Why is the curl the limiting value of circulation density?
$begingroup$
The circulation of vector field around some region is the line integral of the field over closed path surrounding that region so why do we need the curl at all if we have a quantity that represents the amount of circulation and why the curl is the circulation density and not the circulation?
calculus derivatives vector-fields
$endgroup$
add a comment |
$begingroup$
The circulation of vector field around some region is the line integral of the field over closed path surrounding that region so why do we need the curl at all if we have a quantity that represents the amount of circulation and why the curl is the circulation density and not the circulation?
calculus derivatives vector-fields
$endgroup$
1
$begingroup$
This is proved using the mean value theorem, just like any similar result you can imagine.
$endgroup$
– Charlie Frohman
Jan 25 at 18:13
$begingroup$
You might think about units.
$endgroup$
– Ted Shifrin
Jan 25 at 18:28
add a comment |
$begingroup$
The circulation of vector field around some region is the line integral of the field over closed path surrounding that region so why do we need the curl at all if we have a quantity that represents the amount of circulation and why the curl is the circulation density and not the circulation?
calculus derivatives vector-fields
$endgroup$
The circulation of vector field around some region is the line integral of the field over closed path surrounding that region so why do we need the curl at all if we have a quantity that represents the amount of circulation and why the curl is the circulation density and not the circulation?
calculus derivatives vector-fields
calculus derivatives vector-fields
asked Jan 25 at 18:03


gogo okagogo oka
11
11
1
$begingroup$
This is proved using the mean value theorem, just like any similar result you can imagine.
$endgroup$
– Charlie Frohman
Jan 25 at 18:13
$begingroup$
You might think about units.
$endgroup$
– Ted Shifrin
Jan 25 at 18:28
add a comment |
1
$begingroup$
This is proved using the mean value theorem, just like any similar result you can imagine.
$endgroup$
– Charlie Frohman
Jan 25 at 18:13
$begingroup$
You might think about units.
$endgroup$
– Ted Shifrin
Jan 25 at 18:28
1
1
$begingroup$
This is proved using the mean value theorem, just like any similar result you can imagine.
$endgroup$
– Charlie Frohman
Jan 25 at 18:13
$begingroup$
This is proved using the mean value theorem, just like any similar result you can imagine.
$endgroup$
– Charlie Frohman
Jan 25 at 18:13
$begingroup$
You might think about units.
$endgroup$
– Ted Shifrin
Jan 25 at 18:28
$begingroup$
You might think about units.
$endgroup$
– Ted Shifrin
Jan 25 at 18:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given a vector field ${bf F}$ in the plane ${mathbb R}^2$ the circulation is a function $$Phi_{bf F}(gamma):=int_gamma{bf F}({bf z})cdot d{bf z}$$
that produces a scalar value for each closed curve $gammasubset{mathbb R}^2$, in particular for all boundary curves $gamma:=partial B$ of (reasonable) domains $Bsubset{mathbb R}^2$. Note that the set of all such curves is very huge, and a priori it would be difficult to make general statements about this function $Phi$. But this function is important: It measures the nonconservativity of the field ${bf F}$. A conservative field ${bf F}$ (meaning ${bf F}=nabla f$) has $Phi_{bf F}(gamma)=0$ for all closed curves $gamma$.
It is a "wonder of mathematics" that the nonconservativity of ${bf F}$ can be captured in a simple point-function $${rm curl},{bf F}:quad{mathbb R}^2to{mathbb R}, qquad {bf z}mapsto{rm curl},{bf F}({bf z})$$
in such a way that the circulation of ${bf F}$ around the boundary $partial B$ of a domain $B$ appears as integral of the density ${rm curl},{bf F}$ over $B$:
$$Phi_{bf F}(partial B):=int_B{rm curl},{bf F}({bf z})>{rm d}({bf z}) ,$$
whereby ${rm d}({bf z})$ denotes the area element at ${bf z}=(x,y)$. This means that the nonconservativity of ${bf F}$ is in a precise way localized to the points of the domain of ${bf F}$.
$endgroup$
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%2f3087416%2fwhy-is-the-curl-the-limiting-value-of-circulation-density%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
$begingroup$
Given a vector field ${bf F}$ in the plane ${mathbb R}^2$ the circulation is a function $$Phi_{bf F}(gamma):=int_gamma{bf F}({bf z})cdot d{bf z}$$
that produces a scalar value for each closed curve $gammasubset{mathbb R}^2$, in particular for all boundary curves $gamma:=partial B$ of (reasonable) domains $Bsubset{mathbb R}^2$. Note that the set of all such curves is very huge, and a priori it would be difficult to make general statements about this function $Phi$. But this function is important: It measures the nonconservativity of the field ${bf F}$. A conservative field ${bf F}$ (meaning ${bf F}=nabla f$) has $Phi_{bf F}(gamma)=0$ for all closed curves $gamma$.
It is a "wonder of mathematics" that the nonconservativity of ${bf F}$ can be captured in a simple point-function $${rm curl},{bf F}:quad{mathbb R}^2to{mathbb R}, qquad {bf z}mapsto{rm curl},{bf F}({bf z})$$
in such a way that the circulation of ${bf F}$ around the boundary $partial B$ of a domain $B$ appears as integral of the density ${rm curl},{bf F}$ over $B$:
$$Phi_{bf F}(partial B):=int_B{rm curl},{bf F}({bf z})>{rm d}({bf z}) ,$$
whereby ${rm d}({bf z})$ denotes the area element at ${bf z}=(x,y)$. This means that the nonconservativity of ${bf F}$ is in a precise way localized to the points of the domain of ${bf F}$.
