14

Your understanding of 1. is a correct summary of radiation-reaction. For 2, you are correct that if we use ONLY $L = \frac{1}{2} mv^2 -q\phi +q/c \vec{v}\cdot \vec{A},$ then we are assuming that $\vec{A}$ and $\phi$ are fixed and that the motion of the particle doesn't cause $\vec{E}$ and $\vec{B}$ to change. However, the complete electromagnetic Lagrangian ...


9

Certainly there's no reason why you couldn't define the operator $\hat L$ such that this is true. But in general, doing so won't buy you any of the power that comes from a Lagrangian formalism in classical mechanics. For example, you will not have a action minimization principle would this allow you to implement a Lagrangian formulation for Noether's theorem....


5

H is a function of $q$ and $p$ not $q$ and $\dot q$. The "momentum" $p$ is not always $\dot x$ but is defined by $$ p= \frac{\partial L}{\partial \dot q}. $$ The Legendre transform is between $\dot q$ and $p$ just as the usual Legendre transform of a convex function $f(x)$ replaces $f$ by $$ F(p) = x(p)\left.\frac{df}{dx}\right|_{x=x(p)}- f(x(...


5

The gravitational field equations are the equations one obtains varying with respect to the metric field. These equations determine the form of spacetime. Now, if your theory is coupled to some other field(s) say a scalar, variation with respect to the other field(s) yields the equation of motion for the other field(s). These equations (gravitational and ...


5

Here is the action written with explicit sums over the indices \begin{equation} S = \int {\rm d}^4 x \sum_{\dot{\alpha}=1}^2 \sum_{\beta=1}^2 \sum_{\mu=0}^3 \chi^\dagger_{\dot{\alpha}} i \sigma_{\dot{\alpha} \beta}^\mu \partial_\mu \chi_\beta \end{equation} $\chi$ is a two component spinor and $\partial_\mu \chi$ is the gradient of a spinor. At a deeper ...


4

There are no preferred or default coordinates in General Relativity. You can use any coordinates you want. There are generally no “Cartesian” coordinates in which the metric is flat and the volume element is $d^nx$, simply because a general spacetime has curvature. The volume element is $c\,dt\,dx\,dt\,dz$ in flat Minkowski spacetime but not in a general ...


4

Yes it is correct. You can aply the covariant derivatives in $\phi^2$: $$\nabla_{\mu}\nabla_{\nu}\phi^2 = \nabla_{\mu}(2\phi \nabla_{\nu} \phi) = 2 \nabla_{\mu}\phi\nabla_{\nu} \phi + 2\phi\nabla_{\mu}\nabla_{\nu} \phi$$ and the same for the box term, but your calculation is correct!


4

If I understand correctly when the particle accelerates its movement itself affects the magnetic and electric field $\mathbf{B}$ and $\mathbf{E}$ through electromagnetic radiation. So the movement of the particle is affected by $\mathbf{B}$ and $\mathbf{E}$ and the movement of the particle is affecting the field $\mathbf{B}$ and $\mathbf{E}$. Is my ...


4

The variation of the field $B$ vanishes at the boundary thus the total derivative term vanishes. Now regarding the last term, the derivative should also act on the exponential. When you apply Leibniz rule you have: $$\partial(e^{2 \alpha \phi}\delta(B_{np}) [H^{mnp}+H^{pmn}+H^{npm}] )=0 $$ as a total divergence term. The derivative acts on three terms, you ...


4

What are the most important things missing from my interpretation of H and L? I would add two things here. First note that any dynamical system of equations we can always re-write the system as a first order system by the standard trick of defining higher derivatives to be new variables and adding equations to fix the relationships of these new variables to ...


4

Consider the following computation, where a summation is to be understood whenever there is repeated indices: $$ \nabla q_{i}=\frac{\partial q_i}{\partial r_j}\mathbf{e}_j$$ $$\partial_{k}\mathbf{r}=\frac{\partial r_l}{\partial q_k}\mathbf{e}_l$$ $$\nabla q_{i}\cdot\partial_{k}\mathbf{r}=\frac{\partial q_i}{\partial r_j}\frac{\partial r_l}{\partial q_k}\...


3

Maxwell's theory is scale invariant in any 𝑑, but is not conformally invariant (except in 𝑑=4). So, there is no physical scale, but also no conformal invariance. Scale invariance does not imply conformal invariance. One way to see this is to note that if Ω is simply a scale transformation and not the full conformal transformation, then even if it is not ...


3

What are the most important things missing from my interpretation of H and L? Maybe an important feature is that the Hamiltonian is a conserved quantity of the Lagrangian if the later has no explicit dependence on time. Taking the time derivative of the Lagrangian: $$\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot q + \frac{\partial L}{\partial \dot q} \...


3

remeber that the action for the fields also depends on the metric. You have hidden that dependence in the way you write your action. Writing the dependence explicitly gives $$ S_{\rm fields}= \int d^dx \sqrt{-g}L=\int d^dx \sqrt{-g}\left( \frac 12 g_{\mu\nu} \partial^\mu \phi \partial^\nu \phi -V(\phi)\right) $$ so $$ \frac{\delta S_{\rm fields}}{\delta g_{\...


