New answers tagged

0

What you're doing is correct. In actuality, you're solving Poisson's equation, $ \nabla^2\phi = -\rho/\epsilon $, where $\rho$ is the charge density and $\epsilon$ the permittivity. If you had a particular charge distribution then the permittivity would affect the magnitude of the resulting potential. However, by imposing boundary conditions on the ...


0

Well, such form follows from the Noether theorem, which states that if action (in Minkowski space-time) $$ S = \int d^{4}xL $$ of fermions $\Psi$ is invariant under transformation $$ \tag 1 \Psi \to e^{i\alpha}\Psi , \quad \bar{\Psi} \to e^{-i\alpha}\bar{\Psi}, $$ then exists the object $J$ with one vector indice, namely $$ \tag 2 J_{\mu} = i\frac{\partial ...


2

When we say scalar, spinor, vector, and so on, field, we mean which representation of the frame bundle the field belongs to. Or in index notation, which spacetime indices the field has: none, spinor, vector, and so on. We can combine this with internal symmetries which are $G$-bundles for some gauge group $G$, for example $SU(2)$. In indices this is some ...


-1

(In my experience this tends to be a rather controversial subject, so I think this answer might start some arguments!) First of all, given a specific action, it is a purely mathematical result whether or not there exists a local transformation that depends on an arbitrary smooth function $\lambda(x)$ on spacetime and leaves the action invariant. The ...


5

Let us consider an example and take the Weinberg-Salam Lagrangian: $$ \mathscr{L} = i\bar{\psi}\gamma\cdot\partial\psi - m\bar{\psi}\psi $$ and let us adapt it to the case describing electrons and neutrinos as $$ \mathscr{L} = i\bar{\textrm{e}}_R\gamma\cdot\partial\textrm{e}_R + i\bar{\textrm{e}}_L\gamma\cdot\partial\textrm{e}_L + i\bar{\nu}_L\gamma\cdot\...


8

I can't give an answer using fiber bundles, but I don't think it is important as the confusion is at a much simpler level. A field can be in different representations for different symmetry group. The Higgs field is in the trivial representation of the Poincarre group, that is, under Lorentz transformations, $\phi(x)\to \phi(\Lambda x)$, but in non-trivial ...


2

The stress-energy tensor is not symmetric by its definition. It is symmetric only if we require the system to be rotationally invariant and to have no intrinsic spin (all representations of the internal Lorentz algebra must vanish). In particular one can show that: $$ M^{\mu}_{\alpha\beta} = (x_{\alpha}T^{\mu}_{\beta} - x_{\beta}T^{\mu}_{\alpha}) + \textrm{...


1

If I'm understanding correctly, you're asking whether an arbitrary QFT admits an asymptotic Fock basis. If a QFT does admit a Fock basis, you can talk about particles and do scattering theory in momentum space and construct position operators for single particles. But it's not guaranteed that a given QFT admits asympotic Fock bases. Quantum field theory ...


3

The term gauge transformation refers to two related notions in this context. Let $P$ be a principal $G$-bundle over a manifold $M$, and let $\cup_i U_i$ be a cover of $M$. A connection on $P$ is specified by a collection of $\mathfrak{g}=\mathrm{Lie}(G)$ valued 1-forms $\{A_i\}$ defined in each patch $\{U_i\}$, together with $G$-valued functions $g_{ij} : ...


3

In QFT, a single particle does not scatter, hence its (renomalized) wave function in an interacting theory is the same as the corresponding asymptotic wave function in the asymptotic Fock space. However, the multiparticle picture breaks down as the interacting Hilbert space cannot be identified with the asymptotic Fock space, by Haag's theorem. Thus ...


0

A field is usually a tensor field of arbitrary order, so a function from space(-time) to the corresponding tensor space. For example, in your case, $\vec{B}$ is a vector field, a function from time and space to three dimensional vectors. The field strength is in my experience usually used for the norm of a vector field. And usually, the field gives the force ...


