Vector field and Helmholtz Theorem

In summary, the Helmholtz theorem states that the same vector field ##\bf{F}(r)## can be written as the gradient of a scalar field ##\Phi## + the curl of a vector field ##\bf{C}## which can be obtained through volume integrals involving the fields ##B## and ##\bf{A}##.
  • #1
fog37
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TL;DR Summary
Understand the meaning of Helmholtz theorem
Hello,

A generic vector field ##\bf {F} (r)## is fully specified over a finite region of space once we know both its divergence and the curl:

$$\nabla \times \bf{F}= A$$
$$\nabla \cdot \bf{F}= B$$

where ##B## is a scalar field and ##\bf{A}## is a divergence free vector field. The divergence and curl equations are PDEs, i.e. equations applied at all the different spatial points of the region of interest. The region of interest also needs boundary conditions on its boundary. The fields ##B## and ##\bf{A}## are not unique.

Helmholtz theorem states that the same vector field ##\bf{F} (r)## can be written as the gradient of a scalar field ##\Phi## + the curl of a vector field ##\bf{C}## which can be obtained through volume integrals involving the fields ##B## and ##\bf{A}##.

Hope this is correct. This is my understanding of the Helmholtz theorem so far...

Questions:
  • Why does Helmholtz theorem only apply to vector fields that are only space dependent and not time dependent?
  • What must the boundary conditions be on the region of interest? I have read about the vector field needing to decrease fast to zero at infinity. That means the region of interest is free space and has no boundaries...
Thank you!
 
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  • #2
About your first question I believe that all three F,A,B can be functions of time as well and the theorem continues to hold as well.
There is some confusion that arises if you going to use the theorem in conjunction with Maxwell's equations in the time dependent case but tell me if this is what bugs you and where you get confused and I ll try to help you.
 
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  • #3
Although I am primarily a mathematician, I seem to remember that in order to deal with time, you add [itex] ict[/itex] to the spatial coordinates and [itex]ic\frac{\partial}{\partial t} [/itex] to the [itex]\nabla [/itex] operator, creating the space-and-time operator [itex]\square [/itex].
 
  • #4
the symbol ##\square## is usually used for the wave operator which is $$\square=\frac{1}{c^2}\frac{\partial}{\partial t^2}-\nabla^2$$
 
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  • #5
Delta2 said:
the symbol ##\square## is usually used for the wave operator which is $$\square=\frac{1}{c^2}\frac{\partial}{\partial t^2}-\nabla^2$$
You are probably right - I did a class in "Modern Physics" in 1964 and I haven't visited the subject since then. But - I seem to remember an operator of the type [itex]
\begin{matrix}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} & \frac{i}{c}\frac{\partial}{\partial t} \\
\end{matrix}
[/itex]. Am I totally at sea?
 
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  • #6
well I think we can get the (-##\square##) wave operator by taking the dot product of $$u=<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z},\frac{i}{c}\frac{\partial}{\partial t}>$$ with it self.$$\square=-u\cdot u=-u^2$$
 
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  • #7
Delta2 said:
About your first question I believe that all three F,A,B can be functions of time as well and the theorem continues to hold as well.
There is some confusion that arises if you going to use the theorem in conjunction with Maxwell's equations in the time dependent case but tell me if this is what bugs you and where you get confused and I ll try to help you.
Hello Delta2, yes, that is my end game, connecting Helmholtz theorem to Maxwell's four PDE equations. There are 3 distinct areas in electromagnetics:
  • Electrostatics: the ##B## field is zero everywhere, the ##E## field has ##\nabla \times \bf{E} = 0## and nonzero divergence.
  • Magnetostatics: the ##E## field is zero everywhere but the ##B## field has ##\nabla \cdot \bf{B} = \mu_{0} \bf{J}## and zero divergence.
  • Electrodynamics: both ##E## and ##B## are nonzero and each field has nonzero curl and divergence.
 
