find a vector field in $mathbb{R^3}$ with specific properties
$begingroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$
Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$
What I did:
I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$
We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)
and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.
but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$
and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$
and $ dx wedge dy (bpartial_y) = 0$
which means finally
$ ady(.) = bdx(.) $ (II)
Could we find a,b and c from the two results (I) and (II)?
Thank you!
differential-geometry vector-fields exterior-derivative
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
$endgroup$
add a comment |
$begingroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$
Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$
What I did:
I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$
We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)
and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.
but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$
and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$
and $ dx wedge dy (bpartial_y) = 0$
which means finally
$ ady(.) = bdx(.) $ (II)
Could we find a,b and c from the two results (I) and (II)?
Thank you!
differential-geometry vector-fields exterior-derivative
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
$endgroup$
1
$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32
$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41
1
$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05
add a comment |
$begingroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$
Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$
What I did:
I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$
We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)
and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.
but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$
and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$
and $ dx wedge dy (bpartial_y) = 0$
which means finally
$ ady(.) = bdx(.) $ (II)
Could we find a,b and c from the two results (I) and (II)?
Thank you!
differential-geometry vector-fields exterior-derivative
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
$endgroup$
Let $α = dz - ydx in Ω^1 (mathbb{R}^3)$
Find a vector field Z in $mathbb{R^3}$ such that $α(Z) = 1$ and $dα(Z,.) = 0$
What I did:
I computed $dalpha = dx wedge dy $.
Then let $Z=apartial_x + bpartial_y +cpartial_z $ be a vector field in $mathbb{R}^3$
We have $α(Z) = 1 Rightarrow cpartial_z - yapartial_x = 1 $ (I)
and $dα(Z,.) = 0 Rightarrow dx wedge dy (apartial_x + bpartial_y +cpartial_z , .)$.
but we have $ dx wedge dy (apartial_x) = dx(a partial_x)dy(.) -dx(.)dy(a partial_x) = ady(.)$
and $ dx wedge dy (bpartial_y) = dx(b partial_y)dy(.) -dx(.)dy(b partial_y) = -bdx(.)$
and $ dx wedge dy (bpartial_y) = 0$
which means finally
$ ady(.) = bdx(.) $ (II)
Could we find a,b and c from the two results (I) and (II)?
Thank you!
differential-geometry vector-fields exterior-derivative
differential-geometry vector-fields exterior-derivative
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
edited Dec 9 '18 at 23:23
PerelMan
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
asked Dec 9 '18 at 23:04
PerelManPerelMan
524211
524211
1
$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32
$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41
1
$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05
add a comment |
1
$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32
$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41
1
$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05
1
1
$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32
$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32
$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41
$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41
1
1
$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05
$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40
1
$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01
add a comment |
Your Answer
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1 Answer
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$begingroup$
As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40
1
$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01
add a comment |
$begingroup$
As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40
1
$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01
add a comment |
$begingroup$
As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
$endgroup$
As noted in the comment by @Dog_69 , the first equation should read as $c-ya=1$. Now you derived (II) correctly (although in a slightly complicated way), but since $dx$ and $dy$ are linearly independent in each point, (II) implies $a=b=0$, so (I) reduces to $c=1$. So you get the solution $Z=partial_z$ which is readily seen to be correct.
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.1k923
11.1k923
$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40
1
$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01
add a comment |
$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40
1
$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01
$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40
$begingroup$
How would you derive (II) in easier way? I am new to this field of differential forms. Thank you!
$endgroup$
– PerelMan
Dec 10 '18 at 13:40
1
1
$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01
$begingroup$
It dosen't make much difference, but you can say directly that $(dxwedge dy)(X,underline{ })=dx(Z)dy-dy(Z)dx$ so inserting, you readily get $ady-bdx$.
$endgroup$
– Andreas Cap
Dec 11 '18 at 8:01
add a comment |
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$begingroup$
(II) means that $a=b=0$,, while (I) is wrong; you get that $c-ay=1$. Hence the vector field you are looking for is $Z=partial_z$.
$endgroup$
– Dog_69
Dec 9 '18 at 23:32
$begingroup$
Indeed, $alpha(Z) in mathbb{R}$. Thank you!
$endgroup$
– PerelMan
Dec 9 '18 at 23:41
1
$begingroup$
You're welcome :)
$endgroup$
– Dog_69
Dec 10 '18 at 9:05