In the previous article, I preliminarily observed the manifestation of the law of duality from the perspective of the principle of least action. Although that was a convenient and effective method, it left us with some regrets. The previous section started from the action principle in geometric form, rather than discussing it within the framework of the general action principle. This is because difficulties arise when discussing coordinate transformations under the form S=\int Ldt=\int (T-U)dt. These difficulties stem from the transformation d\tau=|z|^2 dt, which leads to the coupling of time and space, meaning the variation cannot be performed simply. However, this is not an unsolvable problem. We can still discuss transformation issues under the fundamental principle of action. This problem will be discussed below.
Variable Substitution in Variation
Consider the action of a general conservative system: S=\int_{t_1}^{t_2} L\left(q,\frac{dq}{dt}\right)dt
Its variation is: \begin{aligned} \delta S&=\int_{t_1}^{t_2} (\delta L) dt\\ &=\int_{t_1}^{t_2} \left[\frac{\partial L}{\partial q} \delta q+\frac{\partial L}{\partial(\frac{dq}{dt})}\delta\left(\frac{dq}{dt}\right) \right]dt\\ &=\int_{t_1}^{t_2} \left[\frac{\partial L}{\partial q} \delta q+\frac{\partial L}{\partial(\frac{dq}{dt})}\left(\frac{d\delta q}{dt}\right) \right]dt \end{aligned}
The last equality utilizes \delta\left(\frac{dq}{dt}\right)=\frac{d\delta q}{dt}, which is crucial because only the variation form on the far right is useful. The subsequent process involves integration by parts to obtain the Euler-Lagrange equations; if it cannot be converted to the form on the far right, integration by parts cannot be performed. As for the process of deriving the Euler-Lagrange equations, it will not be repeated here. We want to consider the form of the action under the transformation dt=f(q,\frac{dq}{dt})d\tau. In this case: S=\int_{t_1}^{t_2} Ldt=\int_{\tau_1}^{\tau_2} (Lf)d\tau
If Lf has already been written in terms of q and \frac{dq}{d\tau}, can we directly substitute Lf into the Euler-Lagrange equations to obtain the transformed equations of motion? Obviously, it is not that simple; the answer is negative. Let us verify it. Suppose we could directly substitute Lf into the Euler-Lagrange equations; then correspondingly we would have: \delta S=\int_{\tau_1}^{\tau_2} \delta(Lf)d\tau=\int_{\tau_1}^{\tau_2} \left[\frac{\partial (Lf)}{\partial q} \delta q+\frac{\partial (Lf)}{\partial(\frac{dq}{d\tau})}\left(\frac{d\delta q}{d\tau}\right)\right]d\tau
At the same time, we can calculate: \delta S=\int_{\tau_1}^{\tau_2} \delta(Lf)d\tau=\int_{\tau_1}^{\tau_2} (f\delta L +L \delta f)d\tau
Let us focus our attention on \delta L. We know that: \delta L=\frac{\partial L}{\partial q} \delta q+\frac{\partial L}{\partial(\frac{dq}{dt})}\delta\left(\frac{dq}{dt}\right)
Based on our assumption, we have \delta\left(\frac{dq}{d\tau}\right)=\frac{d\delta q}{d\tau}, but the equality of \delta\left(\frac{dq}{dt}\right) and \frac{d\delta q}{dt} is unknown. In fact, they are not equal, because: \begin{aligned} \delta\left(\frac{dq}{dt}\right)&=\delta\left(\frac{dq}{d\tau}\frac{1}{f}\right)\\ &=\frac{1}{f}\delta\left(\frac{dq}{d\tau}\right)-\frac{\delta f}{f^2}\frac{dq}{d\tau}\\ &=\frac{1}{f}\left(\frac{d\delta q}{d\tau}\right)-\frac{\delta f}{f^2}\frac{dq}{d\tau}\end{aligned}
So in fact: \begin{aligned} \delta S &= \int_{\tau_1}^{\tau_2} \left[ \left(\frac{\partial L}{\partial q} \delta q+\frac{\partial L}{\partial(\frac{dq}{dt})}\left(\frac{d\delta q}{dt}\right) \right)f d\tau - \left(\frac{\partial L}{\partial (\frac{dq}{dt})}\frac{dq}{dt}-L \right)(\delta f)d\tau\right]\\ &=\int_{\tau_1}^{\tau_2} \left[ \left(\frac{\partial L}{\partial q} \delta q+\frac{\partial L}{\partial(\frac{dq}{dt})}\left(\frac{d\delta q}{dt}\right) \right)f d\tau-H(\delta f)d\tau\right]\end{aligned}
where H is the Hamiltonian of the original system. If the assumption were to hold, it would require H=0, i.e., the energy of the system must be zero, which is not always true. Therefore, the assumption does not hold. However, this does not mean we have reached a dead end. Since the original system is a conservative system, energy is conserved, meaning there exists an integral H=E. We know that under L \mapsto L+E, the equations of motion of the original system remain unchanged, but it allows H \mapsto H-E=0. Thus, we have found a solution: the equations of motion for the Lagrangian L under variables q, t are equivalent to the equations of motion for the Lagrangian \tilde{L}=f(L+E) under variables q, \tau, where dt=f d\tau.
Generalized Duality
The above content was written with reference to the literature The Kustaanheimo-Stiefel transformation in geometric algebra. As can be seen from the above process, when the variable substitution in the variation is a pure time-time or space-space transformation, the conclusion is very simple: one only needs to substitute the new Lagrangian function into the Euler-Lagrange equations. However, when time-space coupling occurs in the variation, it must be treated with care, as it implies additional complexity. Here, we simply refer to the two conservative systems before and after the transformation as being generalized dual systems of each other.
If f explicitly contains only q, a more explicit solution can be obtained. The Hamiltonian of the new system is: \begin{aligned}\tilde{H} &=\frac{\partial f(L+E)}{\partial (\frac{dq}{d\tau})}\frac{dq}{d\tau}-f(L+E)\\ &=\frac{\partial f(L+E)}{\partial (f\frac{dq}{dt})}\frac{fdq}{dt}-f(L+E)\\ &=f(H-E) \end{aligned}
It can be seen that when we perform clever restrictions and transformations, everything can return to a simple form. This is the charm of mathematical physics: simplicity is hidden within complexity, and simplicity also contains various complexities. Incidentally, path integrals can be seen as the quantum version of the principle of least action. Since dual transformations exist for the principle of least action, one might guess that something similar exists for path integrals. In fact, there does exist a class of transformations that can simplify path integrals. However, since the classical version of the variation of the principle of least action is already so complex, the transformation of path integrals will obviously be even more complicated. If I have the opportunity, I will write properly about the content of path integrals.
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