New answers tagged

1

Let us for simplicity consider a 1D system. If the Lagrangian $L(\dot{x},t)$ has a cyclic variable $x$, then the action has an infinitesimal translation quasi-symmetry $$\delta x~=~\epsilon,$$ and it is well-known that the conserved Noether charge $$ Q~=~\frac{\partial L}{\partial \dot{x}} $$ is the conjugate momentum. OP considers next a coordinate ...


0

The answer of Qmechanic says everything can be said. But I think is worth to mention, that there are references where some details of relativistic quantum strings are discussed with the Nambu-Goto action as the starting point. Zwiebach's textbook on string theory is a well known example of this. But I want to recommend the book "Introduction to The ...


0

As we know, the brachistochrone problem is regarded as having started the concept of variational approach to evaluating motion. Still, there is a fundamental difference. In the case of the brachistochrone problem the question is: design a potential profile such that a particle accelerated by that potential traverses a given distance in the least time. In ...


2

You must inspect how the last piece of the Lagrangian transforms, the rest of them are invariant. For example, let's do P. Dirac fields transform as: $$\psi \xrightarrow{\mathcal{P}} \gamma^0 \psi,$$ $$\overline{\psi} \xrightarrow{\mathcal{P}} \overline{\psi}\gamma^0.$$ So the quatity $\overline{\psi}\gamma_5\psi$ transform as $$\overline{\psi}\gamma_5\...


0

Remember the definition of the partial derivative. $$\frac{\partial}{\partial t}L(x,y,t)=\lim_{h\rightarrow 0}\frac{L(x,y,t +h)-L(x,y,t)}{h}$$ Since the Lagrangian doesn't depend on $t$ explicitly we have that $$L(x,y,t')=L(x,y,t)\quad\text{ for all }t'$$ meaning the numerator of this limit is zero and consequently $\frac{\partial L}{\partial t}=0$. In this ...


0

Eq.(*) means that the Lagrangian is invariant under both time and space translation. Since you are assuming that the Lagrangian is only time translation invariant, that equation should've been \begin{equation} \label{eq:time} L(x(t'),\dot x(t'),t')=L(x(t'),\dot x(t'),t) \end{equation} which implies that $$L(x(t'),\dot x(t'),t)+{\partial L \over \partial t}...


3

The $q_{nm}$ are the amplitudes of the normal modes of vibration of the plate with sides $2\pi a$, $2\pi b$. He is using the usual plate wave equation $$ \frac{\partial^2 y}{\partial t^2}= D \left(\frac{\partial^2 y}{\partial t^x}+\frac{\partial^2 y}{\partial z^2}\right)^2 $$ and expanding the displacment as $$ y(x,y,t)= \sum_{n,m} q_{n,m}(t) \cos (mx/a)\...


2

I think q here denotes a position coordinate. There exists a particular form of potential in which the dynamics of the ball work. It might be that, $q_{mn}$ determines the position of the $m^{\text{th}}$ particle of the ball w.r.t to the $n^{\text{th}}$ particle of the plate. At first, the total potential energy of the $m^{\text{th}}$ particle of the ball ...


0

The third term is a gauge fixing term. At this point in time, the Wikipedia article on gauge fixing has has a nice section on it, here: https://en.wikipedia.org/wiki/Gauge_fixing#R.CE.BE_gauges These express the so-called $R_ξ$ gauges - with $\xi=1$ called the Feynman–'t Hooft gauge, and the $\lim\xi\to0$ being the Landau gauge, the limit taken after ...


3

OP is considering a constrained Lagrangian of the form $$ L(x,\theta;\dot{x},\dot{\theta};\lambda)~=~\frac{M}{2}\dot{x}^2+\frac{I}{2}\dot{\theta}^2 - V(x) -\lambda(x-R\theta). \tag{A}$$ The 'reduced' Lagrange equation reads $$ \left(M+\frac{I}{R^2} \right)\ddot{x}~\approx~-V^{\prime}(x). \tag{B}$$ OP's second 'reduced' approach is correct: $$ H_R(x,p_x)~=~\...


1

I don't recommend to use the notation $d_\mu$ since the derivative that appears in the E.L equations of motion is partial derivative $\partial_\mu$. Then if the Lagrangian density depends on the field $y$ and its derivatives $\partial_\mu y$ the E.L equations of motion read $\begin{equation} \partial_\mu\dfrac{\partial \mathcal L}{\partial (\partial_\mu y)}-...


