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What is the vector field on this sphere?
The Next CEO of Stack OverflowTravel vector fieldFinding total flux of vector field across boundary boxVector field on sphereWhat is an irrotational vector field intuitively?Continuous vector field tangent to even dimensional sphere which only vanish at one pointIs this a vector field?Divergence of Vector Field: Maximizing, Normalizing, Singularities, InterpretationPair of Cartesian Cordinates for a line that is tangent to two surfaces.How can a vector field eat a smooth function?A nowhere vanishing vector field $X$ tangent to the leaves of the foliation?
$begingroup$
What is the vector field shown on this sphere?
The source should be at $P(0,0,0)$ and the sink should be at $P(1,1,1).$
How would I derive this vector field?
It looks like the vectors are all tangent to great circles.
So would the vector field be something like: $$ F= 2x+y^2+z^2,2y+x^2,+z^2,2z+x^2+y^2? $$
I'm sure that is wrong though.
multivariable-calculus vector-fields
asked Mar 19 at 20:35
UltradarkUltradark
3681518
$endgroup$
add a comment |
$begingroup$
What is the vector field shown on this sphere?
The source should be at $P(0,0,0)$ and the sink should be at $P(1,1,1).$
How would I derive this vector field?
It looks like the vectors are all tangent to great circles.
So would the vector field be something like: $$ F= 2x+y^2+z^2,2y+x^2,+z^2,2z+x^2+y^2? $$
I'm sure that is wrong though.
multivariable-calculus vector-fields
asked Mar 19 at 20:35
UltradarkUltradark
3681518
$endgroup$
$begingroup$
Why two coordinates instead of 3 for the source and the sink ?
$endgroup$
– Jean Marie
Mar 19 at 20:57
$begingroup$
It should be 3 coordinates
$endgroup$
– Ultradark
Mar 19 at 20:59
$begingroup$
I don't know if this will work, but locally (at the south pole) if you look on the tangent plane this looks just like $fracx$ so perhaps by transferring this by polar projection you can find something ?
$endgroup$
– Max
Mar 19 at 21:06
add a comment |
$begingroup$
What is the vector field shown on this sphere?
The source should be at $P(0,0,0)$ and the sink should be at $P(1,1,1).$
How would I derive this vector field?
It looks like the vectors are all tangent to great circles.
So would the vector field be something like: $$ F= 2x+y^2+z^2,2y+x^2,+z^2,2z+x^2+y^2? $$
I'm sure that is wrong though.
multivariable-calculus vector-fields
asked Mar 19 at 20:35
UltradarkUltradark
3681518
$endgroup$
What is the vector field shown on this sphere?
The source should be at $P(0,0,0)$ and the sink should be at $P(1,1,1).$
How would I derive this vector field?
It looks like the vectors are all tangent to great circles.
So would the vector field be something like: $$ F= 2x+y^2+z^2,2y+x^2,+z^2,2z+x^2+y^2? $$
I'm sure that is wrong though.
multivariable-calculus vector-fields
multivariable-calculus vector-fields
asked Mar 19 at 20:35
UltradarkUltradark
3681518
asked Mar 19 at 20:35
UltradarkUltradark
3681518
edited Mar 19 at 21:00
Ultradark
asked Mar 19 at 20:35
UltradarkUltradark
3681518
asked Mar 19 at 20:35
UltradarkUltradark
3681518
asked Mar 19 at 20:35
UltradarkUltradark
3681518
3681518
$begingroup$
Why two coordinates instead of 3 for the source and the sink ?
$endgroup$
– Jean Marie
Mar 19 at 20:57
$begingroup$
It should be 3 coordinates
$endgroup$
– Ultradark
Mar 19 at 20:59
$begingroup$
I don't know if this will work, but locally (at the south pole) if you look on the tangent plane this looks just like $fracx$ so perhaps by transferring this by polar projection you can find something ?
$endgroup$
– Max
Mar 19 at 21:06
add a comment |
$begingroup$
Why two coordinates instead of 3 for the source and the sink ?
$endgroup$
– Jean Marie
Mar 19 at 20:57
$begingroup$
It should be 3 coordinates
$endgroup$
– Ultradark
Mar 19 at 20:59
$begingroup$
I don't know if this will work, but locally (at the south pole) if you look on the tangent plane this looks just like $fracx$ so perhaps by transferring this by polar projection you can find something ?
$endgroup$
– Max
Mar 19 at 21:06
$begingroup$
Why two coordinates instead of 3 for the source and the sink ?
$endgroup$
– Jean Marie
Mar 19 at 20:57
$begingroup$
Why two coordinates instead of 3 for the source and the sink ?
$endgroup$
– Jean Marie
Mar 19 at 20:57
$begingroup$
It should be 3 coordinates
$endgroup$
– Ultradark
Mar 19 at 20:59
$begingroup$
It should be 3 coordinates
$endgroup$
– Ultradark
Mar 19 at 20:59
$begingroup$
I don't know if this will work, but locally (at the south pole) if you look on the tangent plane this looks just like $fracx$ so perhaps by transferring this by polar projection you can find something ?
$endgroup$
– Max
Mar 19 at 21:06
$begingroup$
I don't know if this will work, but locally (at the south pole) if you look on the tangent plane this looks just like $fracx$ so perhaps by transferring this by polar projection you can find something ?
$endgroup$
– Max
Mar 19 at 21:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The explanation will be simpler if the source is the South Pole and the sink the North pole on the unit sphere centered in $O(0,0,0)$.
We are going to work on spherical coordinates at first : let $theta, phi$ be the longitude and latitude resp.
The depicted vectors have visibly all of the same norm, orthogonal to unit vectors that are their "attach point".
