New answers tagged

1

Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation. $$\begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ...


1

As you noticed, if we use Euler-Lagrange equation on $L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy$ we get $$\ddot x=0$$ $$\ddot y = -g$$ Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ...


3

It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ...


3

The question that you have asked have some vague arguments as well as some partially true facts regarding Standard Model (SM). First, Yes SM describes physics up to some energy scale which is 14 TeV. On the other hand, if we accept Plank energy ($~10^{18}$GeV) as a fundamental energy scale, then we can possibly expect new beyond the SM energy scale. A ...


0

If the Lagrangian is linear in $v^2$, then $\partial_{v^2} L = k$, $k$ some constant. So the difference becomes $$\delta S \propto \vec \varepsilon \cdot \int dt \frac{d}{dt} \vec x(t) = \vec \varepsilon \cdot \int d \vec x(t) = \varepsilon \cdot (\vec x_b - \vec x_a)$$ That term does not depend on the path $x(t)$, since the endpoint of the variation are ...


0

You are possibly confused by the fact that your textbook, when it says we need this to be a total differential, means with respect to time, not speed. We know that, if we perform a change of ${\cal L}$ into ${\cal L}+\delta\!{\cal L}$ such that $\delta\!{\cal L} = df(\vec x)/dt$, the dynamics is unaffected, where $f$ is allowed to be a function of position ...


0

Schaum series Differential Geometry will solve part of your problem. Search "problem book in riemannian geometry" on google and it should bring out something useful. Also see V.I. Arnold's books.


4

Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ...


2

Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor $g_{\mu\nu}$ in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action $S_{\rm m}[\Phi, g]$ should be a general relativistic diffeomorphism-invariant functional. Then the Hilbert stress-energy-momentum ...


0

If you are wondering if $F=\frac{-\partial U}{\partial x}$ holds, you may check the equivalent condition that you posted, $\nabla\times\vec{F}=0$. In this case the curl will come out to be zero, so you may then proceed to write the force as the negative gradient of a potential. This appears to work because the del operator is not taking time into account, ...


3

The conditions about (i) differentiability of the functions and (ii) the maximal rank of the corresponding rectangular Jacobian matrix are regularization conditions imposed to simplify the mathematical analysis of the physical problem, in particular to legitimate the possible future use of the inverse function theorem. In the affirmative case, the ...


5

The generalised Lagrange equations are $$ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial T}{\partial \dot{q}_j} - \frac{\partial T}{\partial q_j}=Q_j \tag{1} $$ where $T$ is the kinetic energy of the system and $Q$ is the generalised force. This is the most general EoM, and is equivalent to Newton's $F_j=m\ddot x_j$. Now, if the generalised force can be ...


1

All of these resources are saying the same thing, but you have to pay extremely close attention to the definitions of their differential operators. Specifically, in the brown.edu link, they define the divergence of a tensor $\mathbf{A}$ as $$ \nabla \cdot \mathbf{A} = \frac{\partial A_{ij}}{\partial x_i} $$ with summation over the first index of ...


0

Symmetry, stability and dimension analysis. You can consider a scalar field theory, for instance. A dynamical action for such a theory must be $S = \int d^4 x \, \, \partial_\mu \phi \partial^\mu \phi $ because i. Lorentz symmetry indicates that all the indices must be properly contracted ii. The field equations must not exceed second ...


0

According to sec. 4 in Calculus of Variations, Gelfand, Dover 1991, a theorem due to Bernstein concerns existence and uniqueness of solutions to the equation $y'' = F(x,y,y')$: If the functions $F$, $\partial_y F$ and $\partial_{y'} F$ are continuous at every finite point $(x,y)$ for any finite $y'$, and if a constant $k>0$ and functions $\alpha\equiv ...


0

The point is that one should distinguish between a total spacetime derivative $$ \frac{d}{dx^{\mu}}~=~ \frac{\partial }{\partial x^{\mu}}+ \phi_{,\mu} \frac{\partial }{\partial \phi}+ \phi_{,\mu\nu} \frac{\partial }{\partial \phi_{,\nu}}+\ldots \tag{1}$$ (where ellipsis denotes contributions in case of higher space-time derivatives), and an explicit ...


2

There are two kinds of derivatives we should differentiate: $$ \frac{\mathrm d\mathcal L}{\mathrm dx}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x+h),\phi'(x+h),x+h)-\mathcal L(\phi(x),\phi'(x),x)\big]\tag{1} $$ and $$ \frac{\partial\mathcal L}{\partial x}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x),\phi'(x),x+h)-\mathcal ...


2

Consider a smooth 2D sphere and a point of positive mass constrained to move on that without friction. The only force is the reactive force normal to the sphere. The trajectories of the motions of the point are geodesics of the spherical surface. Therefore if you fix the north and south poles as boundary conditions you find infinitely many solutions ...