$endgroup$
add a comment |
$begingroup$
Given a vector field ${bf F}$ in the plane ${mathbb R}^2$ the circulation is a function $$Phi_{bf F}(gamma):=int_gamma{bf F}({bf z})cdot d{bf z}$$
that produces a scalar value for each closed curve $gammasubset{mathbb R}^2$, in particular for all boundary curves $gamma:=partial B$ of (reasonable) domains $Bsubset{mathbb R}^2$. Note that the set of all such curves is very huge, and a priori it would be difficult to make general statements about this function $Phi$. But this function is important: It measures the nonconservativity of the field ${bf F}$. A conservative field ${bf F}$ (meaning ${bf F}=nabla f$) has $Phi_{bf F}(gamma)=0$ for all closed curves $gamma$.
It is a "wonder of mathematics" that the nonconservativity of ${bf F}$ can be captured in a simple point-function $${rm curl},{bf F}:quad{mathbb R}^2to{mathbb R}, qquad {bf z}mapsto{rm curl},{bf F}({bf z})$$
in such a way that the circulation of ${bf F}$ around the boundary $partial B$ of a domain $B$ appears as integral of the density ${rm curl},{bf F}$ over $B$:
$$Phi_{bf F}(partial B):=int_B{rm curl},{bf F}({bf z})>{rm d}({bf z}) ,$$
whereby ${rm d}({bf z})$ denotes the area element at ${bf z}=(x,y)$. This means that the nonconservativity of ${bf F}$ is in a precise way localized to the points of the domain of ${bf F}$.
$endgroup$
add a comment |
$begingroup$
Given a vector field ${bf F}$ in the plane ${mathbb R}^2$ the circulation is a function $$Phi_{bf F}(gamma):=int_gamma{bf F}({bf z})cdot d{bf z}$$
that produces a scalar value for each closed curve $gammasubset{mathbb R}^2$, in particular for all boundary curves $gamma:=partial B$ of (reasonable) domains $Bsubset{mathbb R}^2$. Note that the set of all such curves is very huge, and a priori it would be difficult to make general statements about this function $Phi$. But this function is important: It measures the nonconservativity of the field ${bf F}$. A conservative field ${bf F}$ (meaning ${bf F}=nabla f$) has $Phi_{bf F}(gamma)=0$ for all closed curves $gamma$.
It is a "wonder of mathematics" that the nonconservativity of ${bf F}$ can be captured in a simple point-function $${rm curl},{bf F}:quad{mathbb R}^2to{mathbb R}, qquad {bf z}mapsto{rm curl},{bf F}({bf z})$$
in such a way that the circulation of ${bf F}$ around the boundary $partial B$ of a domain $B$ appears as integral of the density ${rm curl},{bf F}$ over $B$:
$$Phi_{bf F}(partial B):=int_B{rm curl},{bf F}({bf z})>{rm d}({bf z}) ,$$
whereby ${rm d}({bf z})$ denotes the area element at ${bf z}=(x,y)$. This means that the nonconservativity of ${bf F}$ is in a precise way localized to the points of the domain of ${bf F}$.
$endgroup$
Given a vector field ${bf F}$ in the plane ${mathbb R}^2$ the circulation is a function $$Phi_{bf F}(gamma):=int_gamma{bf F}({bf z})cdot d{bf z}$$
that produces a scalar value for each closed curve $gammasubset{mathbb R}^2$, in particular for all boundary curves $gamma:=partial B$ of (reasonable) domains $Bsubset{mathbb R}^2$. Note that the set of all such curves is very huge, and a priori it would be difficult to make general statements about this function $Phi$. But this function is important: It measures the nonconservativity of the field ${bf F}$. A conservative field ${bf F}$ (meaning ${bf F}=nabla f$) has $Phi_{bf F}(gamma)=0$ for all closed curves $gamma$.
It is a "wonder of mathematics" that the nonconservativity of ${bf F}$ can be captured in a simple point-function $${rm curl},{bf F}:quad{mathbb R}^2to{mathbb R}, qquad {bf z}mapsto{rm curl},{bf F}({bf z})$$
in such a way that the circulation of ${bf F}$ around the boundary $partial B$ of a domain $B$ appears as integral of the density ${rm curl},{bf F}$ over $B$:
$$Phi_{bf F}(partial B):=int_B{rm curl},{bf F}({bf z})>{rm d}({bf z}) ,$$
whereby ${rm d}({bf z})$ denotes the area element at ${bf z}=(x,y)$. This means that the nonconservativity of ${bf F}$ is in a precise way localized to the points of the domain of ${bf F}$.
answered Jan 25 at 20:20


Christian BlatterChristian Blatter
175k8115327
175k8115327
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.
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%2f3087416%2fwhy-is-the-curl-the-limiting-value-of-circulation-density%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
$begingroup$
This is proved using the mean value theorem, just like any similar result you can imagine.
$endgroup$
– Charlie Frohman
Jan 25 at 18:13
$begingroup$
You might think about units.
$endgroup$
– Ted Shifrin
Jan 25 at 18:28