3

It seems that OP is only considering infinitesimal gauge transformations $$\delta g_{\mu\nu}~=~\nabla_{\{\mu}\xi_{\nu\}}.$$ The stationary action principle for $S_{\rm gravity}+S_{\rm matter}$ holds for arbitrary infinitesimal variations $\delta g_{\mu\nu}$ (that satisfy appropriate boundary conditions). In the latter case, $\delta S_{\rm gravity}$ is only ...


3

Classical mechanics answer A simple example of this would be to impose constraints that can't be solved simultaneously, using Lagrange multipliers. For example, let's take a particle in 2 spatial dimensions and require that it is simultaneously on a circle of radius $R$ and a circle of radius $2R$ \begin{equation} L = \frac{m}{2} \dot{\vec r}^2 + \lambda_1(|\...


3

Some preliminary consideration: In a uniform expanding universe the gravitational potential fulfills in atomic units the equation $$\frac{GM}{R} \propto 1$$ with $c=1$ and $R$ is the reciprocal Hubble parameter and $M$ the total mass of the visible universe. Now the assumption of Brans was that not only the geometry defines gravitation, but also the mass. So ...


3

That seems to be a misunderstanding. The field-theoretic case behaves in essentially the same way as the point mechanical situation. On one hand, the arguments of the Lagrangian density $${\cal L}(\phi,\partial_t\phi,\partial_x\phi,\partial_y\phi,\partial_z\phi,t,x,y,z)$$ are independent variables in the same way as the arguments of the Lagrangian $$L(q,\...


3

A better way to do variations in Lagrangian field theory, and indeed this is essentially the way it is done within high energy theory, is to not worry about writing a generic formula like we are used to in point particle mechanics. Though this can be done, it's often simpler to actually define the functional derivative $$ \frac{\delta}{\delta\phi(x)} $$ ...


2

Well, $\dot{q}$ is defined as $$\dot{q}=\frac{dq}{dt}.$$ Hence when $$q\to q+\delta q,$$ we have $$\dot{q}\to \frac{d}{dt}(q+\delta q)=\dot{q}+\frac{d\delta q}{dt}.$$ Identifying this last expression with $$\dot{q}\to\dot{q}+\delta\dot{q},$$ we see that $$\delta\dot{q}=\frac{d\delta q}{dt}.$$


2

As you state in the question, $\frac{\partial f}{\partial y}=0$ and so it follows that $\frac{\partial f}{\partial y^\prime}$ is a constant. Let's give a suggestive name to this constant and call it $C$. Then we have $$ \frac{y^\prime}{(1+y^{\prime\,2})^{1/2}}=\frac{\partial f}{\partial y^\prime}=C. $$ Squaring both sides of this equality, $$ \frac{y^{\prime\...


2

The Pauli matrices are operating on the implied isospin index that distinguishes the $u$ and $d$ components of $\psi$. The Dirac gamma matrices are operating on the implied Lorentz spinor index that distinguishes the four components of $u$ and the four components of $d$. Thus they commute.


2

I haven't opened the link, but I guess that the field transformation you are writing is a global symmetry (SU(2)) action, which means that Pauli matrices act on a different space than the gamma matrices. In other words, gamma matrices act on the spinorial indexes, while Pauli matrices act on the internal symmetry indexes - so, they commute. In particular, ...


2

Jess Riedel writes in a blog post titled Legendre transform that there is a form of representing the Legrendre transform that makes it transparent what that transform entails: Two convex functions f and g are Legendre transforms of each other when their first derivatives are inverse functions $$ g' = (f')^{-1}$$ That is, the Legendre transform is its own ...


2

You simply derivate every component. You can write (sum over repeated indices implied, where $\alpha,\beta$ run from $1$ to $4$) $$ L = \psi^*_{\alpha} \gamma_0 (i (\gamma_{\mu})_{\alpha \beta} \partial^{\mu} -m) \psi_{\beta}. $$ Now the components $\psi_{\alpha}$ are just complex numbers (or operators in the quantum theory) and so is $$ \pi_{\alpha} \equiv \...


2

There is one Lagrangian for both $x$'s, but there is a separate Euler-Lagrange equation for each of $x_1$, $x_2$.


2

Perhaps this will help. Let $t^a$ be the generators of the Lie algebra is some representation. For example, this could be the adjoint representation as you're asking about and each $t^a$ will be a $3\times 3$ matrix. Then any field $\phi$ valued in the Lie algebra may be written generically as $\phi=\phi_a t^a$. This is a common trick employed when dealing ...


1

Do not confuse multiplication with the idea of a dot product. The problem at hand, of constructing the kinetic term in components, is not equivalent to multiplying $(a+b)(c_d)$. Rather, we have two vectors whose components are $\langle \partial_0,\nabla\psi\rangle$ (the spatial components could be written out as well) and we are computing a dot product as ...


1

In so far as I'm aware, there is not direct connection between the two as you might like, however there are a few things that might be worth mentioning. I will take this back a few steps from the question itself because I feel there's some important context to the point I would like to make in the end. Since the problem at hand is looking for a steady-state ...


1

That space is the tangent bundle $TQ$ of the configuration space $Q$, and it is already in use, see e.g. my Phys.SE answer here.


Only top voted, non community-wiki answers of a minimum length are eligible