      Show that $G=dfrac{p_1-aq_1}{q_2}$ is a first integral (integral of motion) of this system.




      According to the proposed solution, I need to show that the Poisson Bracket $[G,H] = 0$.



      I used the following method, is this valid?



      Other method?



      The Hamiltionian equations are:



      $$dot q_1 = dfrac{partial H}{partial p_1} = q_1 qquadqquad dot q_2 = -p_2$$



      $$dot p_1 = -dfrac{partial H}{partial q_1} = -p_1+2aq_1 qquadqquad dot p_2 = q_2-2bq_2$$



      Then, if $G$ is a integral of motion the total time derivative must equal zero (true?)



      So:



      $$begin{align}
      dfrac{operatorname d G}{operatorname d t} & = dfrac{1}{q_2^2}left[ q_2(dot p_1-adot q_1) - dot q_2(p_1-aq_1)right]\
      &= ldots text{(substitution using the Hamiltionian equations)} \
      &= 0
      end{align}$$






      solution-verification classical-mechanics







      share|cite|improve this question









      $endgroup$





      Let $H=q_1p_1-q_2p_2-aq_1^2+bq_2^2$ (with $a,b$ constant) be a Hamiltionian.



      Show that $G=dfrac{p_1-aq_1}{q_2}$ is a first integral (integral of motion) of this system.




      According to the proposed solution, I need to show that the Poisson Bracket $[G,H] = 0$.



      I used the following method, is this valid?



      Other method?



      The Hamiltionian equations are:



      $$dot q_1 = dfrac{partial H}{partial p_1} = q_1 qquadqquad dot q_2 = -p_2$$



      $$dot p_1 = -dfrac{partial H}{partial q_1} = -p_1+2aq_1 qquadqquad dot p_2 = q_2-2bq_2$$



      Then, if $G$ is a integral of motion the total time derivative must equal zero (true?)



      So:



      $$begin{align}
      dfrac{operatorname d G}{operatorname d t} & = dfrac{1}{q_2^2}left[ q_2(dot p_1-adot q_1) - dot q_2(p_1-aq_1)right]\
      &= ldots text{(substitution using the Hamiltionian equations)} \
      &= 0
      end{align}$$






      solution-verification classical-mechanics




      solution-verification classical-mechanics






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      asked May 15 '15 at 14:45









      dietervdfdietervdf

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      2,6761327






















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          $begingroup$

          Your solution method and the proposed one are equivalent. Integrals of motion are those functions whose total time derivative is zero (by definition), and the total time derivative of a function $G$ dependent only on coordinates is equal to the Poisson bracket: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$






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            1 Answer
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            0












            $begingroup$

            Your solution method and the proposed one are equivalent. Integrals of motion are those functions whose total time derivative is zero (by definition), and the total time derivative of a function $G$ dependent only on coordinates is equal to the Poisson bracket: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$






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              $begingroup$

              Your solution method and the proposed one are equivalent. Integrals of motion are those functions whose total time derivative is zero (by definition), and the total time derivative of a function $G$ dependent only on coordinates is equal to the Poisson bracket: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$






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                $begingroup$

                Your solution method and the proposed one are equivalent. Integrals of motion are those functions whose total time derivative is zero (by definition), and the total time derivative of a function $G$ dependent only on coordinates is equal to the Poisson bracket: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$






                share|cite|improve this answer









                $endgroup$



                Your solution method and the proposed one are equivalent. Integrals of motion are those functions whose total time derivative is zero (by definition), and the total time derivative of a function $G$ dependent only on coordinates is equal to the Poisson bracket: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$







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                answered Jan 22 at 20:03









                Ricardo BuringRicardo Buring

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