Mixing
Does zero curl imply a conservative field?
Clash Royale CLAN TAG#URR8PPP
up vote
10
down vote
favorite
A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9).
vector-spaces
asked Feb 4 at 18:10
John A.
18212
 |Â
show 3 more comments
up vote
10
down vote
favorite
A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9).
vector-spaces
asked Feb 4 at 18:10
John A.
18212
2
Do you know Stokes theorem? The important fact that leads to the existence of a potential is that the line integral just depends on the start and end point and not curve connecting these.
– Fabian
Feb 4 at 18:15
5
You were a bit hasty in accepting an answer. The one you’ve chosen tacitly makes some important assumptions about the domain of the vector field, without which your proposition is false.
– amd
Feb 4 at 19:24
1
There's a factually incorrect answer rated at -2 and a correct answer rated at 7, and the former is the one that's accepted?
– anomaly
Feb 4 at 21:05
2
@anomaly - in fairness to the OP, the argument in the accepted answer becomes correct if one assumes the OP is simply-connected; meanwhile, the other answer is correct in greater generality but doesn't provide an argument.
– Ben Blum-Smith
Feb 4 at 21:12
1
@BenBlum-Smith: That's not a reasonable assumption, though, even in the physics world. The other answer links to a counterexample and states the assumption necessary to make the conclusion valid; actually proving it requires algebraic topology that's beyond the scope of the original question.
– anomaly
Feb 5 at 4:11
 |Â
show 3 more comments
up vote
10
down vote
favorite
up vote
10
down vote
favorite
A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9).
vector-spaces
asked Feb 4 at 18:10
John A.
18212
A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9).
vector-spaces
asked Feb 4 at 18:10
John A.
18212
asked Feb 4 at 18:10
John A.
18212
asked Feb 4 at 18:10
John A.
18212
asked Feb 4 at 18:10
John A.
18212
18212
2
Do you know Stokes theorem? The important fact that leads to the existence of a potential is that the line integral just depends on the start and end point and not curve connecting these.
– Fabian
Feb 4 at 18:15
5
You were a bit hasty in accepting an answer. The one you’ve chosen tacitly makes some important assumptions about the domain of the vector field, without which your proposition is false.
– amd
Feb 4 at 19:24
1
There's a factually incorrect answer rated at -2 and a correct answer rated at 7, and the former is the one that's accepted?
– anomaly
Feb 4 at 21:05
2
@anomaly - in fairness to the OP, the argument in the accepted answer becomes correct if one assumes the OP is simply-connected; meanwhile, the other answer is correct in greater generality but doesn't provide an argument.
– Ben Blum-Smith
Feb 4 at 21:12
1
@BenBlum-Smith: That's not a reasonable assumption, though, even in the physics world. The other answer links to a counterexample and states the assumption necessary to make the conclusion valid; actually proving it requires algebraic topology that's beyond the scope of the original question.
– anomaly
Feb 5 at 4:11
 |Â
show 3 more comments
2
Do you know Stokes theorem? The important fact that leads to the existence of a potential is that the line integral just depends on the start and end point and not curve connecting these.
– Fabian
Feb 4 at 18:15
5
You were a bit hasty in accepting an answer. The one you’ve chosen tacitly makes some important assumptions about the domain of the vector field, without which your proposition is false.
– amd
Feb 4 at 19:24
1
There's a factually incorrect answer rated at -2 and a correct answer rated at 7, and the former is the one that's accepted?
– anomaly
Feb 4 at 21:05
2
@anomaly - in fairness to the OP, the argument in the accepted answer becomes correct if one assumes the OP is simply-connected; meanwhile, the other answer is correct in greater generality but doesn't provide an argument.
– Ben Blum-Smith
Feb 4 at 21:12
1
@BenBlum-Smith: That's not a reasonable assumption, though, even in the physics world. The other answer links to a counterexample and states the assumption necessary to make the conclusion valid; actually proving it requires algebraic topology that's beyond the scope of the original question.
– anomaly
Feb 5 at 4:11
2
2
Do you know Stokes theorem? The important fact that leads to the existence of a potential is that the line integral just depends on the start and end point and not curve connecting these.
– Fabian
Feb 4 at 18:15
Do you know Stokes theorem? The important fact that leads to the existence of a potential is that the line integral just depends on the start and end point and not curve connecting these.
