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There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as $$R = \sum_{i=1}^n p_i\dot{q}_i - L$$ You can learn more about it by clicking this link for wikipedia's description of it.

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It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ...

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Yes, OP is right. In the field-theoretic case, the partial derivatives in OP's first formula (1) should be replaced with functional derivatives $$\delta S~=~\int_{t_1}^{t_2}\!\mathrm{d}t\left(\frac{\delta L}{\delta q}~\delta q+\left. \frac{\delta L}{\delta v}\right|_{v=\dot{q}}~\delta \dot{q}\right),\tag{1'}$$ where the Lagrangian $$L[q(\cdot,t),v(\... 2 This is just supplementing Qmechanic's answer. I think the notations here need to be addressed. OP might be confusing Lagrangian (normal L) with Lagrangian density (\mathcal{L}). Formally, we have three fundamental relations:$$L = \displaystyle\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3xS = \displaystyle\int dt \space L = \...

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Okay, let's give it a try. $SU(2)$ sector of Standard Model Lagrangian is rather involved, so we will take a look at something simpler. Neutron-proton interaction comes to my mind. In low energy limit it is mediated by a massive scalar particle — a pion. We will be very qualitative about this, in reality there are a lot of details. Lagrangian will look ...

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I) The first part of OP's construction is directly related to the covariant Hamiltonian formalism for a real scalar field with Lagrangian density $${\cal L} ~=~ \frac{1}{2}\partial_{\alpha} \phi ~\partial^{\alpha} \phi -{\cal V}(\phi), \tag{CW4}$$ see e.g. Ref. [CW] and this Phys.SE post. [In this answer we use the $(+,-,-,-)$ Minkowski signature ...

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No, many other couplings are possible. For example, in the very simple Lagrangian $$L = \frac{1}{2} mv^2 - mgh$$ we have coupled the particle to the gravitational field $\phi = gh$. This is already in relativistically invariant form, since both $m$ and $\phi$ are scalars. (Of course the real story for coupling to gravity is more complicated, but this works ...

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This might help you out a bit: In what sense is a quantum field an infinite set of harmonic oscillators? From my understanding, most people think it provides a useful way to conceptualize uncoupled quantum fields physically. It doesn't, however, work for coupled quantum fields. The main problem with this seems to be that infinite harmonic oscillators give ...

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