If $vec{nabla} cdot vec{V} neq 0$ at only one point, will this prevent us from saying that $vec{V}=vec{nabla}...












8














This question has an answer in the language of high level mathematics. Can somebody explain this in the language of vector calculus.



Part I: Let us consider Cartesian coordinate system with origin $O$ and axes $x,y,z$. Let:



$$r=sqrt{x^2+y^2+z^2}$$



$$text{and }vec{V}=0 (hat{i})
+dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$$



It is obvious that $dfrac{1}{r}$ is defined everywhere except at the origin.



Now let us take the divergence of $vec{V}$:



$$vec{nabla} cdot vec{V}=0$$



Since $dfrac{1}{r}$ is not defined at the origin, $vec{nabla} cdot vec{V}=0$ is true everywhere except at the origin.



Since $vec{nabla} cdot vec{V} neq 0$ at a point $P (0,0,0)$, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at points other than $P$? Why? Why not?



Part II: If $vec{nabla} cdot vec{V} neq 0$ at points on a one dimensional arbitrary curve in space, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at other points not on the curve? Why? Why not?




NOTE - For both Part I and Part II:



If (Why/Why not) is beyond the scope of vector (multivariable) calculus, just reply as yes/no. However please try your best to explain (Why/Why not) in the language of vector (multivariable) calculus.




SEMI ANSWER: Please point out the limitations



I have stumbled upon a derivation in the language of elementary vector calculus. Please point out if there are any limitations in my derivation. In the context of advanced mathematics (de Rham cohomology or Poincare lemma), it seems to me that there are limitations.



Derivation:



To prove: At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), if $vec{nabla} cdot vec{B}=0$, then $vec{B}=vec{nabla} times vec{A}$



Proof:



At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):
begin{align}
vec{B} &= B_x (hat{i}) + B_y (hat{j}) + B_z (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) + int^{(x,y,z)}_{(x,y,infty)}
dfrac{partial B_z}{partial z} dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( vec{nabla} cdot vec{B} - dfrac{partial B_z}{partial z} right) dz (hat{k})\
&text{{Since $vec{nabla} cdot vec{B}=0$}}\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( dfrac{partial B_x}{partial x} + dfrac{partial B_y}{partial y} right) dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j})
+ left[
dfrac{partial}{partial x}
left(- int^{(x,y,z)}_{(x,y,infty)}B_x dz right)
-dfrac{partial}{partial y}
left( int^{(x,y,z)}_{(x,y,infty)}B_y dz right)
right]
(hat{k})\
&text{{By changing the order of integration and differentiation}}\
end{align}




At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), let's define:



$displaystyle vec{A}=int^{(x,y,z)}_{(x,y,infty)}B_y dz (hat{i}) - int^{(x,y,z)}_{(x,y,infty)}B_x dz (hat{j}) + 0 (hat{k}) + vec{nabla}f$



where $f$ is an arbitrary function of $(x,y,z)$




Therefore at all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):



begin{align}
vec{B} &= left(
dfrac{partial A_z}{partial y}-dfrac{partial A_y}{partial z}
right) (hat{i})
+left(
dfrac{partial A_x}{partial z}-dfrac{partial A_z}{partial x}
right) (hat{j})
+left(
dfrac{partial A_y}{partial x}-dfrac{partial A_x}{partial y}
right) (hat{k})\
&= vec{nabla} times vec{A}
end{align}










share|cite|improve this question
























  • Does the statement say something about how it is defined $vec U$?
    – manooooh
    Nov 27 at 7:54










  • $vec{U} rightarrow vec{U} +vec{nabla}f$ is the vector whose curl gives $vec{V}$......$f$ is an arbitrary function of $x,y,z$
    – Joe
    Nov 27 at 9:03










  • The origin isn't in the domain of $bf V$, so it doesn't even make sense to ask whether $nabla cdot {bf V} = 0$. It's true that $nabla cdot {bf V} = 0$ everywhere on the domain of $V$.
    – Travis
    Nov 27 at 10:43










  • How do we know that the origin isn't in the domain of $bf {V}$?
    – Joe
    Nov 27 at 11:00












  • Because the quantity $frac{1}{r}$ that appears in the definition of $bf V$ isn't defined at the origin.
    – Travis
    Nov 27 at 12:42
















8














This question has an answer in the language of high level mathematics. Can somebody explain this in the language of vector calculus.



