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## Main Question or Discussion Point

How do you solve ((grad(f(x,y,z))))^2?

- Thread starter quantumfoam
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- #1

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How do you solve ((grad(f(x,y,z))))^2?

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Or what I meant was how do you solve (∇σ)^2? Where σ is a scalar function.

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And this should be equal to what, some other scalar function?

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I was hoping you could tell me.

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- #8

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What do you even mean by "solve"? Your problem is not clearly stated...

BiP

BiP

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-∇[itex]\varphi[/itex] = E I want to square both sides.

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[tex](-\nabla \varphi) \cdot (-\nabla \varphi) = E \cdot E[/tex]

Okay...is that all you wanted?

Okay...is that all you wanted?

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How do you simplify (-∇[itex]\varphi[/itex])[itex]\bullet[/itex](-∇[itex]\varphi[/itex])?

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Is it possible to simplify further at all?

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Long answer: sometimes it's useful to use the product rule on [itex]\nabla (\varphi \nabla \varphi) = [\nabla \varphi]^2 + \varphi \nabla^2 \varphi = - \nabla (\varphi E) = E^2 - \varphi \rho/\epsilon_0[/itex]. This is about the only somewhat useful identity I know of in this set of circumstances. I don't know if I'd call it "simpler", but the form [itex]E^2 = \varphi \rho/\epsilon_0 - \nabla (\varphi E)[/itex] can be useful in some integrals, particularly if there is no charge density.

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HallsofIvy

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Thank you very much. I will be more specific next time.

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- #18

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I'm not sure how to interpret the operation of 'squaring a function'. What I needed was a way to expand a certain equation. More specifically, the equation of the square of an electric field.For instance, if E was an electric field and [itex]\phi[/itex] was the electrostatic potential,then the following relationship is true: E=-∇[itex]\phi[/itex]. We know the total energy density of an electromagnetic field to be [itex]\zeta[/itex]=[itex]\iota[/itex]E^2. If E=-∇[itex]\phi[/itex], then what is the expanded form of E^2? From what I got, I understand from using the vector derivative that E^2=∇[itex]\cdot[/itex]([itex]\phi[/itex]∇[itex]\phi[/itex])-[itex]\phi[/itex](∇^2)[itex]\phi[/itex]? Is this true?

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