The Lagrangian \(L(x, \dot{x}, t)\) is a function of the generalized coordinate \(x\), the generalized velocity \(\dot{x}\), and also the time parameter. One is familiar with the Hamiltonian \(H(x, p, t)\), which is a function of the generalized momentum \(p\) and the generalized coordinate. These two functions are related via
\begin{equation}
H = \dot{x} \cdot p - L \label{Ham}
\end{equation}
From the identities
\begin{equation}
\mathrm{d}L = \frac{\partial L}{\partial x} \cdot \mathrm{d}x + \frac{\partial L}{\partial \dot{x}} \cdot \mathrm{d}\dot{x} + \frac{\partial L}{\partial t}\mathrm{d}t
\end{equation}
and
\begin{equation}
\mathrm{d}H = \frac{\partial H}{\partial x} \cdot \mathrm{d}x + \frac{\partial H}{\partial p} \cdot \mathrm{d}p + \frac{\partial H}{\partial t}\mathrm{d}t
\end{equation}
it follows that
\begin{equation}
p = \frac{\partial L}{\partial \dot{x}} \qquad \dot{x} = \frac{\partial H}{\partial p} \qquad {- \frac{\partial L}{\partial x}} = \frac{\partial H}{\partial x} \qquad {- \frac{\partial L}{\partial t}} = \frac{\partial H}{\partial t} \label{Heq}
\end{equation}
If one assumes the Euler-Lagrange equations,
\begin{equation}
\frac{\partial L}{\partial x} = \frac{\mathrm{d}}{\mathrm{d} t} \left( \frac{\partial L}{\partial \dot{x}} \right)
\end{equation}
then the relations in \((\ref{Heq})\) become the Hamilton equations.
Dual Functions
Nothing prevents us from postulating the existence of a dual Lagrangian \(K(\dot{p}, p, t)\), which is a function of the generalized force \(\dot{p}\) and the generalized momentum, and also a dual Hamiltonian \(G(\dot{p}, \dot{x}, t)\), which is a function of the generalized force and the generalized velocity. Obviously, we have
\begin{equation}
\mathrm{d}K = \frac{\partial K}{\partial \dot{p}} \cdot \mathrm{d} \dot{p} + \frac{\partial K}{\partial p} \cdot \mathrm{d}p + \frac{\partial K}{\partial t} \mathrm{d}t
\end{equation}
and
\begin{equation}
\mathrm{d}G = \frac{\partial G}{\partial \dot{p}} \cdot \mathrm{d}\dot{p} + \frac{\partial G}{\partial \dot{x}} \cdot \mathrm{d}\dot{x} + \frac{\partial G}{\partial t} \mathrm{d}t
\end{equation}
In analogy with \((\ref{Ham})\), we postulate the relations
\begin{equation}
H = \dot{x} \cdot p - L \qquad H = -x \cdot \dot{p} + K \qquad G = -x \cdot \dot{p} + L \qquad G = \dot{x} \cdot p - K
\end{equation}
So besides \((\ref{Heq})\), we also have
\begin{equation}
\dot{p} = {-\frac{\partial H}{\partial x}} \qquad x = \frac{\partial K}{\partial \dot{p}} \qquad \frac{\partial K}{\partial p} = \frac{\partial H}{\partial p} \qquad \frac{\partial K}{\partial t} = \frac{\partial H}{\partial t}
\end{equation}
\begin{equation}
\dot{p} = \frac{\partial L}{\partial x} \qquad x = -\frac{\partial G}{\partial \dot{p}} \qquad \frac{\partial L}{\partial \dot{x}} = \frac{\partial G}{\partial \dot{x}} \qquad \frac{\partial L}{\partial t} = \frac{\partial G}{\partial t}
\end{equation}
and
\begin{equation}
p = \frac{\partial G}{\partial \dot{x}} \qquad \dot{x} = \frac{\partial K}{\partial p} \qquad \frac{\partial K}{\partial \dot{p}} = {-\frac{\partial G}{\partial \dot{p}}} \qquad {-\frac{\partial K}{\partial t}} = \frac{\partial G}{\partial t}
\end{equation}
We can collect all the terms with derivatives of \(H\) and \(G\),
\begin{equation}
\dot{x} = \frac{\partial H}{\partial p} \qquad \dot{p} = - \frac{\partial H}{\partial x} \qquad x = {-\frac{\partial G}{\partial \dot{p}}} \qquad p = \frac{\partial G}{\partial \dot{x}} \label{Hderiv}
\end{equation}
and all the terms with derivatives of \(L\) and \(K\),
\begin{equation}
p = \frac{\partial L}{\partial \dot{x}} \qquad x = \frac{\partial K}{\partial \dot{p}} \qquad \dot{p} = \frac{\partial L}{\partial x} \qquad \dot{x} = \frac{\partial K}{\partial p}
\end{equation}
which lead to two Euler-Lagrange equations,
\begin{equation}
\frac{\partial L}{\partial x} = \frac{\mathrm{d}}{\mathrm{d} t} \left( \frac{\partial L}{\partial \dot{x}} \right) \qquad \frac{\partial K}{\partial p} = \frac{\mathrm{d}}{\mathrm{d} t} \left( \frac{\partial K}{\partial \dot{p}} \right)
\end{equation}
We also have the relations among partial time derivatives,
\begin{equation}
\frac{\partial H}{\partial t} = {-\frac{\partial G}{\partial t}} \qquad \frac{\partial L}{\partial t} = {-\frac{\partial K}{\partial t}}
\end{equation}
It appears that assuming the existence of the dual functions and the relationships via Legendre transforms is equivalent to requiring the Euler-Lagrange equations.
