Time derivative of curl
I'm confused about the time derivative of the curl. If the following true? Where do I go wrong?
$$frac { partial } { partial t} vec{nabla} times vec{v} = frac{partial vec nabla }{ partial t} times vec v + vec nabla times frac { partial vec v }{partial t} $$
The reason I'm confused is that the second term looks exactly like the first term, but the second term doesn't necessarily/immediately look like it vanishes.
If it does vanish, is it because $$ frac {partial vec v}{partial t} = vec a = vec nabla Phi $$ and since $mathrm {curl(grad (Phi))} = 0$, my thought that the first term is just like the LHS is true?
Something seems wrong here to me because it seems too strong a hypothesis to consider the time derivative of the vector field, $vec v$ conservative.
calculus vectors vector-analysis
asked Jun 30 '17 at 20:54
jaslibra
260214
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I'm confused about the time derivative of the curl. If the following true? Where do I go wrong?
$$frac { partial } { partial t} vec{nabla} times vec{v} = frac{partial vec nabla }{ partial t} times vec v + vec nabla times frac { partial vec v }{partial t} $$
The reason I'm confused is that the second term looks exactly like the first term, but the second term doesn't necessarily/immediately look like it vanishes.
If it does vanish, is it because $$ frac {partial vec v}{partial t} = vec a = vec nabla Phi $$ and since $mathrm {curl(grad (Phi))} = 0$, my thought that the first term is just like the LHS is true?
Something seems wrong here to me because it seems too strong a hypothesis to consider the time derivative of the vector field, $vec v$ conservative.
calculus vectors vector-analysis
asked Jun 30 '17 at 20:54
jaslibra
260214
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I'm confused about the time derivative of the curl. If the following true? Where do I go wrong?
$$frac { partial } { partial t} vec{nabla} times vec{v} = frac{partial vec nabla }{ partial t} times vec v + vec nabla times frac { partial vec v }{partial t} $$
The reason I'm confused is that the second term looks exactly like the first term, but the second term doesn't necessarily/immediately look like it vanishes.
If it does vanish, is it because $$ frac {partial vec v}{partial t} = vec a = vec nabla Phi $$ and since $mathrm {curl(grad (Phi))} = 0$, my thought that the first term is just like the LHS is true?
Something seems wrong here to me because it seems too strong a hypothesis to consider the time derivative of the vector field, $vec v$ conservative.
calculus vectors vector-analysis
asked Jun 30 '17 at 20:54
jaslibra
260214
I'm confused about the time derivative of the curl. If the following true? Where do I go wrong?
$$frac { partial } { partial t} vec{nabla} times vec{v} = frac{partial vec nabla }{ partial t} times vec v + vec nabla times frac { partial vec v }{partial t} $$
The reason I'm confused is that the second term looks exactly like the first term, but the second term doesn't necessarily/immediately look like it vanishes.
If it does vanish, is it because $$ frac {partial vec v}{partial t} = vec a = vec nabla Phi $$ and since $mathrm {curl(grad (Phi))} = 0$, my thought that the first term is just like the LHS is true?
Something seems wrong here to me because it seems too strong a hypothesis to consider the time derivative of the vector field, $vec v$ conservative.
calculus vectors vector-analysis
calculus vectors vector-analysis
asked Jun 30 '17 at 20:54
jaslibra
260214
asked Jun 30 '17 at 20:54
jaslibra
260214
edited Jun 30 '17 at 21:02
asked Jun 30 '17 at 20:54
jaslibra
260214
asked Jun 30 '17 at 20:54
jaslibra
260214
asked Jun 30 '17 at 20:54
jaslibra
260214
260214
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The notation $nabla times ,cdot,$ is a shorthand for the curl operator, but it should not be taken too literally: In particular, $nabla$ does not signify a vector field, and hence the usual Leibniz (product) rule for the cross product does not apply literally.
Hint If ${bf v} = (v^1, v^2, v^3)$ is a (time-dependent) vector field on $Bbb R^3_{xyz}$ that is sufficiently smooth ($C^2$ will do), then, e.g., the first component of $$frac{d}{dt}(nabla times {bf v})$$ is
$$frac{partial}{partial t}left(frac{partial v^2}{partial z} - frac{partial v^3}{partial y}right) = frac{partial}{partial z}left(frac{partial v^2}{partial t}right) - frac{partial}{partial y}left(frac{partial v^3}{partial t}right) .$$
answered Jun 30 '17 at 21:46
Travis
59.8k767146
What happened to $v^1$? Is this just demonstrating operator commutativity?
– jaslibra
Jul 1 '17 at 13:24
The last display equation just gives the first component of the derivative. And yes, the point is that $frac{partial}{partial t}$ commutes with $nabla times ,cdot,$.
– Travis
Jul 1 '17 at 17:02
Perhaps the downvoter would explain their objection?
