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Thanks to the procedure suggested by Qmechanic, I have clarified myself. I need just to invert the matrix $m$, since it has for the momenta $\sigma_{em}=\Sigma :\nabla u+P^T\nabla \phi$ $d_{em}=P:\nabla u-\epsilon\cdot\nabla\phi$ or $\left(\begin{array}{c}\sigma_{em}\\d\end{array}\right)= \left(\begin{array}{cc}\Sigma & P^T\\ P & ...


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In order to perform the (possibly singular) Legendre transformation, it is necessary to have information about pertinent rank conditions of the structure constants $\rho$, $\omega$, $\Sigma_{ij,k\ell}$, $\epsilon_{ij}$ and $P_{ijk}$. In this answer, we will sketch how the (possibly singular) Legendre transformation is performed in principle: We will use ...


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It is there, it's just hidden by the change of coordinates. Written in Cartesian coordinates, the kinetic energy is $$ T=\frac12m_1\dot{x}_1^2+\frac12m_1\dot{y}_1^2+\frac12m_2\dot{x}_2^2+\frac12m_2\dot{y}_2^2+\frac12I_1\dot\theta_1^2+\frac12I_2\dot\theta_2^2\tag{1} $$ where the last term is the rotational kinetic enregy. If you let \begin{align} ...


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The stress-energy tensor, is up to multiplicative factors, can be defined by $\frac{\delta S}{\delta g^{\mu\nu}}$, where $S$ is the action and $g_{\mu\nu}$ is the metric. When people talk about the graviton, they talk about quantizing the metric around it's classical solution, so we consider field values $g_{\mu\nu} = g^{(c)}_{\mu\nu} + h_{\mu\nu}$, where ...


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OP asks (v1): How one can know the gauge field emerging from the local gauge invariance is actually the EM field? Assuming that OP is pondering about gauging theoretical models (rather than concerned with our actual world and phenomenological inputs) then the answer is: One cannot know. For starters, the gauge group $G$ could be different than $U(1)$. ...


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It is not known yet. Gravitons are from quantum mechanics model, while stress-energy tensor is from General relativity (GR) model. Two models are not connected until quantum gravity created. Also, gravitons were never observed, so they are pretty hypothetical. Simultaneously, it is known, that metric tensor is "generated" by stress-energy tensor. ...


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Read Lagrange's Mécanique analytique (English translation: Analytical Mechanics). The book is split up into two parts: statics and dynamics. The first chapter, "The Various Principles of Statics," is a beautiful historical overview. Lagrange works out many problems; for example, he has a chapter entitled "The Solution of Various Problems of Statics." But, ...


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I have finished reading a great book called "The Theoretical Minimum" by Leonard Susskind (a famous string theorist) and George Hrabovsky. It's about classical mechanics but mainly talks about both the Lagrangian formulation and the Hamiltonian formulation of classical mechanics. It is great for beginners in physics or just about anyone. It also reviews the ...


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I don't think that there would be any more diagrams. Having a total derivative term in the Lagrangian leads to derivative interaction vertex, which after symmetrising gives you something like \begin{equation} ig \sum_i p_i \ , \end{equation} where $g$ is some coupling and $p_i$ the momenta of the particles. This vertex, however, vanishes due to momentum ...


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The error is that $\gamma_5$ doesn't intrinsically change sign under parity. Also, don't forget that under parity the spatial components of $W_\mu$ change sign. And also $\gamma^0 \gamma^\mu \gamma^0 \neq \gamma^\mu$.


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An external force $F_{\rm ext}(t)$ appears as a source term $qF_{\rm ext}(t)$ in the Lagrangian. For example, if the equation of motion is, $$\tag{1} m\ddot{q}~=~-\frac{\partial V(q)}{\partial q} + F_{\rm ext}(t), $$ then the Lagrangian reads $$\tag{2} L(q,\dot{q},t)~=~\frac{m}{2}\dot{q}^2-V(q)+ qF_{\rm ext}(t).$$


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In the case that the force is conservative I would model the force by adding an extra potential term $\psi$ to the Lagrangian such that: $$\vec{F} = -\nabla\psi$$ If the unforced Lagrangian was $$L_{\text{unforced}} = T - V$$ the forced version is now $$L_{\text{forced}} = T - (V + \psi)$$ As far as I am aware modelling non-conservative forces in the ...


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$$L= 1/2(-f(x)\dot{t}^2 + g(x)\dot{x}^2 + 2l(x)\dot{x}\dot{t})$$ I assumed from the form given that $L$ is a function of $x(\tau), \frac{d x}{d\tau}, t(\tau)$ and $\frac{d t}{d\tau}$ for some parameter $\tau$. Therefore we have the Euler-Lagrange equation for $t$ and one for $x$ \begin{equation} \frac{\partial L}{\partial t} = ...


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Short answer: Already "true path" is an ugly choice of wording to say the "path that minimizes the action", taken from Hamilton's principle of least action, intuitively: Mechanical systems favor paths along which the difference between the kinetic and potential energy is as small as possible. More formally Hamilton's principle says: Given the action ...


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Yes, I agree it looks like the second equation is missing an overall factor of $\Delta t$ on the right. The action is a function of all the $x$s for the whole trajectory, but the Lagrangian is not. It is only a function of a position and a velocity. So it makes perfect sense to evaluate it using the position and velocity for a single time. Points 8 and 9 ...


