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

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The discussion centers on the vector identity grad(div(V)) being dropped in the derivation of electromagnetic wave equations. Participants clarify that this is valid because, in the absence of charges or sources, the divergence of the electric field (E) and magnetic field (H) is zero, as stated in Maxwell's equations. This aligns with Gauss's law, which confirms that divergence without sources equals zero in a vacuum. The conversation highlights the importance of understanding the underlying principles of electromagnetism to justify the simplification. The conclusion emphasizes the formal reasoning behind dropping the grad(div(V)) term in these contexts.
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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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