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Proof of equivalence between nabla form and integral form of Divergence

  1. Oct 14, 2014 #1
    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?
  2. jcsd
  3. Oct 15, 2014 #2


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  4. Oct 15, 2014 #3
    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:


    This aproximation improves as radius gets smaller, and in the limit (for r → 0 ⇒ Vr → 0) the equality holds.
    Hope this helps.
  5. Oct 15, 2014 #4
    Wikipedia was one of my first reference (as always) but the equivalence is merely stated, but not demonstrated.

    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: Oct 16, 2014
  6. Oct 16, 2014 #5
    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.

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