10

No it is not. The Weyl tensor vanishes by definition in three dimensions, Einstein's equations (in the absence of matter) impose: $$R_{\mu\nu} = 0 \rightarrow R=0$$ and since $Riemann = Weyl + Ricci$ no geometry can be formed. The solution in three dimensional spacetime is the BTZ black hole (https://arxiv.org/abs/hep-th/9204099v3) which includes a ...


8

The concept of particles also makes sense for non-inertial observers and for observers in curved spacetime, as long as we remember that real observers are local and that the particle concept is only approximate. (In this answer, "local/localized" doesn't mean localized at a point. It only means localized in some small neighborhood.) The ...


8

A manifold is a set - you don't need to put the manifold structure onto anything. Take a look at the first line of the wikipedia page for a manifold: a manifold is defined as a topological space which satisfies certain properties (and a topological space is a set of points). Intuitively: a manifold is a set which looks flat if you zoom in close enough on ...


5

There are two different $dt$s here. One, $dt_{you}$ is the time interval measured on your clock while the other $dt_{me}$ is the time interval measured on my clock. In your coordinates, i.e. your rest frame, you are at rest at the origin and your spacetime looks locally like flat spacetime. So in your coordinates $g_{00}=1$ and the time you measure on your ...


5

The set is not predetermined but arises from physical/mathematical requirements of the given solution. GR is local theory and sufficinetly small region of spacetime is assumed to be isomorphic to open region of $\mathbb{R}^4.$ Globally the set is given by "gluing" these regions together until you arive at global solution you are satisfied with. GR ...


4

A stationary spacetime $M$ is one which admits a complete timelike Killing vector field $\boldsymbol \xi = \xi^a \frac{\partial}{\partial x^a}$. The integral curves of $\boldsymbol \xi$ define a flow which preserves the metric. That is, if you define a map $\phi_t:M\rightarrow M$ which takes some point $p\in M$ and "flows" along the vector field $...


4

Well I'll answer but I would be interested to know if my answer is wrong. I thought we defined $\partial^a$ such that $$ \partial^a f \equiv g^{a\mu} \partial_\mu f . $$ The thing to be careful about is then the fact that $$ \partial_\mu (g^{a\mu} f) = (\partial_\mu g^{a\mu}) f + g^{a\mu} \partial_\mu f $$ so usually $$ \partial_\mu (g^{a\mu} f) \ne \...


4

You should use the definition of the curvature tensor $$[\nabla_\mu, \nabla_\nu]\xi_\kappa = R_{\mu\nu\kappa}^{\ \ \ \ \ \ \lambda}\xi_\lambda.$$ Then by differentiating the Killing condition we obtain $$\nabla_\kappa\nabla_\mu\xi_\nu + \nabla_\kappa\nabla_\nu\xi_\mu = 0.$$ One can then add a term to both sides to get a commutator and insert the above ...


4

The relation $\Lambda^{\mu}_{\ \ \sigma}\,\Lambda^{\nu\sigma}=\eta^{\mu\nu}$, or more properly $$ \Lambda^{\mu}_{\ \ \sigma}\,\eta^{\sigma\tau}\,\Lambda^{\nu}_{\ \ \tau}=\eta^{\mu\nu}\ , $$ actually is the definition of a Lorentz transformation $\Lambda^{\mu}_{\nu}$. As such it cannot be proven true, unless you rest on some alternative definition. In case ...


4

Mathematical conditions such as Hausdorf or paracompactness apply to mathematical models of reality. They are introduced to prove theorems that apply to these models. Do not confuse mathematical models of reality with reality itself. Whether the universe is everywhere Hausdorf or paracompact is something to be decided by experiment. No amount of studying ...


3

As mentioned in ApolloRa's answer, in 2+1 dimensions there exist no asympototically flat black hole solutions. However, you can still solve the Einstein Field Equations to find the metric of a non-spinning point mass $M$. The answer is given by $$ ds^2 = -dt^2 +\frac{1}{(1-4GM)^2}dr^2 + r^2 d\phi^2$$ As you can easily check this metric is flat for all $r>...


3

For a scalar function; $\nabla_{\mu}f=\partial_{\mu}f$, its a definition, we know $\nabla_{\mu}f$ is a $(0,1)$ tensor so $g^{\mu\nu}\nabla_{\mu}f=\nabla^{\nu}f$ but there doesn't exist any similar definition where we identify contravariant version of $\nabla^{\mu}f$ with $\partial^{\mu}f$. So ultimately it boils down to the definition of $\partial^{\mu}$.


3

Yes, your action is of the form \begin{equation} S=\int\text{d}^{4}x\sqrt{-g}(\mathcal{L}_{\text{EH}}+\mathcal{L}_{\text{M}}), \end{equation} where \begin{equation} \mathcal{L}_{\text{EH}}=\frac{R}{2k^{2}} \end{equation} is the part whose variation with respect to $g_{\mu\nu}$ gives you the Einstein tensor in the equations of motion, and \begin{equation} \...


2

Force, in its Newtonian sense, is an integral part of general relativity. General relativity attributes the effects of gravitation and inertia to a dynamic 3D metric and a "dynamic" time lapse. But all other forces remain as they were. In fact we can say that it's force, momentum, and energy – not mass – that cause curvature and are affected by ...


