# Tag Info

1

A general advice: Before trying to understand Hamiltonian field theory, make sure you understand Lagrangian field theory. Before trying to understand Lagrangian field theory, make sure you understand Lagrangian point mechanics. In Lagrangian point mechanics, the functional derivative of the action is $$\tag{1} \frac{\delta S}{\delta q(t)} ... 2 Consider a map$$S \ni\phi \mapsto F[\phi] \in \mathbb R$$defined on a class S of smooth functions \phi defined on the compact set \Omega \subset \mathbb R^n obtained by taking the closure of an open set with regular boundary \partial \Omega. Thus the map F associates a real number F[\phi] to each function \phi\in S. We say that the ... 0 Particles are simply high momentum wave states interacting weakly with matter, i.e. their "existence" is observer dependent. Trying to derive them from some form of free field equation is therefor useless and so is the assumption that they are a general phenomenon. They are a highly likely phenomenon for energies that are much higher than the typical em ... 1 And regarding why it's called a "free" theory, it's not specific to a momentum-space formulation. It's "free" because the Lagrangian is quadratic in the fields, and therefore the equations of motion (what you get from plugging the Lagrangian into the Euler-Lagrange equation) are linear in the fields. Therefore you can superpose different classical ... 2 Yes. You are correct. A non-relativistic theory would be invariant under the Galilean group. Lorentz invariance (specifically, invariance under Lorentz boosts) is what defines a relativistic theory. 0 Take for example q a vector field:$$ q_a=A_\mu $$where a=\mu is a vector index. The conjugate momentum is$$ \frac{\partial\mathcal L}{\partial A_\mu} $$and, as it is an lower index in the denominator, it makes sense to write it as \pi^\mu. Also, you can use the definition of vectors and covectors to prove that \pi^\mu transforms as a vector, and ... 0 The critical observation here is that before the magnet is introduced, there are equal numbers of oppositely charged ions moving in opposite directions, exerting no net force on the water around them. The magnet, however, will deflect a positive ion going one way in the same direction as it will deflect a negative ion going the other way. This produces a ... -1 The electron was the number of electron which has number of charge is 1.6\cdot10^{-18}\,eV. As the electric field is the region in which charged particles experience the force to perpendicular direction between electrical poles.The moving ions consists of positive charge and negative charge therefore the cations and anions which can be moving ions in ... 0 I) Here we will assume that OP is taking about a relativistic point particle with zero spin in a Minkowski spacetime with metric \eta_{\mu\nu} of sign convention (−,+,+,+). Also we put c=1 for simplicity. (OP mentioned that the particle has charge but since it is free that is irrelevant.) Note that the relativistic point particle has world-line ... 0 First of all, electrostatic field is conservative in nature. So this force can be written as the gradient of a scalar potential V E=-∇V The d in your equation is not right. The gradient indicates the rate of change of V with respect to the three coordinates along the three directions. Since the electric field is zero, we can write ∇V=0 which ... 0 "Potential at a point is sort of energy you used to move a point particle."It is work done to move a unit positive charge in a field. Consider a shell with radius r and charge q. Potential at infinity is taken as 0 by convention. When you move a charge from infinity to surface of shell , you did work W = \frac{kq}{r} in present of electric field E. ... -4 Do Weyl fermions carry electric charge? That depends on whether Weyl fermions exist, and whether they are what people say they are. See this from the article mentioned by John Rennie: 'Whereas electrons and all the other known fermions have mass, in 1929, mathematician and physicist Hermann Weyl theorized that massless fermions that carry electric ... 2 We should probably start by pointing out that no Weyl fermion has ever been observed. The recent observations are of quasiparticles that behave like Weyl fermions. Speaking rather loosely (and at the risk of upsetting the QFT experts hereabouts) a Dirac fermion can be viewed as a sum of two Weyl fermions, and the observations are of paired quasiparticles ... 0 You can "derive" the Lagrangian formulation from Shannon entropy arguing Liouville's theorem in reverse. 2 Well, not really. We COULD write hamiltonian as square root - if we know, what is a square root of an operator. Of course we have simple approximation:$$\sqrt{1+x}=1+\frac x2-\frac{x^2}{8}+O(x^3)$$Using this we could write your hamiltonian as:$$\mathcal H=mc^2\sqrt{1+\frac{p^2}{m^2c^2}}=mc^2+\frac{p^2}{2m}+O(p^4).$$The problem is that this form of ... 3 Going from action to EOMs is simple: it is just (functional) differentiation. Going the other way from EOMs to the action is hard: It is (functional) integration, and sometimes impossible! OP is now essentially asking: Can we integrate one more time? Well, not the action itself. But if we replace the EOMs and the Lagrangian L with their dynamical ... 2 The key is: Landau theory doesn't assume the order parameter is small. All it assumes is that the free energy is analytic in the order parameter. One then usually expands this free energy up to some order (which is possibly by definition of 'analytic'). It is key to realize that expanding a function in a variable to some order does not mean this variable has ... 3 Going the way stated in the question's title is easy: The Euler-Lagrange condition is, inherently, a condition on the action -- the statement is that the classical path is the path for which the action takes a minimum value for the path. Since this is a statement about the value of the action, and the action is Lorentz-invariant, then this minimum value is ... 2 I) Assuming that the variational problem for the action S=\int \! d^nx~{\cal L} is well-posed (with appropriate boundary conditions), the field-theoretic Euler-Lagrange (EL) equations read in general$$\tag{1} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial ...