Part I: Let us consider Cartesian coordinate system with origin $O$ and axes $x,y,z$. Let:



$$r=sqrt{x^2+y^2+z^2}$$



$$text{and }vec{V}=0 (hat{i})
+dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$$



It is obvious that $dfrac{1}{r}$ is defined everywhere except at the origin.



Now let us take the divergence of $vec{V}$:



$$vec{nabla} cdot vec{V}=0$$



Since $dfrac{1}{r}$ is not defined at the origin, $vec{nabla} cdot vec{V}=0$ is true everywhere except at the origin.



Since $vec{nabla} cdot vec{V} neq 0$ at a point $P (0,0,0)$, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at points other than $P$? Why? Why not?



Part II: If $vec{nabla} cdot vec{V} neq 0$ at points on a one dimensional arbitrary curve in space, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at other points not on the curve? Why? Why not?




NOTE - For both Part I and Part II:



If (Why/Why not) is beyond the scope of vector (multivariable) calculus, just reply as yes/no. However please try your best to explain (Why/Why not) in the language of vector (multivariable) calculus.




SEMI ANSWER: Please point out the limitations



I have stumbled upon a derivation in the language of elementary vector calculus. Please point out if there are any limitations in my derivation. In the context of advanced mathematics (de Rham cohomology or Poincare lemma), it seems to me that there are limitations.



Derivation:



To prove: At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), if $vec{nabla} cdot vec{B}=0$, then $vec{B}=vec{nabla} times vec{A}$



Proof:



At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):
begin{align}
vec{B} &= B_x (hat{i}) + B_y (hat{j}) + B_z (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) + int^{(x,y,z)}_{(x,y,infty)}
dfrac{partial B_z}{partial z} dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( vec{nabla} cdot vec{B} - dfrac{partial B_z}{partial z} right) dz (hat{k})\
&text{{Since $vec{nabla} cdot vec{B}=0$}}\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( dfrac{partial B_x}{partial x} + dfrac{partial B_y}{partial y} right) dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j})
+ left[
dfrac{partial}{partial x}
left(- int^{(x,y,z)}_{(x,y,infty)}B_x dz right)
-dfrac{partial}{partial y}
left( int^{(x,y,z)}_{(x,y,infty)}B_y dz right)
right]
(hat{k})\
&text{{By changing the order of integration and differentiation}}\
end{align}




At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), let's define:



$displaystyle vec{A}=int^{(x,y,z)}_{(x,y,infty)}B_y dz (hat{i}) - int^{(x,y,z)}_{(x,y,infty)}B_x dz (hat{j}) + 0 (hat{k}) + vec{nabla}f$



where $f$ is an arbitrary function of $(x,y,z)$




Therefore at all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):



begin{align}
vec{B} &= left(
dfrac{partial A_z}{partial y}-dfrac{partial A_y}{partial z}
right) (hat{i})
+left(
dfrac{partial A_x}{partial z}-dfrac{partial A_z}{partial x}
right) (hat{j})
+left(
dfrac{partial A_y}{partial x}-dfrac{partial A_x}{partial y}
right) (hat{k})\
&= vec{nabla} times vec{A}
end{align}










share|cite|improve this question
























  • Does the statement say something about how it is defined $vec U$?
    – manooooh
    Nov 27 at 7:54










  • $vec{U} rightarrow vec{U} +vec{nabla}f$ is the vector whose curl gives $vec{V}$......$f$ is an arbitrary function of $x,y,z$
    – Joe
    Nov 27 at 9:03










  • The origin isn't in the domain of $bf V$, so it doesn't even make sense to ask whether $nabla cdot {bf V} = 0$. It's true that $nabla cdot {bf V} = 0$ everywhere on the domain of $V$.
    – Travis
    Nov 27 at 10:43










  • How do we know that the origin isn't in the domain of $bf {V}$?
    – Joe
    Nov 27 at 11:00












  • Because the quantity $frac{1}{r}$ that appears in the definition of $bf V$ isn't defined at the origin.
    – Travis
    Nov 27 at 12:42














8












8








8


2





This question has an answer in the language of high level mathematics. Can somebody explain this in the language of vector calculus.



