Mixing
Why the Hamiltonian is constant along the integral curves of the hamiltonian vector field?
$begingroup$
Let $H$ the Hamiltonian of a system and $gamma $ an integral curve of the Hamiltonian vector field, i.e. if $gamma (t)=(q(t),p(t))$ and $H(p,q)$ is the Hamiltonian, then $$begin{cases} dot p=-H_q\ dot q= H_pend{cases}.$$
In wikipedia the say that $H$ is constant along $gamma $. Why is this true ? i.e. why $H(gamma (t))$ is constant ?
symplectic-geometry
asked Jan 11 at 15:53
user623855user623855
1507
$endgroup$
add a comment |
$begingroup$
Let $H$ the Hamiltonian of a system and $gamma $ an integral curve of the Hamiltonian vector field, i.e. if $gamma (t)=(q(t),p(t))$ and $H(p,q)$ is the Hamiltonian, then $$begin{cases} dot p=-H_q\ dot q= H_pend{cases}.$$
In wikipedia the say that $H$ is constant along $gamma $. Why is this true ? i.e. why $H(gamma (t))$ is constant ?
symplectic-geometry
asked Jan 11 at 15:53
user623855user623855
1507
$endgroup$
add a comment |
$begingroup$
Let $H$ the Hamiltonian of a system and $gamma $ an integral curve of the Hamiltonian vector field, i.e. if $gamma (t)=(q(t),p(t))$ and $H(p,q)$ is the Hamiltonian, then $$begin{cases} dot p=-H_q\ dot q= H_pend{cases}.$$
In wikipedia the say that $H$ is constant along $gamma $. Why is this true ? i.e. why $H(gamma (t))$ is constant ?
symplectic-geometry
asked Jan 11 at 15:53
user623855user623855
1507
$endgroup$
Let $H$ the Hamiltonian of a system and $gamma $ an integral curve of the Hamiltonian vector field, i.e. if $gamma (t)=(q(t),p(t))$ and $H(p,q)$ is the Hamiltonian, then $$begin{cases} dot p=-H_q\ dot q= H_pend{cases}.$$
In wikipedia the say that $H$ is constant along $gamma $. Why is this true ? i.e. why $H(gamma (t))$ is constant ?
symplectic-geometry
symplectic-geometry
asked Jan 11 at 15:53
user623855user623855
1507
asked Jan 11 at 15:53
user623855user623855
1507
asked Jan 11 at 15:53
user623855user623855
1507
asked Jan 11 at 15:53
user623855user623855
1507
asked Jan 11 at 15:53
user623855user623855
1507
1507
add a comment |
add a comment |
1 Answer
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$begingroup$
For $H(p,q,t) = H(p(t),q(t))$ think that
$$
frac{d}{dt}H(p(t),q(t)) = H_p dot p + H_q dot q
$$
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
$endgroup$
$begingroup$
Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,dot gamma >=0$ ? Because $omega $ is not a scalar product (is not a riemann manifold)
$endgroup$
– user623855
Jan 11 at 16:03
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
$begingroup$
For $H(p,q,t) = H(p(t),q(t))$ think that
$$
frac{d}{dt}H(p(t),q(t)) = H_p dot p + H_q dot q
$$
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
$endgroup$
$begingroup$
Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,dot gamma >=0$ ? Because $omega $ is not a scalar product (is not a riemann manifold)
$endgroup$
– user623855
Jan 11 at 16:03
add a comment |
$begingroup$
For $H(p,q,t) = H(p(t),q(t))$ think that
$$
frac{d}{dt}H(p(t),q(t)) = H_p dot p + H_q dot q
$$
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
$endgroup$
$begingroup$
Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,dot gamma >=0$ ? Because $omega $ is not a scalar product (is not a riemann manifold)
$endgroup$
– user623855
Jan 11 at 16:03
add a comment |
$begingroup$
For $H(p,q,t) = H(p(t),q(t))$ think that
$$
frac{d}{dt}H(p(t),q(t)) = H_p dot p + H_q dot q
$$
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
$endgroup$
For $H(p,q,t) = H(p(t),q(t))$ think that
$$
frac{d}{dt}H(p(t),q(t)) = H_p dot p + H_q dot q
$$
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
edited Jan 11 at 16:11
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
answered Jan 11 at 16:01
CesareoCesareo
8,6793516
8,6793516
$begingroup$
Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,dot gamma >=0$ ? Because $omega $ is not a scalar product (is not a riemann manifold)
$endgroup$
– user623855
Jan 11 at 16:03
add a comment |
$begingroup$
Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,dot gamma >=0$ ? Because $omega $ is not a scalar product (is not a riemann manifold)
$endgroup$
– user623855
Jan 11 at 16:03
$begingroup$
Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,dot gamma >=0$ ? Because $omega $ is not a scalar product (is not a riemann manifold)
$endgroup$
– user623855
Jan 11 at 16:03
$begingroup$
Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,dot gamma >=0$ ? Because $omega $ is not a scalar product (is not a riemann manifold)
$endgroup$
– user623855
Jan 11 at 16:03
add a comment |
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