2

I) When Ref. 1 writes let there be $\ell$ equations (17'') of constraints, it is implicitly assumed that they are independent, as also noted in Peter Diehr's answer. Obviously this implies that $\ell\leq 3N$. Moreover, the rectangular $\ell \times 3N$ matrix $$\tag{A} \left( \frac{\partial f_k}{\partial x_i}\right)_{1\leq k\leq \ell, ~1\leq i \leq ...


1

The equations of constraint must be independent of each other - otherwise they would represent the same constraint; this independence is what guarantees that Lagrange's theorem is satisfied. Joos' method is correct if the conditions are met. An Introduction to Lagrange Multipliers provides a geometric analysis which is often helpful in understanding ...


2

OP is observing that in Minkowski space $g_{\mu\nu}=\eta_{\mu\nu}$, it doesn't matter whether we write $${\cal L} ~=~\sqrt{|g|}\partial\phi\partial\bar{\phi} \tag{1} $$ or $${\cal L}~=~-\sqrt{|g|}\phi\Box\bar{\phi}\tag{2} $$ for the Lagrangian density, if we don't care about total divergence terms. OP is pondering what happens in curved spacetime $(M,g)$? ...


2

The covariant divergence of a vector is $$\nabla_\mu V^\mu = \frac{\partial_\mu (V^\mu \sqrt{-g})}{\sqrt{-g}}$$ Meaning that adding a covariant divergence to the Lagrangian will result in the following change : $$\Delta S = \int d^4x \sqrt{-g} \nabla_\mu V^\mu = \int d^4x \partial_\mu (V^\mu \sqrt{-g})$$ which is once again easy to see that it vanishes ...


2

Yes, there is systematic way called the Noether procedure. Simply you write down all possible 2-derivative fermionic terms with arbitrary coefficients and vary the action using the SUSY transformation rules. Then, you fix the coefficients to obtain the invariance up to a total derivative. When you have 4-derivatives, there are two cases: a. Off-Shell ...


1

We do not start from the gauge fixed path integral in the BRST construction. What you describe (once one adds the missing Faddeev-Popov determinant) is the original Faddeev-Popov trick to get the ghosts, not the systematic BRST construction. The (Hamiltonian) BRST construction crucially first introduces the ghosts as parts of the extended phase space, and ...


1

The BRST symmetry cannot be seen without introducing auxiliary variables. The fastest way to realize the BRST symmetry is to "exponentiate" the delta function $$\delta(G)~=~\int \!{\cal D} B ~\exp\left[iB_{\alpha}G^{\alpha}\right]$$ and the Faddeev-Popov (FP) determinant $$\det\Delta ~=~\int \!{\cal D} c ~{\cal D} \bar{c} ~\exp\left[\bar{c}_{\alpha} ...


0

For the coupling of p-brane Gauge fields to B 2-forms and U(1) Gauge fields check the Chern-Simons effective action for Dp branes. For coupling to fermions check some SUGRA actions, for instance the D=11 SUGRA has the graviton, the gravitino and 3-form Gauge field. Coupling to scalars is omnipresent in dimensional reductions of SUGRA actions.


2

It depends on what you mean by vacuum. If you mean a field configuration that has $F=0$, then the gauge potential is locally pure gauge (cf. this answer by Qmechanic), so there can only be global obstructions to the gauge equivalence class of the $A$ corresponding to $F=0$ being $A=0$ everywhere. On $\mathbb{R}^{1,3}$, there is no such obstruction. If ...


1

Indeed the Lagrangian is independent of $\phi$. However, the partial derivative w.r.t $\dot{\phi}$, \begin{equation} \frac{\partial L}{\partial\dot{\phi}}=mr^2\dot{\phi}, \end{equation} contains $r$ and $\dot{\phi}$, both depending on time $t$. Therefore, you need product rule to compute the total time derivative. That's why you have two terms. They come ...


6

Theorem: let $L$ be a homogeneous function of degree $k$; then the on-shell lagrangian is a total derivative. Proof: according to the Euler's homogeneous function theorem, $$ k\ L(q,\dot q)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\tag{1} $$ On the other hand, because of the Euler-Lagrange equations, $$ (1)=q\frac{\partial ...


0

When a top rotates, it rotates about its centre of mass. The centre of the mass is a point on the axis of rotation. Since the axis is also stationary as is the centre of mass, therefore all the points in the axis are eligible to be considered fixed about which the top is rotating. Besides,I would prefer to use the term axis instead of a fixed point.


3

You should study Newtonian mechanics before Lagrangian mechanics because Newtonian mechanics is more general than Lagrangian mechanics. In other words, while whenever a system allows a Lagrangian formulation it also allows a Newtonian formulation, the converse is not true; the quintessential case is dynamics in the presence of dissipative forces. Lagrangian ...