– Travis
Jul 18 '17 at 12:35
add a comment |
It's best when you're working with not-true-vectors such as curl to calculate from first principles: $frac{partial}{partial t}vec{nabla}timesvec{v}$ has $i$th component $epsilon_{ijk}frac{partial}{partial t}frac{partial}{partial x_j} v_k$, with implicit summation over repeated indices. Since the partial derivatives commute, this expression is $epsilon_{ijk}frac{partial}{partial x_j} frac{partial}{partial t}v_k$, which is the $i$th component of $vec{nabla}timesfrac{partial}{partial t}vec{v}$. In other words, one of the terms expected from a naïve Leibniz rule vanishes. Please don't interpret this as meaning $frac{partial}{partial t}vec{nabla}=vec{0}$, whatever that would mean.
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
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%2f2342322%2ftime-derivative-of-curl%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The notation $nabla times ,cdot,$ is a shorthand for the curl operator, but it should not be taken too literally: In particular, $nabla$ does not signify a vector field, and hence the usual Leibniz (product) rule for the cross product does not apply literally.
Hint If ${bf v} = (v^1, v^2, v^3)$ is a (time-dependent) vector field on $Bbb R^3_{xyz}$ that is sufficiently smooth ($C^2$ will do), then, e.g., the first component of $$frac{d}{dt}(nabla times {bf v})$$ is
$$frac{partial}{partial t}left(frac{partial v^2}{partial z} - frac{partial v^3}{partial y}right) = frac{partial}{partial z}left(frac{partial v^2}{partial t}right) - frac{partial}{partial y}left(frac{partial v^3}{partial t}right) .$$
answered Jun 30 '17 at 21:46
Travis
59.8k767146
What happened to $v^1$? Is this just demonstrating operator commutativity?
– jaslibra
Jul 1 '17 at 13:24
The last display equation just gives the first component of the derivative. And yes, the point is that $frac{partial}{partial t}$ commutes with $nabla times ,cdot,$.
– Travis
Jul 1 '17 at 17:02
Perhaps the downvoter would explain their objection?
– Travis
Jul 18 '17 at 12:35
add a comment |
The notation $nabla times ,cdot,$ is a shorthand for the curl operator, but it should not be taken too literally: In particular, $nabla$ does not signify a vector field, and hence the usual Leibniz (product) rule for the cross product does not apply literally.
Hint If ${bf v} = (v^1, v^2, v^3)$ is a (time-dependent) vector field on $Bbb R^3_{xyz}$ that is sufficiently smooth ($C^2$ will do), then, e.g., the first component of $$frac{d}{dt}(nabla times {bf v})$$ is
$$frac{partial}{partial t}left(frac{partial v^2}{partial z} - frac{partial v^3}{partial y}right) = frac{partial}{partial z}left(frac{partial v^2}{partial t}right) - frac{partial}{partial y}left(frac{partial v^3}{partial t}right) .$$
answered Jun 30 '17 at 21:46
Travis
59.8k767146
What happened to $v^1$? Is this just demonstrating operator commutativity?
– jaslibra
Jul 1 '17 at 13:24
The last display equation just gives the first component of the derivative. And yes, the point is that $frac{partial}{partial t}$ commutes with $nabla times ,cdot,$.
– Travis
Jul 1 '17 at 17:02
Perhaps the downvoter would explain their objection?
– Travis
Jul 18 '17 at 12:35
add a comment |
The notation $nabla times ,cdot,$ is a shorthand for the curl operator, but it should not be taken too literally: In particular, $nabla$ does not signify a vector field, and hence the usual Leibniz (product) rule for the cross product does not apply literally.
Hint If ${bf v} = (v^1, v^2, v^3)$ is a (time-dependent) vector field on $Bbb R^3_{xyz}$ that is sufficiently smooth ($C^2$ will do), then, e.g., the first component of $$frac{d}{dt}(nabla times {bf v})$$ is
$$frac{partial}{partial t}left(frac{partial v^2}{partial z} - frac{partial v^3}{partial y}right) = frac{partial}{partial z}left(frac{partial v^2}{partial t}right) - frac{partial}{partial y}left(frac{partial v^3}{partial t}right) .$$
answered Jun 30 '17 at 21:46
Travis
59.8k767146
The notation $nabla times ,cdot,$ is a shorthand for the curl operator, but it should not be taken too literally: In particular, $nabla$ does not signify a vector field, and hence the usual Leibniz (product) rule for the cross product does not apply literally.
Hint If ${bf v} = (v^1, v^2, v^3)$ is a (time-dependent) vector field on $Bbb R^3_{xyz}$ that is sufficiently smooth ($C^2$ will do), then, e.g., the first component of $$frac{d}{dt}(nabla times {bf v})$$ is
$$frac{partial}{partial t}left(frac{partial v^2}{partial z} - frac{partial v^3}{partial y}right) = frac{partial}{partial z}left(frac{partial v^2}{partial t}right) - frac{partial}{partial y}left(frac{partial v^3}{partial t}right) .$$
answered Jun 30 '17 at 21:46
Travis
59.8k767146
answered Jun 30 '17 at 21:46
Travis
59.8k767146
answered Jun 30 '17 at 21:46
Travis
59.8k767146
answered Jun 30 '17 at 21:46
Travis
59.8k767146
59.8k767146
What happened to $v^1$? Is this just demonstrating operator commutativity?
