Grad operation

  • #1
[SOLVED] grad operation

Homework Statement



Calculate grad([tex]\Phi[/tex]) in terms of r and r, where [tex]\Phi[/tex]=[tex]\frac{1}{r^{3}}[/tex]

r= xi + yj + zk
r = magnitude (r)

The r cubed term is scalar by the way, it comes out looking bold for some reason.

The Attempt at a Solution



[tex]\Phi[/tex] = ( [tex]\sqrt{x^{2} + y^{2} + z^{2}}[/tex] [tex])^{-3}[/tex] = [tex](x^{2} + y^{2} + z^{2})[/tex][tex]^{\frac{-3}{2}}[/tex]

grad[tex]\Phi[/tex] = [tex]\frac{\partial \Phi}{\partial x}[/tex]i + [tex]\frac{\partial \Phi}{\partial y}[/tex]j + [tex]\frac{\partial \Phi}{\partial z}[/tex]k

[tex]\frac{\partial \Phi}{\partial x}[/tex] [tex]\Rightarrow[/tex] let [tex](x^{2} + y^{2} + z^{2})[/tex] = u

[tex]\frac{\partial \Phi}{\partial u}[/tex] = [tex]\frac{-3}{2}u^{\frac{-5}{2}}[/tex]

[tex]\frac{\partial u}{\partial x}[/tex] = 2x

[tex]\frac{\partial \Phi}{\partial x}[/tex] = 2x([tex]\frac{-3}{2}u^{\frac{-5}{2}}[/tex])

= -3x([tex](x^{2} + y^{2} + z^{2})[/tex][tex])^{\frac{-5}{2}}[/tex] = -3x( [tex]\frac{1}{r^{5}}[/tex] )

differentials with respect to y and z will be identical except swapping -3x for -3y and -3z.

These will be the terms for i j and k

Giving a final answer of

-[tex]\frac{3}{r^{5}}[/tex] (r).


EDIT (again the r^5 term is the magnitude not the vector)

Is this the correct solution? can anybody help?
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
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Looks fine to me.

Another way to go about this problem is to use the formula for grad in spherical polar coordinates (see http://www.mas.ncl.ac.uk/~nas13/mas2104/handout5.pdf [Broken] .pdf for example). In this case, it will reduce to [tex]\nabla\left(\frac{1}{r^3}\right)=\hat{\bold{r}}\frac{\partial}{\partial r}(r^{-3})=\hat{\bold{r}}\cdot\frac{-3}{r^4}[/tex] which is the same result.
 
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  • #3
4,254
2
[tex]r= \sqrt{x^{2} + y^{2} + z^{2}}[/tex]

you will need to rethink your unit vector r.
 
  • #4
Thank you very much

EDIT: what do you mean phrak? i used the equation you give for the magnitude of r, and it is not a unit vector in my question.
 
  • #5
cristo
Staff Emeritus
Science Advisor
8,122
74
[tex]r= \sqrt{x^{2} + y^{2} + z^{2}}[/tex]

you will need to rethink your unit vector r.

His r isn't a unit vector, it is [itex]\bold{r}=(x,y,z)[/itex]. The unit vector, which I denoted with a hat, is then [tex]\hat{\bold{r}}=\frac{\bold{r}}{r}[/tex]
 
  • #6
4,254
2
Thanks, Cristo. my bad. I didn't read far enough.
 

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