- #1
jostpuur
- 2,112
- 19
http://en.wikipedia.org/wiki/Vector_potential
The Wikipedia article is mentioning an interesting formula for defining vector potentials. Isn't that the same thing as a formula
<br /> \nabla\times\int d^3y\;(V(y)\times\nabla)\frac{1}{\|x-y\|} = -4\pi V(x),<br />
where V:\mathbb{R}^3\to\mathbb{R}^3 is some test function, and nablas are derivatives with respect to x? I tried to prove this, and reduced it first to a formula
<br /> \sum_{j,k,m=1}^3 \epsilon_{ijk}\epsilon_{lmk} \partial_j \partial_m \frac{1}{\|x-y\|} = -4\pi\delta_{il}\delta^3(x - y).<br />
This formula is to interpreted so that actually there should be an integral with some test function, and at least other one of the derivative operators should be somewhere else, like outside the integral, or acting on the test function with minus sign.
These formulas should be equivalent:
<br /> \Big(\nabla\times\int d^3y\;(V(y)\times\nabla)\frac{1}{\|x-y\|}\Big)_i<br /> =\epsilon_{ijk}\partial_j \int d^3y\; (V(y)\times\nabla)_k \frac{1}{\|x-y\|} = \epsilon_{ijk}\underbrace{\epsilon_{klm}}_{=\epsilon_{lmk}} \partial_j \int d^3y\; V_l(y)\partial_m \frac{1}{\|x-y\|}<br />
But then I encountered problems when trying to prove this. By substituting
<br /> \sum_{k=1}^3 \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}<br />
we get
<br /> \sum_{j,m=1}^3(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\partial_j\partial_m \frac{1}{\|x-y\|} = \delta_{il}\nabla^2 \frac{1}{\|x-y\|}\; -\; \partial_l\partial_i \frac{1}{\|x-y\|},<br />
but now the second term with derivatives \partial_l\partial_i is messing everything. The first term is already all we would want, because it is known result that
<br /> \nabla^2\frac{1}{\|x-y\|} = -4\pi\delta^3(x-y)<br />
again in the sense that there should be a test function, and derivatives outside the integral.
The Wikipedia article is mentioning an interesting formula for defining vector potentials. Isn't that the same thing as a formula
<br /> \nabla\times\int d^3y\;(V(y)\times\nabla)\frac{1}{\|x-y\|} = -4\pi V(x),<br />
where V:\mathbb{R}^3\to\mathbb{R}^3 is some test function, and nablas are derivatives with respect to x? I tried to prove this, and reduced it first to a formula
<br /> \sum_{j,k,m=1}^3 \epsilon_{ijk}\epsilon_{lmk} \partial_j \partial_m \frac{1}{\|x-y\|} = -4\pi\delta_{il}\delta^3(x - y).<br />
This formula is to interpreted so that actually there should be an integral with some test function, and at least other one of the derivative operators should be somewhere else, like outside the integral, or acting on the test function with minus sign.
These formulas should be equivalent:
<br /> \Big(\nabla\times\int d^3y\;(V(y)\times\nabla)\frac{1}{\|x-y\|}\Big)_i<br /> =\epsilon_{ijk}\partial_j \int d^3y\; (V(y)\times\nabla)_k \frac{1}{\|x-y\|} = \epsilon_{ijk}\underbrace{\epsilon_{klm}}_{=\epsilon_{lmk}} \partial_j \int d^3y\; V_l(y)\partial_m \frac{1}{\|x-y\|}<br />
But then I encountered problems when trying to prove this. By substituting
<br /> \sum_{k=1}^3 \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}<br />
we get
<br /> \sum_{j,m=1}^3(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\partial_j\partial_m \frac{1}{\|x-y\|} = \delta_{il}\nabla^2 \frac{1}{\|x-y\|}\; -\; \partial_l\partial_i \frac{1}{\|x-y\|},<br />
but now the second term with derivatives \partial_l\partial_i is messing everything. The first term is already all we would want, because it is known result that
<br /> \nabla^2\frac{1}{\|x-y\|} = -4\pi\delta^3(x-y)<br />
again in the sense that there should be a test function, and derivatives outside the integral.
After replacing the derivative with respect to x into an derivative with respect to y, and integration by parts, the second term gives
