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
679313
$endgroup$
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
679313
$endgroup$
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
679313
$endgroup$
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
679313
asked Dec 9 '18 at 23:04
PerelManPerelMan
679313
edited Dec 9 '18 at 23:23
PerelMan
asked Dec 9 '18 at 23:04
PerelManPerelMan
679313
asked Dec 9 '18 at 23:04
PerelManPerelMan
679313
asked Dec 9 '18 at 23:04
PerelManPerelMan
679313
679313
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
11.3k923
$endgroup$
$begingroup$
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
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033158%2ffind-a-vector-field-in-mathbbr3-with-specific-properties%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
11.3k923
$endgroup$
$begingroup$
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
11.3k923
$endgroup$
$begingroup$
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
11.3k923
$endgroup$
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
11.3k923
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.3k923
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.3k923
answered Dec 10 '18 at 11:30
Andreas CapAndreas Cap
11.3k923
11.3k923
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033158%2ffind-a-vector-field-in-mathbbr3-with-specific-properties%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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