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  • #8
Some other things I know and I think are true:
  • For a generic vector ##\bf{F}(r)##, the decomposition $$\bf{F}(r) = - \nabla \phi + \nabla \times \bf{C}$$ is not unique in the sense that there are infinite pairs ##\phi, \bf{C}## that would work for the same vector field ##\bf{F}(r)##.
  • The specific vector field ##\bf{F}(r)=0## everywhere in space is the only vector field which vanishes at infinity and has both divergence and curl equal to zero everywhere.
  • In electromagnetics, the vector ##\bf{C}## would correspond the famous magnetic vector potential ##\bf{A}## and ##\phi## the scalar electric potential. These two potentials can have different "gauges" (Coloumb, Hamilton, etc.), correct?
  • If both curl and divergence of a vector field ##\bf{F}(r)## go to zero faster than ##\frac {1}{r^2}##, if ##\bf{F}(r)## tends to zero as ##r \rightarrow \infty##, then ##\bf{F}(r)## is UNIQUELY specified by the scalar potential and vector potential. Does that means that the scalar and vector potentials are unique in that case?
All these facts connect to electromagnetism because radiated fields exist and propagate in an unbounded region of space. The only boundary condition on this infinitely large region of space is that the fields must be zero (tend to zero). Is there also a zero derivative involved as boundary condition?

However, electromagnetics exist also in bounded domains (regions of space). I guess the condition "vanish to zero" may or many not be applicable in that case...
 
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  • #9
fog37 said:
Hello Delta2, yes, that is my end game, connecting Helmholtz theorem to Maxwell's four PDE equations. There are 3 distinct areas in electromagnetics:
  • Electrostatics: the ##B## field is zero everywhere, the ##E## field has ##\nabla \times \bf{E} = 0## and nonzero divergence.
  • Magnetostatics: the ##E## field is zero everywhere but the ##B## field has ##\nabla \cdot \bf{B} = \mu_{0} \bf{J}## and zero divergence.
  • Electrodynamics: both ##E## and ##B## are nonzero and each field has nonzero curl and divergence.
some small corrections here
  • Electrostatics: The ##B## field is zero or constant w.r.t to time everywhere (and the ##E## field has zero curl everywhere but non zero divergence as you said)
  • Magnetostatics: The ##E## field is zero or constant w.r.t to time everywhere but the ##B## field has non zero curl ##\mu_0\bf{J}##(and zero divergence as you said).
  • Electrodynamics: both ##E## and ##B## vary w.r.t to time, each field has nonzero curl and the divergence of the ##E## is not necessarily zero everywhere(as you said) , while the divergence of ##B## is zero everywhere
 
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  • #10
I will comment soon enough on your post #8 but while I ponder take a look at the attached pdf file, I do not agree with all the things stated in this paper but I agree to the conclusion written with bold letters at the end of page 3.
 

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  • #11
fog37 said:
Some other things I know and I think are true:
  • For a generic vector ##\bf{F}(r)##, the decomposition $$\bf{F}(r) = - \nabla \phi + \nabla \times \bf{C}$$ is not unique in the sense that there are infinite pairs ##\phi, \bf{C}## that would work for the same vector field ##\bf{F}(r)##.
  • The specific vector field ##\bf{F}(r)=0## everywhere in space is the only vector field which vanishes at infinity and has both divergence and curl equal to zero everywhere.
  • In electromagnetics, the vector ##\bf{C}## would correspond the famous magnetic vector potential ##\bf{A}## and ##\phi## the scalar electric potential. These two potentials can have different "gauges" (Coloumb, Hamilton, etc.), correct?
  • If both curl and divergence of a vector field ##\bf{F}(r)## go to zero faster than ##\frac {1}{r^2}##, if ##\bf{F}(r)## tends to zero as ##r \rightarrow \infty##, then ##\bf{F}(r)## is UNIQUELY specified by the scalar potential and vector potential. Does that means that the scalar and vector potentials are unique in that case?
The first 2 I think are correct.
For the third, we have to be abit more careful, if the Helmholtz decomposition of the electric field is ##\vec{E}=\nabla\phi+\nabla\times \vec{C}## then the ##\phi## is indeed the electric scalar potential (under the coulomb gauge) but ##\vec{C}## is not exactly the famous magnetic vector potential ##\vec{A}##. It is rather $$\nabla\times\vec{C}=-\frac{\partial \vec{A}}{\partial t}$$