Part I: Let us consider Cartesian coordinate system with origin $O$ and axes $x,y,z$. Let:



$$r=sqrt{x^2+y^2+z^2}$$



$$text{and }vec{V}=0 (hat{i})
+dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$$



It is obvious that $dfrac{1}{r}$ is defined everywhere except at the origin.



Now let us take the divergence of $vec{V}$:



$$vec{nabla} cdot vec{V}=0$$



Since $dfrac{1}{r}$ is not defined at the origin, $vec{nabla} cdot vec{V}=0$ is true everywhere except at the origin.



Since $vec{nabla} cdot vec{V} neq 0$ at a point $P (0,0,0)$, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at points other than $P$? Why? Why not?



Part II: If $vec{nabla} cdot vec{V} neq 0$ at points on a one dimensional arbitrary curve in space, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at other points not on the curve? Why? Why not?




NOTE - For both Part I and Part II:



If (Why/Why not) is beyond the scope of vector (multivariable) calculus, just reply as yes/no. However please try your best to explain (Why/Why not) in the language of vector (multivariable) calculus.




SEMI ANSWER: Please point out the limitations



I have stumbled upon a derivation in the language of elementary vector calculus. Please point out if there are any limitations in my derivation. In the context of advanced mathematics (de Rham cohomology or Poincare lemma), it seems to me that there are limitations.



Derivation:



To prove: At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), if $vec{nabla} cdot vec{B}=0$, then $vec{B}=vec{nabla} times vec{A}$



Proof:



At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):
begin{align}
vec{B} &= B_x (hat{i}) + B_y (hat{j}) + B_z (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) + int^{(x,y,z)}_{(x,y,infty)}
dfrac{partial B_z}{partial z} dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( vec{nabla} cdot vec{B} - dfrac{partial B_z}{partial z} right) dz (hat{k})\
&text{{Since $vec{nabla} cdot vec{B}=0$}}\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( dfrac{partial B_x}{partial x} + dfrac{partial B_y}{partial y} right) dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j})
+ left[
dfrac{partial}{partial x}
left(- int^{(x,y,z)}_{(x,y,infty)}B_x dz right)
-dfrac{partial}{partial y}
left( int^{(x,y,z)}_{(x,y,infty)}B_y dz right)
right]
(hat{k})\
&text{{By changing the order of integration and differentiation}}\
end{align}




At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), let's define:



$displaystyle vec{A}=int^{(x,y,z)}_{(x,y,infty)}B_y dz (hat{i}) - int^{(x,y,z)}_{(x,y,infty)}B_x dz (hat{j}) + 0 (hat{k}) + vec{nabla}f$



where $f$ is an arbitrary function of $(x,y,z)$




Therefore at all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):



begin{align}
vec{B} &= left(
dfrac{partial A_z}{partial y}-dfrac{partial A_y}{partial z}
right) (hat{i})
+left(
dfrac{partial A_x}{partial z}-dfrac{partial A_z}{partial x}
right) (hat{j})
+left(
dfrac{partial A_y}{partial x}-dfrac{partial A_x}{partial y}
right) (hat{k})\
&= vec{nabla} times vec{A}
end{align}










share|cite|improve this question















This question has an answer in the language of high level mathematics. Can somebody explain this in the language of vector calculus.



Part I: Let us consider Cartesian coordinate system with origin $O$ and axes $x,y,z$. Let:



$$r=sqrt{x^2+y^2+z^2}$$



$$text{and }vec{V}=0 (hat{i})
+dfrac{partial}{partial z} left( dfrac{1}{r} right) (hat{j})
-dfrac{partial}{partial y} left( dfrac{1}{r} right) (hat{k})$$



It is obvious that $dfrac{1}{r}$ is defined everywhere except at the origin.



Now let us take the divergence of $vec{V}$:



$$vec{nabla} cdot vec{V}=0$$



Since $dfrac{1}{r}$ is not defined at the origin, $vec{nabla} cdot vec{V}=0$ is true everywhere except at the origin.



Since $vec{nabla} cdot vec{V} neq 0$ at a point $P (0,0,0)$, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at points other than $P$? Why? Why not?



Part II: If $vec{nabla} cdot vec{V} neq 0$ at points on a one dimensional arbitrary curve in space, will this prevent us from concluding $vec{V}=vec{nabla} times vec{U}$ at other points not on the curve? Why? Why not?




NOTE - For both Part I and Part II:



If (Why/Why not) is beyond the scope of vector (multivariable) calculus, just reply as yes/no. However please try your best to explain (Why/Why not) in the language of vector (multivariable) calculus.




