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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 ...


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 ...


0

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

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 ...


0

Maybe you can give a rough idea about what the subject is about. You can introduce first for example the Fermat's principle of least time and maybe kind of make an analogy like. "There is a similar principle of minimization in mechanics where you minimize another quantity called action". Maybe if an student is interested you can give him more information. ...


3

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.


0

The Lagrangian, L, of a dynamical system is a mathematical function that summarizes the dynamics of the system. For a simple mechanical system, it is the value given by the kinetic energy of the particle minus the potential energy of the particle but it may be generalized to more complex systems. It is used primarily as a key component in the Euler-Lagrange ...


0

It's the net resultant that a force would react with on application of a certain change in momentum in the body, on which the force is applied.


0

Regarding question 2, in Weinberg's The Quantum Theory of Fields page 295, the author defines the functional derivative of an arbitrary bosonic functional $F[q,p]$: $$\frac{\delta F[q,p]}{\delta q}=i[p,F[q,p]]$$ $$\frac{\delta F[q,p]}{\delta p}=i[F[q,p],q]$$ where $q,p$ are a pair of operators that satisfy canonical conmutation relations. If that is indeed ...


4

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 ...


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 $L_H: TM\times \mathbb{R}\to \mathbb{R}$.


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 ...


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}~=~ ...


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 ...


3

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 ...


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 ...


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( ...


0

What can be frustrating is to start from the idea that we are calculating the general action for any path connecting $x_a$ and $x_b$. Then is when you meet the uncomfortable expression $ S = \frac{m}{2}\int_{t_a}^{t_b} \dot{x}^2 dt =\frac{m}{2}\{[x \dot{x}]_{t_a}^{t_b}-\int_{t_a}^{t_b}x\ddot{x} dt\}$ However, we can still write such general action as $ S = ...


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 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

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 $\frac{\partial}{\partial \dot{q}} F$?


0

I would say that this exercise in particular has nothing to do with constraints, because the existence of a dissipation force is not giving you any type of constraint in the coordinates - I mean, it certainly isn't holonomic since it doesn't eliminate any degree of freedom from the problem. It is just more information on how the system behaves along the only ...


0

All these mathematical objects (scalars, vectors, tensors) are carefully selected to represent real world phenomena and their observable properties. For example, invariance under transformation, means that the variable (or physical object) represented via the transformed, let's say vector or tensor, does not change ,i.e. it is conserved. In general, theories ...


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 ...


0

In this context, it is a change of variables. The variable in the original Lagrangian is $q$, and Goldstein is asking you to use another variable $s$, which is related to the original $q$ via the "transformation": $$s = \exp(\gamma t) \ q$$ and later on, make sense of it (with the later questions). Point transformation in this context refers merely to this ...


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: ...


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 ...


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.


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.


2

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 $$ ...


0

Let us do the RHS first. This just gives us a derivative on the metric: $$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$ The first derivative on the LHS is essentially a derivative of a square, thus $$\frac{\partial L}{\partial \dot x^\lambda}=g_{\mu\lambda}(x(\lambda))\dot x^\mu$$ where we have made the ...


0

I've found the inconsistency. What follows is a derivation of the Hamiltonian for convolutional Lagrangians. Starting with the total variation of a convolutional Lagrangian, we have: $$\delta \mathbb{L}=\int^t_0 \left(\frac{\delta \mathbb{L}}{\delta \dot{q}}\delta\dot{q}(t-\tau)+\frac{\delta \mathbb{L}}{\delta q}\delta q(t-\tau)\right)\,\text{d}\tau ...


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 ...


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 ...


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 ...


0

There is this famous reality: Addition or subtraction of any constant to potential energy doesn't change the equations of motion. In your case $U_0$ is just a constant that one can add or subtract freely. You can assume that system had an initial constant potential energy (independent of your generalized coordinates of course) just before you started to ...


1

Before going to field theory, it seems instructive to first ask the same questions in point mechanics: Can the Lagrangian $L(q,v,t)$ depend on time explicitly? Yes. The Lagrangian $L(q,v,t)$ can depend explicitly on time. E.g. there could be external sources. On the other hand, if the Lagrangian does not depend on time explicitly, then the ...


