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5

You have seen that the substitution $$L\longrightarrow L':= L+\frac{\mathrm{d}F}{\mathrm{d}t}$$ does not change the Euler-Lagrange equations. Now, this happens because the time derivative satisfies the Euler-Lagrange equations identically. Let us consider a concrete example. Take the Lagrangian of a simple harmonic oscillator: ...


4

Let here consider point mechanics (as opposed to field theory) for simplicity, i.e. the generalization to field theory is left as an exercise. I) Bad news. If the Lagrangian action functional $S[q]$ is non-local, the usual definition of Lagrangian momenta as a partial derivative $$p_i~:=~\frac{\partial L}{\partial v^i} \tag{1}$$ does not apply. II) Good ...


4

A negative coupling leads to a Hamiltonian that is unbounded from below, and hence unphysical, since there is no lowest-lying energy state. Similarily, a phase would mean the Hamiltonian is not self-adjoint, and time evolution would not be unitary.


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Comment to the question (v4): Classically, the Lagrangian for a fermion system reads $$ L ~=~ \int\! d^3x~ i\psi^{\dagger}\dot{\psi}-H.\tag{A}$$ The Legendre transformation from the Lagrangian to the Hamiltonian formalism is tricky for at least three reasons: The traditional Dirac-Bergmann analysis leads to constraints. See e.g. my Phys.SE answers here ...


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The most fundamental parts of Lagrangian mechanics involve calculus. The action principle involves an integral and the Euler-Lagrange equation is a partial differential equation. Unless the students are pretty good with calculus it will be quite hard to teach.


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You are right. If $q$ is a generalized coordinate then $\dot{q}$ is the generalized velocity and hence the generalized momentum is \begin{equation} p = \frac{\partial L}{\partial\dot{q}} \end{equation} Therefore, your sequence looks correct. Further, equations (20) and (21) of the article you have referenced also tell that the $p_{\theta_i}$ are indeed ...


3

If $Q$ is configuration space, then the Lagrangian is a function $L: TQ\times \mathbb{R}\to \mathbb{R}$. Let the cotangent bundle $M:=T^{\ast}Q$ be the corresponding phase space. The Hamiltonian/phase space Lagrangian is a function $L_H: TM\times \mathbb{R}\to \mathbb{R}$.


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Disclaimer: Let us here avoid the discussion of how to assign a stress-energy-momentum (SEM) pseudo-tensor $t^{\mu\nu}$ to the gravitational field. The word pseudo here refers to the fact that $t^{\mu\nu}$ is not a tensor wrt. general coordinate transformations; only a rigid subgroup thereof. In other words, the pseudo-tensor ...


3

OP wrote (v3): Is there anything in particular I should be careful of? Yes. Watch out for secondary constraints, cf. e.g. this Phys.SE post. Below follows a brief partial derivation. Let Greek letters $\mu,\nu,\ldots$ denote spacetime indices, while Roman letters $i,j,\ldots$ denote only spatial indices. The Lagrangian density $$ {\cal L}~=~ ...


2

Let $X$ be the phase space. Then $L_\text{ph}(q,p,\dot{q},\dot{p},t)$ is a function on $TX\times \mathbb{R}$1, since the coordinates of $TX\times\mathbb{R}$ are precisely the coordinates of $X$, i.e. $(q,p)$ and their derivatives $(\dot{q},\dot{p})$ (and time $t$). If Hamilton's equations are fulfilled, there are relations among $q,\dot{q},p,\dot{p}$ (the ...


2

Well you just showed ${d \over dt } { \partial L' \over \partial \dot q}- { \partial L' \over \partial q}= {d \over dt } { \partial L \over \partial \dot q}- { \partial L \over \partial q}=0$ right? ${d \over dt } { \partial L \over \partial \dot q}- { \partial L \over \partial q}=0$ is the equation of motion for $q$, in other words this equation means ...


2

With a potential which is proportional to the inverse square of the radius, you can write the energy as $$E=\frac{1}{2}m \dot{r}^2 +\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$ where $L$ is the angular momentum $$L=m r^2 \dot{\phi}$$ You can rewrite the energy in the following way $$E=\frac{1}{2} m \left(\frac{dr}{d\phi} \dot{\phi} \right)^2+\left( ...


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In this answer we apply the general non-local theory developed in my Phys.SE answer here to OP's non-local example. Let us for simplicity assume that time belongs to the unit interval $[t_i,t_f]=[0,1]$. OP's non-local Lagrangian action functional reads (modulo some sign conventions$^1$) $$ \left. S[q,v]\right|_{v=\dot{q}}, \tag{A} $$ where $$ ...


