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Working in terms of velocity variables: $v_x=\dot{x}, v_y=\dot{y}$. The second suggestion redily reduces the order of the system: $$\dot{v}_x + \beta v_x^2\sqrt{v_x^2+v_y^2}=0,\\ \dot{v}_y + \beta v_y^2\sqrt{v_x^2+v_y^2}=0$$ One now has now multiple options available: Switching to polar coordinates: $$v_x=v\cos\phi, v_y=v\sin\phi$$ Solving for the ...

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The relations $$\frac{\partial (L+H)}{\partial t}~=~0\qquad\text{and}\qquad\frac{\partial (L+H)}{\partial q^i}~=~0\tag{1}$$ signify/illustrate the fact that time $t$ and positions $q^i$ are external parameters/passive spectators in the $v\leftrightarrow p$ Legendre transformation $$L(q,v,t)+H(q,p,t)~=~p_iv^i.\tag{2}$$ Eqs. (1) are direct consequences of eq. ...

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Here's a very heuristic argument for how one might come up with such an action. First of all, by looking at the typical Maxwell's equations in terms of $E$ and $B$, we see that the PDEs are linear in the electric and magnetic fields. As such, if we want to encapsulate the information of $E$ and $B$ into the field strength 2-form $F$, then the equations ...

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the position vector of a spherical pendulum is: $$\mathbf R_s=S_z(\theta)\,S_y(\phi)\,\begin{bmatrix} 0 \\ 0 \\ l \end{bmatrix}=l\, \left[ \begin {array}{c} \cos \left( \theta \right) \sin \left( \phi \right) \\ \sin \left( \theta \right) \sin \left( \phi \right) \\ \cos \left( \phi \right) \end {array} \right]$$ you want to ...

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I would say that when external non-conservative forces are present in a system, the model is designed by means of d'Alabert's principle, i.e. the principle of virtual work, which generalizes Euler-Lagrange. D'Alamber's principle (principle of virtual work) states that the variation $$\int_{t_0}^{t_1}\, \Big(\, \delta L\big(q, \, \dot{q}, \, t\big) \, - \,Q(... 2 Any textbook/lecture notes derivation of the correspondence between$$\begin{align} &\text{Operator formalism} \cr & \qquad\qquad \updownarrow \cr &\text{Path integral formalism} \end{align} \tag{1}$$is a formal derivation, which discards contributions in the process. It is naturally to split the derivation into 2 parts:$$\begin{align} &\...

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OP is essentially asking the following. How does the point-mechanical Faddeev-Jackiw formulas from chapter II generalizes to the field-theoretic formulas in section III.B? Well, let's see. The field-theoretic first-order Lagrangian is $$L~=~\int\!d^3{\bf x}~A_{\alpha}({\bf x}) \dot{\xi}^{\alpha}({\bf x})~-~H.\tag{2.1'}$$ An infinitesimal variation reads $$... 1 Firstly, note that S does not represent at all the energy of the system. Indeed,$$ [S]=J\cdot s=[ET]. $$Besides, it contains the Lagrangian but, for a proper definition of the energy, you will need the Hamiltonian and, in general relativity, this is a quite complex and interesting issue. Then, note that the choice of the signature of the metric is an ... 1 Let us start with OP's 2nd point. Yes, in phase space positions q^i and momenta p_i are 2n independent variables, where i\in\{1,\ldots, n\}. Let's introduce a collective notation$$(z^1,\ldots,z^{2n})~=~(q^1,\ldots,q^n,p_1,\ldots p_n).Concerning OP's 1st point, the main issue is that one should use the phase space variables z^I, not the ... 0 H is that thing which is invariant or conserved over time. This follows from Noether's Theorem. That thing is usually called energy.(It is also the generator of time translation. There is a connection between time translation and energy by Noether's Theorem.) But there is a possibly even deeper explanation. This can be found in Leonard Susskind's Theoretical ... 3 You can just define some Hamiltonian H(\vec{q},\vec{p}) and the equations of motion are Hamilton's equations \begin{eqnarray} \frac{{\rm d} q_i}{{\rm d} t} &=& \frac{\partial H}{\partial p^i} \\ \frac{{\rm d} p^i}{{\rm d} t} &=& - \frac{\partial H}{\partial q_i} \end{eqnarray} If you integrate the equations you'll get q_i(t). You might ... 0 You can simply start with a Hamiltonian, but the Hamiltonian has to be written in terms of the correct Hamiltonian variables—the canonical coordinates and their conjugate momenta. In one dimensions, that means that you can just write out a function H=H(x,p) and derive Hamilton's equations from it in the usual way (e.g. \dot{p}=\partial H/\partial x). ... 1 Generally people do not work with operator-valued Lagrangians. In fact, the only source I can think of off the top of my head that does is Weinberg's QFT volume 1, and if I recall correctly he only uses it to work out some things for QED. Working with operator-valued Lagrangians will generically get you into a lot of trouble with ordering issues. However, ... 0 There must be an external force - a torque - to keep the rod rotating at a constant speed when the bead is moving. Your Lagrangian for theta does not allow for this. 1 Regarding the first question, a simple way to convince yourself that multiplying fields in the lagrangian corresponds to coupling is to think back to particle mechanics. Consider the lagrangian for two masses m_1, m_2 coupled by an ideal spring (in one dimension for simplicity): \begin{align} L = \frac{1}{2}m_1 \dot{q}_1^2 + \frac{1}{2}m_2 \dot{q}_2^2 - \... 2 All definitions (1)-(3) are in principle the same. However, the various notations^1 may warrant some explanation: Eq. (2) uses the standard/traditional definition of a functional/variational derivative (FD) of the action functional S=\int\!d^dx ~{\cal L} in d spacetime dimensions. Eq. (1) uses a 'same-spacetime' FD \frac{\delta {\cal L}(x)}{\delta\...

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For example a particle in a newtonian setting has a configuration space 𝑄=ℝ3. The particles trajectory is then a particular curve in this configuration space. Right. A configuration is a point in $Q=\mathbb R^3$, and the particle evolves along a curve $\gamma:t \mapsto \gamma(t)\in Q$. These seem like two different definitions to me. Following the first ...

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They are different definitions. Take your first definition where $Q=\mathbb R^3$ for a Newtonian particle. According to that, the configuration space in classical field theory per your example is the space of functions $$\mathbb R^3\to\mathbb R^n\,.$$ So for each point in time $t$, you get a function that depends on the coordinates $(x,y,z)\in\mathbb R^3$: ...

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I think you are including a spurious minus sign in your summation. The index contraction over $\mu$ should be $$\gamma^{\mu} \partial_{\mu} = \gamma^0 \partial_0 + \gamma^1 \partial_1 + \gamma^2 \partial_2 + \gamma^3 \partial_3 = \gamma^0 \partial_0 + \vec{\gamma} \cdot\nabla$$ The reason the minus sign occurs in a sum like $p \cdot x$ is because the metric ...

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You are correct that time invariance of laws does not require energy conservation. It is a common misunderstanding of the content of Noether's theorem. It is not about "laws of physics being independent of time" implying "physics concept of energy is conserved". Those are simplified colloquial statements that are easy to state and ...

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If $\lambda>0$, the force is repulsive. If it $\lambda<0$, the force is attractive, but the system is unstable to vacuum decay. To be more specific $\lambda \phi^4$ interaction is the relativistic version of the Schrodinger delta-function potential.

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There isn't much to solve actually. The Euler-Lagrange equation is $$\frac{d}{dt}\frac{\partial L(r,\dot r,t)}{\partial \dot r_i} = \frac{\partial L(r,\dot r,t )}{\partial r_i}$$ where $r$ stands for the collection of all coordinates $r_i \in r$. In your case $r=(x,y,z$). The Lagrange function for a free particle is simply $$L(r,\dot r,t) = \frac{1}{2}m(\... 2 Here is an easier way to think about this. Imagine that the laws of physics could be time dependent, and you arrange for the law of gravity to be turned off in the shank of an otherwise dead Thursday afternoon. In anticipation of this useful event, you have a machine set up which uses falling weights to perform useful work, and a pile of weights ready to ... 0 Noether's theorem is completely general and makes it easy to give general results with a minimum effort. However, its generality and its relations with symmetries are sometimes misunderstood. A typical misunderstanding is that it is necessary to explore the relations between symmetries and conservation laws. This is not the case, as shown in the answer by ... 2 It's easy to imagine a universe that has time-symmetry but not energy conservation. It's certainly not easy for me. To me, time-translation symmetry means that if I start a system in some arbitrary initial state x_0 at some initial time t_0, then it will evolve the same way regardless of what t_0 is. But since evolution is generated by the Hamilton ... 1 OP has proven that every infinitesimal transformation of an action is automatically a symmetry on-shell, i.e. the notion of on-shell symmetry is a vacuous notion, and therefore not used in practice. In contrast, a symmetry of an action is conventionally assumed to hold off-shell. See also this related Phys.SE post. 1 The point is that we define the new coupling parameter e as$$e:=g\sin\theta_W=g'\cos\theta_W$$in order to avoid carrying around these longer expressions 0 I would like to add an explanation to @OON's statement that I_{3} is an eigenvalue to T^{3}: If we consider at least an SU(2)_{L}-doublet of the form$$ T^{3}\psi = T^{3}\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix}= \frac{1}{2}\sigma^{3}\begin{pmatrix} \nu_{L} \\ e_{L} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\...

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$I_3$ denotes the eigenvalue of the $T_3$ matrix corresponding to the fermionic field in the sum. In the very first equation you write the sum is not over electrons and neutrinos. Instead you have the left doublets on which $T_3$ acts as the Pauli matrix, $$l_k= \begin{pmatrix}\nu_{Lk}\\e_{Lk}\end{pmatrix}$$ (where $k$ denotes the ...

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I totally concur with J. Murray's explanation, which is excellent, IMO. But because you seem to be somewhat in the darkness about where the need for a matrix comes from in the Lagrangian formalism, let me add an approach which is to be understood as complementary of the previous one. Let's say your mechanical system has $n$ Cartesian coordinates. The ...

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A broad class of Lagrangians is of the form $$L(\mathbf q,\dot{\mathbf q})= \frac{1}{2}\sum_{i,j}g_{ij} \dot q^i \dot q^j - V(\mathbf q)$$ where $g_{ij}$ are the components of a (generally $\mathbf q$-dependent) symmetric bilinear form; this is the "kinetic matrix" to which the other question refers. In this class of Lagrangians, the canonical ...

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As a simple example, imagine rubbing your hand on some wood. Running your hand along the grain of the wood feels different from running your hand against the grain of the wood. Another example would be in your every day life: motion in the plane of the earth is quite different than perpendicular to it due to gravity! Thinking more mathematically, you could ...

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You might be talking at cross purposes. If $x_2$ is the equilibrium position with both springs unstretched, then when the spring on the left is compressed by $x_2$ the spring on the right is stretched by that amount: that's how it is in your current setup. If $x_2$ is defined differently, e.g where $x_2=0$ has one spring unstretched but the other ...

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OP's sought-for commutators $$\left[\frac{\partial}{\partial \dot{q}^j},\frac{\mathrm d}{\mathrm d t}\right]~\stackrel{(3)}{=}~\frac{\partial}{\partial q^j}\tag{1}$$ and $$\left[\frac{\partial}{\partial q^j},\frac{\mathrm d}{\mathrm d t}\right]~\stackrel{(3)}{=}~0\tag{2}$$ follow from the explicit formula $$\frac{\mathrm d}{\mathrm d t} ~=~\frac{\partial}{\... 0 Careful with interchanging total and partial derivatives. In this context, I'm guessing you consider f along a curve q(t), then$$ \dfrac{d}{dt}f(q(t), \dot{q}(t), t) = \dfrac{\partial f}{\partial q}\dot{q}+ \dfrac{\partial f}{\partial \dot{q}}\ddot{q}+\dfrac{\partial f}{\partial t} $$Something similar holds for \tfrac{\partial f}{\partial \dot{q}}(q(t)... 0 This is quite a tedious computation, if you want to perform it this way. I would propose an alternative approach (but if you're really interested in doing the calculation in this particular manner, I'll just erase this answer). My line of reasoning is the following: it is easily provable that \mathcal{L}_\mathrm{kin} is invariant under the gauge ... 0 Let's consider an isolated free particle A with Lagrangian $$\label{kinetic_energy} T(\mathbf{q},\dot{\mathbf{q}})=\frac{1}{2}m\sum_{i,j=1}^{n}\mathcal{G}_{ij}(\mathbf{q})\dot{q}_{i}\dot{q}_{j}$$ and let's bring from infinite distance a system of particles that will interact with A. The nature of the interaction can be seen ... 2 I'll do it step by step. First lower all the indices: $$-\frac{1}{4}F_{\mu \nu}F^{\mu\nu}=-\frac{1}{4}F_{\mu \nu}g^{\mu\rho}g^{\nu\sigma}F_{\rho\sigma}$$ then expand the products, -\frac{1}{4}(\partial_{\mu}A_{\nu}\partial_{\rho}A_{\sigma}-\partial_{\mu}A_{\nu}\partial_{\sigma}A_{\rho}-\partial_{\nu}A_{\mu}\... 1 I think that maybe you are interpreting {\boldsymbol \gamma}\cdot \nabla as \gamma_a \partial _a while P&S probably mean {\boldsymbol \gamma}\cdot \nabla\stackrel{?}{=} \gamma^a \partial_a . The sign then comes from \gamma^a=-\gamma_a in (+,-,-,-) metric. 5 If we are tasked with quantizing a classical gauge theory, we want to try the lightcone (LC) quantization first because it is so much simpler. In this way we can separate out & count physical DOFs, and otherwise familiarize ourselves with the quantum theory at hand. However, LC quantization sacrifices manifest Lorentz symmetry, so ultimately we would ... 0 In the language of the OP, action is a functional, since it is an integral of the Lagrangian... but over an arbitrary path. In other words, it is an abstract mathematical object, which has no counterpart in the real world. This functional is then minimized in respect to all possible trajectories. In quantum mechanical terms the action along the optimal ... 1 Wonderful answer by Alex G! This provides a nice plot of velocity versus position. Follow up question: If I put in a value for M, say 1500 meters so that this black hole is roughly the mass of our sun and initially r = 9000 meters, is it fair to calculate a time of fall from this dr/dt? I get .000064 seconds of proper time by breaking this up into 700 ... 1 A rigorous definition in function spaces can be given via the Legendre-Fenchel transform; see, e.g., H. Attouch and R.J.-B. Wets, Isometries for the Legendre-Fenchel transform, Trans. Amer. Math. Soc. 296 (1986), 33-60. Applications in elasticity theory are given in P.G. Ciarlet, G. Geymonat and F. Krasucki, Legendre–Fenchel duality in elasticity. Comptes ... 1 Yes, the values of the masses are real positive numbers. Recall how you find them out of the complex matrix Y. Note first that Y Y^\dagger is hermitian, and has positive-eigenvalues, and so can be written as$$ Y Y^\dagger = U D ^2 U^\dagger $$for some unitary U and diagonal real D with no zero entries, for simplicity. Take the positive square roots. ... 0 You're integrating over k twice, which is a bad idea. If you introduce a second dummy variable q, then the first term becomes something like$$\int d^4x \int d^4k\int d^4q \left[ \tilde \phi(k)\tilde \phi(q)(-k^2-m^2+i\epsilon) e^{-ix\cdot (k+q)}\right]$$Now the only thing which depends on x is that exponential. Noting that \int d^4x \ e^{-ix\cdot ... 3 You can see it in lots of ways. Here is one of them: Consider the first-order Lagrangian for the Maxwell theory$$L=-\frac{1}{4g^2}F^{\mu\nu}F_{\mu\nu}+\partial_{\mu}\widehat{A}_{\nu}(\star F)^{\mu\nu}.$$The fields F^{\mu\nu},\widehat{A}_\mu are assumed to be independent and F^{\mu\nu} is not the field strength of a 1-form at this level. Varying this ... 1 The statement by Symon is not specific for using the lagrangian formulation of classical dynamics. The statement extends to any use of the concept of kinetic energy in theory of motion. The equivalence class of inertial coordinate systems expresses the physical properties of inertia. In order to formulate a theory of motion one must grant the existence of ... 2 Well, there are a ton of reasons. Let's just mention a few: In the Feynman path integral one should sum over all histories/off-shell configurations, not just insert the stationary values. If we start from the Lagrangian path integral, at the physics level of rigor, we would have to insert by hand the ad-hoc Feynman fudge factor, cf. e.g this Phys.SE post. ... 1 I think that the key issue is what you already pinpointed in your last paragraph. The integrand, that is the e^{iS} part, isn't the only result of a derivation of a path integral formula. Another equally important ingredient that has to be derived is the path measure {\cal D}q or {\cal D}x. It is perfectly possible to go from one formulation to the ... 0 Consider the total derivative of the Langrange density with respect to position,$$ \frac{dL}{dx}=\frac{\partial\Phi}{\partial x}\frac{\partial L}{\partial \Phi} +\frac{\partial d\Phi}{\partial x}\frac{\partial L}{\partial d\Phi}+\frac{\partial L}{\partial x} $$With$$ \delta\Phi = \delta a \frac{\partial\Phi}{\partial x}, d\delta\Phi = d(\delta a \frac{\...

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