0

To understand this f you need to be aware that f can only be a function of q and t, not of q dot. I got tripped up by the same question a long time ago as an undergrad. When you get $a^2 \omega^2 sin^2(\omega t)$ then this is a function of t only (a and omega are not dynamical variables, right?) We know any function of one variable (t) can be written as the ...


2

Yes, you can do that, and people did pretty much exactly that when first trying to canonically quantize gravity. These are what is called ``Gaussian normal coordinates''. For a nice, detailed introduction to the strategy in the 3+1 formalism of canonical gravity, check https://arxiv.org/pdf/gr-qc/0703035v1.pdf (especially section 4.4, where they do exactly ...


1

Yes you can do that. I believe it is called the synchronous gauge.


1

A force of constraint perpendicular to the surface will prevent any momentum gain in the direction perpendicular to the tangent surface. So the particle can only move in the tangent surface and is hence constrained to the surface.


3

If the constraint force was not perpendicular to the surface, then it would have some component tangent to the surface. Suppose for a moment that the particle constrained to the surface is otherwise a free particle. Since it's a free particle, if it's stationary at $t=0$, it will be stationary for all time. But if there's a tangential component of the ...


0

It seems prudent to point out some counterexamples: In general the Lagrangian $L$ does not need to be of the form $T-U$, cf. this Phys.SE post. In general, the Lagrangian $L(q,\dot{q},t)$ could depend explicitly on time $t$, e.g. if there are external forces/sources, c.f. this Phys.SE post. In general the potential $U(q,\dot{q},t)$ can depend on velocity $\...


1

For geometric theories, the Lagrangian $L$, the kinetic term $T$ and the velocity-dependent potential $U$ are scalars, i.e. invariant under change of generalized coordinates. This is e.g. the case for E&M. Similarly, the Lagrange equations are covariant, cf. this & this related Phys.SE posts.


1

Hints: Eq. (3) follows from the boundary condition $$\gamma(t_2, \mathbf{q}_2; t\!=\!t_2)~=~\mathbf{q}_2. \tag{A}$$ Eq. (4) follows by differentiating eq. (A) wrt. $t_2$: $$ \left.\frac{\partial\gamma(t_2, \mathbf{q}_2; t)}{\partial t_2}\right|_{t=t_2} + \left. \frac{\partial\gamma(t_2, \mathbf{q}_2; t)}{\partial t}\right|_{t=t_2}~=~0.\tag{B}$$


3

The answer to the linked question largely answers your question, but it assumes a bit. Contrast the following cases. I italicize your special interest parts, but they should be set in context. Unbroken, not anomalous symmetry. Good symmetry, both classically and quantum, with visible and pleasing effects. Global or local (e.g. color). Symmetry currents ...


3

It really depends on how you interpret these higher order terms. For example, in the case of treating it as a LEEFT, one can simply just add these terms on and the perform perturbative calculations in the low energy limit. In this case we find that at low energies we retain GR results, with the higher derivative terms contributing via a cut-off mass ...


0

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says: $\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$. This ...


3

Technically it is easier to start from the non-square root Lagrangian $$ L_0(x,\dot{x})~:=~ g_{ij}(x) \dot{x}^i \dot{x}^j~\geq~0,\qquad \dot{x}^i~:=~\frac{dx^i}{d\lambda},\tag{1}$$ and consider the new Lagrangian $$ L~:=~f(L_0). \tag{1'}$$ (This is equivalent to OP's set-up, although the notation is a bit different.) The corresponding energy functions become ...


2

Let's start by assuming $q_i$ and $\dot q_i$ are completely separate coordinates. They both have to be varied separately $$q_i\rightarrow q_i'=q_i+\delta q_i\\ \dot q_i\rightarrow \dot q_i'=\dot q_i+\delta\dot q_i$$ Now we know that in reality $\dot q_i$ is the time derivative of $q_i$, so we can impose $\dot q_i=\frac d{dt}q_i$. Plugging this in the varied ...


3

Let us look at the first term in your Lagrangian: $$\partial_\mu \phi \partial^\mu \phi$$ Now, for a general derivative $ \partial_x $ we have: $$\partial_x \left(y \partial_x y\right) = (\partial_x y)^2 + y \partial^2_x y$$ $$\Rightarrow (\partial_x y)^2 = \partial_x \left(y \partial_x y\right) - y \partial^2_x y$$ If you substitute the above for $\...