2

I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer. II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of ...


2

The kinetic term of the Lagrangian is proportional to $$g_{ij}v^iv^j$$ where the $v$s are the generalised velocities. Writing them as the time derivative of the generalised coordinates, i.e. $v^i\dot q^i$, taking the square root, and multiplying by a small time lapse $\epsilon$ you get $$\sqrt{g_{ij}\dot q^i\dot q^j}\epsilon,$$ which is a first order ...


1

The result is not physical, so there must be a mistake! The potential energy is not correct. The negative sign in the gravity term suggests that the lowest potential is at theta=0, whereas this is actually highest potential. I guess with this change you'll have a transition from positive to negative with oscillation to falling over being the two cases.


1

Just as a quick example: say that the dot product were not invariant under transformations. Then let's say that we have two reference frames, A and B, where reference frame B is rotated and displaced with respect to A and which moves at a constant speed w.r.t. A (where $v\ll c$). Then the researcher in A wants to calculate the gravitational attraction ...


1

Hint: $\frac{\partial }{\partial \dot{q}}\dot{F}=\frac{\partial}{\partial \dot{q}}\left(\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial \dot{q}}\ddot{q}+\frac{\partial F}{\partial t}\right)$ What does $F=F(q,t)$ imply about $\frac{\partial}{\partial \dot{q}} F$?


1

The small oscillation approximation considers terms in the Lagrangian to quadratic order in $\theta_i$ and $\dot{\theta}_i$. The reason to only work to zeroth order in $\theta_i$ in one term is because the pertinent term in the Lagrangian is already quadratic in $\dot{\theta}_i$.


1

A simple figure of the system at hand is given below. In writing the kinetic energy we use the velocities of particles. In particular, we square the velocities. And velocity is nothing but the derivative of the displacement. Here we have two particles where we showed their displacements by $x_1$ and $x_2$. That means kinetic energy for the first particle ...


1

The Euler-Lagrange equation holds for Lagrangians that depend on at most first order derivatives of $q$. But when we take the transformation: $$L' = L+\frac{dF(q,\dot q,t)}{dt} $$ we find that our new Lagrangian now has a dependence on the second order derivative of $q$: $$L' = L+\frac{dF(q,\dot q,t)}{dt} = L+\frac{\partial F}{\partial ...


1

Gauss-Ostrogradsky theorem does not actually "care" for a "type" of space or functions that you've got, so it has the same form in curved space and in flat space: $$ \int_D d^4x \frac{\partial}{\partial x^\mu} Anything = \oint_{\partial D} d\sigma_\mu Anything $$ where $d^4 x$ and $d\sigma_\mu$ are constructed from differentials of coordinates the same ...


1

A constraint condition can reduce the DOF of the system if it can be used to express a coordinate in terms of the others. This can always be done in case of holonomic constraints which are basically just algebraic functions of the coordinates and time. This means that you just have to manipulate the constraint equation in such a way that one of the ...


1

Normally, when combining Lagrangians, we often leave the constant multiplying factor to be determined by experiment. For example, if $\mathcal{L}_{k}$ is the kinetic term (for a system of charges and the electromagnetic field), and we choose to describe the electromagnetic coupling by $\mathcal{L}_{int} = A_\mu J^\mu$, then we combine them as ...


1

If all you are looking for is a basic introduction without the calculus of variations, then the following article (which, however, assumes knowledge of elementary calculus as a prerequisite) may be of help: Hanc, Jozef, Edwin F. Taylor, and Slavomir Tuleja. "Deriving Lagrange’s equations using elementary calculus." American Journal of Physics 72.4 (2004): ...


1

The two body problem can be made equivalent to the one body problem. Say you have a mass $m_1$ at $\vec{r}_1$ and a mass $m_2$ at $\vec{r}_2$ interacting gravitationally. Now focus your attention on the center of mass $\vec{r}$ and the difference vector $\vec{r}_{12}=\vec{r}_2-\vec{r}_1$. Then the kinetic energy of the system can be written as ...


1

I'm not familiar with "Modern Analytical Mechanics" by Pellegrini & Cooper so I can't comment on that one but I'm very familiar with the other two books you mentioned. Landau's books are generally excellent but tend to be shorter in length and sometimes very dense. Nearly every paragraph has some profound insight that you'll miss if you don't ponder ...



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