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Square of Gradient!

  1. Sep 26, 2012 #1
    How do you solve ((grad(f(x,y,z))))^2?
     
  2. jcsd
  3. Sep 26, 2012 #2
    Or what I meant was how do you solve (∇σ)^2? Where σ is a scalar function.
     
  4. Sep 26, 2012 #3
    And this should be equal to what, some other scalar function?
     
  5. Sep 26, 2012 #4
    I was hoping you could tell me.:smile:
     
  6. Sep 26, 2012 #5
    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?
     
  7. Sep 26, 2012 #6
    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?
     
  8. Sep 26, 2012 #7
    But yes, when we look at it from the view of the electric field, the electric field squared should equal a scalar.
     
  9. Sep 26, 2012 #8
    What do you even mean by "solve"? Your problem is not clearly stated...

    BiP
     
  10. Sep 26, 2012 #9
    -∇[itex]\varphi[/itex] = E I want to square both sides.
     
  11. Sep 26, 2012 #10
    [tex](-\nabla \varphi) \cdot (-\nabla \varphi) = E \cdot E[/tex]

    Okay...is that all you wanted?
     
  12. Sep 26, 2012 #11
    How do you simplify (-∇[itex]\varphi[/itex])[itex]\bullet[/itex](-∇[itex]\varphi[/itex])?
     
  13. Sep 26, 2012 #12
    Is it possible to simplify further at all?
     
  14. Sep 26, 2012 #13
    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.
     
  15. Sep 26, 2012 #14
    I know exactly what you mean:smile: Thank you very much Muphrid! I am trying to check if my initial steps to "simplifying" were correct. Thank you very much:smile:
     
  16. Sep 27, 2012 #15

    HallsofIvy

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    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.
     
  17. Sep 28, 2012 #16
    Thank you very much. I will be more specific next time.
     
  18. Sep 29, 2012 #17
    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?
     
  19. Sep 29, 2012 #18
    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?
     
    Last edited: Sep 29, 2012
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