Differential Geometry in Hamiltonian Mechanics
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I have following question about the Hamiltonian mechanics from differential geometrical viewpoint:
We start with a physical system parametrized by generalized (position) coordinates $(q^i)$ providing under given restrictions the configuration space $Q$ or more precisely a (smooth) manifold.
The tangent bundle $TQ$ over $Q$ provides for the fixed coordinates $q^i$ the corresponding velocities $dot q ^j$. Geometrically the velocities $dot q ^j$ at $q^i$ belong to the tangent space $TQ_{q_0}$ at fixed point $q_0^i$ of $Q$.
So $q^i$ and $dot q ^j$ provide parameters for the Lagrangian $L[q(t), dot q(t)]$, a function on $TQ$.
On the other hand it's known that the Hamiltionian $H[q(t), p(t)]$ can be interpreted as a function on the cotangent bundle $T^*Q$ where for fixed $q_0^i in Q$ the momenta $(p_0)_j$ are living in cotangent space $T^*Q_{q_0}$.
My question is why (mathematically) the phase space spanned by position $q^i$ and momentum coordinates $p_j$ come from the cotangent bundle $T^*Q$ while the position $q^i$ and velocity coordinates $dot q_j$ come from tangent bundle ?
Or in other words why the cotangent space $T^*Q_{q}$ corresponds to momenta while the tangent space $TQ_{q}$ to velocities from viewpoint of differential geometry?
Remark: I know that elemenary physical approach always associates the velocity to the tangent space of the position but this doesn't answer concretely why the momenta spaces should belong exactly to the cotangents.
Intuitively I guess that the velocities and momenta should behave differently under transformations in sense of co- and contravariant coordinates but I'm not sure if this is the real reason for the problem above...
differential-geometry mathematical-physics hamilton-jacobi-equation
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
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add a comment |
$begingroup$
I have following question about the Hamiltonian mechanics from differential geometrical viewpoint:
We start with a physical system parametrized by generalized (position) coordinates $(q^i)$ providing under given restrictions the configuration space $Q$ or more precisely a (smooth) manifold.
The tangent bundle $TQ$ over $Q$ provides for the fixed coordinates $q^i$ the corresponding velocities $dot q ^j$. Geometrically the velocities $dot q ^j$ at $q^i$ belong to the tangent space $TQ_{q_0}$ at fixed point $q_0^i$ of $Q$.
So $q^i$ and $dot q ^j$ provide parameters for the Lagrangian $L[q(t), dot q(t)]$, a function on $TQ$.
On the other hand it's known that the Hamiltionian $H[q(t), p(t)]$ can be interpreted as a function on the cotangent bundle $T^*Q$ where for fixed $q_0^i in Q$ the momenta $(p_0)_j$ are living in cotangent space $T^*Q_{q_0}$.
My question is why (mathematically) the phase space spanned by position $q^i$ and momentum coordinates $p_j$ come from the cotangent bundle $T^*Q$ while the position $q^i$ and velocity coordinates $dot q_j$ come from tangent bundle ?
Or in other words why the cotangent space $T^*Q_{q}$ corresponds to momenta while the tangent space $TQ_{q}$ to velocities from viewpoint of differential geometry?
Remark: I know that elemenary physical approach always associates the velocity to the tangent space of the position but this doesn't answer concretely why the momenta spaces should belong exactly to the cotangents.
Intuitively I guess that the velocities and momenta should behave differently under transformations in sense of co- and contravariant coordinates but I'm not sure if this is the real reason for the problem above...
differential-geometry mathematical-physics hamilton-jacobi-equation
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
$endgroup$
$begingroup$
Perhaps this answer could be helpful for you.
$endgroup$
– hypernova
Jan 10 at 3:16
add a comment |
$begingroup$
I have following question about the Hamiltonian mechanics from differential geometrical viewpoint:
We start with a physical system parametrized by generalized (position) coordinates $(q^i)$ providing under given restrictions the configuration space $Q$ or more precisely a (smooth) manifold.
The tangent bundle $TQ$ over $Q$ provides for the fixed coordinates $q^i$ the corresponding velocities $dot q ^j$. Geometrically the velocities $dot q ^j$ at $q^i$ belong to the tangent space $TQ_{q_0}$ at fixed point $q_0^i$ of $Q$.
