LH vs Gaussian units in Biot-Savart Law

[avikarto]

I have a question regarding the relationship between the Biot-Savart formula in Gaussian and Lorentz-Heaviside units. In Gaussian, we have a $\frac{1}{c}$ outside the integral, but in LH units we have a $\frac{1}{4\pi}\frac{1}{c}$. This does not make sense, considering the transformation between Gaussian and LH for B is that $B_{LH}=\frac{B_G}{\sqrt{4\pi}}$. Where does the extra $\sqrt{4\pi}$ come from? Thanks.

Source

 

[Dale]

the transformation between Gaussian and LH for B is that $B_{LH}=\frac{B_G}{\sqrt{4\pi}}$. Where does the extra $\sqrt{4\pi}$ come from? Thanks.

Source

it comes from I.

 

[avikarto]

If we take $I=q v$, then using that $q_{LH}=\sqrt{4\pi} q_G$, we would have that $I_{LH}=\sqrt{4\pi} I_G$. This seems to brings an extra $\sqrt{4\pi}$ into the numerator, not the denominator, and would therefore remove the factor all together. I don't understand how this resolves the issue. Besides, the transformation on B should be the transformation on B, without also having to transform its subcomponents additionally. Could you please elaborate?

 

[Dale]

This seems to brings an extra 4π−−√4π\sqrt{4\pi} into the numerator, not the denominator, and would therefore remove the factor all together.

Check your math. Write the equation in one set of units, and then substitute. The factor is correct.