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1

Often we impose regularity conditions on a Lagrangian formulation to simplify the calculations, and/or so that we can work within some mathematical framework, such as e.g., differentiability, that constraints are holonomic, that the rank of the Hessian is maximal, or at least don't jump, other rank conditions, see e.g. this Phys.SE post, locality, etc. ...


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Generally speaking, people do not consider the generic class of Lagrangians, since they tend to be interested in physical systems, but yes, there are examples of "bad Lagrangians". A Lagrangian should certainly be an integrable function, as well as $C^1$ (or at least weakly differentiable) in its variables, but more importantly, it should have an ...


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As you point out: surely the Lagrangian must be capitalizing on the fact that with all interconversion of kinetic energy and potential energy the decrease and increase must match each other. We can restate that as follows: During the entire time the rate of change of kinetic energy is equal to the minus rate of change of potential energy: In mathematical ...


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OP's heuristic explanation certainly contains the right physical ingredients of both dynamical and kinematic nature, although the important relative minus sign between $T$ and $V$ in the Lagrangian $L=T-V$ could preferably use some further elucidation. For a mathematical proof, see e.g. this related Phys.SE post.


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It is easy to construct situations in which light does not take the "fastest" route. For example, set up a laser pointer to send a beam across the room. Put a mirror in its path and deflect it to a second mirror, then tilt the second mirror to direct the beam to the point on the far wall where the beam would hit if all the mirrors were removed. ...


3

Within a homogenous material, light travels in straight lines. To make light curve as you describe, you would have to vary the refractive index (and hence the effective speed of light) continuously from one side of the cube to the other. What you have then invented is a gradient-index lens.


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Correct me if I’m wrong. You try 2 functions $\phi$ and $\phi_2=\phi+\delta\phi$ which both must satisfy boundary condition. Since boundary condition is linear, logically $\delta\phi$ satisfies it too.


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Comments to the post (v5): First of all: an integral where the support of the Dirac delta distribution coincides with one of the integration limits is ill-defined. However, in OP's case this can be avoided altogether. Just add the term $i\lambda(\phi^{\prime}(0)-c\phi(0))$ to the action $S$ instead. For consistency, we need 2 boundary conditions (BCs): 1 ...


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Well, if you don't mind explaining before Hamilton-Jacobi's equation then it's not impossible. From the derivation of the Hamilton-Jacobi equation (see for yourself!) I have that \begin{equation} \mathrm{d}\mathcal{S}=p\ \mathrm{d}q-\mathcal{H}\ \mathrm{d}t \end{equation} Where $\mathcal{S}$ is the action, $p$ the generalized momentum, $\mathcal{H}$ the ...


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A nice analogy can be made with the low-lying states of the ammonia molecule, where the potential and lowest energy levels are illustrated below. The barrier in the middle splits the energy of the low-lying states and the symmetric combination has lower energy than the antisymmetric one. Roughly speaking, the antisymmetric solution must go through $0$ and ...


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Let's graph the potential and your two wavefunctions (I have normalised the wavefunctions): And this all makes sense. We have either the sum or the difference of the two gaussians. Now let's draw the same graph this time of the wavefunctions squared i.e. the probability density: The total energy is the sum of the particle potential and kinetic energy. Of ...


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Why not try? for example with a Lagrangian of the specific form "polynomial of second order" $$\mathcal{L}(x^{\mu},X,\partial_{\mu}X):= \sum_{i,j=1}^6\left( A_{ij} X^i X ^j + \sum_{\mu=0}^3 B^{\mu}_{ij} \partial_{\mu}X^i X ^j +\sum_{\mu,\nu=0}^3 C^{\mu\nu}_{ij} \partial_{\mu} X^i \partial_{\nu} X^j \right) \tag{1}\label{1}$$ in $X=\big(E_x, E_y,E_z,...


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