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Most theories do not have a conserved energy-momentum tensor, regardless of whether they are Lagrangian or not. For example you need locality and Lorentz invariance. When you have those, you can define the energy-momentum tensor via the partition function (which always exists, it basically defines the QFT): $$ \langle T_{\mu\nu}\rangle:=\frac{\delta}{\delta ...


10

Not all non-Lagrangian theories have a stress-energy tensor, an example of this is the critical point of the long-range Ising model, which can be expressed as a "defect" field theory where the action consists of two pieces integrated over spaces of different dimensionality and hence has no single Lagrangian that would describe it. See chapter 6 of &...


5

You are not taking it too literally. The only adjustment that you should make to your mental picture is that the galaxies are a gas with zero temperature and thus zero pressure. This is sometimes referred to as “dust.” As the universe expands, the density of galaxies decreases and the pressure remains at zero$^\dagger$. Note however that there are other ...


3

… why does one simply add the two actions and not add a coupling action? One does add coupling action. When we replace the partial derivatives that are present in the matter action in flat spacetime in Cartesian coordinates with covariant derivatives and integrate using volume form of the curved spacetime (this approach is called minimal coupling), this is ...


3

The $T^{\mu\nu}$ on the right side of the Einstein field equations includes all non-gravitational forms of energy, including mass-energy, kinetic energy, electrostatic potential energy, the energy of static magnetic fields, of electromagnetic waves, of gluons, of the Higgs field, of a possible inflaton field, etc. The curvature of spacetime doesn’t depend on ...


3

This is the Belinfante-Rosenfeld procedure (note that it is not necessary to invoke spin currents for the free electromagnetic theory). Just to spell things out, this involves modifying the canonical energy-momentum tensor by adding a divergenceless term (an alternative method is using the Hilbert definition, see this question for the relation between the ...


3

Your $Q$ still is four independent quantities - one for each independent choice of the translation direction $a^\nu$. Noether's theorem states that for a one-parameter continuous quasi-symmetry \begin{align} x^\mu & \mapsto x^\mu + \epsilon \delta x^\mu \\ \phi^a & \mapsto \phi^a + \epsilon \delta \phi^a \end{align} where the theory is quasi-...


2

The question Given an energy-momentum tensor, $T_{zz}[g]$, in a general CFT with central charge, $c$, defined with respect to a metric, $g$, and conformal frame coordinates, $z$, $$ T_{zz}[g]\quad \textrm{wrt}\quad g=\rho(z,\bar{z})dz d\bar{z},\tag{1}\label{Tg} $$ how does $T_{zz}[g]$ change under a Weyl transformation, $g\rightarrow g'$, while keeping ...


2

First I'll start with the original stress-energy tensor obtained through Noether's theorem, without symmetrisation. Interestingly, you get the (manifestly) correct Hamiltonian density first try: $$ T^{\mu\nu}=-F^{\mu\sigma}\partial^{\nu}A_{\sigma} - \eta^{\mu\nu}\mathcal{L}\\ T^{00} = -F^{0\sigma}\partial^{0}A_{\sigma} - \eta^{00}\mathcal{L}\\ \mathcal{H} = -...


2

I'll share my view of the matter hoping that it helps here. First there is the definition of the stress-energy tensor. We define the stress-energy tensor $T_{\mu\nu}$ from the matter action $I_M$ as $$T_{\mu\nu}=\dfrac{-2}{\sqrt{-g}}\dfrac{\delta I_M}{\delta g^{\mu\nu}}\tag{1}.$$ If you want a definition that is the one. Now you may of course ask about ...


2

Start with $$\frac12\partial_0^2\int_\Sigma T^{00}x^i x^j \ d^3 x = \frac12\partial_0\left(\int_\Sigma \partial_0T^{00}x^i x^j \ d^3 x\right)$$ Since we're integrating over a spatial hypersurface, we can add total spatial derivatives without altering anything, since by the divergence theorem, $\int_\Sigma \partial_iA^{ijk...} \ d^3x = 0$. $$ \frac12\...


2

The Christoffel symbols of the FLRW metric contain the scale factor, that is how the Hubble parameter comes about when you expand $\nabla_\mu T^{\mu\nu} = 0$. The cosmic fluid is implicitly assumed to be at rest, since you can calculate that $R_{0i} = 0$ from the metric, therefore $T_{0i} =0$ from the Einstein equations. There is no three-momentum density, ...


1

To appreciate the general definition, we need a general perspective. The action principle is one of the key principles in classical physics. Loosely speaking, it says that if A influences B, then B must also influence A: all influences must go both ways. The general definition of the stress-energy tensor described below relies on appreciating a precise ...


1

The stress-energy tensor comes from thinking of matter as a fluid - that is, even if you choose to think of matter as a bunch of particles, the stress-energy tensor at a given point in spacetime comes from averaging over all the particles in the vicinity of that point. First consider the Newtonian version. A single particle only has its momentum, $m\vec v$, ...


1

Such a theory disagrees with experiment. There would be no gravitational waves and coupling to a free electromagnetic wave would be impossible, as the trace of its EM tensor is zero.


1

An ideal gas in an otherwise empty universe doesn't actually press on anything, but it would press on an object that you hypothetically placed in it, and you can define the pressure by the effect the gas would hypothetically have on the object. If you insert into the gas a thin barrier that is at rest in the local inertial frame of isotropy of the gas, then ...


1

General relativity is a classical field theory, and so is classical E&M. In these classical field theories, the notion of potential energy isn't particularly useful. For example, when you accelerate an electron from the cathode to the anode of a vacuum tube, what's really happening is that the total energy of the electric field is decreasing, and you're ...


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