Proof of equivalence between nabla form and integral form of Divergence

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Discussion Overview

The discussion revolves around the equivalence between the nabla form and the integral form of the divergence theorem. Participants explore the relationship between these two formulations, seeking to understand how one can be derived from the other and the underlying principles that support this equivalence.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants suggest that the equivalence is a direct consequence of Gauss' divergence theorem, using a solid ball centered at a point in 3D space to illustrate the concept.
  • One participant expresses frustration over the lack of a formal proof for the equivalence, noting that while the divergence theorem seems logical, they seek a rigorous demonstration that does not rely on circular reasoning.
  • Another participant acknowledges the beauty of the explanation provided but critiques it as a "sloppy proof," indicating a desire for a more polished and rigorous approach.
  • There is a mention of the historical context of the development of differential equations versus their integral counterparts, suggesting that this might influence understanding and proof approaches.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of a formal proof for the equivalence. While some agree on the validity of the divergence theorem as a basis for understanding the equivalence, others express the need for a more rigorous demonstration.

Contextual Notes

Participants highlight the challenge of avoiding circular logic in proving the equivalence, indicating that the discussion is limited by the need for a clear, non-circular proof structure.

The_Logos
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Does anybody knows how you can reach one form of the divergence formula from the other? Or in general, why is the equivalence
3d6d3c3e068d3ce679fa3391149e3a84.png
true?
 
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The_Logos said:
Does anybody knows how you can reach one form of the divergence formula from the other? Or in general, why is the equivalence
3d6d3c3e068d3ce679fa3391149e3a84.png
true?
The formula is a direct consequence of Gauss' divergence theorem. You may look at the things in the following manner.
Suppose p is a point in 3D Euclidean space. Let Er be a solid ball centered at p with radius r , and let Sr be the boundary surface of Er with outward pointing normal (and F is the vector field). Using Gauss' theorem we have:

_{r}}\cdot%20\!%20\underset{\!%20\!%20S_{r}}{\int%20\!%20\!%20\int%20}%20\vec{F}\cdot%20d\vec{S}.gif

This aproximation improves as radius gets smaller, and in the limit (for r → 0 ⇒ Vr → 0) the equality holds.
Hope this helps.
 
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mathman said:

Wikipedia was one of my first reference (as always) but the equivalence is merely stated, but not demonstrated.

zoki85 said:
The formula is a direct consequence of Gauss' divergence theorem. You may look at the things in the following manner.
Suppose p is a point in 3D Euclidean space. Let Er be a solid ball centered at p with radius r , and let Sr be the boundary surface of Er with outward pointing normal (and F is the vector field). Using Gauss' theorem we have:

_{r}}\cdot%20\!%20\underset{\!%20\!%20S_{r}}{\int%20\!%20\!%20\int%20}%20\vec{F}\cdot%20d\vec{S}.gif

This aproximation improves as radius gets smaller, and in the limit (for r → 0 ⇒ Vr → 0) the equality holds.
Hope this helps.

That is just beautiful! you should just go ahead and take my user name, you are definitely more the logos than me! I wasted all day trying to make linear transformations thinking that it was my only hope (And managed nothing).
I actually was seeking for the proof of the equivalence before giving deep thought to the divergence theorem, just to find out that ironically I needed it to prove what apparently came before!.
I'm starting to think that the main issue here is that this kind of differential equations where invented much later than their integral equivalent, maybe I should start taking into consideration the historical order of invention?

Anyways, thanks a lot zoki85!

However, I must say that this leads me to another question. While the equivalence of the divergence theorem seem logical on itself (summing all the "lines" that flow trough the surface is going to give us the total flux variation on the surface, and at the same time summing the variations at all points of the volume will give us the total variation too. ) and therefore more acceptable (with "verbal" logic on the previous equivalence I got to very similar "concepts" but didn't quite seemed the same on my mind, that's why I reached out for a more formal proof) the same question inevitably rises, ¿Is there a formal proof of this equivalence? (without of course, falling on a circular logic with the previous equivalence, which would led us to establish one of the two equivalence as a "axiom", which I want to avoid).
 
Last edited:
The_Logos said:
That is just beautiful! you should just go ahead and take my user name, you are definitely more the logos than me! I wasted all day trying to make linear transformations thinking that it was my only hope (And managed nothing).
I actually was seeking for the proof of the equivalence before giving deep thought to the divergence theorem, just to find out that ironically I needed it to prove what apparently came before!.
Meh, my "logos" is quite irrational in everyday life, so I don't think that user name would suit me fine. And, althought short, this is kind of a sloppy proof. It's more a demonstration by Gauss' D.T. why it holds true. I guess a pro-mathematcian should feel a need to "polish" it, to write more rigouros proof. I agree there must be other ways to prove the formula (without G.D.T.) but I didn't try it.

Cheers
 

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