About the fourth statement looks correct to me, except that the scalar and vector potentials are still not unique, we can do the transformations $$\phi'\to\phi+c,\vec{C'}\to\vec{C}+\nabla\chi$$ (where ##c## is any constant and ##\chi## is any scalar function)and the new potentials ##\phi'## and ##\vec{C'}## continue to satisfy the Helmholtz decomposition equation for the same vector field, because (##\nabla c=\vec{0}## and ##\nabla\times(\nabla\chi)=\vec{0}##).
All these facts connect to electromagnetism because radiated fields exist and propagate in an unbounded region of space. The only boundary condition on this infinitely large region of space is that the fields must be zero (tend to zero). Is there also a zero derivative involved as boundary condition?

However, electromagnetics exist also in bounded domains (regions of space). I guess the condition "vanish to zero" may or many not be applicable in that case...
The boundary condition I know (regardless if the fields are in bound or unbounded space) is that $$\lim_{|\vec{r}|\to S}|\vec{r}||F(\vec{r})|=0$$ (where S is the surface that bounds the region or ##S=\infty##) and then the vector field ##\vec{F}## is uniquely determined by its curl and divergence.
 
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  • #12
Delta2 said:
well I think we can get the (-##\square##) wave operator by taking the dot product of $$u=<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z},\frac{i}{c}\frac{\partial}{\partial t}>$$ with it self.$$\square=-u\cdot u=-u^2$$
Your statement looks OK, but there are several problems with it:
  • Is the scalar product of operators meaningful or even defined?
  • If you concatenate the operators you would get a lot of mixed partial derivatives
  • And even if the scalar product makes sense, you would get components like [itex] \frac{\partial}{\partial x}\cdot \frac{\partial}{\partial x}[/itex] which is not equal to [itex]\frac{\partial^{2}}{\partial x^{2}} [/itex]
Operator arithmetic is tricky!
 
  • #13
I don't think I am say something too exotic, the dot product of ##u## with itself is defined pretty much the same way as the dot product of ##\nabla## with itself is defined and it is ##\nabla^2=\nabla\cdot\nabla## at least that's how I was being taught back at 1998.
 
  • #14
Delta2 said:
The first 2 I think are correct.
For the third, we have to be abit more careful, if the Helmholtz decomposition of the electric field is ##\vec{E}=\nabla\phi+\nabla\times \vec{C}## then the ##\phi## is indeed the electric scalar potential (under the coulomb gauge) but ##\vec{C}## is not exactly the famous magnetic vector potential ##\vec{A}##. It is rather $$\nabla\times\vec{C}=-\frac{\partial \vec{A}}{\partial t}$$

About the fourth statement looks correct to me, except that the scalar and vector potentials are still not unique, we can do the transformations $$\phi'\to\phi+c,\vec{C'}\to\vec{C}+\nabla\chi$$ (where ##c## is any constant and ##\chi## is any scalar function)and the new potentials ##\phi'## and ##\vec{C'}## continue to satisfy the Helmholtz decomposition equation for the same vector field, because (##\nabla c=\vec{0}## and ##\nabla\times(\nabla\chi)=\vec{0}##).

The boundary condition I know (regardless if the fields are in bound or unbounded space) is that $$\lim_{|\vec{r}|\to S}|\vec{r}||F(\vec{r})|=0$$ (where S is the surface that bounds the region or ##S=\infty##) and then the vector field ##\vec{F}## is uniquely determined by its curl and divergence.

Thank you!

!) When you mention that "the vector field ##\vec{F}## is uniquely determined by its curl and divergence".
Do you mean that the scalar potential and vector potential in the decomposition become unique under that boundary condition and therefore they uniquely define the vector ##E##?

