I'm ok with the following expression for Biot-Savart: $$ {\mathbf B}({\mathbf r}) = \frac{{\mu _0}}{{4\pi }}\int\frac{{{\mathbf J}({\mathbf r'}) \times {(\mathbf{r-r'})}}}{{|\mathbf{r-r'}|^3 }}dV'. $$ Many sources give a slightly different formula for a curve C: $$ {\mathbf B}({\mathbf r}) = \frac{{\mu _0 I}}{{4\pi }}\int_C\frac{{d{\mathbf r'} \times {\mathbf{(r-r')}}}}{{|\mathbf{r-r'}|^3 }}. $$ That the general formula implies the one for a curve seems reasonable, and even intuitive. But how would one go about showing it properly?

My idea was to write the current density as a product of delta functions whose arguments have uncountably many roots (those on the curve C), thus the triple integral would be annihilated by the delta functions, while a new integral would arise because of the delta functions. However this is quite complicated, and it doesn't seem to work out. Anyone has any ideas?


Yes, the strategy is right. If one has a one-dimensional current $I$ in the thin wire, the volume density of the current is $$ \vec J(\vec r) = I\cdot \delta^{(2)}(\vec r - \vec R_{\rm nearest}) \cdot \vec n $$ where $\vec R_{\rm nearest}$ is the point on the wire that is closest to the point $\vec r$. There are other ways to write the current but this is probably the simplest one. Also, $\vec n$ is the normal direction along the wire at the point $\vec R_{\rm nearest}$.

Now, substitute this $\vec J$ to the first formula which involves a volume integral. In some local coordinates, the two transverse integrals cancel against the two-dimensional delta-function, $I$ survives as a factor, and what's left from $\vec J \cdot dV$ except for the parts we have already accounted for is simply $\vec n\cdot d z$ where $z$ is a local coordinate along the wire, and that's what is called $d\vec r'$ in the second form of the integral. Because of the delta-function, the three-dimensional integral gets reduced to the one-dimensional integral along the wire because this is where the two-dimensional delta-function vanishes.

I've been a bit sloppy about another detail that would make the notation less comprehensible if I used it immediately. The two-dimensional delta-function should have an argument that is a two-vector. So the argument shouldn't be just $\vec r-\vec R_{\rm nearest}$ but the projection of this 3-vector to the 2-plane orthogonal to $\vec n$ (the wire).

It may be right to say that to write the exact delta-function-containing formula for $\vec J$ is more complicated than to guess the second form of the law immediately. But a virtue of this comment is the correct assertion that a formula for $\vec J$ that uses the delta-functions does exist and the second form may indeed be derived from the first if things are done right.


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