2

Yes, your equations aren't quite right. The main issue is that you're assuming a certain form for the normal force that isn't correct. What follows should illuminate why this is so in some detail. When using forces and Newton's Laws to solve this problem, it is overwhelmingly helpful to work in spherical coordinates, not just for locating the position of ...


1

A simpler method would be to use spherical coordinates. To clear the notation, I define: $$\begin{align} z&=R \cos \theta (t)\\ y&=R \sin \theta (t) \sin \phi(t)\\ x&=R \sin \theta (t) \cos \phi(t) \end{align}$$ The kinetic energy is then $$T=\frac{m}{2}\left(\dot x^2+\dot y^2+\dot z^2\right) =\frac{mR^2}{2}\left(\dot ...


2

The laws of physics are discovered through a mixture of heuristics, modelling and inference. In case of momentum, the story goes like this: It is possible to 'transfer motion' from one body to another. However, experiment shows that it is not velocity that is conserved during such transfers, but another 'quantity of motion'. We give that quantity the name ...


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In particle mechanics you integrate along a path, which is bounded by points, but in field theory you integrate over a spacetime volume, so your boundary is a hypersurface, not just points. For a typical quantum field theory process (at least the way it's formulated for calculations), there is some initial state consisting of noninteracting wavepackets, ...


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In field theory the values of the field at every point in space are independent degrees of freedom, just like the positions of different particles in a multi-particle system. So, AFAIK to specify the initial and final configurations for an action integral you have to give the values of the field at every point in space at the initial and final times. The ...


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The author assumed separation of variables, $$ L(u,v)=E(u)B(v) $$ which leads to $$ \frac{\partial L}{\partial u}\frac{\partial L}{\partial v}=B\frac{\partial E}{\partial u}E\frac{\partial B}{\partial v}=-1 $$ Rearranging, $$ E\frac{\partial E}{\partial u}=-\left(B\frac{\partial B}{\partial v}\right)^{-1}\tag{1} $$ Note that $$ E\frac{\partial E}{\partial ...


1

I'll get you started. It's not as bad as you might think at first glance. I think it's easiest to see how to proceed using vector notation. Let $\mathbf r_1, \mathbf r_2, \mathbf r_3$ denote the positions of the three masses in the plane containing the hoop. Let $\ell_{ij}$ denote the length of the spring attaching mass $i$ to mass $j$, then ...


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The Lagrangian is what is integrated over spacetime in the action, i.e. has to be a 4-form. As such, it is necessarily a (pseudo-)scalar under Lorentz transformations. When wondering about Lorentz transformations and such, the Hamiltonian is, as a non-Lorentz-covariant object, not a good starting point, by the way. It is often better to start with the ...


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Yes, a projectile confined to the $xz$-plane is a holonomic system, since the constraint $y=0$ is a holonomic constraint.


1

We start with the action $$ S_{\Lambda,F} = \int d^4 x \left[ \frac{1}{4g^2} F_{\mu\nu} F^{\mu\nu} + a \Lambda_\mu \partial_\nu \ast F^{\mu\nu}\right] $$ This action is equivalent to $S_A$ in the sense that the equation of motion for $\Lambda_\mu$ when plugged into $S_{\Lambda,F}$ gives $S_A$. This is legit since $\Lambda_\mu$ appears linearly and without ...


0

No, the principle of least action started well before quantum mechanics. It is a variational principle that, when applied to the action of a mechanical system, can be used to obtain the equations of motion for that system. The principle led to the development of the Lagrangian and Hamiltonian formulations of classical mechanics. this new formulation of ...


2

Comments to the question (v4): The question formulation seems to talk about affine parametrizations before applying the principle of stationary action. In the context of Riemannian$^1$ geometry, an affine parametrization of a (not necessarily geodesic) curve means by definition that the arc-length $s$ and the curve parameter $\lambda$ are affinely related ...


3

What Qmechanic said in comments is pretty solid, "Lagrangian (2) is not bounded from below because the kinetic term of $A_0$ field has the wrong sign, and hence the theory is not physical in the first place", but I think your Question needs a change of emphasis. Your Lagrangian allows us to construct four equations of motion for four non-interacting fields. ...


0

After thinking about this a bit I realize the question was due to a rather silly misconception,the converse was not mentioned in the usual references because it's trivial from the proof of the usual direction. For future reference I will include a quick version here. $$F = L^2/2$$ $$ EL[F] \equiv \frac{d}{ds}(\frac{\partial}{\partial \dot{x^\mu}}F)- ...


3

Orientation. Let a lagrangian $L$ that is a local function of position and velocity be given. For a parameterized path $x$ on $M$, define a corresponding action $S$ as follows: \begin{align} S[x] = \int_a^b ds\,L(x(s), \dot x(s)). \end{align} Let $\delta x$ be a fixed-endpoint variation, then a standard computation shows that the corresponding variation ...


0

Field $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ with a given spin and mass (i.e. field which transforms under irrep of the Poincare group) must satisfy some determined conditions called irreducibility conditions: $$ \tag 1 \hat{W}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = -m^{2}\frac{n + m}{2}\left(\frac{n + m}{2} + ...


0

Hmm, cool problem. At each collision with the wall, $v_{r}=\dot{r}$ flips sign but $\dot{\theta}$ stays the same, and thus so does the angular momentum $L$. No idea if this works, but it seems worth a shot - I was inspired by quantum mechanics. You basically have a particle in a well. So \begin{align} L &= T-V\\ &= \frac{1}{2}mv^{2} - V\\ &= ...



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