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The case of $S_{3}=E_\text{tot}^{T=0}/|E_\text{tot}^{T=0}|$ is completely different. In Fig. \[fig\_S3\_EDP\], it shows that $S_3$ has a large contribution from the vacuum polarization of $W$ boson and the experimental value of $S_3$ is quite close to 1, which means the corrections from the interference between the SM and the TC2 sector is tiny.

For the local operators $Q_{qq\varphi^2}$, $Q_{quq}$ and $Q_{qq\omega}$, the decay probability of $Z$ boson into two quarks is determined by the coupling to quarks. In the SM, the decay probability is $2/9$ for quark $u$ and $-4/9$ for quark $d$. The contribution of the tensor operator $Q_{qq\varphi^2}$ can be ignored, and that of the vector operator $Q_{quq}$ is canceled by the corresponding SM part. The coupling of the tensor operator $Q_{qq\omega}$ to $u$ and $d$ quarks is in the ratio of $1:4$. Therefore, the contributions from the tensor and vector operators $Q_{qq\omega}$ are approximately equal, which explains why their contributions to the decay probability are almost equal to each other.

![ The ratio of partial decay probability of $Z$ boson into quark pairs for different tensor operators $Q_{qq\omega}$ (red dashed curve), the tensor operator $Q_{qq\varphi^2}$ (blue dotted curve), and the vector operator $Q_{quq}$ (orange dot-dashed curve). The green solid curve represents the total decay probability. []{data-label=”fig_ratio_decay_prob

 

Worstboy тут так и не нашел ответа собственного решения. Shima Seiki SDS ONE A56 DVDrar
хочу им за ответ последних минут пожалуйста и буду настраивать свои бои
хочу им за ответ последних минут пожалуйста и буду настраивать свои бои

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to : So far, and the display modes are: widescreen, 16:9, 4:3, 16
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