– Fabian
Feb 4 at 18:15
5
5
You were a bit hasty in accepting an answer. The one you’ve chosen tacitly makes some important assumptions about the domain of the vector field, without which your proposition is false.
– amd
Feb 4 at 19:24
You were a bit hasty in accepting an answer. The one you’ve chosen tacitly makes some important assumptions about the domain of the vector field, without which your proposition is false.
– amd
Feb 4 at 19:24
1
1
There's a factually incorrect answer rated at -2 and a correct answer rated at 7, and the former is the one that's accepted?
– anomaly
Feb 4 at 21:05
There's a factually incorrect answer rated at -2 and a correct answer rated at 7, and the former is the one that's accepted?
– anomaly
Feb 4 at 21:05
2
2
@anomaly - in fairness to the OP, the argument in the accepted answer becomes correct if one assumes the OP is simply-connected; meanwhile, the other answer is correct in greater generality but doesn't provide an argument.
– Ben Blum-Smith
Feb 4 at 21:12
@anomaly - in fairness to the OP, the argument in the accepted answer becomes correct if one assumes the OP is simply-connected; meanwhile, the other answer is correct in greater generality but doesn't provide an argument.
– Ben Blum-Smith
Feb 4 at 21:12
1
1
@BenBlum-Smith: That's not a reasonable assumption, though, even in the physics world. The other answer links to a counterexample and states the assumption necessary to make the conclusion valid; actually proving it requires algebraic topology that's beyond the scope of the original question.
– anomaly
Feb 5 at 4:11
@BenBlum-Smith: That's not a reasonable assumption, though, even in the physics world. The other answer links to a counterexample and states the assumption necessary to make the conclusion valid; actually proving it requires algebraic topology that's beyond the scope of the original question.
– anomaly
Feb 5 at 4:11
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
15
down vote
accepted
Not necessarily. Look at the following potential $A%$ defined on some region:
The associated vector field $F=mathrmgrad(A)$ looks like this:
Since it is a gradient, it has $mathrmcurl(F)=0$. But we can complete it into the following still curl-free vector field:
This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero.
This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $Bbb R^3$, then you are safe: $Bbb R^3$ is simply connected and every curl-free vector field is conservative.
answered Feb 5 at 9:44
M. Winter
17.7k62764
Thanks for the help!
– John A.
Feb 5 at 21:18
Great illustrations! How did you make them?
– Emily
May 6 at 16:53
@Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case).
– M. Winter
May 6 at 16:54
add a comment |Â
up vote
35
down vote
Any conservative vector field $F :U to mathbbR^3$ is irrotational, i.e. $mathbfcurl (F)=0$, but the converse is true only if the domain $U$ is simply connected (see here for a classical example).
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
Thanks for your help!
– John A.
Feb 5 at 21:18
you are welcome! :)
– Emilio Novati
Feb 5 at 21:28
add a comment |Â
up vote
1
down vote
You have to keep in mind that a vector field is not just a set of functions, but also a domain. For instance, the vector field $mathbfF = left<-fracyx^2+y^2,fracxx^2+y^2right>$ on the set $U = left(x,y) neq (0,0)right$ has a curl of zero. But it's not conservative, because integrating it around the unit circle results in $2pi$, not zero as predicted by path-independence.
On the other hand, the same vector field restricted to $U' = leftx>0right$ is conservative. A potential function is $f(x,y) = arctanleft(fracyxright)$.
The difference is that $U'$ is simply connected, while $U$ is not. In fact, this is an symbolic version of M. Winter's graphical example.
answered Feb 5 at 15:29
Matthew Leingang
15k12143
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
Not necessarily. Look at the following potential $A%$ defined on some region:
The associated vector field $F=mathrmgrad(A)$ looks like this:
Since it is a gradient, it has $mathrmcurl(F)=0$. But we can complete it into the following still curl-free vector field:
This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero.
This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $Bbb R^3$, then you are safe: $Bbb R^3$ is simply connected and every curl-free vector field is conservative.
answered Feb 5 at 9:44
M. Winter
17.7k62764
Thanks for the help!
– John A.
Feb 5 at 21:18
Great illustrations! How did you make them?
– Emily
May 6 at 16:53
@Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case).
– M. Winter
May 6 at 16:54
add a comment |Â
up vote
15
down vote
accepted
Not necessarily. Look at the following potential $A%$ defined on some region:
The associated vector field $F=mathrmgrad(A)$ looks like this:
Since it is a gradient, it has $mathrmcurl(F)=0$. But we can complete it into the following still curl-free vector field:
This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero.