1

OP wrote (v1): Why the Lagrangian density becomes a function of the spatial derivatives? Well, one intuitive answer is, that if the theory is supposed to be relativistic, and if the Lagrangian density has temporal field derivatives, then it must also contain spatial field derivatives. Another answer is that if the theory is supposed to be local, this ...


1

To make the correct answer clearer, allow me to introduce the canonical momentum $\vec{p}$, given by: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}$$ This way we can rewrite the Hamiltonian as: $$H=\vec{p}\cdot\vec{\dot{x}}-L$$ Let's start by computing $\vec{p}$: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}=m\vec{\dot{x}}+\dfrac{e}{c}\vec{A}(\vec{x},t)$$ ...


1

What happened with $V\left(\sqrt{x^2+y^2+z^2}\right)$? You mean, why does V(r) disappear from the $\frac{\partial }{\partial \dot q_j}$ term, right? It's because V(r) is a function only of $q_j$ not $\dot q_j$. Those variables are treated as independent and so $\frac{\partial V}{\partial \dot q_j}=0$. and why $\partial\dot q_{j} = \partial\dot ...


1

Comments to the question (v7): The director $\vec{n}(\vec{r},t)$ is a vector-valued field. Ericksen-Leslie theory is a field theory. Before studying variational calculus in field theory, and asking which variables are independent, and which are not, it is highly recommended to understand the corresponding problem in point mechanics, see e.g. this Phys.SE ...


2

The extra derivative in Polchinski comes from the following version of the Fundamental Lemma of Calculus of Variation (FLCV): $$\tag{1} \left[ \forall g : ~~\int_a^b\! dx~ g(x) ~=~0 \quad\Rightarrow \quad \int_a^b\! dx ~f(x) g(x) ~=~0\right]\quad\Rightarrow \quad f^{\prime}~=~0.\quad $$ FLCV (1) states in words: If it is true that for all functions ...


0

Finding the potential radii is actually quite simple. I already have: $$V_{eff}(r) = \frac{L^2}{2mr^2}-\frac{GMm}{r} - \frac{GMmL^2}{r^3(mc)^2} $$ I was mistakenly graphing $1/r^2 - 1/r - 1/r^3$, when in actuality it makes more sense to take $1/r^2 - 1/r - 0.1/r^3$, since $1/r^3$ is sufficiently small. Since the orbits are circular, the potential radii ...


1

Comments to the question (v3): On one hand, traditionally, the Batalin-Vilkovisky (BV) operator $\Delta$ in Lagrangian BRST formulation encodes geometric data of the antisymplectic phase space for the model, specifically the antisymplectic structure [i.e. the so-called antibracket $(\cdot,\cdot)$, or odd Poisson bracket] and a path integral volume density ...


0

Your equation has the form $$ V_\text{eff}(r) = \frac{\alpha}{r^2} - \frac{\beta}{r} - \frac{\gamma}{r^3} $$ If you set $\alpha=\beta=\gamma=1$, then you're overestimating the $r^{-3}$ term, which is supposed to be a small correction. You will only find two extrema if the derivative has two roots: $$ V_\text{eff}'(r) = -\frac{2\alpha}{r^3} + ...


0

$$\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{q}_j}=\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{r}_i}\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}=2\dot{r}_i^T\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}$$ (A simple chain rule, while bearing in mind that $\dot{r}_i^T\dot{r}_i$ is essentially $\dot{r}_i^2$. You just need a transposed vector on the ...


2

I'll do all calculations assuming the lagrangian $\mathcal{L}$ acts on a 1-dimensional manifold $M$. I believe you'll find the generalization absolutely trivial, and this will spare me of writing tons of sums. Let \begin{equation} \mathcal{L}: \mathbb{R} \times T M \rightarrow \mathbb{R} \end{equation} be a lagrangian over $T M$, with time in ...


5

(a) Write down the potential energy of the rope as the function $y(x)$. You're almost right, up to a minus sign in the limits of the integral: $$V=\dfrac{mg}{l}\int_{-x_0}^{d-x_0}y(x)\sqrt{1+y'^2}\ dx$$ (b) Since the problem is static, interpret the potential energy as the Lagrangian and find the Lagrangian density. The Lagrangian is usually given ...



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