Or what I meant was how do you solve (∇σ)^2? Where σ is a scalar function.
#3
Muphrid
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And this should be equal to what, some other scalar function?
#4
quantumfoam
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I was hoping you could tell me.
#5
quantumfoam
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For example, the electric field is equal to -∇[itex]\varphi[/itex] where [itex]\varphi[/itex] is the electrostatic potential I believe. Now what I want to do is square the electric field. How can this be done on the gradient and the electrostatic potential?
#6
Muphrid
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I can't tell you anything. You have an expression. I assumed you wished to solve an equation. Right now you have an expression. You can't solve anything with that. What do you want to solve for? What is this expression supposed to be equal to?
#7
quantumfoam
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But yes, when we look at it from the view of the electric field, the electric field squared should equal a scalar.
#8
Bipolarity
775
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What do you even mean by "solve"? Your problem is not clearly stated...
BiP
#9
quantumfoam
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-∇[itex]\varphi[/itex] = E I want to square both sides.
#10
Muphrid
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[tex](-\nabla \varphi) \cdot (-\nabla \varphi) = E \cdot E[/tex]
Okay...is that all you wanted?
#11
quantumfoam
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How do you simplify (-∇[itex]\varphi[/itex])[itex]\bullet[/itex](-∇[itex]\varphi[/itex])?
#12
quantumfoam
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Is it possible to simplify further at all?
#13
Muphrid
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Short answer: you don't.
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.
#14
quantumfoam
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I know exactly what you mean Thank you very much Muphrid! I am trying to check if my initial steps to "simplifying" were correct. Thank you very much
There are two different vector products defined in three dimensions, the "scalar product" and the "cross product". So you can't simply say "square both sides" and expect people to know which product you mean.
#16
quantumfoam
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Thank you very much. I will be more specific next time.
#17
laserblue
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I would like to suggest perusing the literature of Dr. David Hestenes et al. regarding Geometric Algebra and Geometric Calculus. It soumds like you want to talk about the D' Alembertian and possibly Dirac's derivation of his equations as the inverse of 'squaring'. In what way would you interpret the operation of 'squaring' a function?
#18
quantumfoam
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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?