Mixing
How to decompose the vector field by the parameter t
i have a fields:
$X=X_1 dfrac {partial }{partial q_1}+...+X_n dfrac {partial }{partial q_n}$
$Y=Y_1 dfrac {partial }{partial q_1}+...+Y_n dfrac {partial }{partial q_n}$
$Z=Y_1(X_t(q)) dfrac {partial }{partial X^1_{t}(q)}+...+Y_n(X_t(q)) dfrac {partial }{partial X^n_{t}}$,
where $X_t(q)$ - vector flow.
All i want is to find $dot{Z}_{t=0}$.
I tried to do next calculations:
$Z(X_t(q))=Z_0(X_t(q))+dot{Z}t+o(t)$;
easy to get that $Z_0(X_t(q))=Y(X_t(q))=Y(q)+Y(X(q))t+o(t)$,
then $Z(X_t(q))=Y(q)+Y(X(q))t+dot{Z}t+0(t)$.
I'm not very sure, but it turns out that on the other hand, $Z(X_t(q))=d(X_t(q))[Y(q)]$;
since $X_t(q)=q+X(q)t+o(t)$ then $d(X_t(q))approx Id+d(X(q)t)$,
then $Z(X_t(q))=Y(q)+d(X(q)t)[Y(q)]+o(t)=Y(q)+Y(X(q))t+dot{Z}t+o(t)$
but i don’t really know what is $d(X(q)t)[Y(q)]$, it looks like $Y(X(q))$ or $X(Y(q))$..
so the answer is $[X,Y](q)$ or 0..
can you help me with this?
vector-fields lie-derivative
asked Nov 21 '18 at 10:12
Ilya
285
add a comment |
i have a fields:
$X=X_1 dfrac {partial }{partial q_1}+...+X_n dfrac {partial }{partial q_n}$
$Y=Y_1 dfrac {partial }{partial q_1}+...+Y_n dfrac {partial }{partial q_n}$
$Z=Y_1(X_t(q)) dfrac {partial }{partial X^1_{t}(q)}+...+Y_n(X_t(q)) dfrac {partial }{partial X^n_{t}}$,
where $X_t(q)$ - vector flow.
All i want is to find $dot{Z}_{t=0}$.
I tried to do next calculations:
$Z(X_t(q))=Z_0(X_t(q))+dot{Z}t+o(t)$;
easy to get that $Z_0(X_t(q))=Y(X_t(q))=Y(q)+Y(X(q))t+o(t)$,
then $Z(X_t(q))=Y(q)+Y(X(q))t+dot{Z}t+0(t)$.
I'm not very sure, but it turns out that on the other hand, $Z(X_t(q))=d(X_t(q))[Y(q)]$;
since $X_t(q)=q+X(q)t+o(t)$ then $d(X_t(q))approx Id+d(X(q)t)$,
then $Z(X_t(q))=Y(q)+d(X(q)t)[Y(q)]+o(t)=Y(q)+Y(X(q))t+dot{Z}t+o(t)$
but i don’t really know what is $d(X(q)t)[Y(q)]$, it looks like $Y(X(q))$ or $X(Y(q))$..
so the answer is $[X,Y](q)$ or 0..
can you help me with this?
vector-fields lie-derivative
asked Nov 21 '18 at 10:12
Ilya
285
add a comment |
i have a fields:
$X=X_1 dfrac {partial }{partial q_1}+...+X_n dfrac {partial }{partial q_n}$
$Y=Y_1 dfrac {partial }{partial q_1}+...+Y_n dfrac {partial }{partial q_n}$
$Z=Y_1(X_t(q)) dfrac {partial }{partial X^1_{t}(q)}+...+Y_n(X_t(q)) dfrac {partial }{partial X^n_{t}}$,
where $X_t(q)$ - vector flow.
All i want is to find $dot{Z}_{t=0}$.
I tried to do next calculations:
$Z(X_t(q))=Z_0(X_t(q))+dot{Z}t+o(t)$;
easy to get that $Z_0(X_t(q))=Y(X_t(q))=Y(q)+Y(X(q))t+o(t)$,
then $Z(X_t(q))=Y(q)+Y(X(q))t+dot{Z}t+0(t)$.
I'm not very sure, but it turns out that on the other hand, $Z(X_t(q))=d(X_t(q))[Y(q)]$;
since $X_t(q)=q+X(q)t+o(t)$ then $d(X_t(q))approx Id+d(X(q)t)$,
then $Z(X_t(q))=Y(q)+d(X(q)t)[Y(q)]+o(t)=Y(q)+Y(X(q))t+dot{Z}t+o(t)$
but i don’t really know what is $d(X(q)t)[Y(q)]$, it looks like $Y(X(q))$ or $X(Y(q))$..
so the answer is $[X,Y](q)$ or 0..
can you help me with this?
vector-fields lie-derivative
asked Nov 21 '18 at 10:12
Ilya
285
i have a fields:
$X=X_1 dfrac {partial }{partial q_1}+...+X_n dfrac {partial }{partial q_n}$
$Y=Y_1 dfrac {partial }{partial q_1}+...+Y_n dfrac {partial }{partial q_n}$
$Z=Y_1(X_t(q)) dfrac {partial }{partial X^1_{t}(q)}+...+Y_n(X_t(q)) dfrac {partial }{partial X^n_{t}}$,
where $X_t(q)$ - vector flow.
All i want is to find $dot{Z}_{t=0}$.
I tried to do next calculations:
$Z(X_t(q))=Z_0(X_t(q))+dot{Z}t+o(t)$;
easy to get that $Z_0(X_t(q))=Y(X_t(q))=Y(q)+Y(X(q))t+o(t)$,
then $Z(X_t(q))=Y(q)+Y(X(q))t+dot{Z}t+0(t)$.
I'm not very sure, but it turns out that on the other hand, $Z(X_t(q))=d(X_t(q))[Y(q)]$;
since $X_t(q)=q+X(q)t+o(t)$ then $d(X_t(q))approx Id+d(X(q)t)$,
then $Z(X_t(q))=Y(q)+d(X(q)t)[Y(q)]+o(t)=Y(q)+Y(X(q))t+dot{Z}t+o(t)$
but i don’t really know what is $d(X(q)t)[Y(q)]$, it looks like $Y(X(q))$ or $X(Y(q))$..
so the answer is $[X,Y](q)$ or 0..
can you help me with this?
vector-fields lie-derivative
vector-fields lie-derivative
asked Nov 21 '18 at 10:12
Ilya
285
asked Nov 21 '18 at 10:12
Ilya
285
edited Nov 21 '18 at 10:17
asked Nov 21 '18 at 10:12
Ilya
285
asked Nov 21 '18 at 10:12
Ilya
285
asked Nov 21 '18 at 10:12
Ilya
285
285
add a comment |
add a comment |
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