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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

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 ...


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

No, the Lagrangian density is different: $$ \mathcal{L} = \pm \frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi. $$ The Hamiltonian density is actually the same in both conventions. However, this has no physical meaning. The choice of the signature is purely conventional.


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".


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

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

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 ...


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

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

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

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

Conventions do not change physics. If they would, we would not call them conventions. When studying Lagrangian mechanics, you may have noticed that you can multiply a lagrangian by any constant, and receive the same dynamics. Thus, we often (Or always) choose the constant such that the term $(\partial_0\phi)^2$ appears with a positive sign. (And often with ...


1

Hints: Then potential term $\frac{1}{2}(\nabla\phi)^2$ is semipositive definite and is only zero for a $x$-independent configuration $\phi$. If one completes the square of the potential $$V(\phi)~=~\frac{\lambda}{4}\phi^4-\frac{\mu^2}{2}\phi^2~=~ \frac{\lambda}{4} \left(\phi^2-\frac{\mu^2}{\lambda}\right)^2-\frac{\mu^4}{4\lambda},$$ then it becomes clear ...


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 $F_{\mu\nu}$).



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