So $q^i$ and $dot q ^j$ provide parameters for the Lagrangian $L[q(t), dot q(t)]$, a function on $TQ$.
On the other hand it's known that the Hamiltionian $H[q(t), p(t)]$ can be interpreted as a function on the cotangent bundle $T^*Q$ where for fixed $q_0^i in Q$ the momenta $(p_0)_j$ are living in cotangent space $T^*Q_{q_0}$.
My question is why (mathematically) the phase space spanned by position $q^i$ and momentum coordinates $p_j$ come from the cotangent bundle $T^*Q$ while the position $q^i$ and velocity coordinates $dot q_j$ come from tangent bundle ?
Or in other words why the cotangent space $T^*Q_{q}$ corresponds to momenta while the tangent space $TQ_{q}$ to velocities from viewpoint of differential geometry?
Remark: I know that elemenary physical approach always associates the velocity to the tangent space of the position but this doesn't answer concretely why the momenta spaces should belong exactly to the cotangents.
Intuitively I guess that the velocities and momenta should behave differently under transformations in sense of co- and contravariant coordinates but I'm not sure if this is the real reason for the problem above...
differential-geometry mathematical-physics hamilton-jacobi-equation
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
$endgroup$
I have following question about the Hamiltonian mechanics from differential geometrical viewpoint:
We start with a physical system parametrized by generalized (position) coordinates $(q^i)$ providing under given restrictions the configuration space $Q$ or more precisely a (smooth) manifold.
The tangent bundle $TQ$ over $Q$ provides for the fixed coordinates $q^i$ the corresponding velocities $dot q ^j$. Geometrically the velocities $dot q ^j$ at $q^i$ belong to the tangent space $TQ_{q_0}$ at fixed point $q_0^i$ of $Q$.
So $q^i$ and $dot q ^j$ provide parameters for the Lagrangian $L[q(t), dot q(t)]$, a function on $TQ$.
On the other hand it's known that the Hamiltionian $H[q(t), p(t)]$ can be interpreted as a function on the cotangent bundle $T^*Q$ where for fixed $q_0^i in Q$ the momenta $(p_0)_j$ are living in cotangent space $T^*Q_{q_0}$.
My question is why (mathematically) the phase space spanned by position $q^i$ and momentum coordinates $p_j$ come from the cotangent bundle $T^*Q$ while the position $q^i$ and velocity coordinates $dot q_j$ come from tangent bundle ?
Or in other words why the cotangent space $T^*Q_{q}$ corresponds to momenta while the tangent space $TQ_{q}$ to velocities from viewpoint of differential geometry?
Remark: I know that elemenary physical approach always associates the velocity to the tangent space of the position but this doesn't answer concretely why the momenta spaces should belong exactly to the cotangents.
Intuitively I guess that the velocities and momenta should behave differently under transformations in sense of co- and contravariant coordinates but I'm not sure if this is the real reason for the problem above...
differential-geometry mathematical-physics hamilton-jacobi-equation
differential-geometry mathematical-physics hamilton-jacobi-equation
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
asked Jan 9 at 23:02
KarlPeterKarlPeter
3541315
3541315
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Perhaps this answer could be helpful for you.
$endgroup$
– hypernova
Jan 10 at 3:16
add a comment |
$begingroup$
Perhaps this answer could be helpful for you.
$endgroup$
– hypernova
Jan 10 at 3:16
$begingroup$
Perhaps this answer could be helpful for you.
$endgroup$
– hypernova
Jan 10 at 3:16
$begingroup$
Perhaps this answer could be helpful for you.
$endgroup$
– hypernova
Jan 10 at 3:16
add a comment |
1 Answer
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Let $T(q, dot q)$ denote the kinetic energy of the system at configuration $q$ and velocity $dot q$. On the tangent plane of some fixed point $q_0$, the kinetic energy induces a norm $frac{1}{2}|dot q|^2 = T(q_0, dot q)$ on tangent vectors, which can be extended to an inner product via the polarization identity, and from there to a kinetic energy metric $g_T$ on the entire tangent bundle $TQ$. In this way kinetic energy gives $Q$ its geometric structure.