2

The two Lagrangians are the same modulo total spacetime divergence terms. If you vary with respect to the scalar field the action constructed from the lagrangian you will obtain the Klein-Gordon equation. People usually write the kinetic term of the scalar field as $(\partial_{μ} \phi)^2$. You can write it as $\phi \partial^2 \phi$. Both these two ...


0

Non-linear systems of odinary differential equations are pretty well studied. In particular, a conservative system with an arbitrary potential energy (so-called nonlinear oscillator) will have only one type of stable states, corresponding to its potential minina, with periodic oscillations around each of them. There may be also periodic trajectories in ...


0

Can you write it as a total divergence? If so, then you can use the divergence theorem to argue that it goes to 0 at the surface, which you can take to infinity.


1

Sterile neutrinos do not have any gauge interaction (electroweak or strong) but only gravitational.


1

with : $$r(t)=\sqrt{x(t)^2+y(t)^2}$$ and $$\tan(\varphi(t))=\varphi(t)=\frac{y(t)}{x(t)}$$ $$L\mapsto L(x,y,\dot{x},\dot{y})$$ $$L=1/2\,m \left( 1/4\,{\frac { \left( 2\,x{\it \dot{x}}+2\,y{\it \dot{y}} \right) ^ {2}}{{x}^{2}+{y}^{2}}}+ \left( {x}^{2}+{y}^{2} \right) \left( {\frac {{\it \dot{y}}}{x}}-{\frac {y{\it \dot{x}}}{{x}^{2}}} \right) ^{2} \right)...


1

Hint: Compute $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot \theta}\right)$ and check if any of the terms is of degree greater than one.


3

I am not sure I understand OP's problem, but aside from what Qmechanic already said, the standard Poincaré's lemma does not always correspond to the variational Poincaré's lemma when it is applied to functional/variational forms. In particular, consider the horizontal row of the variational bicomplex $$ 0\rightarrow\mathbb R\rightarrow\Omega^{0,0}(J^\infty E)...


3

As far as I know, in continuum mechanics a body is discribed by a fixed orientable Riemannian manifold $B$ together with a (time-dependent) embedding in space, e.g. the $\mathbb{R}^3$: $$X_t:B\rightarrow\mathbb{R}^3.$$ You could then define a free Lagrangian as a integral over the model body $B$: $$L(X)=\int_B \frac{1}{2} \rho \langle\partial_t X, \partial_t ...


2

If you are only interested in the dynamics of the matter field (that is, if you can approximate gravity by a fixed background metric and ignore back-reaction), then both expressions are correct. Remember that you can always add a constant to the Hamiltonian without changing the equations of motion (Hamilton's equations). Here this constant is simply $$ \int ...


2

If you already know point mechanics, then one way to build intuition for field theory is to consider the space variable $\vec{x}$ of spacetime as a continuous index $j$. Even better: discretize space $\mathbb{R}^3$ with some lattice parameter $a$ that eventually is taken to zero. (But keep the time interval $[t_i,t_f]$ continuous.) Then the field $\phi(t,\...


1

In field theory, action us defined as $ S[x] = \int L(φ_a, ∂_µφ_a)\, \text dt $ = $\int L'd^3x dt$ = $\int L'd^4x $, where L is Lagrangian and L' is Lagrangian density. Then we vary action and require that $ \delta S=0 $ by the principle of least action. This variation is taken under the boundary condition that $\delta \phi(x, t_1)= \delta \phi(x,t_2)=0$. It ...


1

The specific example you give of a spring-mass system is a natural system for which $2T=\sum \dot{q}p$ and, if the coordinates $q_i$ measure a departure from equilibrium, then $q_i=0$ are the equilibrium positions, just as you find. The equation of motion for $q_i$ should not depend on the velocities $\dot{q}_k$ unless you have friction since $\frac{\...


1

Lets look at this example: Double Pendulum with the position vector to mass one and two $$\vec{R}_1= \left[ \begin {array}{c} \rho\,\cos \left( q_{{1}} \right) \\ \rho\,\sin \left( q_{{1}} \right) \end {array} \right] $$ $$\vec{R}_2=\left[ \begin {array}{c} \rho\,\cos \left( q_{{1}} \right) +\rho\, \cos \left( q_{{2}} \right) \\ \rho\,\sin \left( q_{...