0

For non-hermitian products of Dirac field operators the parity is not well defined and depends on the phase $\eta=\pm1,\,\pm i$ of the parity transformation $\eta \gamma^0$. For example, $I_P = -\eta^2 I$, where $I = \overline{\Psi_C}\Phi$. In the $S$-matrix elements, however, all phases go away eventually, because creation and annihilation operators come ...

4

Hint: Eq. (6) in its current form (v4) is meaningless since the lhs. depends on $x$, while the rhs. is integrated over $x$. The functional derivative $$\tag{A}\frac{\delta F}{\delta u(x^{\prime})}~\stackrel{(B)}{=}~\frac{\delta u(x)}{\delta u(x^{\prime})}~=~\delta(x\!-\!x^{\prime})$$ in eq. (6) for the functional $$\tag{B} F[u]~:=~u(x)~=~\int \! ... 4 It is the celebrated spin connection on the tangent space, gauging Lorentz rotations so you can take Lorentz covariant derivatives on spinors---you would not be able to do Supergravity without it. As you see, however, \omega_\mu^{ab} is a composite gauge field, that is, it is is an elaborate function of Vierbeine (or Vielbeine) and their derivatives, ... 0 I was able to find an answer to my question in literature. The reference is: A.S. Wightman, "Introduction to some aspects of quantizes fields", in "Lectures notes, Cargese Summer School, 1964". At the bottom of p. 204 Wightman writes ".. there is no such mathematical object as a free scalar field of mass zero in two-dimensional space-time unless one of the ... 2 I cannot quite vouch for exhaustive panoramas, but the crucial point is that GL(N), SU(N) matrices are representable in a nonhermitean basis discovered by Sylvester in 1882, the clock and shift matrices which he called nonions for N=3 (long before the Gell-Mann basis!), sedenions, etc. Their braiding relations, and maximal grading, and hence commutators, ... 3 There most definitely is, and your text should have used it in defining the unitary gauge more conventionally: the SU(2) group element parameterization of physics, that is the rotation matrix for spinors R. Absorb v into the definition of σ, where it belongs and from where it can re-emerge at will.$$R=\exp (i\theta ~\hat{n}\cdot\vec{\sigma})=I \cos \theta ...

0

There is something very essential to keep in mind for electric field and that is although the electric field lines are continuous but as you increase your distance from the electric charge which makes the field appear, the strength of the field decrease, and what you have mentioned about the size of seeds is also right, since when the seeds have a small mass ...

2

The standard way to put in a temperature is to go to imaginary time (euclidean space) and impose periodic/anti-periodic boundary conditions on bosonic/fermionic fields $$\phi(x,\tau)=\phi(x,\tau+\beta) , \;\;\;\; \psi(x,\tau)=-\psi(x,\tau+\beta),$$ where $\beta=1/T$. This ensures that the path integral represents the partition function \$Z=Tr[\exp(-\beta ...

Top 50 recent answers are included