2) If the electric field is time independent, it is automatically conservative and only equal to the gradient of the scalar field ##\phi##. Apparently, conservative vector fields can never be time-dependent. I always forget the full reasoning behind that. I guess the line integral must be always path-independent and if the vector field was time-dependent, the line integral would be path-dependent? What would be the problem with that?
You mention that all our discussion and Helmholtz theorem should equally apply to time-dependent vector fields as they do to position-only dependent vector fields...
 
  • #15
fog37 said:
Thank you!

!) When you mention that "the vector field ##\vec{F}## is uniquely determined by its curl and divergence".
Do you mean that the scalar potential and vector potential in the decomposition become unique under that boundary condition and therefore they uniquely define the vector ##E##?
No, the scalar and vector potential of the Helmholtz decomposition are never unique for a given vector field(regardless of the boundary conditions for the field, or of the conditions for its curl and divergence if we are given such conditions instead of being given the vector field) cause you can always do the transformations outlined in post #11 and the vector field remains the same under those transformations of potentials. Take for a example the vector field (in a cartesian coordinate system)
$$\vec{F}=\nabla x+\nabla\times (y\hat x)$$
so we can immediately see that the scalar potential ##\phi=x## and vector potential ##\vec{C}=y\hat x## are its Helmholtz decomposition. Now try for ##\phi'=x+5## and ##\vec{C'}=y\hat x+\nabla x## and you ll see that ##\phi'## and ##\vec{C'}## also satisfy the Helmholtz decomposition of this field, that is it holds that $$\vec{F}=\nabla\phi'+\nabla\times\vec{C'}$$.

2) If the electric field is time independent, it is automatically conservative and only equal to the gradient of the scalar field ##\phi##. Apparently, conservative vector fields can never be time-dependent. I always forget the full reasoning behind that. I guess the line integral must be always path-independent and if the vector field was time-dependent, the line integral would be path-dependent? What would be the problem with that?
You mention that all our discussion and Helmholtz theorem should equally apply to time-dependent vector fields as they do to position-only dependent vector fields...
We can have conservative vector fields that can be time dependent (take for example the field ##\vec{F}=2x\sin(2\pi t)\hat x=\nabla x^2\sin(2\pi t)##, but they will not correspond to an electric or magnetic field, because for an electric or magnetic field the Maxwell's equations hold and from Maxwell's equations we can conclude that if the electric or magnetic field are time dependent then they are not conservative. The vector field ##\vec{F}## as it is given above does NOT satisfy Maxwell's equations (try it with ##\vec{E}=\vec{F},\vec{B}=\vec{0}## you ll find it out that in free space it does not satisfy Gauss's law or Maxwell- Ampere law and I don't think you can find a choice of ##\vec{B}## such that all of the Maxwell's equations (in free space) are satisfied for ##\vec{E}=\vec{F}##. I think you will have to define unbounded, unrealistic charge and current densities in order for Maxwell's equations to hold for this choice of ##\vec{E}##).

And yes helmholtz decomposition can be applied to time dependent vector fields as well. The time coordinate is something like a constant in helmholtz decomposition.
 
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  • #16
Delta2 said:
No, the scalar and vector potential of the Helmholtz decomposition are never unique for a given vector field(regardless of the boundary conditions for the field, or of the conditions for its curl and divergence if we are given such conditions instead of being given the vector field) cause you can always do the transformations outlined in post #11 and the vector field remains the same under those transformations of potentials. Take for a example the vector field (in a cartesian coordinate system)
$$\vec{F}=\nabla x+\nabla\times (y\hat x)$$
so we can immediately see that the scalar potential ##\phi=x## and vector potential ##\vec{C}=y\hat x## are its Helmholtz decomposition. Now try for ##\phi'=x+5## and ##\vec{C'}=y\hat x+\nabla x## and you ll see that ##\phi'## and ##\vec{C'}## also satisfy the Helmholtz decomposition of this field, that is it holds that $$\vec{F}=\nabla\phi'+\nabla\times\vec{C'}$$.We can have conservative vector fields that can be time dependent (take for example the field ##\vec{F}=2x\sin(2\pi t)\hat x=\nabla x^2\sin(2\pi t)##, but they will not correspond to an electric or magnetic field, because for an electric or magnetic field the Maxwell's equations hold and from Maxwell's equations we can conclude that if the electric or magnetic field are time dependent then they are not conservative. The vector field ##\vec{F}## as it is given above does NOT satisfy Maxwell's equations (try it with ##\vec{E}=\vec{F},\vec{B}=\vec{0}## you ll find it out that in free space it does not satisfy Gauss's law or Maxwell- Ampere law and I don't think you can find a choice of ##\vec{B}## such that all of the Maxwell's equations (in free space) are satisfied for ##\vec{E}=\vec{F}##. I think you will have to define unbounded, unrealistic charge and current densities in order for Maxwell's equations to hold for this choice of ##\vec{E}##).

