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楼主 |
发表于 2026-4-13 19:38
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本帖最后由 谢芝灵 于 2026-4-14 08:40 编辑
\(t_q = \Delta \tau_{m_1} = \Delta T = t_P = \sqrt{\frac{\hbar G}{c^5}} \approx 5.39 \times 10^{-44} \text{s} \quad \text{(5)}
\)
\(\frac{dT}{d\tau_m} = 1 \quad \text{(1)}\)
\( \frac{dT}{d\tau_m} = 1 \quad \forall m \in \mathcal{M}\)
\(\tau_m\)
\(\tau dτ/dT\)
\(\mathcal{M}\)
\(\zeta(v, \Phi, \mathbf{M}) = \frac{\nu}{\nu_0} = f\left(v, \Phi, {\lambda_i}\right) \)
\(\mathbf{M} = {\lambda_i}\)
\( \zeta(v, \Phi, \mathbf{M}) = \exp\left[-\frac{1}{2}\left(\alpha(\mathbf{M})\frac{v^2}{c^2} + \beta(\mathbf{M})\frac{\Phi}{c^2}\right)\right] \)
\(\frac{dT_m}{dt} = \frac{dT_M}{dt} = 1\)
\(i\hbar\frac{\partial}{\partial T}\psi(\mathbf{x}, T) = \hat{H}\psi(\mathbf{x}, T) \)
\(\frac{d\tau}{dT} = f(v, \Phi, \text{m})dTdτ=f(v,Φ,m)\)
\(\zeta(v) = \exp\left[-\frac{1}{2}\left(\alpha\frac{v^2}{c^2} + \gamma\frac{v^4}{c^4} + \cdots\right)\right] \)
\(\mathcal{E} = \sum_{i=1}^{N} \left[\tau^{\text{exp}}_i - \zeta(v_i, \Phi_i, \mathbf{M}_i)T_i\right]^2
\)
\(\mathcal{E}\) |
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