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Since magnetic moment can be view as a small electric current circle. Pictorially, when apply inversion operation, the current direction is reversed, so I think the $\vec{m}\to -\vec{m}$ under inversion symmetry operation.

On the other hand, the formula for the magnetic moment is $\vec{m}=\int_V \vec{r}\times\vec{j}\mathrm{d}V$. Under the inversion symmetry operation $\vec{r}\to -\vec{r}$ and $\vec{j}\to -\vec{j}$, therefore $\vec{m}$ is unchanged.

The above two reasoning must have one being wrong, which one and what is the flaw of the reasoning?

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    $\begingroup$ Which inversion symmetry? $\endgroup$ – Qmechanic Jul 31 at 15:53
  • $\begingroup$ @Qmechanic I think any inversion should be same for a circular loop current $\endgroup$ – an offer can't refuse Aug 1 at 5:40
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Under a full parity inversion, the magnetic moment is unchanged. In your first paragraph, the current is reversed, but the location of the current is also changed.

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  • $\begingroup$ Do you change the coordinate system from right hand to left hand together with the current loop reversed also? $\endgroup$ – an offer can't refuse Aug 2 at 23:34
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$\vec m$ and $\vec B$ are indeed "axial vectors" and invariant under space inversion. Imagine a circular current. Under inversion the current changes direction but also the circle is inverted. This leaves $\vec m$ unchanged. (just noticed Ben Crowell's ear lier answer which is identical to mine.

$\vec E$ also is not a true 3D vector as it changes sign under time reversal.

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  • $\begingroup$ but how do you draw the current loop after inversion, will it look the same as I described in the first part? $\endgroup$ – an offer can't refuse Aug 2 at 23:33
  • $\begingroup$ I also don't understand why E change sign under time reversal, I see in many places E is invariant... $\endgroup$ – an offer can't refuse Aug 8 at 0:00

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