Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question in Helmholtz Theorem

  1. Aug 9, 2013 #1
    In page 555, Appendix B of Intro to electrodynamics by D Griffiths:
    [tex]\nabla\cdot\vec F=D,\;\nabla\times\vec F=\vec C\;\Rightarrow\;\nabla\cdot \vec C=0[/tex]
    Griffiths let ##\vec F=-\nabla U+\nabla\times \vec W##
    [tex]\nabla\cdot \vec F=-\nabla^2U=-\frac{1}{4\pi}\int D\nabla^2\left(\frac{1}{\vec{\vartheta}}\right)d\tau'=\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=D(\vec r) \;\hbox { where }\;\;\vec{\vartheta}=\vec r-\vec r'[/tex]

    My questions are:
    (1)Why the theorem use ##D(\vec r),\; and \;\vec C(\vec r)## where ##\vec r## is a position vector pointing at the OBSERVATION point ##P##?
    In electrostatic, ##\nabla\cdot\vec E=\frac{\rho(\vec r')}{\epsilon}## where ##\rho(\vec r')## represented the charge density at the SOURCE point pointed by position vector ##\vec r'##. ##\nabla\cdot\vec F## is the divergence of the field at the OBSERVATION point ##P(\vec r)## where position vector ##\vec r## pointing at the point ##P##. Why in Helmholtz Theorem using ##D(\vec r)##?

    (2) Why ##D(\vec r)## and ##\vec C(\vec r)## goes to zero when ##|\vec r| \rightarrow \;\infty##? Which, pretty much back to question (1)......why use ##D(\vec r)##?

    Thanks
     
    Last edited: Aug 9, 2013
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Question in Helmholtz Theorem
Loading...