Actions
Note that
\begin{equation}
K = \frac{\mathrm{d}}{\mathrm{d} t}(x \cdot p) - L \qquad G = \dot{x} \cdot p - x \cdot \dot{p} - H \qquad L + H = K + G
\end{equation}
This suggest that the action \(S\) and the dual action \(R\) defined by
\begin{equation}
S = \int\limits_{t_{i}}^{t_{f}}\mathrm{d}t \, L \qquad R = \int\limits_{t_{i}}^{t_{f}}\mathrm{d}t \, K
\end{equation}
are related via
\begin{equation}
R = x_{f} \cdot p_{f} - x_{i} \cdot p_{i} - S
\end{equation}
In some way, this relation can be viewed as a Legendre transform. If \(R\) is a function of \(p_{i}\) and \(p_{f}\), and \(S\) is a function of \(x_{i}\) and \(x_{f}\), it follows that
\begin{equation}
p_{f} = \frac{\partial S}{\partial x_{f}} \qquad p_{i} = - \frac{\partial S}{\partial x_{i}} \qquad x_{f} = \frac{\partial R}{\partial p_{f}} \qquad x_{i} = - \frac{\partial R}{\partial p_{i}}
\end{equation}
In principle, only two pieces of the temporal boundary data are independent. One can imagine working with \(x_{f}\) and \(p_{f}\) as the independent pieces of data. Then \(x_{i}\) and \(p_{i}\) are functions of \(x_{f}\) and \(p_{f}\). If we want to change the independent pieces of data to be \(x_{i}\) and \(x_{f}\), we would have to compute the Jacobian and inverse Jacobian matrices
\begin{equation}
\frac{\partial x_{i}}{\partial p_{f}} = - \frac{\partial^{2} R}{\partial p_{f} \partial p_{i}} \qquad \frac{\partial p_{f}}{\partial x_{i}} = \frac{\partial^{2} S}{\partial x_{i} \partial x_{f}}
\end{equation}
Similarly, if we work with \(x_{i}\) and \(p_{i}\) as independent pieces of data and then switch to \(x_{i}\) and \(x_{f}\) as the independent pieces of data, the Jacobian and inverse Jacobian matrices are
\begin{equation}
\frac{\partial x_{f}}{\partial p_{i}} = \frac{\partial^{2} R}{\partial p_{i} \partial p_{f}} \qquad \frac{\partial p_{i}}{\partial x_{f}} = - \frac{\partial^{2} S}{\partial x_{f} \partial x_{i}}
\end{equation}
If \(S\) and \(R\) are well-behaved functions, then it follows that
\begin{equation}
\frac{\partial x_{i}}{\partial p_{f}} + \frac{\partial x_{f}}{\partial p_{i}} = 0 \qquad \frac{\partial p_{f}}{\partial x_{i}} + \frac{\partial p_{i}}{\partial x_{f}} = 0
\end{equation}
Gauge Transformations
Shifting \(L\) by the total time derivative of a function of \(x\) and \(t\) leaves the Euler-Lagrange equations invariant. The action \(S\) aquires a shift too:
\begin{equation}
L \rightarrow L + \frac{\mathrm{d} \Lambda}{\mathrm{d} t} \quad \Longrightarrow \quad S \rightarrow S + \Lambda(x_{f}, t_{f}) - \Lambda(x_{i}, t_{i})
\end{equation}
If we write \(L\) in terms of \(p\) and \(H\), it follows that
\begin{equation}
L \rightarrow L + \frac{\mathrm{d} \Lambda}{\mathrm{d} t} \quad \Longrightarrow \quad p \rightarrow p + \frac{\partial \Lambda}{\partial x} \qquad H \rightarrow H - \frac{\partial \Lambda}{\partial t}
\end{equation}
These are the familiar gauge transformations of the electromagnetic potentials.