2

Expanding on my comment, I think the Rarita Schwinger field (spin 3/2) has exactly the gauge symmetry you want: https://books.google.be/books?id=KFUhAwAAQBAJ&lpg=PA96&ots=vh0WtWM5rg&dq=rarita%20schwinger%20fermionic%20gauge%20symmetry&pg=PA95#v=onepage&q&f=false This gauge symmetry removes the spin 1/2 component of the field so only ...


-2

Imposing local gauge symmetry on the Dirac equation produces the electromagnetic field interacting with it. See http://www.physics.rutgers.edu/~steves/613/lectures/Lec06.pdf Before any down voting please see my comments below. The question was not about we whether Dirac's equation can be used to represent a spin 3/2 or higher fermion, though you could ...


0

I consider the potential (slightly modifying your notation for the sake of clarity): $$V= m_1^2\,\phi_1^2+m_2^2\,\phi_1^2 + \tfrac{1}{2}\lambda_1\,\phi_1^4+\tfrac{1}{2}\lambda_2\, \phi_2^4+\lambda_{12}\,\phi_1^2\phi_2^2\,\,,$$ then the conditions for boundedness from below are: \begin{eqnarray} \lambda_{1} &\geq& 0 \\ \lambda_{2} &\geq& 0 \\...


4

You already got your answer, all right, several times over, but I will emphasize the central puzzle of your question which you only got indirect answers for, connected to the peculiar special structure of SO(4). Any self-respecting text introducing the standard model more or less has it. I'll skip all superfluous issues like lagrangian terms, the U(1)s, etc....


1

Answer of this question is quite subtle. First let us consider the most general Higgs potential which is renormalizable and invariant under $SU(2)_{L}\otimes U(1)_{Y}$ gauge transformations, which has the form \begin{equation} V = \lambda(\phi^{\dagger}\phi-\mu^{2})^{2} \end{equation} Where \begin{equation} \phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi_{1}+...


0

After a bit of discussion I believe there is actually a $SU(2)\times SU(2)$ symmetry in a sense. So in principle there is a $U(2)$ symmetry if $\phi=(\phi_1,\phi_2)^T$, $\phi^\dagger=(\phi_1^*,\phi_2^*)$ and the lagrangian $$\mathscr{L}=\partial_\mu \phi^\dagger\partial^\mu \phi-m\phi^\dagger\phi-\lambda(\phi^\dagger\phi)^2,$$ simply sent $\phi\to U\phi$, ...


-1

If the field is a simple complex scalar field, than the symmetry is just $U(1)$. For a higher symmetry, $\phi$ need to be higher dimensional too, for instance you can add a vector index $\phi_i$ with $i=1,2$ for simplicity, which means that you add an additional complex field. If these two fields interact, you can have two cases now: Each field has a $U(1)$ ...


2

This is just supplementing Qmechanic's answer. I think the notations here need to be addressed. OP might be confusing Lagrangian (normal $L$) with Lagrangian density ($\mathcal{L}$). Formally, we have three fundamental relations: $$L = \displaystyle\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3x$$ $$S = \displaystyle\int dt \space L = \...


3

Yes, OP is right. In the field-theoretic case, the partial derivatives in OP's first formula (1) should be replaced with functional derivatives $$ \delta S~=~\int_{t_1}^{t_2}\!\mathrm{d}t\left(\frac{\delta L}{\delta q}~\delta q+\left. \frac{\delta L}{\delta v}\right|_{v=\dot{q}}~\delta \dot{q}\right),\tag{1'}$$ where the Lagrangian $$L[q(\cdot,t),v(\...


6

There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as $$R = \sum_{i=1}^n p_i\dot{q}_i - L$$ You can learn more about it by clicking this link for wikipedia's description of it. Reading more in regards to the routhian because I was bored, I realized it is ...


3

It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ...



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