And yes helmholtz decomposition can be applied to time dependent vector fields as well. The time coordinate is something like a constant in helmholtz decomposition.

Thank you. So your point is that admissible electric and magnetic fields that both satisfy Maxwell's equations and are also time-dependent cannot conservative. But, as you mention, it is possible for a time-dependent field to be conservative.

In one of my calculus books, I read that if a force vector field is conservative conservation of energy must hold and that is only true for conservative, position-dependent only fields. Conservative means that energy is conserved and that line integrals only depend on the initial and final positions (not on the path). I can see that if the vector changes with time it is possible for the line integral between point ##A## and ##B# to possibly vary along the same path or along different paths.

The static electric field is conservative. The dynamic electric field, the dynamic magnetic field, the static magnetic field are not conservative.
 
  • #17
fog37 said:
Thank you. So your point is that admissible electric and magnetic fields that both satisfy Maxwell's equations and are also time-dependent cannot conservative. But, as you mention, it is possible for a time-dependent field to be conservative.
That is absolutely correct. The vector field ##\vec{F}## I give above is time dependent and conservative
Its curl is zero at any time instant t and the line integral around a closed curve is zero (or its path independent) at any time instant t. But the field F as it is given , probably it cannot correspond to any field of the physical reality, because fields of the physical reality obey some additional equations (e.g Maxwell's equations) that make the field non conservative (its curl becomes non zero) if it is time-dependent.

In one of my calculus books, I read that if a force vector field is conservative conservation of energy must hold and that is only true for conservative, position-dependent only fields. Conservative means that energy is conserved and that line integrals only depend on the initial and final positions (not on the path). I can see that if the vector changes with time it is possible for the line integral between point ##A## and ##B# to possibly vary along the same path or along different paths.
I am not sure how conservation of energy comes into play in order for physical vector fields to be conservative only if they are time independent.
The static electric field is conservative. The dynamic electric field, the dynamic magnetic field, the static magnetic field are not conservative.
Correct with one little correction, I believe the static magnetic field from natural magnets is conservative.
 
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  • #18
Thank you again.

  • You are right, a static magnetic field generated by electric currents (i.e. by an electromagnet) and a time-dependent B field are NOT conservative (only a static B field generated by permanent magnet is conservative).

  • As far as the conservation of energy and time-dependent fields, a generic force vector field ##F(t)## cannot be conservative (i.e. cannot keep mechanical energy constant) because it would be the negative gradient of a time-dependent scalar potential ##U(t)##: $$ F(t) = - \nabla U(t)$$ and the derivation below shows how the total mechanical energy ##ME = KE + U## would stay constant only if ##U##, hence the vector field ##F##, is not time-dependent.

1590009848386.png
 
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  • #19
Thanks, nice derivation which I was eluding. It shows (in my opinion) that the notion of conservative in mathematics(defined as a vector field being conservative when it is equal to the gradient of a scalar field or equivalently the curl of the vector field being zero) does not necessarily coincides with the notion of conservative in physics (defined as a force vector field that conserves mechanical energy) . The force field ##\vec{F}=-\nabla U(x,t)## in this derivation has curl zero for any time instant t however it does not necessarily conserves mechanical energy.
 
