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I'm going to use c.g.s. units for the derivation, and I will show that ## B=H+4 \pi M ##.

Let ## a=-\nabla \frac{1}{|x-x'|}=\frac{x-x'}{|x-x'|^3} ##.

Let ## b=M(x') ##. Then ## \nabla' \times b=\nabla' \times M(x')=\frac{J_m(x')}{c} ##.

Biot-Savart gives ## B(x)=\int (\nabla' \times b) \times a \, d^3x'=-\int a \times (\nabla' \times b) ##.

By a vector identity ## B=-\int [\nabla'(a \cdot b)-(b \cdot \nabla') a-(a \cdot \nabla') b-b \times \nabla' \times a] \, d^3 x' ##.

The first term integrates to zero because of a vector identity similar to Gauss' law that ## \int \nabla \psi \, d \tau=\int \psi \, dA ##, and for a finite sample size, ## b ## vanishes on the large outer surface that surrounds the magnetized sample.

The 4th term is zero because it is the curl of a gradient. This leaves the two middle terms.

The derivation is a little more lengthy than I anticipated, so let me just add a couple of key steps here, and I will try to fill in the pieces at a later time:

## \nabla' \cdot a=-4 \pi \delta(x-x') ##, and ## \nabla \cdot a=-\nabla' \cdot a=4 \pi \delta(x-x') ##.

Now ## b \cdot \nabla' a=\nabla' \times (a \times b)+b(\nabla' \cdot a)-a(\nabla' \cdot b)+(a \cdot \nabla')b ##.

Again, I hope to fill in all the steps=there are quite a number of them=but in any case, the result of all of this vector calculus is that ## B=H+4 \pi M ##.

It is worth mentioning one more step in this derivation: ## (a \cdot \nabla') b=(b \cdot \nabla')a-\nabla' \times (a \times b) -b (\nabla' \cdot a)+a (\nabla' \cdot b) ##.

This last substitution seems to be necessary, because in a later step ## (b \cdot \nabla')a=-(b \cdot \nabla)a ## (without a prime on the last gradient operator ) is used to evaluate this term, while evaluation of ## (a \cdot \nabla')b ## as is would be problematic.

Much of the rest is just evaluating the terms, and also determining that certain integrals equal to zero, because they equate to surface integrals, etc.

The editor is giving me problems again, so I think I will go to post 27 to add additional steps.

Let ## a=-\nabla \frac{1}{|x-x'|}=\frac{x-x'}{|x-x'|^3} ##.

Let ## b=M(x') ##. Then ## \nabla' \times b=\nabla' \times M(x')=\frac{J_m(x')}{c} ##.

Biot-Savart gives ## B(x)=\int (\nabla' \times b) \times a \, d^3x'=-\int a \times (\nabla' \times b) ##.

By a vector identity ## B=-\int [\nabla'(a \cdot b)-(b \cdot \nabla') a-(a \cdot \nabla') b-b \times \nabla' \times a] \, d^3 x' ##.

The first term integrates to zero because of a vector identity similar to Gauss' law that ## \int \nabla \psi \, d \tau=\int \psi \, dA ##, and for a finite sample size, ## b ## vanishes on the large outer surface that surrounds the magnetized sample.

The 4th term is zero because it is the curl of a gradient. This leaves the two middle terms.

The derivation is a little more lengthy than I anticipated, so let me just add a couple of key steps here, and I will try to fill in the pieces at a later time:

## \nabla' \cdot a=-4 \pi \delta(x-x') ##, and ## \nabla \cdot a=-\nabla' \cdot a=4 \pi \delta(x-x') ##.

Now ## b \cdot \nabla' a=\nabla' \times (a \times b)+b(\nabla' \cdot a)-a(\nabla' \cdot b)+(a \cdot \nabla')b ##.

Again, I hope to fill in all the steps=there are quite a number of them=but in any case, the result of all of this vector calculus is that ## B=H+4 \pi M ##.

It is worth mentioning one more step in this derivation: ## (a \cdot \nabla') b=(b \cdot \nabla')a-\nabla' \times (a \times b) -b (\nabla' \cdot a)+a (\nabla' \cdot b) ##.

This last substitution seems to be necessary, because in a later step ## (b \cdot \nabla')a=-(b \cdot \nabla)a ## (without a prime on the last gradient operator ) is used to evaluate this term, while evaluation of ## (a \cdot \nabla')b ## as is would be problematic.

Much of the rest is just evaluating the terms, and also determining that certain integrals equal to zero, because they equate to surface integrals, etc.

The editor is giving me problems again, so I think I will go to post 27 to add additional steps.

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