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

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SUMMARY

The discussion centers on the vector identity grad(div(V)) = 0, specifically in the context of electromagnetic (EM) wave equations. Participants confirm that this identity is valid under the assumption of no sources or charges, leading to a divergence of zero. This conclusion is supported by Gauss's law from Maxwell's equations, which states that the divergence of electric field E and magnetic field H is zero in a vacuum. Thus, the grad(div(V)) term can be dropped in these scenarios.

PREREQUISITES
  • Understanding of vector calculus, specifically divergence and gradient operations.
  • Familiarity with Maxwell's equations, particularly Gauss's law.
  • Knowledge of electromagnetic wave theory.
  • Basic concepts of fields in physics, including electric and magnetic fields.
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  • Study the implications of Gauss's law in various physical scenarios.
  • Explore the derivation of electromagnetic wave equations from Maxwell's equations.
  • Investigate the role of sources and charges in electromagnetic theory.
  • Learn about vector identities and their applications in physics and engineering.
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Physicists, electrical engineers, and students studying electromagnetism or vector calculus will benefit from this discussion, particularly those interested in the mathematical foundations of electromagnetic wave propagation.

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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