Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's have the equality $$ \int \limits_{-\infty}^{\infty} \left[ [\nabla_{\mathbf r'} \times \mathbf A (\mathbf r' )] \times \frac{\mathbf r' - \mathbf r}{|\mathbf r' - \mathbf r |^{3}}\right]d^{3}\mathbf r' = 4 \pi\mathbf A (\mathbf r), \qquad (.0) $$ where curl operator acts only on $\mathbf A$. How to prove it?

Addition. The equality is proven by me. [:))]. $$ \int \limits_{-\infty}^{\infty} \left[ [\nabla_{\mathbf r'} \times \mathbf A (\mathbf r' )] \times \frac{\mathbf r' - \mathbf r}{|\mathbf r' - \mathbf r |^{3}}\right]d^{3}\mathbf r' = -\int \left[ \left[ \nabla ' \times \mathbf A ' \right] \times \nabla \left( \frac{1}{|\mathbf r' - \mathbf r|}\right)\right]d^{3}\mathbf r' = $$ $$ =\left[ \nabla \times \int \frac{[\nabla ' \times \mathbf A' ]d^{3}\mathbf r' }{|\mathbf r' - \mathbf r |} \right] = $$ $$ = \left[ \nabla \times \int \nabla' \times \left( \frac{\mathbf A'}{|\mathbf r' - \mathbf r|}\right)d^{3}\mathbf r'\right] - \left[ \nabla \times \left[ \int \nabla' \left(\frac{1}{|\mathbf r' - \mathbf r|}\right) \times \mathbf A ' \right]d^{3}\mathbf r '\right] = $$ $$ \left[ \nabla \times \left[ \int \nabla \left(\frac{1}{|\mathbf r' - \mathbf r|}\right) \times \mathbf A ' \right]d^{3}\mathbf r '\right] = $$ $$ \left( \nabla \cdot \left( \nabla \int \frac{\mathbf A' d^{3}\mathbf r'}{|\mathbf r' - \mathbf r |} \right)\right) - \Delta \int \frac{\mathbf A' d^{3}\mathbf r'}{|\mathbf r' - \mathbf r|} = 4 \pi \mathbf A - \left( \nabla \cdot \left( \nabla \int \frac{\mathbf A' d^{3}\mathbf r'}{|\mathbf r' - \mathbf r |} \right)\right) = $$ $$ = 4\pi\mathbf A - \left(\nabla \cdot \left( \int \nabla'\left( \frac{\mathbf A'd^{3}\mathbf r'}{|\mathbf r' - \mathbf r|}\right) - \int \frac{(\nabla ' \cdot \mathbf A')d^{3}\mathbf r' }{|\mathbf x' - \mathbf x |}\right) \right) = 4 \pi\mathbf A $$ only if $(\nabla ' \cdot \mathbf A') = 0$.

share|improve this question
    
The first formula cannot be true unless you impose some condition on $A$, such as $\nabla\cdot A=0$. There are many $A$'s such that $\nabla\times A=0$, and the formula would give for them $A=0$. –  user8268 Sep 16 '13 at 16:15
    
@user8268 . Why I can impose condition $(\nabla \cdot \mathbf A ) = 0$? From which it follows that the field is not uniquely defined (in sense of additional summand of gradient of some function)? –  John Taylor Sep 16 '13 at 16:17
    
I'm just saying that if $\nabla\times A=0$ then the formula gives $0$, i.e. it is wrong. It might be true under some conditions imposed on $A$; I'm not saying that you can impose them (though your notation looks electomagnetic, and there $A$ has a gauge freedom) –  user8268 Sep 16 '13 at 16:22
    
@user8268 . I can impose a condition on $\mathbf A$, according to which its curl must be non-zero. Unfortunately, task does not involve the imposition of calibration of Coulomb type: it is not associated with the physics. –  John Taylor Sep 16 '13 at 16:26
2  
But you need to re-ask the question: if you add to $A$ an $A'$ s.t. $\nabla\times A'=0$, the LHS doesn't change, but the RHS does. As it is, it is not true. –  user8268 Sep 16 '13 at 16:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.