# Helmholtz Decomposition is wrong

There is a theorem called Helmholtz Decomposition that is believed among meteorologists and hydrodynamicists.

That theorem says that any flow can be decomposed into a curl-free flow and a divergence-free flow.

That is,**F**=+

Here, **F** shows any flow, shows differential operator, shows velocity potential, **A** shows vector potential.

So, the first term of right hand is considered as a curl-free flow, and the second term is considered as a divergence-free flow. But,I have found some problems on this theorem. This theorem has been established on many mistakes.

**mistake 1**

We can calculate the distribution of () and () from any flow **F**.

It is not needed to divide this flow to calculate those distribution.

If we have not proved that Helmholtz Decomposition is right yet, we should assume that any flow has curl component, divergence component and the component which play both role of curl and divergence as shown in right illustration. You can not brush aside the last component, because that is to use preliminarily Helmhltz Decomposition theorem.

They have mistook these distributions of and to be decomposed. And then, they think to be able to calculate **uniquely** and .

Here, I want to define **G**=+.

As you can see in upper illustration, aparently

For example, someone has explained about Helmholtz Decomposition in Wikipedia as follows,

He preliminarily beleave Helmholtz Decomposition is right as follow

If you can prove that is perpendicular to , the original flue has not such component which play both role of curl and divergence . Therefor Helmholtz Decomposition is right.

But, we sometime see actual “divergence wind” is not perpendicular to “the wind from steam function”. So, we should consider any flow has the component which play both roles of curl and divergence.

Even if there is a component which paly both roles, he can calculate the distributions of the fields of and .

And, if he thinks those fields are separated, he may calculate and

He (or She) has confused G with F as shown in next illustration.

**mistake 2**

There are two of basically mathematical theorems on any flow(vector function).

One of them says that if a flow has culr(voticity) component, there is no velocity potential , and if there are velocity potential in a flow, the flow has not curl. That is, not to have curl component is the necessary and sufficient condition for existing of velocity potential .

And other says that if a flow has divergent component in the flow, there is no vector potential **A**, and if there are vector potentials **A** in the flow, the flow has no divergence. That is, not to have divergence in the flow is the necessary and sufficient condition for exiting of vector potential **A**.

So, I can definitively say that there is no and **A** in the flow which has curl and divergence.

But, almost of authorities of Meteorology and Hydrodynamics don’t think so. They think that any flow can be divided into two kinds of flow. They are a curl-free flow and a divergence-free flow.

I can show you some good example.

I would like to liken all kinds of flows to container boxes.

Then, there are just two kinds of shape of container, one of them is spherical shape, and other is cubic. A cubic shape container means a curl-free flow, and a spherical shape container means a flow including curl.

Then, I check all of containers(flow), and if it’s a cubic container(curl-free), I put cotton( ) into it. And if it’s a spherical container(including curl), I confirm that there is no cotton( ) in it.

And then, all container is painted with only two colors, red or black. Red one means a divergence-free flow, and black one means a flow including divergence.

Then, I check all of containers(flow), and if it’s a red container(divergence-free), I put star ornaments(**A**) into it. And if it’s a black container(including divergence), I confirm that there is no star ornament(**A**) in it.

Aren’t you sure that there is no cotton nor star ornament in any spherical black container.

I can’t believe that those clever persons think that there are cotton and star ornament in a spherical black container.

Why do they think so? I think they confuse G with F in < mistake 1 >

The spherical black container is **F**, not **G**. There is not nor **A** in **F**.

**mistake 3**

If you want to prove that any vector **F** can be devided into a irrotational vector **V**e and a solenoidal vector **V**r, you need to find some identity which can be described by like **F**=**V**e + **V**r.

As far as I know, the only equation is the vector triple product identity.

For example, “The vector triple product identity” is posted in the next homepage.

(reference:http://en.wikipedia.org/wiki/Triple_product#Proof)

If you replace **U** and **V** with ▽, you may get

So,you can get

In above equation, the first term of the right hand shows a curl-free flow, and the second one shows a divergence-free flow.

So, you might say that Helmholtz decomposition is perfectly proved.

But, the third mistake is that they take as any flow.

Any flow **F** certainly exist. But **W** is not guaranteed to exist. You need to make sure that there exists **W** for any flow **F**.

And if **W** exists, at the same time, and are decided **uniquely** with **W**, and they need to appear together.

I can show you that there are many flows which do not include **W**.

You have got following expression from the vector triple product identity.

**F**=-+

It is given on terms and conditions as required by

That is, and **A** are deriven from the common function **W**.

If is decided from **W**, then **A** should be decided uniquely at the same instance.

And, the flow given by is supposed to exist independently from other flows.

It is called solenoidal flow. That means a flow like in the tube. According to Helmholtz Decomposition there exists such a flow.

I do not think that such a solenoidal flow exists in the real world.

If there were such solenoidal flows, I would be able to show the collapses of Helmholtz decomposition.

Assuming that Helmholtz decomposition theorem is correct. you can consider two flows as following.

_{1} and **A**_{1} are functions which are derived from **W**_{1}. and, _{2}, and **A**_{2} are derived from **W**_{2}.

Here, because an arbitrary flow (vector function) must be possible, you can consider the flow **F**_{3} which includes divergent component of (-_{1}) and rotational component of (_{2}).

