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9

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 $F_{\mu\nu} F^{\mu\nu}$.


8

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 ...


6

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 ...


5

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 $\mathrm{U}(1)$ ...


4

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 ...


4

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} ...


4

"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 ...


3

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 ...


3

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 = ...


3

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 ...


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 ...


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 ...


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 $\frac{d q^k}{dt}$ ...


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

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 ...


2

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 ...


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 ...


2

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 ...


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} ...


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

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 ...


1

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 ...


1

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 ...


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 ...


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

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 $\Gamma^r_{\phi\phi}$.


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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