You are analyzing a constrained dynamical system, and it is commonly said that 'first class constraints are generators of gauge transformations', e.g. section 1.4 of Henneaux 'Quantization of Gauge Systems'. On a classical level, they are not commonly written in exponential form like that...
The particle-in-a-box problem from the pov of matrix mechanics seems to be discussed in section 7.5 of Razavy's 'Heisenberg's Quantum Mechanics' which cites this paper, it looks very non-trivial.
Your article is a discussion of the GSW equation (2.1.5) I mentioned above, however GSW introduce it out of thin air and the article introduces the einbein basically via magic.
The fool-proof way to arrive at the einbein form of the action, the way that generalizes, is discussed from equations...
It's just the einbein form of the action for a point particle, e.g. eq. (2.1.5) of GSW (note the action you posted is technically zero for massless particles, one adds in the constraint p^2 = - m^2 as a Lagrange multiplier then eliminates momentum to get the action as in (7.1)).
The GL paper
http://www.jetpletters.ac.ru/ps/1584/article_24309.shtml
one of the first papers on what became known as supersymmetry, introduced the idea by noting that only a fraction of the possible interactions invariant under the Poincare group are realized in nature, hence it may be that...
The Hamiltonian is introduced in the calculus of variations as part of breaking up a 2nd order ode into a system of first order ode's, and of course one knows the Hamiltonian is related to the Lagrangian by a Legendre transform, ##L = p \dot{x} - H##, where now the derivatives are at most of...
The result, summarized in
Eisenhart - Separable Systems in Euclidean 3-Space
seems to be proven in
https://www.jstor.org/stable/1968433
and seems to depend on
https://www.jstor.org/stable/2306278
If anyone goes through these and gets a sense of the idea of the theorem, proof and where it...
Peeling ##\Omega^{-2} g^{bd}## off ##I## and ##V## easily gave me the ##\Omega^{-2}## terms in Zee's Ricci scalar, the other three should give the ##\Omega^{-1}## part which I can check another day but there should be no issue.
One can add potentials in SR: to add scalar potentials one considers the action ##S = - mc \int ds## we and on adding a scalar potential ##-\int V(x^{\mu}(\tau)) d \tau##, with ##\tau## proper time, this leads to ##S = - mc \int ds - \int V \frac{d s}{c} = - c \int (m + \frac{V}{c^2}) ds##. The...
I believe you are right, in exponential form those factors from the Christoffels cancel so unfortunately I'm not sure where you're going wrong without basically re-doing it in this notation.
I have a write-up of the conformal transformation of the Ricci scalar in Zee's notation attached in a...
Your Christoffel symbols do not account for the fact that ##g_{ab} \to e^{\Omega} g_{ab}## implies ##g^{ab} \to e^{-\Omega} g^{ab}##, so the Christoffel symbols (compare to those in the wiki link), and especially the derivatives of the Christoffel symbols, are off. Taking the conformal factor to...
The point is very simple: velocity is described by a vector, a vector points in a certain direction - if the Lagrangian is to be a direction-independent function of velocity the Lagrangian can only depend on the velocity vector in a way that eliminate the notion of directionality associated to...
The anti-symmetric tensor is built from the direct product of a dotted spin ##k = \tfrac{1}{2}## and an un-dotted spin ##l=\tfrac{1}{2}## ##\mathrm{SL}(2,\mathbb{C})## spinor, in which case it has ##(2k + 1)(2l + 1) = 4## components - an anti-symmetric tensor built from the direct product of two...
A four vector ##x^{\mu}## transforms under a Lorentz transformation as ##x^{\mu} \to x'^{\mu} = (e^{-\frac{i}{2}\omega_{\rho \sigma} J^{\rho \sigma}})^{\mu} \, _{\nu} x^{\nu}## where ##J^{\rho \sigma}## generate the vector representation of the Lorentz algebra, ##(J^{\rho \sigma})^{\mu} \...