SEMI ANSWER: Please point out the limitations



I have stumbled upon a derivation in the language of elementary vector calculus. Please point out if there are any limitations in my derivation. In the context of advanced mathematics (de Rham cohomology or Poincare lemma), it seems to me that there are limitations.



Derivation:



To prove: At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), if $vec{nabla} cdot vec{B}=0$, then $vec{B}=vec{nabla} times vec{A}$



Proof:



At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):
begin{align}
vec{B} &= B_x (hat{i}) + B_y (hat{j}) + B_z (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) + int^{(x,y,z)}_{(x,y,infty)}
dfrac{partial B_z}{partial z} dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( vec{nabla} cdot vec{B} - dfrac{partial B_z}{partial z} right) dz (hat{k})\
&text{{Since $vec{nabla} cdot vec{B}=0$}}\
&= B_x (hat{i}) + B_y (hat{j}) - int^{(x,y,z)}_{(x,y,infty)}
left( dfrac{partial B_x}{partial x} + dfrac{partial B_y}{partial y} right) dz (hat{k})\
&= B_x (hat{i}) + B_y (hat{j})
+ left[
dfrac{partial}{partial x}
left(- int^{(x,y,z)}_{(x,y,infty)}B_x dz right)
-dfrac{partial}{partial y}
left( int^{(x,y,z)}_{(x,y,infty)}B_y dz right)
right]
(hat{k})\
&text{{By changing the order of integration and differentiation}}\
end{align}




At all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$), let's define:



$displaystyle vec{A}=int^{(x,y,z)}_{(x,y,infty)}B_y dz (hat{i}) - int^{(x,y,z)}_{(x,y,infty)}B_x dz (hat{j}) + 0 (hat{k}) + vec{nabla}f$



where $f$ is an arbitrary function of $(x,y,z)$




Therefore at all points where $vec{B}$ is defined (whatever be the domain of $vec{B}$):



begin{align}
vec{B} &= left(
dfrac{partial A_z}{partial y}-dfrac{partial A_y}{partial z}
right) (hat{i})
+left(
dfrac{partial A_x}{partial z}-dfrac{partial A_z}{partial x}
right) (hat{j})
+left(
dfrac{partial A_y}{partial x}-dfrac{partial A_x}{partial y}
right) (hat{k})\
&= vec{nabla} times vec{A}
end{align}







multivariable-calculus vector-fields divergence curl






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edited Dec 1 at 9:53

























asked Nov 27 at 7:26









Joe

292113




292113












  • Does the statement say something about how it is defined $vec U$?
    – manooooh
    Nov 27 at 7:54










  • $vec{U} rightarrow vec{U} +vec{nabla}f$ is the vector whose curl gives $vec{V}$......$f$ is an arbitrary function of $x,y,z$
    – Joe
    Nov 27 at 9:03










  • The origin isn't in the domain of $bf V$, so it doesn't even make sense to ask whether $nabla cdot {bf V} = 0$. It's true that $nabla cdot {bf V} = 0$ everywhere on the domain of $V$.
    – Travis
    Nov 27 at 10:43










  • How do we know that the origin isn't in the domain of $bf {V}$?
    – Joe
    Nov 27 at 11:00












  • Because the quantity $frac{1}{r}$ that appears in the definition of $bf V$ isn't defined at the origin.
    – Travis
    Nov 27 at 12:42


















  • Does the statement say something about how it is defined $vec U$?
    – manooooh
    Nov 27 at 7:54










  • $vec{U} rightarrow vec{U} +vec{nabla}f$ is the vector whose curl gives $vec{V}$......$f$ is an arbitrary function of $x,y,z$
    – Joe
    Nov 27 at 9:03










  • The origin isn't in the domain of $bf V$, so it doesn't even make sense to ask whether $nabla cdot {bf V} = 0$. It's true that $nabla cdot {bf V} = 0$ everywhere on the domain of $V$.
    – Travis
    Nov 27 at 10:43










  • How do we know that the origin isn't in the domain of $bf {V}$?
    – Joe
    Nov 27 at 11:00












  • Because the quantity $frac{1}{r}$ that appears in the definition of $bf V$ isn't defined at the origin.
    – Travis
    Nov 27 at 12:42
