2

The divergence of the energy-momentum-stress tensor is zero. The time component of this expresses energy conservation, and the spatial components express momentum conservation. This means that the time-time component is energy density, the time-space components are energy flow / momentum density, and the space-space components are momentum flow. The ...


2

There are two possibilities to look at this: Either the exact form of $V(B^2)$ is known and you can simply take the variation with respect to the $B^2$-dependency in that potential. More general, you can use use the chain rule as follows \begin{equation} \frac{\delta V(B^2)}{\delta B^2} \frac{\delta B^2}{\delta g^{\mu \nu}} \end{equation} The same ...


2

Yes. You have to vary $g^{ab}$ everywhere it appears.


2

You forgot to lower the $\mu$ index in ${(\mathcal{J}^{\rho\sigma})^\mu}_\nu$ before using its definition. Doing it makes $g$ appear as you need.


2

This expression: $$(g^{\mu\nu} \nabla_\mu \nabla_\nu f) g_{\mu\nu} = \square f g_{\mu\nu} = (\nabla_\mu \nabla_\nu f)g^{\mu\nu}g_{\mu\nu} =\nabla_\mu \nabla_\nu f,$$ is wrong. This does not hold. The Box operator is a scalar quantity. It is defined as: $$\Box = g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}$$ and for a four dimensional diagonal metric this is: $$\Box = ...


2

General Relativity is a mathematical framework within which we can construct Lorentzian manifold models of reality. In general, structures (e.g. the spacetime manifold) of a given model taken to represent some aspect of observable reality need not be physically real aspects of nature, whatever that might mean. Indeed, they're almost certainly not-- what are ...


1

Yes. The field strength tensor is $F_{\mu \nu}$. The only two fundamental fields in your action are $g_{\mu \nu}$ and $A_\mu$ (unless you use Palatini variation and treat the connection as independent of the metric). So before preforming the variation, you should write the action in terms of these fundamental fields, and vary w.r.t each of them. The only ...


1

The corresponding action $S=\int_{\sigma_i}^{\sigma_f} \mathrm{d}\sigma~L$ is the arc length between 2 spacetime events. The principle of stationary action (with Dirichlet boundary conditions) therefore leads to geodesics. See also e.g. this related Phys.SE post.


1

Can it have a consistent explanation in Relativity? If I understand it right, what you talking about is a single point $p$ in spacetime, may call it an event. So yeah, it is well defined as $p\in M$. However, it is unintresting as it does not posess any dynamics per se. As you said, it remains at the same point of the manifold. Would it be described as ...


1

Having one temporal dimension and three spatial dimensions means that the metric tensor has signature $(1,3)$ or $(3,1)$ depending on your convention. Two temporal dimensions and two spatial dimensions would mean that the metric has signature $(2,2)$. Signature is a concrete property of a metric, so there is no sense in which a dimension could be both ...


1

I've eventually found the answer. So we have : $$ \mathcal{L} = \frac{1}{2} m g_{ij}(\vec{x}(t))\dot{q}^i\dot{q}^j = - U(q,t) $$ Computing the different terms in Euler-Lagrange equation for a particle $k$: $$ \frac{\partial \mathcal{L}}{\partial \dot{q}^k} = \frac{1}{2}mg_{ij}\dot{q}^i\delta^j_k + \frac{1}{2}mg_{ij}\delta^i_k\dot{q}^j = \frac{1}{2} m g_{ik} \...


1

Does General Relavity also make the introduction of other forces (i.e. coriolis- , centripetal- and normal force) obsolete by explaining its related phenomena through curved spacetime as well? You're mixing together unlike things. A normal force is not a fictitious force. A coriolis force is a fictitious force, and is described that way both by Newtonian ...


1

OP has a point: Eq. (9.91) should read $$\pi~=~-\sqrt{-g}g^{0\mu}\nabla_{\mu}\phi~=~-\sqrt{-g}\nabla^0\phi,\tag{9.91'}$$ i.e. the 0-index should be upstairs.


1

The definition $$ \partial^{\mu} = g^{\mu \nu} \, \partial_{\nu} \tag{1} $$ doesn't have more sense than $$ x_{\mu} = g_{\mu \nu} \, x^{\nu}. \tag{2} $$ Using (1) may lead to many troubles that you want to avoid at all cost. Don't use it! However, you may use the same for the covariant derivative: $$ \nabla^{\mu} = g^{\mu \nu} \, \nabla_{\nu}, \tag{3} $$ ...


1

If $\xi^a$ is a Killing vector then Killing's equation says that $\xi_{a;b}$ is antisymmetric. It follows that, when completely contracted with a symmetric tensor, it will give zero. But $\xi^a \xi^b$ is a symmetric tensor. Thus I do both steps in one go. But if you want to separate the two steps, then the argument is as follows. Step 1: any second rank ...


1

The metrics, $g_{ab}$ and $\bar{g}_{ab}$ are conformally equivalent if we can write $$ \bar{g}_{ab} = \Omega^2g_{ab}\,, $$ where $\Omega\equiv \Omega(x)$ is a non-zero differentiable function of the space-time coordinates. In this case, their Weyl tensors are equivalent, i.e., $\bar{C}^{a}_{\,\,bcd} = {C}^{a}_{\,\,bcd}$. Setting $g_{ab}$ equal to the ...


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