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  • #20
Good points!
Delta2 said:
No, the scalar and vector potential of the Helmholtz decomposition are never unique for a given vector field(regardless of the boundary conditions for the field, or of the conditions for its curl and divergence if we are given such conditions instead of being given the vector field) cause you can always do the transformations outlined in post #11 and the vector field remains the same under those transformations of potentials. Take for a example the vector field (in a cartesian coordinate system)
$$\vec{F}=\nabla x+\nabla\times (y\hat x)$$
so we can immediately see that the scalar potential ##\phi=x## and vector potential ##\vec{C}=y\hat x## are its Helmholtz decomposition. Now try for ##\phi'=x+5## and ##\vec{C'}=y\hat x+\nabla x## and you ll see that ##\phi'## and ##\vec{C'}## also satisfy the Helmholtz decomposition of this field, that is it holds that $$\vec{F}=\nabla\phi'+\nabla\times\vec{C'}$$.We can have conservative vector fields that can be time dependent (take for example the field ##\vec{F}=2x\sin(2\pi t)\hat x=\nabla x^2\sin(2\pi t)##, but they will not correspond to an electric or magnetic field, because for an electric or magnetic field the Maxwell's equations hold and from Maxwell's equations we can conclude that if the electric or magnetic field are time dependent then they are not conservative. The vector field ##\vec{F}## as it is given above does NOT satisfy Maxwell's equations (try it with ##\vec{E}=\vec{F},\vec{B}=\vec{0}## you ll find it out that in free space it does not satisfy Gauss's law or Maxwell- Ampere law and I don't think you can find a choice of ##\vec{B}## such that all of the Maxwell's equations (in free space) are satisfied for ##\vec{E}=\vec{F}##. I think you will have to define unbounded, unrealistic charge and current densities in order for Maxwell's equations to hold for this choice of ##\vec{E}##).

And yes helmholtz decomposition can be applied to time dependent vector fields as well. The time coordinate is something like a constant in helmholtz decomposition.

Delta2,

I am re-reading your post #15 and I am still not clear on your statement that a vector field ## \bf{F}## is UNIQUELY determined by its curl and divergence if that specific boundary condition is satisfied.

I think that:
  • Both the curl and divergence of the vector field ## \bf{F}## are "connected" (i.e. can be expressed) to the pair of potentials ##(\phi, \bf{A})##.
  • The vector field ##\bf{F}(r)## can be expressed using different, but valid, pairs ##(\phi, \bf{A})## which means that both ##\phi## and ## \bf{A}## are never a unique pair for the decomposition of vector ## \bf{F}##. Curl and divergence of ## \bf{F}##, being connected to ##(\phi, \bf{A})##, are therefore never unique either.
  • When we choose a particular and arbitrary pair ##(\phi, \bf{A})## we are essentially choosing a gauge (Coloumb, Hamilton, etc.), correct? For example, the Coulomb gauge is when we choose a vector potential ##\bf{A}##, among the many possible ones, whose divergence is zero, i.e. ##\nabla \cdot /bf{A}=0##, correct? What does the Coloumb gauge impose on the scalar potential ##\phi##?
Finally, what does "uniquely" exactly mean when you state that the vector field ##\bf{F}(r)## is UNIQUELY determined by its curl and divergence when that specific boundary condition you mention is satisfied? There is clearly one vector field ##\bf{F}(r)## but different pairs of potentials ##(\phi, \bf{A})## that always work. Same goes for different divergences ##\nabla \cdot \bf{F}## and curls ##\nabla \times \bf{F}##...

thanks!
 