Does the statement say something about how it is defined $vec U$?
– manooooh
Nov 27 at 7:54




Does the statement say something about how it is defined $vec U$?
– manooooh
Nov 27 at 7:54












$vec{U} rightarrow vec{U} +vec{nabla}f$ is the vector whose curl gives $vec{V}$......$f$ is an arbitrary function of $x,y,z$
– Joe
Nov 27 at 9:03




$vec{U} rightarrow vec{U} +vec{nabla}f$ is the vector whose curl gives $vec{V}$......$f$ is an arbitrary function of $x,y,z$
– Joe
Nov 27 at 9:03












The origin isn't in the domain of $bf V$, so it doesn't even make sense to ask whether $nabla cdot {bf V} = 0$. It's true that $nabla cdot {bf V} = 0$ everywhere on the domain of $V$.
– Travis
Nov 27 at 10:43




The origin isn't in the domain of $bf V$, so it doesn't even make sense to ask whether $nabla cdot {bf V} = 0$. It's true that $nabla cdot {bf V} = 0$ everywhere on the domain of $V$.
– Travis
Nov 27 at 10:43












How do we know that the origin isn't in the domain of $bf {V}$?
– Joe
Nov 27 at 11:00






How do we know that the origin isn't in the domain of $bf {V}$?
– Joe
Nov 27 at 11:00














Because the quantity $frac{1}{r}$ that appears in the definition of $bf V$ isn't defined at the origin.
– Travis
Nov 27 at 12:42




Because the quantity $frac{1}{r}$ that appears in the definition of $bf V$ isn't defined at the origin.
– Travis
Nov 27 at 12:42










2 Answers
2






active

oldest

votes


















7














You seem to know that a divergence free field $vec V$ can be regarded as curl of some other field: There is a field $vec U$ such that $vec V={rm curl}(vec U)$. This is a consequence of the so-called Poincaré Lemma.



But there is a catch: The Poincaré Lemma guarantees the existence of such a vector potential $vec U$ only if the domain of $vec V$ is, e.g., a ball or star shaped. For your field $vec V$ this is not the case. Therefore we only can say the following: Each point ${bf p}$ in the punctured space $dot{mathbb R}^3:={mathbb R}^3setminus{{bf 0}}$ is the center of an open ball $B_r({bf p})subset dot{mathbb R}^3$ such that within $B_r({bf p})$ the field $vec V$ can be written in the form $vec V={rm curl}(vec U)$ for some $vec U$ defined in $B_r({bf p})$ only. These local fields $vec U$ are not uniquely determined, and it is not at all sure whether the implied "integration constants" can be chosen in a coherent way such that we obtain a single field $vec U_*$, which then would be defined on all of $dot{mathbb R}^3$.



Of course, it could be that "by coincidence" the $vec V$ in your example possesses a global vector potential $vec U_*$ nevertheless: not by the Poincaré Lemma per se, but because a certain integrability condition (say, the flux of $vec V$ across a sphere around ${bf 0}$ should be $=0$) is fulfilled. Consider as an analogue the functions $zmapsto{1over z}$ and $zmapsto{1over z^2}$ in the punctured complex plane $dot{mathbb C}$. Both have local primitives. But one of them does not have a global primitive in $dot{mathbb C}$, the other does, the reason being that $$int_{partial D}{1over z}>dz=2pi ine0>, qquad int_{partial D}{1over z^2}>dz=0 .$$



Update: I suggest you look at the entry Helmholtz decomposition in Wikipedia and apply Helmholtz' theorem to a large ball minus a tiny ball around the origin.






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  • Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand.
    – Joe
    Nov 27 at 11:59










  • You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand.
    – Christian Blatter
    Nov 27 at 14:22










  • "The flux of $vec{V}$ across a sphere around $bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere....
    – Joe
    Nov 27 at 16:37












  • Full answer. What is the meaning of $dot{Bbb R}^3$?
    – manooooh
    Nov 27 at 16:51












  • I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${mathbb R}^3$ or balls.
    – Christian Blatter
    Nov 27 at 16:52



















0














It is a consequence of the Kelvin-Stokes theorem:



https://en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem



If you take a potential $U=frac{1}{r}$ with associated "electric field" $E:=nabla U$ then you know that the flux of $E$ through a unit sphere $Sigma$ centered around $(0,0,0)$ is 1.



