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6

Momentum is defined via the Noether theorem. It is the conserved charge correspoding to translations. If you consider $\delta x^{\mu} = a^{\mu}$ where $a^{\mu}$ is just a constant four-vector with real components, you can run the Noether procedure to compute $$ \delta S = \int d^{4} x \, a^{\mu} \partial^{\nu} T_{\mu \nu} $$ Now given a conserved current $...


0

The standard procedure to find the components of metric tensor has the following steps. Guess a trial formula for dynamic evolution of the metric tensor components ($g_{\mu\nu}$), considering the symmetries your given system possesses. The trial formula should include enough unknown parameters. Calculate the connection coefficients from your trial formula. ...


0

$\nabla^{α}(\nabla_{α}φ\nabla_{β}φ) -\cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ\nabla_{ν}φ) - \nabla^{α}g_{αβ}V(φ) = 0 \Rightarrow $ $\nabla^{α}\nabla_{α}φ\nabla_{β}φ +\nabla_{α}φ \nabla^{α}\nabla_{β}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ)\nabla_{ν}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla_{μ}φ\nabla^{α}\nabla_{ν}φ - \nabla^{α}g_{αβ}V(φ) = 0 \...


4

You are not going to find such a table. There are a several problems with this concept. (1) You're talking about expressing the stress-energy in terms of its components in a certain coordinate system. But the same spacetime can be described using infinitely many different coordinate systems, none of which is preferred. (2) This also isn't going to work ...


1

Diffeomorphism invariance implies (via Noether's 2nd theorem) that $$\nabla_{\mu} T^{\mu\nu}~\approx~0.\tag{1}$$ Perhaps surprisingly, there are no conserved quantities associated with the identity (1) per se for generic spacetime metric $g_{\mu\nu}$. However, if the metric $g_{\mu\nu}$ has a Killing vector field $K^{\mu}$, then the $4$-current $$J^{\mu}~:=...


1

This isn't a fact about general relativity, it's a fact about special relativity. Based on hundreds of years worth of experiments, we know that the energy-momentum four-vector is conserved locally. This means that the stress-energy tensor has zero divergence. (The stress-energy tensor isn't conserved. What's conserved is the energy-momentum four-vector.) If ...


0

There are various definitions of the total global mass-energy contained in a spacetime: the ADM mass, Bondi mass, etc. In present understanding, these require specific conditions such as symmetry in time, or asymptotic flatness. Then there are an awful lot of quasi-local proposals. The idea is that, since you can't define the gravitational energy at a point,...


7

The traditional derivation of the Navier-Stokes equations starts by looking at a fluid parcel and the different fluxes over the surface in the integral form. The integral form is preferred as it is more general than the differential form: For the latter one has to assume differentiability and thus it is not valid for flow discontinuities such as shocks in ...


7

Well, first of all you should ask yourself what do you expect from the concept of energy and momentum. Or in other words, what is energy and momentum, really? You have a set of intuitions in mind, but the minimal requirement is that these are quantities that are 1) conserved, and 2) reduce to our "usual" definitions of energy and momentum in suitable limits. ...


10

I came across this passage in Misner, Thorne & Wheeler (20.4) where they first talk about the stress-energy pseudotensor of the gravitational field and how one might calculate a contribution to the local momentum vector... and then memorably state: Right? No, the question is wrong. The motivation is wrong. The result is wrong. The idea is wrong. ...


6

The simplest way to see this is that by the equivalence principle, we can always let the gravitational field at any point have any value we like, including zero. Therefore there is no possibility of defining an energy density of the gravitational field at one point.


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