# On the components in Helmholtz Decomposition Theorem

There are two kinds of vector decompositions. One is a normal vector decomposition, and the second appears in Helmholtz Decomposition.

**1. The normal decomposition of vector**

Talking of the decomposition, you generally think about projected components of a given vector on the reference axes or rectangular coordinates.

And the vector projected on the reference axes is called the components of a given vector.

Fig.1 a normal decomposition and components

But you can also think about a decomposition into non orthogonal directions.

If a given vector is equal to the vector sum of other two vectors, these two vectors are called as components of a given vector.

Fig.2 a decomposition into non-right angle components

The basic concept of vector is that two vectors which are not perpendicular to each other have the components of each other. The dimension of it is given by the length from the foot of perpendicular to the point of intersection of two vectors.

Fig.3 a vector component of a vector on another vector direction

When the directions of two vectors are perpendicular to each other, the foot of perpendicular to other axis is on the foot of other vectors. So, in these situations, you can say that these two vectors are independent of each other.

Fig.4 the component on orthogonal axis

In other words, when two vectors are perpendicular to each other, one does not influence the other.

The normal vector decomposition is thinkable at a point.

**2. The decomposition in Helmholtz Decomposition theorem**.

Generally you can’t think about the components of vectors which can’t be decomposed.

But the components that come up by the decomposition of Helmholtz Decomposition are not simply the projections of the flow vectors. They should be called “Roles”. They are a curl-free divergence component and a divergence-free curl component. Those components are given as “Roles” of a flow at a point in the set of adjacent flows.

These components are not thinkable at just a point. It is given in a region.

**3. It is not needed to get a curl distribution and divergence**

Helmholtz Decomposition is not needed to get distributions of curl and divergence.

You can calculate “a distribution of curl” and “a distribution of divergence” by calculating F and F respectively.

It is not necessary to use Helmholtz Decomposition theorem to get them. Therefor to be able to uniquely obtain these distributions can not be the proof of Helmholtz Decomposition.

You just be able to calculate these distributions, so you should not think that any flow is decomposed into two flows.

**4. The component which plays the both roles of curl and divergence**

Because a decomposition in Helmholtz Decomposition can be thought as roles of any flow at a point in the neighborhood, you should think about a component which plays both roles of curl and divergence.

If you don’t think of it, you preliminary use the Helmholtz Decomposition theorem without the proof.

If you would prove that there is no component which plays the both role of curl and divergence in any flows, you can say that Helmholtz Decomposition is correct.

At first you should think about a component which plays the both roles of curl and divergence.

Fig.5 If any flow has the component which play the both roles

In Fig.5, **F** is any flow which has both of curl and divergence. And **F**l shows a divergence component as a curl-free component, and **F**t shows a curl component as a divergence-free component, and **F**b shows a component which plays the both roles of curl and divergence.

These **F**l and **F**t are borroeing character from the article of ”Longitudinal and transverse fields” in Wikipedia.

The substitute character “l” means as shown in Fig.6,

Fig.6 the longitudinal component

and “t” means as shown inFig.7,

Fig.7 the transversal component

I might have mistook “**k**” what he/she wrote, but I beleave that “k” should be the gradient directin of potential.

These illustrations are made by myself, so if there is any fault in these illustrations, it is my fault.

As we know, they can calculate “a distribution of curl” and “a distribution of divergence” by calculate**F** and**F** .

Here, if they had mistook a flow had been decomposed into a curl-free flow and a divergence-free flow, they could calculate a velocity potential from the distribution of divergence, and a stream function from distribution of curl.

But actually they just calculated the distributions of divergence by (**F**l+**F**b）, not by （**F**l’）, and the distributions of curl by **(****F**t+**F**b）, not by （**F**t’）.

Therefore the composition **G** of these two flows does not match the original flow **F**.

If there is no component which plays the both roles of curl and divergence , Helmholtz Decomposition theorem is correct as showing in Fig.8.

Fig.8 If any flow has not the component which play the both roles

In fact, there is an example in real world. I may say that an electromagnetic wave has an electric field as a scalar potential φ, and a magnetic field as a vector potential **A** as shown in Fig.9. But I have to say that I don’t know exactly the electromagnetic wave.

Fig.9 is drawn from Wiki.

Fig.9 The electromagnetic wave

In the article of Wikipedia, an electromagnetic wave has an electric field as a scalar potential and a magnetic field as a vector potential, and they are perpendicular to each other.

In this case, the force given in an electric field and the force in a magnetic field are perpendicular to each other.

In electromagnetic wave case, you can confirm to exist a set of a scalar potential and a vector potential. But in fluid case, we can not confirm to exist these potential.

