Grad(div(V)) = 0: Why is this Vector Identity Dropped?

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Discussion Overview

The discussion revolves around the vector identity grad(div(V)) and its application in the context of electromagnetic wave equations. Participants explore why this term is often dropped in derivations, particularly in relation to the absence of sources such as charges.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions the reasoning behind dropping the grad(div(V)) term, suggesting it may relate to the absence of sources, leading to a divergence of zero.
  • Another participant proposes that the identity is specifically used for the electric field (E) and magnetic field (H), referencing Maxwell's equations which indicate that the divergence of both fields is zero in the absence of charges.
  • A third participant agrees with the reasoning provided, affirming that the divergence is indeed zero due to Gauss's law in vacuum.
  • A later reply acknowledges the formalism behind the divergence being zero without charges, expressing a realization of the connection to Maxwell's equations.

Areas of Agreement / Disagreement

Participants generally agree on the reasoning that the divergence is zero in the absence of sources, as supported by Maxwell's equations. However, the discussion does not reach a consensus on the broader implications or formal proofs regarding the dropping of the term.

Contextual Notes

Some assumptions regarding the conditions under which the divergence is considered zero may not be fully explored, and the discussion relies on interpretations of Maxwell's equations without delving into formal proofs.

DoobleD
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This is closely related to this thread I posted yesterday, but the question is different so I created another thread. There is a vector identity often used when deriving EM waves equation :

d0e4740eaf9a820b14f267ae70cf9bca.png


Then the grad(div(V)) part of it is simply dropped, assuming it equals 0. And I wonder why.

Is it because, since there is no "sources" here (no charges), any divergence is 0 ? Can this be proven more formally ?
 
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Isn't it because the identity is used for ##V=E## and ##V=H##, and according to Maxwell's equations (see your Wikipedia link):
##\nabla.{E}=0##
##\nabla.{H}=0##
?
 
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Yes, that's the reason.
 
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Samy_A said:
Isn't it because the identity is used for ##V=E## and ##V=H##, and according to Maxwell's equations (see your Wikipedia link):
##\nabla.{E}=0##
##\nabla.{H}=0##
?

Oh ! Of course ! Thank you. Well, I formulated that divergence without charges/sources is 0, that is indeed Gauss's law from Maxwell's in vacuum...There is the obvious formalism I was looking for, I should have seen it. -_-'
 

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