0

There's nothing in principle impossible about a system being in equilibrium at every point given that the initial conditions are specified such that the system is initially not in motion. If the Euler-Lagrange equations are satisfied in all such cases, it means that at least up to first order, these represent legitimate extremal points of the action ...


1

The action $S[\phi]$ is a functional of the map $\phi:{\mathbb R}^4\to {\mathbb R}$, which we seek to make stationary under variations $\delta \phi$ that vanish at infinity. We can think of as a history $\phi(x,t):{\mathbb R}^3\times {\mathbb R}\to {\mathbb R}$. In principle we should do the same as in classical mechanics: choose a beginning and end time $...


5

There are two notions which tend to be a bit confused in Lagrangian mechanics for fields. First, there is the functional derivative. As the name implies, this is a derivative for functionals, such as the action functional. Functionals are maps from function spaces to $\mathbb{R}$ (or some other field). If we consider, say, a scalar field $\phi$, the action ...


-1

How you choose new generalized coordinates From Euler Lagrange equation $$ \frac d{dt}\left(\frac{\partial L}{\partial \dot q}\right) - \frac{\partial L}{\partial q} = Q$$ you get: $$M_q(\vec{q})\,\vec{\ddot{q}}=\vec{Q}-\vec{f}(\vec{q},\vec{\dot{q}})\tag 1$$ if you chose new set of generalized coordinate ($\vec{w}$) thus $$\vec{q}=\vec{q}(\vec{w})\tag ...


1

When we eliminate/integrate out the auxiliary field $F$, the SUSY transformation for $F$ is rendered moot, and the appearance of $F$ on the RHSs of the other SUSY transformations is replaced with its algebraic EOM. It's not true that we do not know anything about the model -- we assume that the action $S$ is SUSY-invariant. In particular, the EOM for $F$ is ...


0

First, that the Lagrangian will have a term containing $F(t)$, or you will not get $(M+m)\ddot{x}=F(t)$. Second that if $F(t)$ is an explicit function of $t$, then Lagrangian will also be the explicit function of $t$ and then you have to consider more general form of Euler-Langrangian equation. For that refer to https://physics.stackexchange.com/a/437198/...


8

OP's 2 Lagrangians $L_1$ & $L_2$ are (up to normalization and total time derivative terms) just $$\overbrace{L_H~=~ p\dot{q}-\underbrace{\frac{p^2}{2m}}_{\text{Hamiltonian}}}^{\text{Hamiltonian Lagrangian}^1}\qquad\stackrel{p~\approx~ m\dot{q}}\longrightarrow\qquad \overbrace{L~=~\frac{m}{2}\dot{q}^2}^{\text{Lagrangian}}$$ for a free non-relativistic 1D ...


4

Yes you can! Indeed, observe that the two Lagrangians basically give rise to the same action (modulo boundary terms and an overall normalization which do not impact the equations of motion). First of all, write $$ S_{1}=\int_{t_{i}}^{t_{f}}dt\ \left\{q\frac{d\alpha}{dt}+\alpha^{2}\right\}=[q\alpha]^{t_{f}}_{t_{i}}-\int_{t_{i}}^{t_{f}}dt\ \alpha(\dot{q}-\...


2

The scalars in $D$ dimensions consist of $(11-D)$ dilatons: $\phi_i, \ i = 1 \cdots 11-D$ Axions $A_{(0)ijk}$ (call them Type A axions for now) Axions $A^i_{(0)j}$ (call them Type B axions for now) If you go through the appendix in a bit of detail, you see that the dilaton vectors $\vec b_{ij}$ corresponding to Type B axions and $\vec a_{ijk}$ ...


0

The main problem with the Nambu-Goto path integral is how to obtain a consistent path integral measure (PIM). When people say that it is ill-defined (and not just difficult to work with), they presumably mean that various naive choices of PIMs are inconsistent. The easiest way to obtain a consistent PIM is to go to the Hamiltonian formulation, cf. e.g. this ...


1

The last step uses the Fundamental Lemma of Calculus of Variation: $$ \left[ \forall \eta : \quad \int_a^b\! dx ~f(x) \eta(x) ~=~0\right]\quad\Rightarrow \quad f~=~0. $$


4

It can be an odd function, but we are trying to find the case where the integral is 0 for any arbitrary function $\eta(t)$. This means that the stuff in the bracket must be 0.


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