I must say that again **F**_{3} should have (_{1}) as divergent component, and have (_{2}) as rorational component.

But according to Helmholtz decomposition, **F**_{3} can be decomposed into two flows only by the vector triple product identity.

Then,

Therfor, **F**_{3} has (_{3}) as divergent component, and has (_{3}) as rorational component.

So,

Here, you must say that **F**_{3} which has combined with divergent component of (_{1}) and rotational component of (_{2}) can not decomposed into divergent component of (_{1}) and rotational component of (_{2}).

Or we may have to say that there is no **W** functions in **F**_{3}.

In real world, “The vector triple product identity” just means that there exit many electromagnetic waves such as X-rays and radio waves. They have their own electric fields(corresponding to velocity potential χ) and their own magnetic fields（corresponding to vector potential **A**）.

That does not mean there exist any electromagnetic wave which has an electric field radiated from “CNN” antenna and a magnetic field radiated from “ABC” antenna.

As I showed above, any vector function in Helmholtz Decomposition can not be able to have arbitrary potential velocity and arbitrary vector potential **A**. Their partners are definitive with same **W**.

That is, “the function **F** in Helmholtz Decomposition” is not “any function”, but the very special function with their own and their own **A**.

Simply, “The vector triple product” does not give proof of Helmholtz Decomposition.

Helmholtz Decomposition theorem is believed even in electromagnetics, but they must not need this theorem. They should be enough to have Maxwell’s equations.

They does not needed to divide any vector function.

I think many articles on Helmholtz Decomposition are written by authority of electromagnetics. They believe the two components are perpendicular to each other, and theoretically they should be so.

This theorem is for fluid dynamics. The authorities of Meteorologist roughly aplly this theorem to the real winds in some plane, and make “the potential velocity” and “the stream function”. If the two components are perpendicular to each other, isolines of these two function should be parallel to each other. But as we can see in NASA home page, they are not parallel.

postscript

Electromagnetics is out of my hands.

But, I may say that the vector triple product identity is applicable just in electromagnetics.

Because, there exists as electric field, and **A** as magnetic field in real world. There is not such phisical potential in hydrodynamics.

In Electromagnetics, there is dynamics in the electric field and in the magnetic field. But there is not dynamics in velocity potential and in vector potential.

If you use ageostrophic winds as substitute for the winds from , there is dynamics, and you can calculate the divergence of the winds.

If you agree with me, please mail me.

( e－mail : taka19440606@yahoo.co.jp )

Dear Takaaki Mitsufuji!

Thanks for interesting work.

The Helmholtz decomposition also known as the Fundamental theorem of vector calculus is more 100 years old. This theorem is used by thousand physicists, mechanics and mathematicians, covered in hundreds textbooks, and taught to many thousands students all over the world. Mathematicians proved an Uniqueness of Helmholtz decomposition. The Helmholtz decomposition used for transformations of the Navier–Stokes and Lame (also called Navier) equations. These transformations included in encyclopedias and textbooks of mathematical physics as counterexamples (for instance Pressure-free velocity formulation ). “This theorem is of great importance in electrostatics, since Maxwell’s equations for the electric and magnetic fields in the static case are of exactly this type” http://en.wikipedia.org/wiki/Helmholtz_decomposition#Fields_with_prescribed_divergence_and_curl Therefore it is very difficult to believe that this theorem is wrong. For the majority of physicists and mathematicians I and you are mistaking odd fellows. Read, for example http://dxdy.ru/topic54334.html .

However this well-known theorem has appeared wrong. We should convince to physicists and mathematicianson simple examples. Therefore I suggest to submit your comment here: http://en.wikipedia.org/wiki/Talk%3AHelmholtz_decomposition#Helmholtz_decomposition_is_wrong

This problem, very unpleasant for modern physics and mathematics, were discussed here http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics/Archive/2012/Mar#Helmholtz_decomposition_is_wrong

Read more here:

http://continuum-paradoxes.narod.ru/Hel … tionCP.pdf

If you are still interested I think I could explain why the theorem is correct and why there is a misconceptions in the theorem’s interpretation.

1) When you get components in the area where both components are nonzero, you cancel the other component so when you sums again the fields in the “mixed area” each component add only her contribution. The components are always orthogonal even if are both nonzero in the same area.

2) I don’t understand but the theorems you talk about seems specific for hydrodynamic flow, not general math.

3) theorem only apply to two times derivable fields, so you are right that not all flow are decomposable but two times derivable yes and two times derivable are writable as two derivable function.

Hope I’ll be useful.

Sorry for my bad English… I’m Italian…

Thank you for giving some comments, Alessandro.

For example, in electromagnetic world, it is right. But irrotatinal fieald(electric field) and nondivergent fieald( magnetic field) in electromagnetic wave do not exist in the “mixed area”. they are orthogonal to each other.

In meteorology, they apply Helmholtz Decomposition to actual wind, and calculate “velocity potential” and “stream function”.

They seem as if they can get them without big problems. But that is big problem.

If Helmholtz Decomposition is correct, they can do divergence analysis by only the velocity potentials. This idea brings big mistakes in the “general (atmospheric) circulation”

You are right. In the real winds, especially about winds in some layer(as 2 dimensions phenomenon), they are not two times derivable. So, they could not apply Helmholtz Decomposition, but they bleave it is right, and they are publishing “velocity potentials” and “stream function”.