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3

First, the classical and semiclassical adjectives are not quite synonyma. "Semiclassical" means a treatment of a quantum system whose part is described classically, and another part quantum mechanically. Fields may be classical, particle positions' inside the fields quantum mechanical; metric field may be classical and other matter fields are quantum ...


2

I) In Palatini $f(R)$ gravity, the Lagrangian density is $$ {\cal L}~=~ \sqrt{-g} f(R), $$ with $$R~:=~ g^{\mu\nu} R_{\mu\nu}(\Gamma),$$ and where $\Gamma^{\lambda}_{\mu\nu}=\Gamma^{\lambda}_{\nu\mu}$ is an arbitrary torsionfree$^1$ connection. II) As OP mentions, the word Palatini refers to that the metric $g_{\mu\nu}$ and the connection ...


4

I'll write here a list of my personal favorites plus some commonly used books. I wouldn't be surprised if your teacher chose either one of the books below as a textbook: i) Mechanics, the first volume of the Landau course on Theoretical Physics; ii) Goldstein's book "Classical Mechanics"; iii) Taylor's book "Classical Mechanics"; iv) Marion's book ...


1

There are results that are mathematically rigorous concerning the semiclassical limit of quantum theories. It is in fact an ongoing and interesting theme of research in mathematical physics. However you need to be rather well versed in analysis to understand the results. The bibliography is quite huge, but I would like to mention the following (some quite ...


1

There seems to be a slight confusion about the meaning of solution: The principle of least action leads to the equation of motion (Euler-Lagrange equation), which correspond to a minimum of the action functional. These equations can have multiple solutions, so there is no contradiction in the formalism. There can multiple solutions that minimize the energy, ...


8

The point is that eq. (1.35) should hold off-shell to have a symmetry, while eq. (1.37) may only hold on-shell. [The term on-shell (in this context) means that the Euler-Lagrange equations are satisfied. See also this Phys.SE post.] In other words: On-shell, the action will only change with at most a boundary term for any infinitesimal variation, whether ...


2

Your last expression (4) is equal to (2), you just have to realize what does it say. $\lambda$ isn't $\tau$ and $$u^c = \frac{d\xi^c}{d \tau} = \frac{d\xi^c}{d \lambda} \frac{d\lambda}{d \tau}$$ If you look back to your Lagrangian and how it was derived, you should be able to say what is $d\lambda/d\tau$. To be very explicit, the action of a free ...


4

The most simplest way is to look at arbitrary process amplitude (S-matrix), to expand it in a series of some constant, then - to use Wick theorem and, finally, to get that n-th amplitude consists of sum over multiplications of all possible numbers of propagators and normal ordering field operators. Sometimes it's convenient to use nonperturbative ...


0

Here is one way to derive the geodesic equations from the Euler-Lagrange equations. First consider a natural Lagrangian system $(M,L)$, where $L\in C^\infty(TM)$. Let $g$ be a Riemannian metric. Suppose in our mechanical system the net force is zero. That is, the Lagrangian is just equal to the kinetic energy, $$L(p,V_p)=\frac{1}{2}mg_p(V_p,V_p)$$In ...


3

Arriving at the Lagrangian Density The Maxwell Lagrangian density in terms of the field-strength tensor $F_{\mu\nu}$ (which has the interpretation of the curvatature of a $U(1)$ connection) is given by, $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} = -\frac{1}{4}\left( \partial_\mu A_\nu-\partial_\nu A_\mu \right)\left( \partial^\mu A^\nu -\partial^\nu ...


0

your Lagrangian is almost correct. But you also need to use a mass term that is conserved, which won't be the case of the mass term with your Lagrangian. If you use: $$ \mathcal{L}_m = \sum_p m_p \gamma_p^{-1}(-g)^{-1/2}\delta^{(3)}(x^j-x^j_p(\tau_p)),$$ where $\gamma_p=dx^0/cd\tau_p$ the Lorentz factor, $u^\mu_p=dx^\mu_p/cd\tau$ the 4-velocity of the ...


6

The strategy is to recall the geodesic equation, $$ \frac{d^2x^\lambda}{dt^2}+\Gamma^\lambda_{\,\mu\nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}=0\tag{1} $$ From your Lagrangian, you'll end up with equations of the form \begin{align} \ddot{\psi}&=f(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\ ...


6

I'll show you how to do this for the 2-plane in polar coordinates. Once you work this out, it should be doable to work it out in your case. You start with the metric $$ds^{2} = dr^{2} + r^{2}d\theta^{2}$$ Since the geodesics of this metric (i.e., straight lines) minimizes distance, we know that the geodesics are an extremum of: $$I = \frac{1}{2}\int ds ...


1

Here we assume that OP is mostly interested in the Eulerian fluid picture (as opposed to the Lagrangian fluid picture). Both fluid pictures are discussed in great detail in Ref. 1. Note however that in the methods of Ref. 1, the mass density $\rho$ is a dynamical variable. The variation of $\rho$ is important in order to obtain a full set of eoms. But OP ...


1

Hint: The canonical stress-energy tensor from Noether's theorem is not necessarily symmetric, and often needs to be improved with appropriate improvements terms. This is e.g. the case for EM. See also e.g. this Phys.SE post and links therein. References: Landau and Lifshitz, Vol.2, The Classical Theory of Fields, $\S$33.


2

Hints: The einbein $e(\lambda)$ is not a dynamical field because there is no $\dot{e}(\lambda)$ present. It is a so-called auxiliary field or generalized Lagrange multiplier. Its EL eq. simplifies to $$\tag{1} (me)^2~\approx~-g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu} . $$ [Here the $\approx$ symbol means equality modulo eom.] Here $m$ is the restmass of the ...


2

As Wikipedia says, a great circle is a circle formed by the intersection of a sphere and a plane that passes through the center of the sphere. The great circles parametrized by $\theta =\tau$ and $\phi =\text{const}$ are not all the great circles. They are only the `vertical' great circles, that is, the great circles formed by intersection with a plane ...


1

Consider an element $g$ of the symmetry group. Say $g$ is represented by a unitary operator on the Hilbertspace $$ T_g = \exp(tX) $$ with generator $X$ and some parameter $t$. It acts on an operator $\phi(y)$ by conjugation $$ (g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other ...


1

The generalized coordinates are the 2 polar coordinates $r$ and $\phi$. There is 1 constraint. This means we have 2 Lagrange equations of first kind with a constraint force term: Lagrange equation for $r$ is the Newton's 2nd law in the radial direction with a centrifugal force and no constraint force. Lagrange equation for $\phi$ is the angular Newton's ...


1

You seem to realize that the rotational analog for Newton's law is important here. This law states that the net torque $\tau$ on an object and its angular momentum $L$ are related by $\tau = \dot{L}$. If I read you question correctly you seem to think that because $\ddot{\phi}=0$, that the angular momentum $L$ must be constant. However, this isn't true. ...


4

Here is one line of motivation: On one hand, in the Lagrangian formalism, the Lagrangian energy function $$\tag{1} h(q,v,t)~:=~v^i \frac{\partial L(q,v,t)}{\partial v^i}- L(q,v,t)$$ is defined as the Noether charge for time translations. Noether's theorem states that if the Lagrangian is invariant under time translations, which implies that ...


0

Once you have the Lagrangian (1) (there is a mistake in your equation, look at the book), just express the momemtum as $p = \partial L / \partial \dot{q} = \dot{q}/\omega$. Insert this expression in the lagrangian expression, and you have the result.



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