– jaslibra
Jul 1 '17 at 13:24
The last display equation just gives the first component of the derivative. And yes, the point is that $frac{partial}{partial t}$ commutes with $nabla times ,cdot,$.
– Travis
Jul 1 '17 at 17:02
Perhaps the downvoter would explain their objection?
– Travis
Jul 18 '17 at 12:35
add a comment |
What happened to $v^1$? Is this just demonstrating operator commutativity?
– jaslibra
Jul 1 '17 at 13:24
The last display equation just gives the first component of the derivative. And yes, the point is that $frac{partial}{partial t}$ commutes with $nabla times ,cdot,$.
– Travis
Jul 1 '17 at 17:02
Perhaps the downvoter would explain their objection?
– Travis
Jul 18 '17 at 12:35
What happened to $v^1$? Is this just demonstrating operator commutativity?
– jaslibra
Jul 1 '17 at 13:24
What happened to $v^1$? Is this just demonstrating operator commutativity?
– jaslibra
Jul 1 '17 at 13:24
The last display equation just gives the first component of the derivative. And yes, the point is that $frac{partial}{partial t}$ commutes with $nabla times ,cdot,$.
– Travis
Jul 1 '17 at 17:02
The last display equation just gives the first component of the derivative. And yes, the point is that $frac{partial}{partial t}$ commutes with $nabla times ,cdot,$.
– Travis
Jul 1 '17 at 17:02
Perhaps the downvoter would explain their objection?
– Travis
Jul 18 '17 at 12:35
Perhaps the downvoter would explain their objection?
– Travis
Jul 18 '17 at 12:35
add a comment |
It's best when you're working with not-true-vectors such as curl to calculate from first principles: $frac{partial}{partial t}vec{nabla}timesvec{v}$ has $i$th component $epsilon_{ijk}frac{partial}{partial t}frac{partial}{partial x_j} v_k$, with implicit summation over repeated indices. Since the partial derivatives commute, this expression is $epsilon_{ijk}frac{partial}{partial x_j} frac{partial}{partial t}v_k$, which is the $i$th component of $vec{nabla}timesfrac{partial}{partial t}vec{v}$. In other words, one of the terms expected from a naïve Leibniz rule vanishes. Please don't interpret this as meaning $frac{partial}{partial t}vec{nabla}=vec{0}$, whatever that would mean.
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
add a comment |
It's best when you're working with not-true-vectors such as curl to calculate from first principles: $frac{partial}{partial t}vec{nabla}timesvec{v}$ has $i$th component $epsilon_{ijk}frac{partial}{partial t}frac{partial}{partial x_j} v_k$, with implicit summation over repeated indices. Since the partial derivatives commute, this expression is $epsilon_{ijk}frac{partial}{partial x_j} frac{partial}{partial t}v_k$, which is the $i$th component of $vec{nabla}timesfrac{partial}{partial t}vec{v}$. In other words, one of the terms expected from a naïve Leibniz rule vanishes. Please don't interpret this as meaning $frac{partial}{partial t}vec{nabla}=vec{0}$, whatever that would mean.
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
add a comment |
It's best when you're working with not-true-vectors such as curl to calculate from first principles: $frac{partial}{partial t}vec{nabla}timesvec{v}$ has $i$th component $epsilon_{ijk}frac{partial}{partial t}frac{partial}{partial x_j} v_k$, with implicit summation over repeated indices. Since the partial derivatives commute, this expression is $epsilon_{ijk}frac{partial}{partial x_j} frac{partial}{partial t}v_k$, which is the $i$th component of $vec{nabla}timesfrac{partial}{partial t}vec{v}$. In other words, one of the terms expected from a naïve Leibniz rule vanishes. Please don't interpret this as meaning $frac{partial}{partial t}vec{nabla}=vec{0}$, whatever that would mean.
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
It's best when you're working with not-true-vectors such as curl to calculate from first principles: $frac{partial}{partial t}vec{nabla}timesvec{v}$ has $i$th component $epsilon_{ijk}frac{partial}{partial t}frac{partial}{partial x_j} v_k$, with implicit summation over repeated indices. Since the partial derivatives commute, this expression is $epsilon_{ijk}frac{partial}{partial x_j} frac{partial}{partial t}v_k$, which is the $i$th component of $vec{nabla}timesfrac{partial}{partial t}vec{v}$. In other words, one of the terms expected from a naïve Leibniz rule vanishes. Please don't interpret this as meaning $frac{partial}{partial t}vec{nabla}=vec{0}$, whatever that would mean.
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
answered Nov 3 '18 at 14:06
J.G.
23.1k22137
23.1k22137
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f2342322%2ftime-derivative-of-curl%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
