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Peter Kravchuk has already given a good answer. Here we will follow the programming hint given in the Exercise 1.6. How would one program this minimization problem? By discretization. So the positions ${\bf r}_n$ live on discrete times $$t_n~=~n\Delta t,\qquad\Delta t ~:=~\frac{T}{N},\qquad n\in\{0,\ldots,N\};$$ and velocities are discretized as e.g. ...


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Hints: The Majorana spinor is real. For instance $\bar{\psi}=\psi^T\rho^0$ without complex conjugation. The SUSY transformation $\delta{\cal L}$ of the Lagrangian density ${\cal L}$ does not have to vanish. It is enough if it is a total divergence. See the notion of quasi-symmetry, cf. e.g. this and this Phys.SE posts.


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If your action only depends on first time derivatives, it is then not required for the trajectory to have second time derivative -- i.e. an abrupt change in velocity does not by itself give a contribution to the action. In other words, there is no penalty for changing your speed instantaneously. It then means that you can ignore the boundary values for ...


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1) Configuration space is, in a sense, the possible "positions" of the mechanical system. The states of motion, eg. velocities/momenta are not part of the configuration space. The configuration space (especially when constraits are in the picture) is modelled as some real, $n$-dimensional differential manifold, which I'll denote as $\mathcal{C}$. The ...


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Gauss's principle of least constraint Principles of Least Action and of Least Constraint (a review paper by E.Ramm) If I remember correctly, this principle has been used to derive equations of motion for Gaussian isokinetic thermostat (i.e., a computational algorithm for maintaining a fixed temperature of the system). Please see, for example, Statistical ...


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Point in configuration space represents configuration of the system, i.e. positions of the constituent particles. Point in phase space represents state of the system, i.e. positions and velocities of the constituent particles together. No. Liouville's theorem has no simple analogue in the configuration space. Depends on what is the task at hand and what are ...


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You may know about it already, but you can find an excellent account of Lagrangian Mechanics on manifolds in the book Mathematical Methods of Classical Mechanics by V. Arnold. Also to specifically address your question: $L:TM \rightarrow \mathbb{R}$ so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate over a ...


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Your professor is telling you something that is absolutely fundamental to a proper understanding of relativity. Suppose we draw out the trajectory of some object on a space time graph, we may get something like this: The path traced out by the object(the blue curve) is called the world line. The length of the world line, $s$, is equal to $c\tau$, where ...


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The shortest world line (geodesic path) is given by the GR coordinate system. GR has $x_0$ (ct), $x_1$ (x), $x_2$ (y), $x_3$ (z). Often in GR, c is set to 1 (c = 1). Distance squared ($ds^2$) is given by $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$. If one second passes, the shortest distance is to stay in the same spatial location, that is $$ds^2 = -dt^2$$ ...


1

With a strong grasp of Lie Algebra and Calculus of variations, "Invariante Variationsprobleme" should provide all the foundation one needs to build Newtonian Mechanics (and so much more). The deeper reason that we use either of these formalism is that they agree with experiment; that either formalism predicts the other is far less valuable than that they ...


1

Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective: The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian density is of the form "Lagrange multipliers times constraints"$^1$ $$ {\cal H}~=~\lambda^{\alpha} ...


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"The number of degrees of freedom can be defined as the MINIMUM number of independent coordinates that can specify the position of the system completely" (wikipedia) In your case the number is ONE, because you only need to know the position of the particle along the curve. It doesn't matter if the curve is not a line, or even contained on a plane, because ...


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I'm not altogether sure what you are asking, but I suspect the following may help. To represent rotations, spins and vectors in $SU(2)$ we work as follows. Rotations live in $SU(2)$. Vectors (in the physicist's sense) live in the algebra $\mathfrak{su}(2)$. The position vector $(x,\,y,z)$ is: $$X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = ...


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I) In this alternative answer we resolve the singular Hessian $H_{\mu\nu}$ of the Nambu-Goto string action by introducing two auxiliary variables from the onset, thereby indirectly showing that the Hessian $H_{\mu\nu}$ must have co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. Consider the extended Nambu-Goto ...


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As the counter example given by Herr_Mitesh shows it is not true and this is because the lagrangian is not uniquely determined. In physics sometimes you don't have to think like in mathematics and in this case you must content yourself thinking that if the lagrangian does not contain x as a variable that is enough for the condition of homogeneity to be ...


