Helmholtz Theorem: What It Is & How It Works

[golfingboy07]

Hi everyone!

This question has me a little stumped.

 

[quasar987]

Let's work on this together. On my side, I "simplified" the 2 original integrals to

$$=\frac{1}{4\pi}\int_V \frac{\nabla _s \cdot \vec{F}(\vec{x}_s)}{R^2}\hat{R} \ d^3x_s - \frac{1}{4\pi}\int_V \frac{\nabla _s \times \vec{F}(\vec{x}_s)}{R^2} \times \hat{R} \ d^3x_s$$

Anyone else care to contribute?

 

[quasar987]

On the other hand, take just the first integral. Isn't it just 0?

$$\nabla_t \int_V \nabla_s \cdot \left( \frac{\vec{F}(\vec{x}_s)}{R} \right) \ d^3x_s = \nabla_t \int_{\partial V} \left( \frac{\vec{F}(\vec{x}_s)}{R} \right)\ \cdot \ \hat{n} \ d^2x_s \rightarrow \nabla_t 0=0$$

just by taking V sufficiently large.

?!

 

[pmb_phy]

See the attached file.

Pete

 

[quasar987]

Did you write this Pete?

There's something I don't get at all:

Equation (10c): " $\nabla$ and $\nabla '$ are related by $\nabla = -\nabla '$ "

How can they be related since the primed coordinates and the non-primed are not related?

 

[quasar987]

Also, in the last paragraph of page 3,

"Regarding the first integral on the right; as the radius of the surface increases as r then the area of the surface increases as r². However the integrand decreases as r³. Therefore as we let the radius go to infinity we see that the first integral vanishes."

The author seems to be making the assumption that F decreases as r², something that was not mentionned in the original statement of the theorem.

It's a detail important to mention imo.

 

[siddharth]

Did you write this Pete?

There's something I don't get at all:

Equation (10c): " $\nabla$ and $\nabla '$ are related by $\nabla = -\nabla '$ "

How can they be related since the primed coordinates and the non-primed are not related?

In this case, it's because the operator is being used on $$\frac{1}{|r - r'|}$$. So, the derivates of $r - r'$ wrt to primed coordinates, are the negative of the derivates wrt to unprimed.

I don't see how it how they are related like that in any other case, except when they operate on functions of $r - r'$.

"Regarding the first integral on the right; as the radius of the surface increases as r then the area of the surface increases as r². However the integrand decreases as r³. Therefore as we let the radius go to infinity we see that the first integral vanishes."

The author seems to be making the assumption that F decreases as r², something that was not mentionned in the original statement of the theorem.

It's a detail important to mention imo.

Yeah, I think that the argument should be the other way around. If that integral should converge, then F should decrease as 1/r^2 or faster.