However by Kelvin-Stokes if you could write $E=curl(B)$ for a certain vector field $B$ then the flux of $E$ through $Sigma$ would be equal to 0 since $Sigma$ is closed surface.






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    7














    You seem to know that a divergence free field $vec V$ can be regarded as curl of some other field: There is a field $vec U$ such that $vec V={rm curl}(vec U)$. This is a consequence of the so-called Poincaré Lemma.



    But there is a catch: The Poincaré Lemma guarantees the existence of such a vector potential $vec U$ only if the domain of $vec V$ is, e.g., a ball or star shaped. For your field $vec V$ this is not the case. Therefore we only can say the following: Each point ${bf p}$ in the punctured space $dot{mathbb R}^3:={mathbb R}^3setminus{{bf 0}}$ is the center of an open ball $B_r({bf p})subset dot{mathbb R}^3$ such that within $B_r({bf p})$ the field $vec V$ can be written in the form $vec V={rm curl}(vec U)$ for some $vec U$ defined in $B_r({bf p})$ only. These local fields $vec U$ are not uniquely determined, and it is not at all sure whether the implied "integration constants" can be chosen in a coherent way such that we obtain a single field $vec U_*$, which then would be defined on all of $dot{mathbb R}^3$.



    Of course, it could be that "by coincidence" the $vec V$ in your example possesses a global vector potential $vec U_*$ nevertheless: not by the Poincaré Lemma per se, but because a certain integrability condition (say, the flux of $vec V$ across a sphere around ${bf 0}$ should be $=0$) is fulfilled. Consider as an analogue the functions $zmapsto{1over z}$ and $zmapsto{1over z^2}$ in the punctured complex plane $dot{mathbb C}$. Both have local primitives. But one of them does not have a global primitive in $dot{mathbb C}$, the other does, the reason being that $$int_{partial D}{1over z}>dz=2pi ine0>, qquad int_{partial D}{1over z^2}>dz=0 .$$



    Update: I suggest you look at the entry Helmholtz decomposition in Wikipedia and apply Helmholtz' theorem to a large ball minus a tiny ball around the origin.






    share|cite|improve this answer























    • Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand.
      – Joe
      Nov 27 at 11:59










    • You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand.
      – Christian Blatter
      Nov 27 at 14:22










    • "The flux of $vec{V}$ across a sphere around $bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere....
      – Joe
      Nov 27 at 16:37












    • Full answer. What is the meaning of $dot{Bbb R}^3$?
      – manooooh
      Nov 27 at 16:51












    • I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${mathbb R}^3$ or balls.
      – Christian Blatter
      Nov 27 at 16:52
















    7














    You seem to know that a divergence free field $vec V$ can be regarded as curl of some other field: There is a field $vec U$ such that $vec V={rm curl}(vec U)$. This is a consequence of the so-called Poincaré Lemma.



    But there is a catch: The Poincaré Lemma guarantees the existence of such a vector potential $vec U$ only if the domain of $vec V$ is, e.g., a ball or star shaped. For your field $vec V$ this is not the case. Therefore we only can say the following: Each point ${bf p}$ in the punctured space $dot{mathbb R}^3:={mathbb R}^3setminus{{bf 0}}$ is the center of an open ball $B_r({bf p})subset dot{mathbb R}^3$ such that within $B_r({bf p})$ the field $vec V$ can be written in the form $vec V={rm curl}(vec U)$ for some $vec U$ defined in $B_r({bf p})$ only. These local fields $vec U$ are not uniquely determined, and it is not at all sure whether the implied "integration constants" can be chosen in a coherent way such that we obtain a single field $vec U_*$, which then would be defined on all of $dot{mathbb R}^3$.



    Of course, it could be that "by coincidence" the $vec V$ in your example possesses a global vector potential $vec U_*$ nevertheless: not by the Poincaré Lemma per se, but because a certain integrability condition (say, the flux of $vec V$ across a sphere around ${bf 0}$ should be $=0$) is fulfilled. Consider as an analogue the functions $zmapsto{1over z}$ and $zmapsto{1over z^2}$ in the punctured complex plane $dot{mathbb C}$. Both have local primitives. But one of them does not have a global primitive in $dot{mathbb C}$, the other does, the reason being that $$int_{partial D}{1over z}>dz=2pi ine0>, qquad int_{partial D}{1over z^2}>dz=0 .$$



