Question on Coulomb and Lorenz Gauge

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  • #1
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This is Problem 10.6 in page 422 of "Introduction to Electrodynamics" Edition 2 by David Griffiths. The question is:

Which of the potentials in the following is in the Coulomb gauge? Which is in Lorenz Gauge?( Notice that these gauges are not mutually exclusive.)

This is what the solution gave:
[tex]V)(\vec r, t)=0,\; and\; \vec A(\vec r,t)=-\hat r \frac{qt}{4\pi\epsilon_0r^2}[/tex]
[tex]\nabla\cdot \vec A=-\frac{qt}{4\pi\epsilon_0)} \nabla\left(\frac{\vec r}{r^2}\right)=-\frac{qt}{\epsilon_0)}\delta^3(\vec r)\Rightarrow\; \nabla\cdot\vec A\neq 0[/tex]



I disagree:
[tex]\nabla\cdot\vec A=\frac{1}{r^2}\frac{\partial (r^2A_r)}{\partial r}+\frac{1}{r\sin\theta}\frac{\partial (A_{\theta}\sin\theta)}{\partial \theta}+\frac{1}{r\sin\theta}\frac{\partial A_{\phi}}{\partial \phi}[/tex]
[tex]\Rightarrow\;\nabla\cdot\vec A=-\frac{1}{r^2}\frac{\partial \left(r^2\frac{qt}{4\pi\epsilon_0 r^2}\right)}{\partial r}= -\frac{1}{r^2}\frac{\partial \left(\frac{qt}{4\pi\epsilon_0}\right)}{\partial r}=0[/tex]

Please help
Thanks
 

Answers and Replies

  • #2
TSny
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[tex]\Rightarrow\;\nabla\cdot\vec A=-\frac{1}{r^2}\frac{\partial \left(r^2\frac{qt}{4\pi\epsilon_0 r^2}\right)}{\partial r}= -\frac{1}{r^2}\frac{\partial \left(\frac{qt}{4\pi\epsilon_0}\right)}{\partial r}=0[/tex]

This result holds for all points except at the origin. For ##r = 0## the expression ##-\frac{1}{r^2}\frac{\partial \left(\frac{qt}{4\pi\epsilon_0}\right)}{\partial r}=0## is not well defined. See section 1.5.1 on page 45 and equation (1.99) on page 50 of Griffiths.
 
  • #3
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This result holds for all points except at the origin. For ##r = 0## the expression ##-\frac{1}{r^2}\frac{\partial \left(\frac{qt}{4\pi\epsilon_0}\right)}{\partial r}=0## is not well defined. See section 1.5.1 on page 45 and equation (1.99) on page 50 of Griffiths.

thanks

So you mean it's not the differentiation part, it's the ##\frac {1}{r^2}## part that goes to infinity at r+0. But if the differentiation part is zero, everything time zero is zero!!!
 
  • #4
TSny
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So you mean it's not the differentiation part, it's the ##\frac {1}{r^2}## part that goes to infinity at r+0. But if the differentiation part is zero, everything time zero is zero!!!

Any finite quantity times zero is zero. But 1/r2 is not a finite quantity at the origin. So, you sort of end up with ∞##\cdot##0 at the origin. What if I say the answer should be ∞ because "infinity times anything is infinity"?
 
  • #5
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Any finite quantity times zero is zero. But 1/r2 is not a finite quantity at the origin. So, you sort of end up with ∞##\cdot##0 at the origin. What if I say the answer should be ∞ because "infinity times anything is infinity"?

Thanks, I since read page 46 to 50. I forgot all about this.
 

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