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Gradient of a scalar field

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the scalar field

    [tex] V = r^n , n ≠ 0[/tex]

    expressed in spherical coordinates. Find it's gradient [itex]\nabla V[/itex] in
    a.) cartesian coordinates
    b.) spherical coordinates

    2. Relevant equations

    cartesian version:
    [tex]\nabla V = \frac{\partial V}{\partial x}\hat{x} + \frac{\partial V}{\partial y}\hat{y} + \frac{\partial V}{\partial z}\hat{z} [/tex]

    spherical version:
    [tex] \nabla V = \frac{\partial V}{\partial r}\hat{r} + \frac{1}{r}*\frac{\partial V}{\partial \phi}\hat{\phi} + \frac{1}{r*sin(\phi)}*\frac{\partial V}{\partial \theta}\hat{\theta} [/tex]

    conversion:
    [tex] r = (x^2+y^2+z^2)^\frac{1}{2} [/tex]

    3. The attempt at a solution

    a.) using the third equation...

    [tex] V = r^n = (x^2+y^2+z^2)^\frac{n}{2} [/tex]

    using the first equation and skipping some steps involving the chain rule...
    [tex] \nabla V = \frac{n(x\hat{x}+y\hat{y}+z\hat{z})}{(x^2+y^2+z^2)^\frac{n}{2}} [/tex]

    b.)Using the second equation
    [tex] \nabla V = nr^m \hat{r} [/tex]
    [tex]m = n-1[/tex]


    Those are my two solutions to this problem. Are these right? Are they wrong? If so where did I go wrong?

    Thanks!
    hover
     
  2. jcsd
  3. Jan 31, 2012 #2

    Dick

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    How did the power n/2 wind up in the denominator?? You should be able to check that the solutions are the same pretty easily. What's [itex]\hat r[/itex] in cartesian coordinates?
     
  4. Jan 31, 2012 #3
    Small mistake there... the denominator should be (n-2)/2 from the chain rule. I think that [itex]\hat r[/itex] in Cartesian coordinates is

    [itex]\hat r[/itex] = sin([itex]\phi[/itex])[itex][/itex]*cos(θ)[itex]\hat{x}[/itex]+sin([itex]\phi[/itex])*sin(θ)[itex]\hat{y}[/itex]+cos([itex]\phi[/itex])[itex]\hat{z}[/itex]

    Is that right or am I completely off the mark?
     
  5. Jan 31, 2012 #4

    Dick

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    The power of r should be (n-2)/2=n/2-1 but it should be in the numerator. [itex]\hat r=\frac{x \hat x + y \hat y + z \hat z}{r}[/itex] is probably a more useful expression.
     
  6. Jan 31, 2012 #5
    I plugged [itex]\hat r *r={x \hat x + y \hat y + z \hat z}[/itex] into my first equation(with r on top and to the power of n/2-1) and I got the value from part b so I guess that these answers are correct. I just have one question. Where exactly did your version of [itex]\hat r=\frac{x \hat x + y \hat y + z \hat z}{r}[/itex] come from? I know this is probably simple but I want to understand where you got that equation from or how you derived it.
     
  7. Jan 31, 2012 #6

    Dick

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    [itex]\hat r[/itex] is a unit vector that points away from the origin. I took the point [itex] x \hat x + y \hat y + z \hat z[/itex] and subtracted [itex] 0 \hat x + 0 \hat y + 0 \hat z[/itex] to get a vector pointing away from the origin and divided by its length, [itex]r=\sqrt(x^2+y^2+z^2)[/itex].
     
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