    Update: I suggest you look at the entry Helmholtz decomposition in Wikipedia and apply Helmholtz' theorem to a large ball minus a tiny ball around the origin.






    share|cite|improve this answer























    • Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand.
      – Joe
      Nov 27 at 11:59










    • You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand.
      – Christian Blatter
      Nov 27 at 14:22










    • "The flux of $vec{V}$ across a sphere around $bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere....
      – Joe
      Nov 27 at 16:37












    • Full answer. What is the meaning of $dot{Bbb R}^3$?
      – manooooh
      Nov 27 at 16:51












    • I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${mathbb R}^3$ or balls.
      – Christian Blatter
      Nov 27 at 16:52














    7












    7








    7






    You seem to know that a divergence free field $vec V$ can be regarded as curl of some other field: There is a field $vec U$ such that $vec V={rm curl}(vec U)$. This is a consequence of the so-called Poincaré Lemma.



    But there is a catch: The Poincaré Lemma guarantees the existence of such a vector potential $vec U$ only if the domain of $vec V$ is, e.g., a ball or star shaped. For your field $vec V$ this is not the case. Therefore we only can say the following: Each point ${bf p}$ in the punctured space $dot{mathbb R}^3:={mathbb R}^3setminus{{bf 0}}$ is the center of an open ball $B_r({bf p})subset dot{mathbb R}^3$ such that within $B_r({bf p})$ the field $vec V$ can be written in the form $vec V={rm curl}(vec U)$ for some $vec U$ defined in $B_r({bf p})$ only. These local fields $vec U$ are not uniquely determined, and it is not at all sure whether the implied "integration constants" can be chosen in a coherent way such that we obtain a single field $vec U_*$, which then would be defined on all of $dot{mathbb R}^3$.



    Of course, it could be that "by coincidence" the $vec V$ in your example possesses a global vector potential $vec U_*$ nevertheless: not by the Poincaré Lemma per se, but because a certain integrability condition (say, the flux of $vec V$ across a sphere around ${bf 0}$ should be $=0$) is fulfilled. Consider as an analogue the functions $zmapsto{1over z}$ and $zmapsto{1over z^2}$ in the punctured complex plane $dot{mathbb C}$. Both have local primitives. But one of them does not have a global primitive in $dot{mathbb C}$, the other does, the reason being that $$int_{partial D}{1over z}>dz=2pi ine0>, qquad int_{partial D}{1over z^2}>dz=0 .$$



    Update: I suggest you look at the entry Helmholtz decomposition in Wikipedia and apply Helmholtz' theorem to a large ball minus a tiny ball around the origin.






    share|cite|improve this answer














    You seem to know that a divergence free field $vec V$ can be regarded as curl of some other field: There is a field $vec U$ such that $vec V={rm curl}(vec U)$. This is a consequence of the so-called Poincaré Lemma.



    But there is a catch: The Poincaré Lemma guarantees the existence of such a vector potential $vec U$ only if the domain of $vec V$ is, e.g., a ball or star shaped. For your field $vec V$ this is not the case. Therefore we only can say the following: Each point ${bf p}$ in the punctured space $dot{mathbb R}^3:={mathbb R}^3setminus{{bf 0}}$ is the center of an open ball $B_r({bf p})subset dot{mathbb R}^3$ such that within $B_r({bf p})$ the field $vec V$ can be written in the form $vec V={rm curl}(vec U)$ for some $vec U$ defined in $B_r({bf p})$ only. These local fields $vec U$ are not uniquely determined, and it is not at all sure whether the implied "integration constants" can be chosen in a coherent way such that we obtain a single field $vec U_*$, which then would be defined on all of $dot{mathbb R}^3$.



    Of course, it could be that "by coincidence" the $vec V$ in your example possesses a global vector potential $vec U_*$ nevertheless: not by the Poincaré Lemma per se, but because a certain integrability condition (say, the flux of $vec V$ across a sphere around ${bf 0}$ should be $=0$) is fulfilled. Consider as an analogue the functions $zmapsto{1over z}$ and $zmapsto{1over z^2}$ in the punctured complex plane $dot{mathbb C}$. Both have local primitives. But one of them does not have a global primitive in $dot{mathbb C}$, the other does, the reason being that $$int_{partial D}{1over z}>dz=2pi ine0>, qquad int_{partial D}{1over z^2}>dz=0 .$$