We have an analogous situation with \(K\). Shifting \(K\) by the total time derivative of a function of \(p\) and \(t\) leaves the Euler-Lagrange equations invariant. The dual action \(R\) aquires a shift too:
\begin{equation}
K \rightarrow K + \frac{\mathrm{d} \Omega}{\mathrm{d} t} \quad \Longrightarrow \quad R \rightarrow R + \Omega(p_{f}, t_{f}) - \Omega(p_{i}, t_{i})
\end{equation}
If we write \(K\) in terms of \(x\) and \(H\), it follows that
\begin{equation}
K \rightarrow K + \frac{\mathrm{d} \Omega}{\mathrm{d} t} \quad \Longrightarrow \quad x \rightarrow x + \frac{\partial \Omega}{\partial p} \qquad H \rightarrow H + \frac{\partial \Omega}{\partial t}
\end{equation}
Can we introduce dual potentials in momentum space such that these transformations compensate some gauge transformations of the dual potentials? What does this dual gauge theory look like? In the quantum theory, do \(Z\) and \(\psi\) correspond to Fourier transforms of \(A\) and \(\phi\), or are they unrelated?
Brackets
Consider a function $ h(x, p, t) $. The total time derivative of $ h $ is
\begin{equation}
\frac{\mathrm{d} h}{\mathrm{d} t} = \frac{\partial h}{\partial x} \cdot \dot{x} + \frac{\partial h}{\partial p} \cdot \dot{p} + \frac{\partial h}{\partial t}
\end{equation}
Using $ (\ref{Hderiv}) $ we find
\begin{equation} \frac{\mathrm{d} h}{\mathrm{d} t} = \frac{\partial h}{\partial x} \cdot \frac{\partial H}{\partial p} - \frac{\partial H}{\partial x} \cdot \frac{\partial h}{\partial p} + \frac{\partial h}{\partial t}
\end{equation}
This involves the familiar Poisson bracket,
\begin{equation} [ h , H ] = \frac{\partial h}{\partial x} \cdot \frac{\partial H}{\partial p} - \frac{\partial H}{\partial x} \cdot \frac{\partial h}{\partial p}
\end{equation}
Harmonic Oscillator
During a course on classical mechanics, one is rarely introduced to the dual Lagrangian \(K\) or the dual Hamiltonian \(G\). Indeed, they are mentioned in Goldstein, Poole and Safko as problems at the end of chapter 8 (alas, with a different sign convention). I suspect the dual functions can be used to study magnetic degrees of freedom. By magnetic, I mean heavy or strongly-coupled.
As an example, let us consider the classical harmonic oscillator. The Hamiltonian for this system is
\begin{equation}
H = \frac{p^{2}}{2m} + \frac{1}{2}k x^{2} \label{HarmOsci}
\end{equation}
Assuming that \(m \geq 0\) and \(k \geq 0\), we have \(H \geq 0\). The partial derivatives of \(H\) are
\begin{equation}\frac{\partial H}{\partial x} = k x \qquad \frac{\partial H}{\partial p} = \frac{p}{m} \qquad \frac{\partial H}{\partial t} = 0
\end{equation}
Using \((\ref{Hderiv})\) leads to
\begin{equation}\dot{x} = \frac{p}{m} \qquad \dot{p} = -k x
\end{equation}
The Lagrangian \(L\) is a function of \(x\) and \(\dot{x}\), so we find
\begin{equation}L = \dot{x} \cdot p - H = \frac{1}{2}m \dot{x}^{2} - \frac{1}{2}kx^{2}
\end{equation}
This should be familiar. Note that in the limit \(m \rightarrow 0\) and \(k \rightarrow 0\) we have \(L \rightarrow 0\). Also, \(k\) appears with a positive power, so we can take the \(k \rightarrow 0\) limit.
The dual Lagrangian \(K\) is a function of \(p\) and \(\dot{p}\), so we find
\begin{equation}K = x \cdot \dot{p} + H = \frac{p^{2}}{2m} - \frac{\dot{p}^{2}}{2k}
\end{equation}
Unlike \(L\), in the limit \(m \rightarrow 0\) or \(k \rightarrow 0\) we find \(K \rightarrow \infty\). However, we can see that in the \(m \rightarrow \infty\) and \(k \rightarrow \infty\) we find \(K \rightarrow 0\). The constant \(k\) appears in the denominator, so we cannot take \(k \rightarrow 0\).
We also have the dual Hamiltonian \(G\), which is a function of \(\dot{x}\) and \(\dot{p}\). This gives
\begin{equation}G = -x \cdot \dot{p} + L = \frac{1}{2}m \dot{x}^{2} + \frac{\dot{p}^{2}}{2k}
\end{equation}
Just like for \(H\), we have \(G \geq 0\) if \(m \geq 0\) and \(k \geq 0\). Both \(H\) and \(G\) contain mixed powers of \(m\), but \(G\) contains a negative power of \(k\) while \(H\) contains a positive power of \(k\). In some ways, \(k\) is the coupling parameter for this system, so we can argue that \(H\) and \(L\) are more compatible with weak-coupling dynamics while \(G\) and \(K\) are more compatible with strong-coupling dynamics.
