Question about mathematical equality

  • #1
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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]
 
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Answers and Replies

  • #2
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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
Meir Achuz
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That equality depends on div J=0 (or d\rho/dt=0).
 
  • #5
vanhees71
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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.
 
  • #6
Meir Achuz
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Sorry about that. I was thinking cases where j appears with other variables.
 

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