This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $Bbb R^3$, then you are safe: $Bbb R^3$ is simply connected and every curl-free vector field is conservative.
answered Feb 5 at 9:44
M. Winter
17.7k62764
Thanks for the help!
– John A.
Feb 5 at 21:18
Great illustrations! How did you make them?
– Emily
May 6 at 16:53
@Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case).
– M. Winter
May 6 at 16:54
add a comment |Â
up vote
15
down vote
accepted
up vote
15
down vote
accepted
Not necessarily. Look at the following potential $A%$ defined on some region:
The associated vector field $F=mathrmgrad(A)$ looks like this:
Since it is a gradient, it has $mathrmcurl(F)=0$. But we can complete it into the following still curl-free vector field:
This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero.
This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $Bbb R^3$, then you are safe: $Bbb R^3$ is simply connected and every curl-free vector field is conservative.
answered Feb 5 at 9:44
M. Winter
17.7k62764
Not necessarily. Look at the following potential $A%$ defined on some region:
The associated vector field $F=mathrmgrad(A)$ looks like this:
Since it is a gradient, it has $mathrmcurl(F)=0$. But we can complete it into the following still curl-free vector field:
This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero.
This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $Bbb R^3$, then you are safe: $Bbb R^3$ is simply connected and every curl-free vector field is conservative.
answered Feb 5 at 9:44
M. Winter
17.7k62764
edited Feb 5 at 12:18
answered Feb 5 at 9:44
M. Winter
17.7k62764
answered Feb 5 at 9:44
M. Winter
17.7k62764
answered Feb 5 at 9:44
M. Winter
17.7k62764
17.7k62764
Thanks for the help!
– John A.
Feb 5 at 21:18
Great illustrations! How did you make them?
– Emily
May 6 at 16:53
@Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case).
– M. Winter
May 6 at 16:54
add a comment |Â
Thanks for the help!
– John A.
Feb 5 at 21:18
Great illustrations! How did you make them?
– Emily
May 6 at 16:53
@Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case).
– M. Winter
May 6 at 16:54
Thanks for the help!
– John A.
Feb 5 at 21:18
Thanks for the help!
– John A.
Feb 5 at 21:18
Great illustrations! How did you make them?
– Emily
May 6 at 16:53
Great illustrations! How did you make them?
– Emily
May 6 at 16:53
@Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case).
– M. Winter
May 6 at 16:54
@Emily Thanks :). These are made with Mathematica and some post-processing in some standard image editing software (Paint.NET in my case).
– M. Winter
May 6 at 16:54
add a comment |Â
up vote
35
down vote
Any conservative vector field $F :U to mathbbR^3$ is irrotational, i.e. $mathbfcurl (F)=0$, but the converse is true only if the domain $U$ is simply connected (see here for a classical example).
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
Thanks for your help!
– John A.
Feb 5 at 21:18
you are welcome! :)
– Emilio Novati
Feb 5 at 21:28
add a comment |Â
up vote
35
down vote
Any conservative vector field $F :U to mathbbR^3$ is irrotational, i.e. $mathbfcurl (F)=0$, but the converse is true only if the domain $U$ is simply connected (see here for a classical example).
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
Thanks for your help!
– John A.
Feb 5 at 21:18
you are welcome! :)
– Emilio Novati
Feb 5 at 21:28
add a comment |Â
up vote
35
down vote
up vote
35
down vote
Any conservative vector field $F :U to mathbbR^3$ is irrotational, i.e. $mathbfcurl (F)=0$, but the converse is true only if the domain $U$ is simply connected (see here for a classical example).
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
Any conservative vector field $F :U to mathbbR^3$ is irrotational, i.e. $mathbfcurl (F)=0$, but the converse is true only if the domain $U$ is simply connected (see here for a classical example).
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
answered Feb 4 at 18:39
Emilio Novati
50.2k43170
50.2k43170
Thanks for your help!
– John A.
Feb 5 at 21:18
you are welcome! :)
– Emilio Novati
Feb 5 at 21:28
add a comment |Â
Thanks for your help!
– John A.
Feb 5 at 21:18
you are welcome! :)
– Emilio Novati
Feb 5 at 21:28
Thanks for your help!
– John A.
Feb 5 at 21:18
Thanks for your help!
– John A.
Feb 5 at 21:18
you are welcome! :)
– Emilio Novati
Feb 5 at 21:28
you are welcome! :)
– Emilio Novati
Feb 5 at 21:28
add a comment |Â
up vote
1
down vote
You have to keep in mind that a vector field is not just a set of functions, but also a domain. For instance, the vector field $mathbfF = left<-fracyx^2+y^2,fracxx^2+y^2right>$ on the set $U = left(x,y) neq (0,0)right$ has a curl of zero. But it's not conservative, because integrating it around the unit circle results in $2pi$, not zero as predicted by path-independence.
On the other hand, the same vector field restricted to $U' = leftx>0right$ is conservative. A potential function is $f(x,y) = arctanleft(fracyxright)$.
The difference is that $U'$ is simply connected, while $U$ is not. In fact, this is an symbolic version of M. Winter's graphical example.
answered Feb 5 at 15:29
Matthew Leingang
15k12143
add a comment |Â
up vote
1
down vote
You have to keep in mind that a vector field is not just a set of functions, but also a domain. For instance, the vector field $mathbfF = left<-fracyx^2+y^2,fracxx^2+y^2right>$ on the set $U = left(x,y) neq (0,0)right$ has a curl of zero. But it's not conservative, because integrating it around the unit circle results in $2pi$, not zero as predicted by path-independence.
On the other hand, the same vector field restricted to $U' = leftx>0right$ is conservative. A potential function is $f(x,y) = arctanleft(fracyxright)$.
The difference is that $U'$ is simply connected, while $U$ is not. In fact, this is an symbolic version of M. Winter's graphical example.
answered Feb 5 at 15:29
Matthew Leingang
15k12143
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have to keep in mind that a vector field is not just a set of functions, but also a domain. For instance, the vector field $mathbfF = left<-fracyx^2+y^2,fracxx^2+y^2right>$ on the set $U = left(x,y) neq (0,0)right$ has a curl of zero. But it's not conservative, because integrating it around the unit circle results in $2pi$, not zero as predicted by path-independence.
On the other hand, the same vector field restricted to $U' = leftx>0right$ is conservative. A potential function is $f(x,y) = arctanleft(fracyxright)$.
The difference is that $U'$ is simply connected, while $U$ is not. In fact, this is an symbolic version of M. Winter's graphical example.
answered Feb 5 at 15:29
Matthew Leingang
15k12143
You have to keep in mind that a vector field is not just a set of functions, but also a domain. For instance, the vector field $mathbfF = left<-fracyx^2+y^2,fracxx^2+y^2right>$ on the set $U = left(x,y) neq (0,0)right$ has a curl of zero. But it's not conservative, because integrating it around the unit circle results in $2pi$, not zero as predicted by path-independence.
On the other hand, the same vector field restricted to $U' = leftx>0right$ is conservative. A potential function is $f(x,y) = arctanleft(fracyxright)$.
The difference is that $U'$ is simply connected, while $U$ is not. In fact, this is an symbolic version of M. Winter's graphical example.
answered Feb 5 at 15:29
Matthew Leingang
15k12143
edited Feb 5 at 17:58
answered Feb 5 at 15:29
Matthew Leingang
15k12143
answered Feb 5 at 15:29
Matthew Leingang
15k12143
answered Feb 5 at 15:29
Matthew Leingang
15k12143
15k12143
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2635850%2fdoes-zero-curl-imply-a-conservative-field%23new-answer', 'question_page');
);
Post as a guest
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
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
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
2
Do you know Stokes theorem? The important fact that leads to the existence of a potential is that the line integral just depends on the start and end point and not curve connecting these.
– Fabian
Feb 4 at 18:15
5
You were a bit hasty in accepting an answer. The one you’ve chosen tacitly makes some important assumptions about the domain of the vector field, without which your proposition is false.
– amd
Feb 4 at 19:24
1
There's a factually incorrect answer rated at -2 and a correct answer rated at 7, and the former is the one that's accepted?
– anomaly
Feb 4 at 21:05
2
@anomaly - in fairness to the OP, the argument in the accepted answer becomes correct if one assumes the OP is simply-connected; meanwhile, the other answer is correct in greater generality but doesn't provide an argument.
– Ben Blum-Smith
Feb 4 at 21:12
1
@BenBlum-Smith: That's not a reasonable assumption, though, even in the physics world. The other answer links to a counterexample and states the assumption necessary to make the conclusion valid; actually proving it requires algebraic topology that's beyond the scope of the original question.
– anomaly
Feb 5 at 4:11