  • #21
fog37 said:
Good points!Delta2,

I am re-reading your post #15 and I am still not clear on your statement that a vector field ## \bf{F}## is UNIQUELY determined by its curl and divergence if that specific boundary condition is satisfied.
It simply means that if we are looking for a vector field ##\bf{F}## whose we are given its curl, its divergence and also given that boundary condition, then this field is one and only (that satisfies the boundary condition and has this curl and this divergence). If we are not given that boundary condition then there might be many different fields (that have this curl and this divergence).
I think that:
  • Both the curl and divergence of the vector field ## \bf{F}## are "connected" (i.e. can be expressed) to the pair of potentials ##(\phi, \bf{A})##.
  • Yes that's correct it will be $$\nabla\cdot \bf{F}=\nabla\cdot(\nabla\phi)=\nabla^2\phi$$ and $$\nabla\times\bf{F}=\nabla\times(\nabla\times\bf{A})$$
  • The vector field ##\bf{F}(r)## can be expressed using different, but valid, pairs ##(\phi, \bf{A})## which means that both ##\phi## and ## \bf{A}## are never a unique pair for the decomposition of vector ## \bf{F}##. Curl and divergence of ## \bf{F}##, being connected to ##(\phi, \bf{A})##, are therefore never unique either.

Correct up to the point about the curl and divergence. The curl and divergence will be unique as long as the vector field ##\bf{F}## is specified. To see this use the transformations that give ##\phi'## and ##\bf{A'}## at post #15 and then calculate $$\nabla\cdot(\nabla\phi')$$ you ll find it to be equal to $$\nabla\cdot(\nabla\phi)=\nabla\cdot\bf{F}$$ and similar for the vector potential ##\bf{A'}## you ll find that $$\nabla\times\nabla\times \bf{A'}=\nabla\times\nabla\times \bf{A}=\nabla\times\bf{F}$$.
  • When we choose a particular and arbitrary pair ##(\phi, \bf{A})## we are essentially choosing a gauge (Coloumb, Hamilton, etc.), correct? For example, the Coulomb gauge is when we choose a vector potential ##\bf{A}##, among the many possible ones, whose divergence is zero, i.e. ##\nabla \cdot /bf{A}=0##, correct? What does the Coloumb gauge impose on the scalar potential ##\phi##?
When we are choosing a particular pair of potentials (the potentials as defined in electromagnetism and not as defined in Helmholtz decomposition) we are choosing a strong or complete or total gauge (not sure about the terminology used here). I know that Coulomb and Lorentz gauge are not complete but partial which means that there can be many pairs of potentials that satisfy the gauge. The paper shows that the potentials defined by Helmholtz decomposition satisfy the coulomb gauge. (BTW I have reasons to believe that the paper analysis is correct only in the electro static and quasi static case, in the full electrodynamic case the paper's conclusion at page 3 does not hold)
Finally, what does "uniquely" exactly mean when you state that the vector field ##\bf{F}(r)## is UNIQUELY determined by its curl and divergence when that specific boundary condition you mention is satisfied? There is clearly one vector field ##\bf{F}(r)## but different pairs of potentials ##(\phi, \bf{A})## that always work. Same goes for different divergences ##\nabla \cdot \bf{F}## and curls ##\nabla \times \bf{F}##...

thanks!
I think the above should be clear be now.
 
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Related to Vector field and Helmholtz Theorem

1. What is a vector field?

A vector field is a mathematical concept that assigns a vector to each point in a given space. This vector represents the direction and magnitude of a physical quantity, such as velocity or force, at that point.

2. How is a vector field represented?

A vector field is represented graphically using arrows, with the direction and length of the arrow indicating the direction and magnitude of the vector at a particular point. Alternatively, it can be represented algebraically using mathematical equations.

3. What is the significance of Helmholtz Theorem in vector fields?

Helmholtz Theorem states that any well-behaved vector field can be decomposed into two components: a solenoidal component (divergence-free) and an irrotational component (curl-free). This allows for a simpler analysis and understanding of complex vector fields.

4. What are some real-world applications of vector fields?

Vector fields have many applications in physics, engineering, and other sciences. Some examples include fluid dynamics, electromagnetism, and weather forecasting. They are also used in computer graphics to create realistic visual effects.

5. How is Helmholtz Theorem used in practical applications?

Helmholtz Theorem is used in practical applications to simplify and solve complex vector field problems. It allows for the separation of different physical phenomena, making it easier to analyze and understand the behavior of a vector field. It is also used in the development of numerical methods for solving vector field equations.

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