## 3 Magnetic Negative Stiffness in QZS Isolators
### Mathematical Modeling of Permanent Magnet Negative Stiffness Elements
#### 3.2.1 Mechanical Modeling of Magnetic Springs with Negative Stiffness of Magnetic Rings
Initially, the negative stiffness spring, comprised of the axially magnetized magnetic ring, undergoes mechanical analysis. The established theoretical analysis model, illustrated in Figures 14a and 15, is employed as an exemplar to analyze the magnetic force acting on the inner magnetic ring end face 2 and the outer magnetic ring end face 3. The magnetic charge of the point P on the end face 2 of the inner magnetic ring is:  
\[q_{2}=\sigma r_{2}dadr_{2} \tag{1}\]  
where \(r_{2}\) and \(\alpha\) are the polar coordinates of any point P on the end face 2 of the inner magnetic ring; \(\sigma\) is the surface density of the magnetic charge on the end face of the ring,  
Figure 15: Schematic diagram of a two-ring permanent magnet unit with radial magnetization: **(a)** side view of negative stiffness elements; and **(b)** section view of negative stiffness elements. assuming that the magnetic charge is uniformly distributed on the end face. According to the theory of electromagnetism, it can be seen that the outer magnetic ring end face 3 at point Q by the P point of the electromagnetic charge generated by the magnetic field strength is:  
\[H=\frac{1}{4\pi\mu_{0}}\frac{q_{2}}{r^{2}}r_{0} \tag{2}\]  
where \(\mu_{0}\) is the vacuum permeability; \(r\) is the separation between points P and Q on end face 2 and end face 3; and \(r_{0}\) is the unit direction vector between points P and Q. Therefore, the magnetic force on the point charge at point Q is:  
\[dF=\textbf{H}q_{3}=H\sigma r_{3}d\beta dr_{3}=\frac{\sigma^{2}}{4\pi\mu_{0}} \frac{r_{2}r_{3}dad\beta dr_{2}dr_{3}}{r^{3}}r \tag{3}\]  
where \(r_{3}\) and \(\beta\) are the polar coordinates of a point Q on the end face 3 of the outer ring. According to the theoretical equation of static magnetism, it is shown as follows:  
\[ \begin{cases}\sigma=jn=\mu_{0}\textbf{M}\textbf{n}\\ \textbf{B}=\mu_{0}(\textbf{H}+\textbf{M})\end{cases}  \tag{4}\]  
where \(j\) is the magnetic polarization intensity vector; \(\textbf{n}\) is the unit vector in the direction normal to the end face of the magnetic ring. \(M\) represents the magnetic induction intensity. When \(H=0\), \(B=B_{r}\), so \(B_{r}=\mu_{0}M\). Since the magnetization direction of the magnetic ring is axial magnetization, which coincides with the normal direction of the end face of the ring, the combined above equations can be obtained:  
\[\sigma=\mu_{0}M=B_{r} \tag{5}\]  
where \(B_{r}\) is the residual magnetic induction. Therefore, Equation (3) can be expressed as:  
\[dF=\frac{B_{r}^{2}}{4\pi\mu_{0}}\frac{r_{2}r_{3}dad\beta dr_{2}dr_{3}}{r^{3}} r=C\frac{r_{2}r_{3}dad\beta dr_{2}dr_{3}}{r^{3}}r \tag{6}\]  
where \(C=\frac{B_{r}^{2}}{4\pi\mu_{0}}\) is the coefficient related to the magnetic properties of the permanent magnet material. By taking Equation (6) and projecting it along the X-direction as shown in Figure 15a, the axial force along the end face of the ring generated by the point magnetic charge at any point P on end face 2 and any point Q on end face 3 is obtained:  
\[dF_{32}=C\frac{r_{2}r_{3}dad\beta dr_{2}dr_{3}}{r^{3}}ri \tag{7}\]  
where \(i\) is the unit vector in the axial X-coordinate direction of the magnetic ring. From Figure 15b, the geometrical relationship between point P and point Q can be established, then  
\[r=r^{\prime}-x_{1}=r_{3}-r_{2}-x_{1} \tag{8}\]  
where \(x_{1}\) is the relative axial displacement between the inner and outer ring end surfaces 2, 3. The unit vectors in the axial coordinate direction of the magnetic ring form angles \(\gamma\) and \(\varphi\) with the x-axis and y-axis, respectively. Substituting equation \(ri\) into, we can obtain  
\[ri=r_{3}i-r_{2}i-x_{1}i \tag{9}\]  
Due to  
\[\left\{ \begin{array}{l}r_{3}i=r_{3}\sin\gamma\\ r_{2}i=r_{2}\sin\varphi\\ x_{1}i=x_{1}\end{array} \right. \tag{10}\]So, Equation (9) can be further simplified as  
\[ri=r_{3}\sin\gamma-r_{2}\sin\varphi-x_{1} \tag{11}\]  
where \(\sin\gamma=\frac{x_{1}}{\sqrt{r_{3}^{2}+x_{1}^{2}}}\), \(\sin\varphi=\frac{x_{1}}{\sqrt{r_{2}^{2}+x_{1}^{2}}}\).