Momenta live on the cotangent space because they are dual to velocities, in the sense of the musical isomorphisms: a configurational velocity $dot q$ and its corresponding momentum $p = dot q^{flat}$ satisfy, for any other tangent vector $v$,
$$pv = langle dot q, vrangle_{g_T}$$
and in particular, $pdot q = 2T(q,dot q).$
answered Jan 10 at 1:51
user7530user7530
34.9k760113
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1 Answer
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1 Answer
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active
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votes
active
oldest
votes
$begingroup$
Let $T(q, dot q)$ denote the kinetic energy of the system at configuration $q$ and velocity $dot q$. On the tangent plane of some fixed point $q_0$, the kinetic energy induces a norm $frac{1}{2}|dot q|^2 = T(q_0, dot q)$ on tangent vectors, which can be extended to an inner product via the polarization identity, and from there to a kinetic energy metric $g_T$ on the entire tangent bundle $TQ$. In this way kinetic energy gives $Q$ its geometric structure.
Momenta live on the cotangent space because they are dual to velocities, in the sense of the musical isomorphisms: a configurational velocity $dot q$ and its corresponding momentum $p = dot q^{flat}$ satisfy, for any other tangent vector $v$,
$$pv = langle dot q, vrangle_{g_T}$$
and in particular, $pdot q = 2T(q,dot q).$
answered Jan 10 at 1:51
user7530user7530
34.9k760113
$endgroup$
add a comment |
$begingroup$
Let $T(q, dot q)$ denote the kinetic energy of the system at configuration $q$ and velocity $dot q$. On the tangent plane of some fixed point $q_0$, the kinetic energy induces a norm $frac{1}{2}|dot q|^2 = T(q_0, dot q)$ on tangent vectors, which can be extended to an inner product via the polarization identity, and from there to a kinetic energy metric $g_T$ on the entire tangent bundle $TQ$. In this way kinetic energy gives $Q$ its geometric structure.
Momenta live on the cotangent space because they are dual to velocities, in the sense of the musical isomorphisms: a configurational velocity $dot q$ and its corresponding momentum $p = dot q^{flat}$ satisfy, for any other tangent vector $v$,
$$pv = langle dot q, vrangle_{g_T}$$
and in particular, $pdot q = 2T(q,dot q).$
answered Jan 10 at 1:51
user7530user7530
34.9k760113
$endgroup$
add a comment |
$begingroup$
Let $T(q, dot q)$ denote the kinetic energy of the system at configuration $q$ and velocity $dot q$. On the tangent plane of some fixed point $q_0$, the kinetic energy induces a norm $frac{1}{2}|dot q|^2 = T(q_0, dot q)$ on tangent vectors, which can be extended to an inner product via the polarization identity, and from there to a kinetic energy metric $g_T$ on the entire tangent bundle $TQ$. In this way kinetic energy gives $Q$ its geometric structure.
Momenta live on the cotangent space because they are dual to velocities, in the sense of the musical isomorphisms: a configurational velocity $dot q$ and its corresponding momentum $p = dot q^{flat}$ satisfy, for any other tangent vector $v$,
$$pv = langle dot q, vrangle_{g_T}$$
and in particular, $pdot q = 2T(q,dot q).$
answered Jan 10 at 1:51
user7530user7530
34.9k760113
$endgroup$
Let $T(q, dot q)$ denote the kinetic energy of the system at configuration $q$ and velocity $dot q$. On the tangent plane of some fixed point $q_0$, the kinetic energy induces a norm $frac{1}{2}|dot q|^2 = T(q_0, dot q)$ on tangent vectors, which can be extended to an inner product via the polarization identity, and from there to a kinetic energy metric $g_T$ on the entire tangent bundle $TQ$. In this way kinetic energy gives $Q$ its geometric structure.
Momenta live on the cotangent space because they are dual to velocities, in the sense of the musical isomorphisms: a configurational velocity $dot q$ and its corresponding momentum $p = dot q^{flat}$ satisfy, for any other tangent vector $v$,
$$pv = langle dot q, vrangle_{g_T}$$
and in particular, $pdot q = 2T(q,dot q).$
answered Jan 10 at 1:51
user7530user7530
34.9k760113
answered Jan 10 at 1:51
user7530user7530
34.9k760113
answered Jan 10 at 1:51
user7530user7530
34.9k760113
answered Jan 10 at 1:51
user7530user7530
34.9k760113
34.9k760113
add a comment |
add a comment |
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– hypernova
Jan 10 at 3:16