7

No, I would highly recommend studying Newtonian mechanics before Lagrangian mechanics. While, yes it is 'possible' to learn about Lagrangian mechanics before Newtonian, a lot of intuition would be lost beginning with one instead of the other which will, in the long run, do no more than harm you or, at best, possibly confuse you. But there are, indeed, many ...


12

It is necessary to study Newtonian mechanics to truly understand Lagrangian mechanics since its underlying foundation is Newtonian mechanics. It is essentially a different formulation of the same thing. In a way when doing Lagrangian mechanics you are still doing Newtonian mechanics just in the way of energy. For example, under Lagrangian mechanics, say we ...


2

$$ F^2 \to F^2 + \alpha_i f^{ijk} F_{j\mu\nu} F_k^{\mu\nu} $$ $f^{ijk}$ is completely antisymmetric in all its indices and is being contracted with something that is symmetric in $jk$. Thus, the action is invariant. We can see that $f^{ijk}$ is totally antisymmetric as follows $$ [T^i , T^j] = f^{ij}{}_k T^k \implies \text{tr} \left( [ T^i , T^j ] T^k ...


4

OK, as per your request…. My sense is you want to learn everything about integrability from here, and combine issues which confuses them, instead of separating them…. How about you supplement L&L with Arnold’s book? The seven additive integrals of L&L are the additive conservation laws of the isolated center of mass system, and standard center of ...


3

There are no contradiction at all and your questions can be answered at once. An isolated system with $s$ degrees of freedom has $2s-1$ integrals of motion since the solutions for the coordinates $q_i$ involve $2s-1$ constants (determined by the initial conditions). The $2s$th can be solved in terms of $t-t_0$ where $t_0$ can be arbitrarily chosen. However ...


2

The virial theorem is valid if few simple conditions are satisfied (system remains bound to finite region) and is thus quite general. Ergodic theorem requires evolution to be ergodic, i.e. evolution of state must be ergodic transformation: http://encyclopedia2.thefreedictionary.com/ergodic+transformation This is a special requirement that is not ...


2

As already mentioned, the action corresponds to massive electrondynamics, including external sources, in Minkowski spacetime. This is also known under the term Proca action. As you mention, the corresponding equations of motion can be found using Euler-Lagrange equation, that is $$0 = \frac{\partial \mathcal{L}}{\partial A_\mu} - \partial_\nu \frac{\partial ...


1

by the Euler-Lagrange method you would simply get the following field equation: $(\square-\nu^2)A_\alpha=-\beta J_\alpha$, which is the Proca equation. You can read up about the Proca action online. I'm new to field theory, and someone should correct me if I am wrong.


0

This is just the lagrangian for electromagnetism. The A is the vector potential and the expressions in parenthesis are the F tensor. You can read about it here


1

While I see the 'use' of a Lagrange multiplier in this question, I don't really see the 'need' for it, so let's make it a little simpler. The Lagrangian for this system is found from: $$ T = \frac{1}{2}[M(r_1\dot\theta)^2 + m_2(\dot{r_2}^2 +(r_2\dot\theta)^2)] $$ $$ V = -(Mr_1+mr_2)g\cos\theta + \frac{1}{2}k(r_2-l_0)^2$$ $$ L = ...


3

123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means ...


-1

The real answer is: it doesn't really. Or rather: we can still extract some physics out of it! Lets derive $$ S[\phi] = \int d^d x \mathcal{L}[\phi,\partial\phi]\\ \delta S[\phi] = \int d^dx \delta \mathcal{L} = \int d^dx \left( \frac{\partial\mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta ...


4

The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does ...


10

Indeed the problem with boundary conditions, generally speaking, is not well-posed. There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations. Examples. (1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the ...


0

Lagrangian mechanics can be derived directly from Newton's second law using only algebraic manipulation and a some calculus. This includes both the general form of the Euler-Lagrange equation and the specific form Langangian $L = T - V$. No assumptions of stationarity, use of the calculus of variations, or even any reference to the concept of action are ...


4

You are asking four questions, whose answers are routinely provided in textbooks. We consider it in QCD since there is no fundamental reason to exclude it, and topological configurations such as instantons, etc.. might well generate it in an effective low energy theory: the rule of thumb is that anything that is not prohibited has to emerge out of the ...


1

And regarding why it's called a "free" theory, it's not specific to a momentum-space formulation. It's "free" because the Lagrangian is quadratic in the fields, and therefore the equations of motion (what you get from plugging the Lagrangian into the Euler-Lagrange equation) are linear in the fields. Therefore you can superpose different classical ...


2

Yes. You are correct. A non-relativistic theory would be invariant under the Galilean group. Lorentz invariance (specifically, invariance under Lorentz boosts) is what defines a relativistic theory.



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