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The following might help: $H = \frac{1}{2}(mv^2 + kx^2) + \gamma mkvx$ decays exponentially with time along the solution of the damped system. Check by integrating $H$ with respect to $t$ and using the equations of the system. So the "energy" $H$ decays exponentially instead of remaining constant.


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Perhaps the most enlightening is just to show how it goes in OP's example. If the Lagrangian reads $${\cal L}_1(A,\phi)~:= ~{\cal F}(A)- Ay(\phi),\qquad F~=~{\cal F}^{\prime}(A),\tag{1}$$ then the eom for the "auxiliary" variable $A$ reads $$ F(A)~\approx~ y(\phi) \qquad \Leftrightarrow\qquad A~\approx~ F^{-1}(y(\phi)),\tag{2}$$ where we have assume that ...


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I) In this answer we will consider the standard Nambu-Goto (NG) string and show that the Hessian has co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. The NG Lagrangian density is $${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}}, $$ $$ {\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} ...


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Can Lagrangian Mechanics be justified without referring back to Newtonian Mechanics? Sure; one can deduce Newtons Laws from it. The question is should one? By deducing Newtons Laws one is missing the crucial aspect of induction; the reverse procedure and in a sense more difficult; that is the discovery and invention of a theory that covers a wider ...


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Since the Lagrangian results in exactly the same equations of motion as Newton's laws, I'd say that based on their agreement with experiment both are on equal footing. Of course, to get the right equations of motion from Lagrange's equation you have to pick the right Lagrangian, so then you ask how we systematically pick the right Lagrangian. The recipe in ...


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We do if there is a color trace. The term $D_{\mu}D_{\nu}F^{\mu\nu}=\frac{1}{2}[D_{\mu},D_{\nu}]F^{\mu\nu}$ is proportional to $F_{\mu\nu} F^{\mu\nu}$.


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As the more upvoted answer said, if there are color indices then the covariant derivative doesn't commute with itself and the expression you wrote makes sense. If not, symmetry arguments about the symmetric nature of the derivative and the connection and the anti-symmetric nature of the curvature tensor are enough to reason like I did below that the ...


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It helps to write the full action: $$S = \int \frac{-mc^2}{\gamma}dt - \int U dt $$ The first term can be put in a much better form by noting that $d\tau = \frac{dt}{\gamma}$ represents the proper time for the particle. The action is then: $$S = -mc^2\int d\tau - \int U dt$$ The first term is Lorentz invariant, being only the distance between two points ...


1

It seems, based on the comments above, that you have figured it out. Just for closure, I am writing the steps out. If you had just a parabolic potential well, $V(x) = ax^2$, you could get the period quite easily - for a given mass $m$, the frequency would be $$\omega = \sqrt{\frac{2\alpha}{m}}\\ T = \frac{2\pi}{\omega} = \pi \sqrt{\frac{2m}{\alpha}}$$ For ...


2

Actually the equations of motion one ends up with are not manifestly the same: If I let $$L_1 = \sqrt{g_{\mu \nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}}$$ one finds that the Euler-Lagrange equation is $$ \frac{d^2 x^{\mu}}{dt^2} + \Gamma^{\mu}_{\nu\sigma} \frac{dx^{\nu}}{dt} \frac{dx^{\sigma}}{dt} = f(t) \frac{dx^{\mu}}{dt}, $$ for a suitable function ...


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The state of the universe is not homogeneous and isotropic, but the laws of physics are. For example, the speed of light propagation is the same in all directions, and the mass of the electron is not a function of position. Writing down a Lagrangian requires an assumption about the laws of physics (or more precisely, an assumption about the dynamics). There ...


2

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$ $\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$ $\newcommand{\l}[0]{\mathcal L}$ $\newcommand{\q}[0]{\dot q}$ $\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$ A word of advice: Landau's book, don't read it if you haven't studied that subject before! Your first question has ...


1

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$ $\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$ As indicated in the comments, the reason why you get a term of the kind $\pdv{A_x}{x}\pdv{x}{t}$ is the chain rule. Notice that the magnetic potential can be written in components as $A_i(x,y,z,t)$. However the components $x,y \text{ and } z$ ...


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The geodesic equation is $$ \frac{d^2x^\mu}{dt^2} + \Gamma^\mu_{\nu \rho} \frac{dx^\nu}{dt}\frac{dx^\rho}{dt} = 0$$ The coefficient of $\dot{\phi}^2$ you're seeing corresponds to $\Gamma^r_{\phi\phi}$.