Consider the generic point $P$ of the sphere characterized by $(theta, phi, r=1)$, i.e., with cartesian coordinates
$$P=(cos theta cos phi, sin theta cos phi, sin phi).tag1$$
The great circles passing through the north and south poles being integral lines of the vector fiels, the generic vector belonging to the vector field at point $P$ is proportional to the derivative of (1) with respect to $phi$ ($theta $ being kept constant).
Thus the general formula for this vector field is :
$$vecV=k(-cos theta sin phi, -sin theta sin phi, cos phi).tag2$$
Please note that $|vecV|=k$. Nothing give us information on the real value of $k$.
One can pass from this particular case to the general case by a rotation and an homothety (the sphere you are dealing with has radius $1/(2 sqrt3)$).
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
The explanation will be simpler if the source is the South Pole and the sink the North pole on the unit sphere centered in $O(0,0,0)$.
We are going to work on spherical coordinates at first : let $theta, phi$ be the longitude and latitude resp.
The depicted vectors have visibly all of the same norm, orthogonal to unit vectors that are their "attach point".
Consider the generic point $P$ of the sphere characterized by $(theta, phi, r=1)$, i.e., with cartesian coordinates
$$P=(cos theta cos phi, sin theta cos phi, sin phi).tag1$$
The great circles passing through the north and south poles being integral lines of the vector fiels, the generic vector belonging to the vector field at point $P$ is proportional to the derivative of (1) with respect to $phi$ ($theta $ being kept constant).
Thus the general formula for this vector field is :
$$vecV=k(-cos theta sin phi, -sin theta sin phi, cos phi).tag2$$
Please note that $|vecV|=k$. Nothing give us information on the real value of $k$.
One can pass from this particular case to the general case by a rotation and an homothety (the sphere you are dealing with has radius $1/(2 sqrt3)$).
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
$endgroup$
add a comment |
$begingroup$
The explanation will be simpler if the source is the South Pole and the sink the North pole on the unit sphere centered in $O(0,0,0)$.
We are going to work on spherical coordinates at first : let $theta, phi$ be the longitude and latitude resp.
The depicted vectors have visibly all of the same norm, orthogonal to unit vectors that are their "attach point".
Consider the generic point $P$ of the sphere characterized by $(theta, phi, r=1)$, i.e., with cartesian coordinates
$$P=(cos theta cos phi, sin theta cos phi, sin phi).tag1$$
The great circles passing through the north and south poles being integral lines of the vector fiels, the generic vector belonging to the vector field at point $P$ is proportional to the derivative of (1) with respect to $phi$ ($theta $ being kept constant).
Thus the general formula for this vector field is :
$$vecV=k(-cos theta sin phi, -sin theta sin phi, cos phi).tag2$$
Please note that $|vecV|=k$. Nothing give us information on the real value of $k$.
One can pass from this particular case to the general case by a rotation and an homothety (the sphere you are dealing with has radius $1/(2 sqrt3)$).
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
$endgroup$
add a comment |
$begingroup$
The explanation will be simpler if the source is the South Pole and the sink the North pole on the unit sphere centered in $O(0,0,0)$.
We are going to work on spherical coordinates at first : let $theta, phi$ be the longitude and latitude resp.
The depicted vectors have visibly all of the same norm, orthogonal to unit vectors that are their "attach point".
Consider the generic point $P$ of the sphere characterized by $(theta, phi, r=1)$, i.e., with cartesian coordinates
$$P=(cos theta cos phi, sin theta cos phi, sin phi).tag1$$
The great circles passing through the north and south poles being integral lines of the vector fiels, the generic vector belonging to the vector field at point $P$ is proportional to the derivative of (1) with respect to $phi$ ($theta $ being kept constant).
Thus the general formula for this vector field is :
$$vecV=k(-cos theta sin phi, -sin theta sin phi, cos phi).tag2$$
Please note that $|vecV|=k$. Nothing give us information on the real value of $k$.
One can pass from this particular case to the general case by a rotation and an homothety (the sphere you are dealing with has radius $1/(2 sqrt3)$).
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
$endgroup$
The explanation will be simpler if the source is the South Pole and the sink the North pole on the unit sphere centered in $O(0,0,0)$.
We are going to work on spherical coordinates at first : let $theta, phi$ be the longitude and latitude resp.
The depicted vectors have visibly all of the same norm, orthogonal to unit vectors that are their "attach point".
Consider the generic point $P$ of the sphere characterized by $(theta, phi, r=1)$, i.e., with cartesian coordinates
$$P=(cos theta cos phi, sin theta cos phi, sin phi).tag1$$
The great circles passing through the north and south poles being integral lines of the vector fiels, the generic vector belonging to the vector field at point $P$ is proportional to the derivative of (1) with respect to $phi$ ($theta $ being kept constant).
Thus the general formula for this vector field is :
$$vecV=k(-cos theta sin phi, -sin theta sin phi, cos phi).tag2$$
Please note that $|vecV|=k$. Nothing give us information on the real value of $k$.
One can pass from this particular case to the general case by a rotation and an homothety (the sphere you are dealing with has radius $1/(2 sqrt3)$).
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
answered Mar 19 at 21:25
Jean MarieJean Marie
31.1k42155
31.1k42155
add a comment |
add a comment |
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$begingroup$
Why two coordinates instead of 3 for the source and the sink ?
$endgroup$
– Jean Marie
Mar 19 at 20:57
$begingroup$
It should be 3 coordinates
$endgroup$
– Ultradark
Mar 19 at 20:59
$begingroup$
I don't know if this will work, but locally (at the south pole) if you look on the tangent plane this looks just like $fracx$ so perhaps by transferring this by polar projection you can find something ?
$endgroup$
– Max
Mar 19 at 21:06