The Hamiltonian $ (\ref{HarmOsci}) $ is compatible with the $ k \rightarrow 0 $ limit (the free system), and the $ m \rightarrow \infty $ limit (the localized system). Usually we define an angular frequency $ \omega $ via
\begin{equation} \omega^{2} = \frac{k}{m}
\end{equation}
Note that both taking $ k \rightarrow 0 $ with $ m $ fixed, and $ m \rightarrow \infty $ with $ k $ fixed lead to $ \omega \rightarrow 0 $, which can be interpreted as large period limits.
Free System
Taking the $ k \rightarrow 0 $ limit while keeping $ m $ fixed in $ (\ref{HarmOsci}) $ gives the Hamiltonian for a single-body free system:
\begin{equation} H = \frac{p^{2}}{2m}
\end{equation}
The Lagrangian becomes
\begin{equation} L = \frac{1}{2} m \dot{x}^{2}
\end{equation}
which is invariant under $ x \rightarrow x + a $. Note that $ K $ and $ G $ are undefined in this regime. The classical solution gives
\begin{equation}x(t) = x_{0} + \left(\frac{p_{0}}{m}\right)t \qquad p(t) = p_{0}
\end{equation}
That is, the position is a linear function of time and the momentum is constant.
It is worth mentioning now that the Hamiltonian for a charged system is found by adding electric and magnetic potentials to the Hamiltonian for a free system:
\begin{equation}\frac{1}{2m}p^{2} \longrightarrow \frac{1}{2m}[p - e A(x, t)]^{2} + e \phi(x, t)
\end{equation}
Using \((\ref{Hderiv})\) we find
\begin{equation}m \dot{x} = p - e A(x, t)
\end{equation}
The Lagrangian becomes
\begin{equation}L = \dot{x} \cdot p - H = \frac{1}{2}m \dot{x}^{2} + e \dot{x} \cdot A(x, t) - e \phi(x, t)
\end{equation}
The gauge transformation
\begin{equation}A \rightarrow A + \frac{\partial \Lambda}{\partial x} \qquad \phi \rightarrow \phi - \frac{\partial \Lambda}{\partial t}
\end{equation}
corresponds to a shift in the Lagrangian
\begin{equation}L \rightarrow L + e \frac{\mathrm{d} \Lambda}{\mathrm{d} t}
\end{equation}
which, as we saw above, does not change the Euler-Lagrange equation.
Localized System
Taking the $ m \rightarrow \infty $ limit while keeping $ k $ fixed in $ (\ref{HarmOsci}) $ gives the Hamiltonian for a localized system:
\begin{equation} H = \frac{1}{2} k x^{2}
\end{equation}
The dual Lagrangian becomes
\begin{equation} K = - \frac{\dot{p}^{2}}{2k}
\end{equation}
which is invariant under $ p \rightarrow p + b $. In this regime $ L $ and $ G $ are undefined. The classical solution is now given by
\begin{equation} x(t) = x_{0} \qquad p(t) = p_{0} - k x_{0} t
\end{equation}
The position is now constant but the momentum is a linear function of time.
In analogy with the free system, we introduce a dual charged system by adding dual electric and magnetic potentials to the Hamiltonian for a localized system:
\begin{equation} \frac{1}{2}k x^{2} \longrightarrow \frac{1}{2}k [x - g Z(p, t)]^{2} + g \psi(p, t)
\end{equation}
Using $ (\ref{Hderiv}) $ we find
\begin{equation} -\frac{\dot{p}}{k} = x - g Z(p, t)
\end{equation}
The dual Lagrangian becomes
\begin{equation} K = x \cdot \dot{p} + H = -\frac{\dot{p}^{2}}{2k} + g \dot{p} \cdot Z(p, t) + g \psi(p, t)
\end{equation}
Now the gauge transformation
\begin{equation} Z \rightarrow Z + \frac{\partial \Omega}{\partial p} \qquad \psi \rightarrow \psi + \frac{\partial \Omega}{\partial t}
\end{equation}
corresponds to a shift in the dual Lagrangian
\begin{equation} K \rightarrow K + g \frac{\mathrm{d} \Omega}{\mathrm{d} t}
\end{equation}
which again, as we saw above, does not change the Euler-Lagrange equation.
General System
Both the free system and the localized system are extreme cases of the harmonic oscillator system. In both cases the Hamiltonian becomes a function of a single mechanical variable. After adding potentials, we have implicit dependence on the second variable.
