35

This was something that confused me for awhile as well until I found this great set of notes: homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf Let me just briefly summarize what's in there. The free Klein-Gordon field satisfies the field equation $$(\partial_{\mu} \partial^{\mu} +m^2) \phi(x) = 0$$ the most general solution to this equation ...


18

I) Let there be given a local action functional $$\tag{1} S[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}, $$ with the Lagrangian density $$\tag{2} {\cal L}(\phi(x),\partial\phi(x),x). $$ [We leave it to the reader to extend to higher-derivative theories. See also e.g. Ref. 1.] II) We want to study an infinitesimal variation$^1$ $$\tag{3} \delta x^{\mu}~=~\...


16

Currently all fundamental fields are quantum, except for gravity. For this reason Quantum Gravity is a hot area of research, but the full Quantum Gravity theory has not been developed yet. Why not? The challenge is not just technical, but conceptual. On one hand, the Quantum Field Theory cannot consistently co-exist with any classical theory. If the Quantum ...


13

When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right. Let's say we have the function: \begin{equation} f(\theta_1, \theta_2)=f_0+f_1\theta_1+f_2\theta_2+f_3\theta_1\theta_2 \end{equation} Then the left derivative ...


12

There is a very easy way to see this and it is through an $\hbar$ series. This claim can be traced back to Sydney Coleman and states that in the ultraviolet one is doing an expansion with $\hbar$ going to zero. A previous answer cited these lectures on classical fields but I would like to start from the generating functional of the scalar field theory and ...


10

The notation $q$ in the functional $S[q]$ stands for the whole parametrized curve/path $q:[t_i,t_f]\to \mathbb{R}$, not just a single position. In particular, the parametrized path already carries all the information about the derivative $\dot{q}\equiv\frac{dq}{dt}$, so there is no need to include it as an extra argument in the action. In contrast, the ...


9

Sometimes, diagrams are more useful than equations! Note that $p^2 = - (p^0)^2 + |\vec{p}|^2$ remains unchanged under any Lorentz transformation. Suppose that $p^2 < 0$ (say $p^2 = -1$). A plot of this hypersurface is shown below Lorentz transformations that are connected to the identity are continuous transformations take a single point on this line to ...


9

The $v$ you write is itself a spinor, not a scalar. A non-zero spinor is obviously not invariant under Lorentz transformations, so a non-zero spinorial VEV breaks Lorentz invariance of the 1-point function.


8

What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid. In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of ...


8

No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. ...


8

The classical limit of bosonic quantum mechanical systems with both finite and infinite degrees of freedom is pretty well understood from a mathematical standpoint (with complete rigour, and for quite general quantum states; see the references in the end). With fermions on the other hand, the situation is more involved. The point is essentially that hinted ...


8

Well, one can reason as follows: One wants a diffeomorphism-invariant action, which must be of the form $$ S=\int d^4x\sqrt{-g}L, $$ where $L$ is a scalar in terms of transformation properties. Gravity is assumed to be purely metric, so $g_{\mu\nu}$ is the only dynamical variable that can appear in the action. So we must look for scalars that can be ...


8

A clean way to make the concept of a classical field precise it to phrase things in terms of a quantum effective action: given a generating function of connected and renormalised Green functions, $W(J)$, with $$ e^{W(J)} = \int \mathcal{D}\phi\,e^{-I[\phi]+\int d^dx\,J\phi} $$ the quantum effective action, $\Gamma[\varphi]$, is the Legendre transform (when ...


8

Perhaps the most clear way to see what's going on is to compare the action $$ S_1[\phi]=\int_{\mathbb R^d} \phi(x)^2\mathrm dx\tag1 $$ to the action $$ S_2[\phi]=\int_{\mathbb R^d} x^2\phi(x)^2\mathrm dx\tag2 $$ The first one should be invariant under translations, while the second one should not. We define the translation operator $T_a\in\mathrm{End}(\...


7

This question (v1) asks many questions. Let us here make some general remarks, which OP hopefully will find useful. Noether's theorem only needs infinitesimal transformations to work. Hence the important object is not the set $G$ of finite transformations, but rather the set $\mathfrak{g}$ of infinitesimal transformations. In general, the set $\mathfrak{g}$...