3

The metric times the Kronecker delta gives $$g_{ab} \delta^a_c = g_{cb}$$ Since the Kronecker delta tells us to replace the $a$ indice with $c$. We do this for both terms in your equation, $$\frac{\partial L}{\partial \dot{x}^c} = \frac{1}{2} g_{cb} \dot{x}^b + \frac{1}{2} g_{ac} \dot{x}^a$$ and then rename the dummy indices (the indices that are summed ...


0

Why do you say it is not a soliton solution? I did not verify your answer but assuming its true, then it seems like a soliton to me. Your vacuum consists of $\phi=0,\pm 1$. This solution obviously interpolates between the two vacua $\phi(+\infty)\rightarrow +1$ and $\phi(-\infty)\rightarrow -1$, and is therefore topologically stable. Moreover the region ...


3

It was pointed out by @Peter Anderson in the comment that you forgot the transformation of the derivative, which in infinitesimal form should read $$\delta \partial_n = - g^{lm} \Lambda_{mn}\partial_l$$ which comes from the Lorentz transformation $$\partial_n \to g^{lm}(L^{-1})_{mn} \partial_l$$ (the metric is there to keep the indices in agreement with OP's ...


0

Actually, you can do without complex fields, at least in some general and important cases, and I don't mean replacing a complex field with two real fields. Schroedinger noted that, in the case of a scalar field interacting with electromagnetic field (the klein-Gordon-Maxwell electrodynamics, or scalar electrodynamics), you can use the so-called unitary ...


0

Actually, you don't have to use proper time for a parametrisation of the Euler-Lagrange-Equations / geodesic equations in GR. Just take any parametrisation you want. However, if you solve the equations and use initial conditions for a time/light/space-like-path, that geodesic will stay time/light/space-like with $$ g_{ij}V^iV^j = \text{const.} $$ over the ...


0

What type of fields are you using? If you are working with spinor fields, the representation of Lorentz transformations is complex. So even if the field is real in some reference frame, if you switch to another reference frame it will become complex. There's no way to avoid complex spinor fields.


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There is no non-trivial one-dimensional representation of $\mathrm{U}(1)$ on a scalar field $\mathbb{R}^4\to\mathbb{R}$, but on complex fields $\mathbb{R}^4\to\mathbb{C}$, we have the one-dimensional "phase" representations by $$\phi\mapsto\mathrm{e}^{e\mathrm{i}\chi}\phi$$ for $e\in\mathbb{Z},\chi\in\mathfrak{u}(1)\cong\mathbb{R}$ for $\mathrm{U}(1)$ ...


2

The OP has a point. If a dot denotes time-differentiation $$\dot{q}~\equiv~ \frac{dq}{dt},$$ and if we add a total time derivative to the Lagrangian $$\tilde{L}(q,\dot{q},t)-L(q, \dot{q}, t) ~=~ \frac{dF(q,t)}{dt}~\equiv~\frac{\partial F(q,t)}{\partial q}\dot{q} + \frac{\partial F(q,t)}{\partial t},$$ and if we want to view position $q$ and velocity ...


2

Rigorously speaking, yes, you are right if dealing with the Lagrangian function. Indeed E.-L. equations should be more properly written $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^k}\right) - \frac{\partial L}{\partial {q}^k}= 0\:, \quad \frac{d q^k}{dt} = \dot{q}^k\quad k=1,\ldots, n\:.$$ In other words $\dot{q}^k$ becomes $\frac{d q^k}{dt}$ ...


0

p dot in d'Alembert's formula represents the derivative of the object's momentum with respect to time. delta * r represents what would have been the displacement of the object during the infinitesimal interval of p dot. So what d'Alembert is saying is that Force minus the effect of Force = zero. If you analyze a system this way, you can treat it as though ...


1

As it happens, it is not necessary that two Lagrangians that have the same equation of motion have the same functional form. Consider the Lagrangians $L_1 = T-V$ and $$L_2 = \frac{1}{3}T^2 + 2TV - V^2$$ where $T = \frac{1}{2}m\dot{x}^2$ and $V(x)$ is the potential energy. They both lead to the same equation of motion: $$m\ddot{x} = -\frac{dV}{dx}$$ This is ...


1

If by 'equivalent' you mean equal, then no. They can clearly differ by a constant, but they moreover can differ by a total time derivative. So if two lagrangians $L_{1}$ and $L_{2}$ are such that $L_{1} - L_{2} = \frac{\mathrm{d} \Phi}{\mathrm{d}t}$ for some function $\Phi$, then they lead to the same equations of motion. You can find a proof of this in Jose ...



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