Why is grad(div(V)) dropped in EM wave derivation?

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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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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. -_-'