**5. How to verify the Helmholtz Decomposition**

As I said before, if Helmholtz Decomposition is right, there is not a component which plays the both roles of curl and divergence.

Then

1) The composition of a curl component and a divergence component is equal to the original flow(vector)

2) The two of components, a curl component flow and a divergence flow are perpendicular to each other.

And a curl component flow is directed along the line of stream functions, and a divergence flow is directed toward perpendicular to iso-velocity potential lines.

Therefore at any point, the lines of stream function and the line of iso-velocity potential line are on a parallel with each other.(Please see Fig.6 and Fig.7)

Stream function is expressed by the dimension of “vector potentials” in the horizontal plane as vector potential. The direction of the stream function as a “vector potential” stands up perpendicular to the plane.(Please see Fig.7)

**6. Current status**

I will show you some current status of “the velocity potential” and “stream function”.

The natural wind in our atmosphere generally blow horizontally, because the air is generally in the condition of hydrostatic equilibrium. So we can think the natural wind as a flow in a horizontal plane.

NOAA(National Oceanic and Atmospheric Administration) and JMA(Japan Meteorological Agency) etc. beleave Helmholtz Decomposition is correct, and they publish their data of “the velocity potential” and “the stream function”

**6.1) Do the compositions of “the divergent flow” and “curl flow” match the original winds?**

JMA had been publishing their data on the net until October 2011.

By using those data, I will show you how “the velocity potentials” and “the stream functions” are going. The following examples are about on 20 Jun 2011.

Fig.10 shows the distributions of the velocity potentials and divergent winds on the water vapor imagery.

Fig.10 distribution of the velocity potentials and divergent wind

And Fig.11 shows the distribution of stream functions and curl winds.

Fig.11 distribution of stream functions and curl winds

If Helmholtz Decomposition theorem is correct, the vector sum of these divergent winds and curl winds should match the original winds.

Fig.12 shows the original winds(black) and the vector sums of two kinds of components(purple).

Fig.12 The comparison between analyzed winds(black) and composed winds

I first drew analyzed wind with black arrows, and after that I drew composed wind with purple arrows. So if the composed winds perfectly match the analyzed winds, all of arrows should be purple.

You may think that analyzed winds( original winds) match the vector sum of divergent winds and curl winds. But I can see the difference between these two kinds of winds.

Fig.13 shows the differences between original winds and the vector sum of divergent winds and curl winds.

Fig.13 the differences between analyzed winds and

vector sum of two components

There are some differences about 5m/sec in some area.

Here, what do you think with Fig.13. Permissible? or Impermissible?

I do not know how to make these potentials, but I want to applaud the efforts of them. Good jobs!

**6.2) Are “the divergent wind” and “the curl wind” perpendicular to each other?**

If “Helmholtz Decomposition” is correct, “the divergent wind” and “the curl wind” should be perpendicular to each other.

But these two kinds of winds are not perpendicular to each other as we can see in Fig.10 and Fig.11.

And if “Helmholtz Decomposition” is correct, two kinds of isolines of “the velocity potential” and “the stream function” should be parallel to each other as we have seen in Fig.6 and Fig.7.

But these two kinds of iso-lines are not parallel to each other as we can see in Fig.13.

<by NOAA data>

We can also see the same consequences in NOAA data. Fig.14 and Fig.15 show the velocity potential and stream function respectively. They were published in NOAA home page(http://www.cpc.ncep.noaa.gov/products/hurricane/).

Fig.14 The example of Velocity potential distribution

(30days mean from 16 May to 14 June in 2014)

Fig.15 The example of Stream function distribution

(30days mean from 16 May to 14 June in 2014)

Fig.14 shows 30days means of 200hPa Velocity potential and divergent wind from 16 May to 14 June in 2014(top of them).

Fig.15 shows 30days mean of 200hPa stream function as the same term as Fig.14. The bottoms of them show anomalies of velocity potential, but now it is out of our argument.

To see how to go on the two potentials, I draw stream function on velocity potential in Fig.16. In Fig.16, blue lines show stream functions, and red lines show velocity potential. And red arrows show divergent wind.

Fig.16 is made by Fig.14 and Fig.15.

Fig.16 velocity potential on stream function at 200hPa as same term as Fig.14,15

The curl wind(=divergence-free wind) blow parallel to blue line(stream function). So we can see that divergent wind and curl wind are not orthogonal to each other.

**As far as has been hitherto seen, we can say Helmholtz Decomposition is not correct**.

**I think Helmholtz Decomposition has been preventing the progress of Meteorology**.