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0

I'd say the problem doesn't make much sense as it stands. Without terms that include a time derivative, the system would just immediately jump the the configuration with the smallest $L$, as the speed with which it does so is not "punished" (does not increase the action). In other words, the solution with the smallest action is the one with ...


2

Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. OP is in this terminology essentially asking Q: What is the meaning of horizontal transformations? A: It is a (horizontal) flow ...


2

Lorentz invariance refers to the action $S=\int\mathcal{L}(x)\,\mathrm{d}x$, not to the Lagrangian. To determine the condition on the Lagrangian which we must have, we make the coordinate change $x\to \Lambda x=:x'$ (a Lorentz transformation) and use the general fact that the Jacobian of a Lorentz transformation is unity, so ...


1

When you integrate the Lagrangian density over a certain region $\Omega$, this is in principle allowed to change and this gives you a "boundary" term in the variation. This is well discussed in, e.g., the book of Goldstein (3rd edition), where the correct proof of the Noether theorem is given.


1

Let there be given a general covariant matter action $$S~=~ \int \! d^4x~ {\cal L}, \qquad {\cal L}~=~e L, \qquad L~=~L(\Phi,\nabla_a\Phi). \tag{1}$$ The main strategy will be to demand that the matter fields $\Phi^A$ carry flat rather than curved indices$^1$. This is achieved with the help of a vielbein $e^a{}_{\mu}$, where $$g_{\mu\nu}~=~e^a{}_{\mu} ...


1

The Einstein-Hilbert lagrangian coupled to a matter action $$ S_m[\varphi,g] = \int d^Dx\, \sqrt{-g}\mathcal L_m(\varphi,\partial_\mu\varphi), $$ i.e. $$ S[g,\varphi] = \frac{1}{16\pi G}\int d^Dx\,\sqrt{-g} R + \int d^Dx\,\sqrt{-g} \mathcal L_m(\varphi,\partial_\mu\varphi), $$ satisfies $$ \delta S=-\frac{1}{16\pi G} \int d^Dx\, \sqrt{-g} \mathcal ...


2

Let $\mathcal{M}$ be our spacetime. Then, a gauge theory is given by a connection form $A$ on a principal bundle over it (that locally projects onto the spacetime in a way compatible with gauge transformations), which is the gauge potential. Maxwell's equations1 (in vacuum) are the equations of motion for the gauge field for the Yang-Mills action coupled to ...


1

Comment to the question (v2): The association (2) is not correct. To find the Hilbert SEM tensor, one varies the action wrt. the metric $g_{\mu\nu}$; not wrt. the gauge potential $A_{\mu}$ (or the field strength $F_{\mu\nu}$).


5

The energy momentum tensor is found by varying the metric and holding all other fields constant. Since clearly $$\frac{\partial F}{\partial g}=0\longleftrightarrow \delta_gF=0$$ we end up with $$\delta_g S=\frac{1}{2}\int\mathrm{d}v\,\left(F^2g_{\mu\nu}/4-F^\tau{}_\mu F_{\tau\nu}\right)\delta g^{\mu\nu}$$ and comparison with ...


2

For quantum field theories the vacuum state is a scalar with respect to lorentz transformations (i.e. $ M \psi = \psi $ for any boost $M$), and the angular momentum 4-vector transforms as a vector (i.e. $M J^{\mu} M^{-1} = M^{\mu}_{\nu} J^{\nu}$ where the boost $M$ sends $ x^{\mu} \to {x^{\prime}}^{\mu} = M^{\mu}_{\nu} x^{\nu} $ ) From this we see the ...


2

The vacuum $\Psi_0$ is the only vector in the Fock representation that is Lorentz invariant. The consequence of this fact can be interpreted as "the vacuum is not polarised, so that any vector must be the zero vector, or otherwise it would determine a privileged direction in space, thus breaking its relativistic invariance".


1

Does a field have any physical meaning or significance? Yes. See Einstein talking about field theory in 1929 and note this: "The two types of field are causally linked in this theory, but still not fused to an identity. It can, however, scarcely be imagined that empty space has conditions or states of two essentially different kinds, and it is natural ...


-1

Field is a mathematical notion. It is possible to formulate electromagnetic theory with particles in such a way that the field is not needed. This is due to Fokker and Tetrode, I think. More accessible reference is the paper (the part about the absorber is additional, not needed for the theory to work without fields): J. A. Wheeler, R. P. Feynman, Classical ...


0

You are mixing up two different types of field. The electron, Higgs and quark fields are quantum fields. The excitations of these fields appear as (real) particles. The electric and magnetic fields are different aspects of the electromagnetic field, and this is a gauge field. It describes the interaction between charged particles as the exchange of virtual ...


1

Tips: 1) Remember that $\mu$ and $\nu$ are dummy indices. It will be easier to see if you lower all indexes, but with pratice this won't be necessary anymore. 2) For terms like $(\partial_ \mu A^\mu)^2$, write them as $g^{\mu \nu} g^{\sigma \rho} (\partial_\mu A_\nu) (\partial_\sigma A_\rho$) and use Leibniz's rule . 3) For terms like $A^\mu A_\mu$ just ...


1

Yes, you are correct. $\mathcal{N}=2$ supersymmetry in 3d can be obtained by dimensionally reducing 4d $\mathcal{N}=1$. The 4d chiral superfield contains a complex scalar, a Weyl fermion, and a complex auxiliary field. Reducing this to 3d we get a compex scalar, a Dirac fermion, and the auxiliary. You can also impose reality conditions to rewrite the 4d ...



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