Time derivative of a differential form Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Computing the restriction of a differential formHow to proceed this computation with differential forms?Covariant and contravariant bases on a diffeomorphismHow to find the derivative of the flow of an autonomous differential equation with respect to $x$Exterior derivative = infinitesimal change in differential form?Exterior derivarive dependent only on pointHow to take this exterior derivative of the expression $du - sum_i p_i dx_i$?covariantly constant one formOn the “exterior derivative” for not-necessarily-differential formsI don't understand understand this differential equality
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Time derivative of a differential form
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Computing the restriction of a differential formHow to proceed this computation with differential forms?Covariant and contravariant bases on a diffeomorphismHow to find the derivative of the flow of an autonomous differential equation with respect to $x$Exterior derivative = infinitesimal change in differential form?Exterior derivarive dependent only on pointHow to take this exterior derivative of the expression $du - sum_i p_i dx_i$?covariantly constant one formOn the “exterior derivative” for not-necessarily-differential formsI don't understand understand this differential equality
$begingroup$
Maybe a very silly question, but I've found the following equation:
$$fracddt omega(p+tV|_p) = sum_ifracddt (p_i+tV_i(p)) partial_i omega (p)$$
I suppose it's a kind of chain rule, and that $partial_i omega (p)$ is the directional derivative of $omega$ in the direction $i$ at $p$, but i can not get full light on it.
calculus differential-geometry differential-forms
$endgroup$
add a comment |
$begingroup$
Maybe a very silly question, but I've found the following equation:
$$fracddt omega(p+tV|_p) = sum_ifracddt (p_i+tV_i(p)) partial_i omega (p)$$
I suppose it's a kind of chain rule, and that $partial_i omega (p)$ is the directional derivative of $omega$ in the direction $i$ at $p$, but i can not get full light on it.
calculus differential-geometry differential-forms
$endgroup$
1
$begingroup$
it's covariant derivative along a curve
$endgroup$
– deb
Mar 27 at 18:43
add a comment |
$begingroup$
Maybe a very silly question, but I've found the following equation:
$$fracddt omega(p+tV|_p) = sum_ifracddt (p_i+tV_i(p)) partial_i omega (p)$$
I suppose it's a kind of chain rule, and that $partial_i omega (p)$ is the directional derivative of $omega$ in the direction $i$ at $p$, but i can not get full light on it.
calculus differential-geometry differential-forms
$endgroup$
Maybe a very silly question, but I've found the following equation:
$$fracddt omega(p+tV|_p) = sum_ifracddt (p_i+tV_i(p)) partial_i omega (p)$$
I suppose it's a kind of chain rule, and that $partial_i omega (p)$ is the directional derivative of $omega$ in the direction $i$ at $p$, but i can not get full light on it.
calculus differential-geometry differential-forms
calculus differential-geometry differential-forms
asked Mar 27 at 17:51
Lo ScrondoLo Scrondo
192111
192111
1
$begingroup$
it's covariant derivative along a curve
$endgroup$
– deb
Mar 27 at 18:43
add a comment |
1
$begingroup$
it's covariant derivative along a curve
$endgroup$
– deb
Mar 27 at 18:43
1
1
$begingroup$
it's covariant derivative along a curve
$endgroup$
– deb
Mar 27 at 18:43
$begingroup$
it's covariant derivative along a curve
$endgroup$
– deb
Mar 27 at 18:43
add a comment |
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$begingroup$
it's covariant derivative along a curve
$endgroup$
– deb
Mar 27 at 18:43