    Update: I suggest you look at the entry Helmholtz decomposition in Wikipedia and apply Helmholtz' theorem to a large ball minus a tiny ball around the origin.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 28 at 17:27

























    answered Nov 27 at 11:20









    Christian Blatter

    172k7112325




    172k7112325












    • Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand.
      – Joe
      Nov 27 at 11:59










    • You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand.
      – Christian Blatter
      Nov 27 at 14:22










    • "The flux of $vec{V}$ across a sphere around $bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere....
      – Joe
      Nov 27 at 16:37












    • Full answer. What is the meaning of $dot{Bbb R}^3$?
      – manooooh
      Nov 27 at 16:51












    • I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${mathbb R}^3$ or balls.
      – Christian Blatter
      Nov 27 at 16:52


















    • Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand.
      – Joe
      Nov 27 at 11:59










    • You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand.
      – Christian Blatter
      Nov 27 at 14:22










    • "The flux of $vec{V}$ across a sphere around $bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere....
      – Joe
      Nov 27 at 16:37












    • Full answer. What is the meaning of $dot{Bbb R}^3$?
      – manooooh
      Nov 27 at 16:51












    • I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${mathbb R}^3$ or balls.
      – Christian Blatter
      Nov 27 at 16:52
















    Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand.
    – Joe
    Nov 27 at 11:59




    Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand.
    – Joe
    Nov 27 at 11:59












    You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand.
    – Christian Blatter
    Nov 27 at 14:22




    You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand.
    – Christian Blatter
    Nov 27 at 14:22












    "The flux of $vec{V}$ across a sphere around $bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere....
    – Joe
    Nov 27 at 16:37






    "The flux of $vec{V}$ across a sphere around $bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere....
    – Joe
    Nov 27 at 16:37














    Full answer. What is the meaning of $dot{Bbb R}^3$?
    – manooooh
    Nov 27 at 16:51






    Full answer. What is the meaning of $dot{Bbb R}^3$?
    – manooooh
    Nov 27 at 16:51














    I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${mathbb R}^3$ or balls.
    – Christian Blatter
    Nov 27 at 16:52




    I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${mathbb R}^3$ or balls.
    – Christian Blatter
    Nov 27 at 16:52











    0














    It is a consequence of the Kelvin-Stokes theorem:



    https://en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem



    If you take a potential $U=frac{1}{r}$ with associated "electric field" $E:=nabla U$ then you know that the flux of $E$ through a unit sphere $Sigma$ centered around $(0,0,0)$ is 1.



    However by Kelvin-Stokes if you could write $E=curl(B)$ for a certain vector field $B$ then the flux of $E$ through $Sigma$ would be equal to 0 since $Sigma$ is closed surface.






    share|cite|improve this answer


























      0














      It is a consequence of the Kelvin-Stokes theorem:



      https://en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem



      If you take a potential $U=frac{1}{r}$ with associated "electric field" $E:=nabla U$ then you know that the flux of $E$ through a unit sphere $Sigma$ centered around $(0,0,0)$ is 1.



      However by Kelvin-Stokes if you could write $E=curl(B)$ for a certain vector field $B$ then the flux of $E$ through $Sigma$ would be equal to 0 since $Sigma$ is closed surface.






      share|cite|improve this answer
























        0












        0








        0






        It is a consequence of the Kelvin-Stokes theorem:



        https://en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem



        If you take a potential $U=frac{1}{r}$ with associated "electric field" $E:=nabla U$ then you know that the flux of $E$ through a unit sphere $Sigma$ centered around $(0,0,0)$ is 1.



        However by Kelvin-Stokes if you could write $E=curl(B)$ for a certain vector field $B$ then the flux of $E$ through $Sigma$ would be equal to 0 since $Sigma$ is closed surface.






        share|cite|improve this answer












        It is a consequence of the Kelvin-Stokes theorem:



        https://en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem



        If you take a potential $U=frac{1}{r}$ with associated "electric field" $E:=nabla U$ then you know that the flux of $E$ through a unit sphere $Sigma$ centered around $(0,0,0)$ is 1.



        However by Kelvin-Stokes if you could write $E=curl(B)$ for a certain vector field $B$ then the flux of $E$ through $Sigma$ would be equal to 0 since $Sigma$ is closed surface.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 at 4:52









        Ezy

        54429




        54429






























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