Mixing
Proving that a certain function is an integral of motion for a Hamiltonian
$begingroup$
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
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
$endgroup$
add a comment |
$begingroup$
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
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
$endgroup$
add a comment |
$begingroup$
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
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
$endgroup$
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
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
asked May 15 '15 at 14:45
dietervdfdietervdf
2,6761327
2,6761327
add a comment |
add a comment |
1 Answer
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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: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
add a comment |
$begingroup$
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: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
add a comment |
$begingroup$
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: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
$endgroup$
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: $$begin{align*}dot{G} &= frac{dG}{dt}\ & = frac{partial G}{partial q_1}dot{q_1} + frac{partial G}{partial q_2}dot{q_2} + frac{partial G}{partial p_1}dot{p_1} + frac{partial G}{partial p_2}dot{p_2}\ &= frac{partial G}{partial q_1}frac{partial H}{partial p_1} + frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_1}frac{partial H}{partial q_1} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}\ &=bigg(frac{partial G}{partial q_1}frac{partial H}{partial p_1} - frac{partial G}{partial p_1}frac{partial H}{partial q_1}bigg) + bigg(frac{partial G}{partial q_2}frac{partial H}{partial p_2} - frac{partial G}{partial p_2}frac{partial H}{partial q_2}bigg) \& = {G,H}end{align*}$$
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
answered Jan 22 at 20:03
Ricardo BuringRicardo Buring
1,4711334
1,4711334
add a comment |
add a comment |
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