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Let me write the Nambu-Goto action as \begin{equation} S_{NG} \propto \int d^2 \sigma \sqrt{-h} = \int d^2 \sigma \mathcal{L} \end{equation} The Euler-Lagrange equations read \begin{equation} \partial_{\alpha}\frac{\partial \mathcal{L}}{\partial x^{\mu},_{\alpha}} = 0 \end{equation} with the shorthand $\partial x^{\mu},_{\alpha} = \partial_{\alpha} x^{\... 0 The wave functions do not exist! They (alongside with the Hilbert space and its axioms) are mathematical tools for describing the reality. Moreover, this is not the only description possible: from the very beginning the matrix formulation of quantum mechanics of Heisenberg was competing with the wave mechanics of Schrödinger, until their equivalence was ... 3 Such a scenario is metastable. It has several time evolutions possible. The single time evolution for the system is not a defined thing. These scenarios are easy to construct. A simple example would be a roller coaster at the top of its highest hill, with zero velocity as an initial state, and its final state at the bottom of the coaster. It could ... 2 So, let us say you are looking for a trajectory$x = x(t) = \big(\, x^1(t), \, x^2(t), ...,\, x^n(t)\, \big)$that connects two fixed points and times$x_0, t_0$and$x_1, t_1$. Such a trajectory is an optimal trajectory for the action: $$S[x] = \int_{t_0}^{t_1} L\Big(\,x, \,\frac{dx}{dt}\,\Big) \, dt = \int_{t_0}^{t_1} L\Big(x(t), \,\frac{dx}{dt}(t)\Big)... 0 It's true that the specific set of coordinates that we choose is not Lorentz invariant, but when you integrate over all space, it doesn't matter. One however has to check that the measure is invariant. It's a nice excercise to show that, under a Lorentz transformation \Lambda the integral transforms are$$ \int \mathrm{d}^4x\, f(x) \mapsto \int \mathrm{d}^... 0 Action is constructed as Lorentz invariant dencity integrated over all Minkowski space.$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$is Lorentz invariant dencity.$d^4x$is Lorentz invariant measure of integration in rectangular coordinate system. If you wanna to choose some another coordinate system (curvilinear coordinate system), in general you need ... 0 The work$W_{\rm nc}(t)$at instant$t$of non-conservative (nc) forces in eq. (4.6) is ill-defined. Given a path$t\mapsto q^j(t)$, it is unclear what work$W_{\rm nc}(t)$refers to. This dooms a variational principle of some action functional for nc systems, in agreement with standard lore. On the other hand, the infinitesimal virtual work $$\delta W_{\rm ... 1 Recall how work is defined via the non-conservative force: W_{nc}[x, \dot{x}, t]=\int_{\vec{x}(t_1)}^{\vec{x}(t)}\vec{F}_{nc}(x', \dot{x}', t)\cdot d\vec{x'} Requiring \int_{t_1}^{t_2}(\delta L + \delta W_{nc})dt=0 gives, by a completely analogous argument to ordinary Euler-Lagrange, (-\frac{\partial}{\partial\vec{x}} +\frac{d}{dt}\frac{\partial}{\... 0 Many empirical facts about the nature can be reduced to set of symmetries. For example conservation of momentum is equivalent to homogenity of space. Lagrangian is then great method to seek available theories from symmetry. For example in the case of classical mechanics, you can start by free particle and general lagrangian L=L(q,\dot{q},t), where q is ... 0 The Ads-Schwarzschild metric reads (I am working by setting b\equiv 1)$$ds_{AdS-Sch}^{2} = -\left(1-\frac{2M}{r} + r^{2}\right)dt^{2} + \left(1-\frac{2M}{r} + r^{2}\right)^{-1}dr^{2} + r^{2}d\Omega^{2}_{2}.$$The gravitational action with cosmological constant \Lambda=-d(d-1)/2b^{2} is (see equation (4) here)$$I=-\frac{1}{2\kappa}\int_{\mathcal{M}}d^... 4 By defining the lagrangian as $$L = T-V$$ you're restricting yourself. This definition works only in classical mechanics. A lagrangian is a vastly more general concept which I'll summatrize:$L(q(t),\eta(t),t)$is an arbitrary differentiable map $$L:A\times\mathbb{R}\to\mathbb{R}\qquad A\subseteq\mathbb{R}^n\times\mathbb{R}^n$$ where$q(t)$are generalized ... 2 It depends on what you take as 'fundamental reason'. From a mathematical point of view, the Lagrangian$L= T - V$, with$q A\cdot v$included in$V$in the electromagnetic case, can be derived from the equations of Motion. These are in principle the Newton's law you mentioned. However, the Lagrangian$L=T-V$is by far not unique. For example, if you add a ... 0 Not only can we derive Newton's Laws and Classical Electromagnetism from the Lagrangians, but, in the first instance, we actually find these Lagrangians from Newton's Laws and Maxwell's equations together with the Lorentz force law. Then, from the point of view of mathematics, the assumptions are equivalent and any decision which to take as more fundamental ... 3 Lagrangian theories are indeed obiquitous because those are the ones we can understand better and ultimately those with which we can do computations. However, sadly, they only make sense when the theory admits a weakly coupled description. The reason is the very existence of the couplings. The fact that your theory has couplings means that there is a fine ... 2 Dijkgraaf-Witten theories (Chern-Simons theory for a finite gauge group) are an example of a quantum theory without a Lagrangian, which nevertheless has an action (an exponentiated action to be precise). It is neither odd nor difficult to deal with, in fact, toy models like these are often used precisely because they are easier to handle than the quantum ... 1 Since the Lagrangian density (which is confusingly also referred as a Lagrangian) is defined as a function which is integrated on, we may always think the$\mathcal{L}$in inside a 4D integral, since the action is defined via $$S = \exp \left( \int d^4x \mathcal{L} \right).$$ Now, we may decompose the term to two identical parts and integrating one of ... 2 The Euler-Lagrange (EL) equations are not affected by total derivative terms, cf. e.g. this Phys.SE post. In OP's concrete example the Lagrangian density (2) is preferred as it is manifestly real. See also this related Phys.SE post. 1 The energy of an electric dipole moment$\bf{p}$/magnetic dipole moment$\bf{m}$in the external field is proportional to it,$W = -\bf{p\cdot E}$or$W = -\bf{m\cdot B}$. In a ferromagnetic sample the local magnetic dipole moment is propotional to element of volume$d^3x$. This is just the same as you have in the expression for$I'\$. Then, if you consider ...

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In a dynamical variational principle like the principle of stationary action, it is necessary to impose boundary conditions (BCs). Without BCs, the action is not bounded from below nor from above, cf. OP's observations. With appropriate BCs, e.g Dirichlet BCs $$x(t_i)~=~x_i\quad\text{and}\quad x(t_f)~=~x_f,$$ one can determine the stationary path(s) via the ...

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The short answer is that the naive, physicist, and de Rham way are all equivalent. Specifically the de Rham way is defined precisely so that it is the "physicist" way when expressed in terms of a specific coordinate system (atlas). (I believe something is up with your volume forms in the "physicist way" in the version of your post I see.) The physicist way ...

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