New answers tagged

0

You'll want to start with the electric field, or better, the electromagnetic field: this is the abstraction of a force, which according to a force law originates at a source, and is applied to an object. With a field we simply ignore the object, removing it from the equation, and consider the situation for a hypothetical test object dropped into the field ...


4

In quantum field theory bosonic fields $\varphi_\alpha\left(x\right)$ satisfy $\left[\varphi_\alpha\left(x\right),\,\varphi_\beta\left(y\right)\right]=0$, unless noncommutative geometry is incorporated in which case we achieve the moral general case in the first equation you've asked about. This is analogous to the fact that discrete quantum mechanics ...


1

The result you described says that the projection of antisymmetric rank 2 $SO(4)$ tensors onto self-dual and anti-self-dual subspaces commutes with the action of $SO(4)$. This just implies that the space of antisymmetric rank 2 tensors of $SO(4)$ is reducible. To show that the self-dual and anti-self-dual subspaces are themselves irreducible, think of ...


0

Note that $\int^{\infty}_{-\infty}f(t)dt=\int^{\infty}_{-\infty}f(-t)dt$ and also $\int^{\infty}_{-\infty}\frac{df(t)}{dt}dt=\int^{\infty}_{-\infty}\frac{df(-t)}{-dt}dt$


7

This has nothing to do with "bras" or "kets" and more with the elementary observation that a complex number has two real degrees of freedom, and that derivatives are with respect to one real degree of freedom. The $\frac{\partial}{\partial\phi}$ and $\frac{\partial}{\partial\phi^\ast}$ are the Wirtinger derivatives, which in particular fulfill ...


1

That $g^{-1}\mathrm{d} g$ is Liealgebra-valued for a Lie group-valued function $g$ has nothing to do with the particular model or with physics, it is true for all matrix groups. Write $g(x) = \exp(k(x))$, where $k(x)$ is now Lie algebra-valued and $\exp$ is the usual power series in the case of a matrix group. Then $\partial_\mu g = \partial_\mu k\exp(k)$ by ...


3

OP's proposal (v2) is a special case of Finsler geometry with $n=3$. The main idea is to replace the quadratic metric tensor $g^{(2)}_{\mu_1\mu_2}$ for pseudo-Riemannian manifolds, which defines (infinitesimal, possibly imaginary) distance on the manifold via $$ds ~=~ \sqrt[2]{g^{(2)}_{\mu_1\mu_2}dx^{\mu_1}dx^{\mu_2}},$$ with (possibly a sequence of) ...


0

Complex fields naturally lend themselves to an associated charge and current density, and this is the main reason for their introduction in physical theories. Consider the Lagrangian density $L = c^2\phi\phi^*-\mu_0^2c^2\phi\phi^*$. A transformation of the type $\phi' = \phi e^{i\epsilon}$, $\phi^{*'} = \phi^* e^{-i\epsilon}$ corresponds in infinitesimal ...


3

Two real scalar fields $\phi_1$ and $\phi_2$ satisfying an $SO(2)$ symmetry and one complex scalar field $\psi$ are equivalent. However, the latter is more convenient because $\psi$ and $\psi^\dagger$ form the antiparticle pair, while in the real case, you need to change basis from $\phi_1$ and $\phi_2$ to $\phi_1 \pm i\phi_2$. Once you do this, you just get ...


2

In my viewpoint, the free complex field theory $$\tag 1 \mathcal{L}=\partial_\mu\phi\partial^\mu\phi*-\frac{1}{2}m^2\phi\phi*$$ is actually equivalent to the free double real field theory $$\tag 2 \mathcal{L}=\partial_\mu\phi_i\partial^\mu\phi_i-\frac{1}{2}m^2\phi_i\phi_i,$$ where $i=1,2$. To see this, simply write $\phi=\phi_1+i\phi_2$ and you can obtain ...


2

The nature of field $a$ - scalar or pseudoscalar - isn't relevant for the existence of invariance of action under continuous Lorentz transformations. Really, under continuous Lorentz transformation both scalar and pseudoscalar fields are transformed trivially, $$ a(x) \to a'(x) = a(\Lambda^{-1} x), $$ However, there is a big difference - scalar or ...


9

Comments to the question (v2): First of all, let us stress that OP is correct, that a given set of equations of motion does not necessarily have a variational/action principle, cf. e.g. this Phys.SE post and links therein. On one hand, if there exists a Lagrangian formulation, then one may in principle obtain a Hamiltonian formulation via a (possible ...


7

The field equations must be conservative in a fairly precise sense in order that this can be done in a physically appropriate sense. Then there are several Hamiltonian approaches to field theory: the De Donder-Weyl formalism and the multisymplectic formalism. Although both formalisms can accommodate Lagrangians, the can also be understood without any ...


2

In a comment you say (I fixed a few words in this quote) "If it was a neutron star an atom would lose its electrons and protons before becoming a part of that star"... And in the question you say, "Are photons absorbed by atoms compressed out by gravity". This reminds me of Feynman's father. If you Google "feynman father photon", you should find the story ...


3

Notice the photons are reduced around the smaller one. Is that happening before the photon sphere? The photon sphere by definition does not send any photons in our direction, as it is a spherical region of space where gravity is strong enough that photons are forced to travel in orbits. So the photons seen come from the region before, ...


0

This is happening because your field depends on $t$, $\Psi = \Psi (x,t)$. Therefore, when you perform the Wick rotation $t = i\tau$, you also Wick rotate to your field, and obtain an action for $\Psi(x,i\tau)$. In the second case, you obtain an action for $\Psi(x,-i\tau)$. Those are not the same $\Psi$'s.


1

A massless vector is the gauge field $A_\mu$ and the massless symmetric traceless tensor is the metric perturbation $h_{\mu\nu}$. They can couple only to conserved currents, namely the $U(1)$ vector current $j_\mu$ and the stress tensor $T_{\mu\nu}$ respectively. The requirement for this arises from gauge invariance. Coupling a gauge field to a current ...


0

You are performing an on-shell counting for an off-shell multiplet. The off-shell vector multiplet has $\sigma [1], A_{\mu} [2], D[1], \lambda [4]$ thus $4+4$ degrees of freedom. The on-shell vector multiplet consists of the scalar $\sigma$, the vector $A_\mu$ and the Dirac fermion $\lambda$. In that case, the counting is $\sigma [1], A_{\mu} [1], \lambda ...


3

The electromagnetic potential $A^\mu$ is a four-vector, and hence transforms in the fundamental representation of $\mathrm{SO}(1,3)$, i.e. $A^\mu\mapsto \Lambda^\mu_\nu A^\nu$ where $\Lambda$ is the usual 4x4 matrix associated to a Lorentz transformation. Your question seems fundamentally confused about the difference between the field and the particle. The ...


1

Have you read the whole page, not just this sentence? I really don't know how to explain that differently. It's analog to the general relativity. You can perform coordinate diffeomorphisms, $$x^\mu\mapsto \tilde{x}^\alpha(x^\mu)$$ and they all get compensated by change of the spacetime metrics $g_{\mu\nu}$, $$g_{\mu\nu}\mapsto ...



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