1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about mathematical equality

  1. Dec 5, 2014 #1
    Hi there, im reading Chapter 9 of Jackson Classic Electrodynamics 3rd edition, and I don't see why this equality is true, it says "integrating by parts", but I still don't know... any help?

    http://imageshack.com/a/img673/9201/4WYcXs.png [Broken]
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 6, 2014 #2


    User Avatar
    Gold Member

    Write [itex] \mathbf J [/itex] as the sum of its Cartesian components and then write the integral as the sum of three integrals of the three Cartesian components. Then integrate by parts each integral, but the [itex] J_x [/itex] w.r.t. to x and so on! If you consider the integrals to be from [itex] -\infty [/itex] to [itex] \infty [/itex] and take [itex] \mathbf J [/itex] to be zero at infinities, you'll get what you want.
  4. Dec 6, 2014 #3
  5. Dec 9, 2014 #4


    User Avatar
    Science Advisor

    That equality depends on div J=0 (or d\rho/dt=0).
  6. Dec 10, 2014 #5


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Just my derivation. For an arbitrary constant vector ##\vec{n}## we have
    $$\vec{\nabla} \cdot [(\vec{n} \cdot \vec{x}) \vec{j}]=(\vec{n} \cdot \vec{x}) \vec{\nabla} \cdot \vec{j} + \vec{n} \cdot \vec{j}.$$
    Now integrate this over the whole space and assume that ##\vec{j}## goes to 0 quickly enough at infinity (or most realistically that it has compact support). Then the left-hand side vanishes, because it's a divergence, and thus can be transformed to a surface integral, which vanishes at infinity, if ##\vec{j}## vanishes quickly enough. This implies
    $$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \vec{n} \cdot \vec{j} = -\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; (\vec{n} \cdot \vec{x} ) \vec{\nabla} \cdot \vec{j}.$$
    Now use for ##\vec{n}## the three basis vectors of a cartesian reference frame, and you see that the vector equation stated by Jackson holds.

    It's independent of whether ##\vec{j}## is a solenoidal field or not. Of course in the case of stationary currents it must be one, but the integral identity is generally valid.
  7. Dec 13, 2014 #6

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Sorry about that. I was thinking cases where j appears with other variables.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook