# Tag Info

10

General remarks. The momentum you define in the first equation, namely \begin{align} p = \frac{\partial L}{\partial \dot q} \end{align} is not necessarily the same momentum that appears in Newton's Second Law. This momentum is called the canonical momentum conjugate to $q$, and it can be quite different from the momentum you're used to (the one ...

6

Let's compare classical mechanics and GR to attempt to get at the intuition you're looking for. Classical mechanics. Recall that in the classical mechanics of a system of $N$ particles, the configuration of the system at every point in time is represented by a point $x\in\mathbb R^{3N}$. The configuration manifold $\mathcal Q$, namely the set of possible ...

6

Symmetry is present when something $x$ doesn't change under some transformation $T$: $$T(x)=x$$ In an infinite cylinder, there is radial symmetry because if you move at constant height and radius, you see the same figure. In the Lagrangian case, if you change coordinates, the Lagrangian doesn't change. $L(x') =L(x)$ In group theory, group elements will ...

5

Think of a two dimensional horizontal elastic sheet in three dimensions. Suppose we specify the height everywhere with $\phi(x,y)$. Then if the sheet moves up and down there is kinetic energy. This kinetic energy is proportional to $\dot{\phi}^2$ because $\dot{\phi}$ is telling you the velocity of that point on the sheet. Also, suppose we want to pull ...

5

What is a physical theory/model? A given physical theory is typically mathematically modeled by some set $\mathscr O$ of mathematical objects, and some rules that tell us how these objects correspond to a physical system and allow us to predict what will happen to that system. For example, many systems in classical mechanics can be described by a pair ...

4

Yes, provided one uses the correct notions of symmetry for the action and the lagrangian. The setup. We assume throughout that the action can be written as the integral of a local Lagrangian. Namely, let $\mathcal C$ be the configuration space of the system, then for any admissible path $q:[t_a, t_b]\to \mathcal C$, there exists a local function $L$ of ...

4

Well, if I read the problem correctly then your kinetic energy is wrong. Your $\dot{x}_m$ is the $x$-component of the velocity of the ball, but you're missing the $y$-component and also the velocity of the cart, not to mention you're multiplying $\dot{x}_m^2$ by $m$ when you should be multiplying it by $M$. The $\cos^2\phi$ goes away because the ...

3

Why don't you just write the equations of motion? You will find that you get a second order lienar homogeneous DE with constant coefficients. Such an equation always has a closed form solution in terms of exponentials, so you can solve it; I guarantee the solution is not too bad. You might even recognize the equation without having to solve it. Edit: OK, ...

3

Let's compare Landau's Lagrangian and the one given by $\mathcal L_{Me}$ in your question: $$\mathcal{L}_{Landau}-\mathcal{L}_{Me}=mal\gamma^2\sin(\phi-\gamma t)-alm\gamma\dot\phi\sin(\phi-\gamma t)=\\ =mal\gamma\left((\gamma-\dot\phi)\sin(\phi-\gamma t)\right)=\\ =mal\gamma \frac{\text{d}}{\text{d}t}\cos(\phi-\gamma t)$$ Now the difference is obviously a ...

3

Yes. Though the energy will not be unbounded, but bounded from above, if my calculation is correct. For real scalar field under $(+---)$ metric, besides the negative classical kinetic energy for the Lagrangian $$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi - \frac{1}{2} m^2 \phi^2 \tag{1}$$, the classical equation of motion will be $$... 3 The equation of motion corresponding with \mathcal{L}_{kin} is$$(∂_{t}^2-{∂_{\mathbf{x}}}^2)φ=0,$$the Klein-Gordon equation, which has its origin in relativistic field theory. The minus sign is essential for relativistic invariance and leads to propagating solutions (waves). With$$φ(\mathbf{x},t)=\text{exp}[iωt]ψ(\mathbf{x}).$$we ... 3 First let's rewrite the kinetic term \mathcal{L_{kin}} such that the metric \eta^{\mu \nu} is explict$$\mathcal{L_{kin}}=\frac 12 \eta^{\mu \nu}\partial_\mu \phi\partial_\nu\phi$$In this answer we will restrict our attention to 4-dimensional Minkowski space-time M^4, the flat "arena" of special relativity, and set the speed of light c=1. The ... 3 It depends on how you "derive" Lagrange's equations, whether taking Newton's laws as fundamental or by assuming an action integral and minimizing it. However, there is no such requirement that you be in an inertial frame of reference. Thus, to look at your pendulum problem, you could start with the Lagrangian L = \frac{1}{2} I ... 3 I) In e.g. Ref. 1 is shown that there exist (possibly velocity-dependent) generalized potentials for all the fictitious forces, such as, e.g., the centrifugal force, the Coriolis force and the Euler force. So Yes, there exist Lagrangian formulations for non-inertial accelerated reference frames. II) OP's image shows Kapitza's pendulum. Kapitza's pendulum ... 3 First some terminology: In general an infinitesimal transformation of a field theory consists of a so-called horizontal infinitesimal transformation$$ \delta x^i ~=~x^{\prime i}- x^i$$of the base manifold, and a so-called vertical infinitesimal transformation$$ \delta_0\phi^{\alpha}(x)~=~\phi^{\prime \alpha}(x)-\phi^{\alpha}(x) $$of the fields. The ... 2 Symmetries indeed have a broad and powerful impact in physics, and I will only be able to scratch the surface of the subject in this answer, but I will try to give you a glimpse of the subject. In the most simple framework, you mention an electrostatic problem. In such a problem, the key factor is the geometric symmetries which apply to the charged ... 2 Hamilton was guided by a hunch that since a minimum principle worked for optics, then perhaps a similar principle worked for mechanics. From his principle of least action he was then able to derive the Euler-Lagrange equations from his paper: W.R. Hamilton, "On a General Method in Dynamics.", Philosophical Transaction of the Royal Society, 1834 2 Finding an action that gives you dust when varied is actually kind of tricky. Part of the reason it is hard is that for scalar matter, the action is usually proportional to the pressure,and the pressure vanishes for dust. I'm not sure where you found your action for dust, but I think you didn't make any calculational errors, it just isn't the right action. ... 2 The use of inertial frames in Lagrangian mechanics is by no means compulsory and everything can be done in any reference frame provided one takes all forces, real and inertial, into account. Actually there are two possibilities in interpreting the question. We work in a non inertial frame R' (instead of an inertial one R) because we are adopting ... 2 The lagrangian of the gauge field is independent of that of the scalar field. You have to "guess" it. The reason we pick this one is twofold: 1-it is the one which gives the Maxwell equations, so when you try to describe E&M, that looks like a good guess; 2- if you think of all the terms that are both Lorentz invariant, parity invariant and gauge ... 2 I assume you're thinking about Minkowski space, i.e. the metric \eta_{\mu\nu}=\text{diag}(c^2,-1,-1,-1). You should be aware that the dot notation is purely a notational shorthand, and has no other information contained in it. In particular, by definition we have$$\dot{A}\equiv\partial_0A=\frac{1}{c}\frac{\partial A}{\partial t}$$Thus, there is no ... 1 It is, in a sense, just semantics but I'd say the natural choice is \mathcal{L}=\sqrt{-g}\times \text{something}. If you take this definition, the general form of the equations of motion is the same as when doing QFT in Minkowski, with the appropriate generalizations to account for curvature. Furthermore, I think it is standard practice to define the ... 1 The special feature of contact geometry is the contact 1-form \lambda, which satisfies \lambda\wedge d\lambda\ne0 (let's restrict to 3-dimensions). In our Lagrangian mechanics example, \lambda = dq-vdt. You want this to pull-back to zero on the permissible'' curves in phase space -- these curves represent the motions of your system. For a more ... 1 I) Let there be given a local action functional$$\tag{1} S[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}, $$with the Lagrangian density$$\tag{2} {\cal L}(\phi(x),\partial\phi(x),x). $$[We leave it to the reader to extend to higher-derivative theories.] II) Assume that a variation of S for arbitrary x-dependent infinitesimal \epsilon(x) takes the ... 1 If \phi is a function of t, then x, for example, is written as$$ x(t)=R\cos(\omega t)+\ell\sin(\phi(t)). $$Applying the chain rule gives$$ \begin{align} \frac{d}{dt}x(t)&=\frac{d}{dt}R\cos(\omega t)+\frac{d}{d\phi}\ell\sin(\phi)\frac{d}{dt}\phi(t)\\ &=-R\omega\sin(\omega t)+\ell\cos(\phi)\dot{\phi}. \end{align} $$1 Define F(u):= \int_0^u f(s) ds, so the equations for the field u(t,\vec{x}) can be re-written as$$\frac{\partial^2 u}{\partial t^2}-\Delta_{\vec x} u + \frac{dF}{du}=0\::$$If defining$${\cal L}:= \frac{1}{2}(-\partial_t u\partial_t u + \nabla u \cdot \nabla u) + F(u)\:.$$this Lagrangian density leads to your field equations. Moreover, as you can ... 1 The technical answer is No. Surprisingly I think Wikipedia gives the better definition, though I think both authors are trying to say the same thing. Let the action be defined as S[\varphi]=\int d^4x\ \mathcal L(\varphi(x),\partial_\mu\varphi(x)) A differentiable symmetry is a symmetry of the functional that does not change the action ... 1 The numbers of deegres of freedom for a planar body in the plane is$$\text{2 (coordinates of CM)}+1\text {(angles to determine the body orientation)}=3.$$As you correctly recognize, the constraint is one:$$|\vec {OA}|=R, so you need two numbers $\varphi, \theta$ to localize the body in the plane. So, you have the first, $\varphi$, and I gave you a big ...

1

Regardless of the form of whichever (holonomic) constraints you may have, non-autonomous systems are most naturally understood from a field theoretical viewpoint. More precisely, one should understand Lagrangian mechanics as Lagrangian field theory in $0+1$ dimensions (that is, the space-time manifold is just the real time line). There, coordinates over each ...

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