7

The classical reference is Landau & Lifshitz, The Classical Theory of Fields, from the Course of Theoretical Physics. As all Landau & Lifshitz books, masterpieces [in my opinion] full of content but sometimes a little difficult to grasp for beginners.


7

First, I'll note that unless the non-inertial frame has a changing acceleration, there is some doubt as to whether it radiates at all. https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field Assuming there is changing acceleration, yes the particle radiates, and this can be observed in both frames. Seeing it radiate would be a way of ...


7

If we linearize the equations of motion about a $\phi = 0$ background (and assume the usual sort of kinetic term), we will find a non-relativistic limit of something like $\nabla^2 \phi \propto \rho^2$, which is inconsistent with Newtonian gravity.


7

Yes, classically, we can unify gravity with electromagnetism. The theories that do so are the famous Kaluza-Klein theories. They are theories of pure gravity in $4+1$ dimensions rather than our usual $3+1$ dimensions. When such theories are viewed from a $3+1$ dimensional perspective, the effects of gravity in the fourth unseen dimension appear in the ...


6

The classical analogue of quantum $\Phi^4$ theory is classical $\Phi^4$ theory, with the same action. There are no particles, but there is still scattering of waves! The correspondence between tree-level QFT and classical fields is on the level of fields only. (Particles make their appearance in classical field theory only in the limit where geometric optics ...


6

People spent a lot of time trying to do this kind of thing ca. 1910, i.e., after SR but before quantum mechanics. To make the electrostatic self-energy no greater than the observed mass of the electron, you have to create some kind of model of an electron as an extended object, with a size that's at least on the order of the classical electron radius. You ...


6

The two quantities don't correspond because they are conserved quantities corresponding to different symmetries. One is a symmetry from shifting your field, the other from shifting space-time itself. Here is what is going on precisely: Let us do a simpler case first: In a particle mechanics system, let's say a free particle with $L = \frac{1}{2}m\dot{x}^2$, ...


6

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 the particles made by $\psi$ and $\psi^\dagger$ are each others' antiparticles. In the real case, the fields that have this property are $\phi_1 \pm i \phi_2$, so once you change basis ...


6

Given a dynamical system $\dot x=F(x)$ with continuously differentiable $F$, there is a canonical bijection between initial conditions at a fixed time and solution trajectories. If the dynamical system is Hamiltonian, the phase space is the space of initial conditions, hence is canonically isomorphic to the space of solution trajectories. Considering the ...


6

That formula is for bosons. In the large particle limit the boson quantum field can be approximated by the classical field. Electrons are fermions and by the Pauli exclusion principle you can have only one particle per state. So you cannot go to the large particle limit in the same way as you can for a boson field which can have many particles in each state. ...


6

In quantum field theories, the fields are there to make locality and causality manifest. This fields are suitable linear combinations of creation and annihilation operators of one-particles states. They are not necessarily hermitian , witch means (in QM) that they are not necessarily observables. They are something as: $$ \phi_{\alpha}(x)=\sum_{p,\sigma,n}...


6

Your confusion resides in the notation. To make it clear I'll use a hat over the operators. The Noether symmetry is $$ \hat\phi'\equiv\mathrm e^{i\theta\hat Q}\hat\phi\mathrm e^{-i\theta\hat Q}\equiv \mathrm e^{i\theta Q}\hat\phi $$ where $Q\in\mathbb R$ is a scalar. The generator of this symmetry is $\hat U(\theta)=\mathrm e^{i\theta\hat Q}$. The ...


6

In classical mechanics there is no reason to identify $m$ with a mass. It is just an inverse of a length, in which case there are no factors of $\hbar$ nor $c$ in the KG equation.


6

A gauge symmetry is simply a symmetry transformation of the action that depends non-trivially on spacetime. You can ask for all physical theories whether such transformations exist. For the case of particle mechanics of finitely many degrees of freedom (where the gauge symmetry then only depends on time), it is known that the existence of a gauge symmetry is ...


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