Tag Info

57

Classical electromagnetic fields carry energy and momentum and therefore cause spacetime curvature. For example, the EM field around a charged black hole is taken into account when finding the Reissner-Nordstrom and Kerr-Newman metrics. The question of whether photons cause spacetime curvature is a question about quantum gravity, and we have no accepted ...

47

The short answer for why gravity is unique is that it is the theory of a massless, spin-2 field. To contrast with the other forces, the strong, weak and electromagnetic forces are all theories of spin-1 particles. Although it's not immediately obvious, this property alone basically fixes all of the essential features of gravity. To begin with, the fact ...

24

In classical general relativity, electromagnetic fields do bend spacetime. They have a nonvanishing stress-energy tensor, and the Einstein field equations relate the stress-energy to the curvature. We even have fairly direct experimental proof that electromagnetic fields interact gravitationally in this way, from Cavendish-like experiments. See Kreuzer, ...

23

I believe it can be useful to define the following concepts (I won't be very formal here for pedagogical reasons): Any event can be described through four real numbers, which we take to be: the moment in time it happens, and the position in space where it takes place. We call this four numbers the coordinates of the event. We collect these numbers in a ...

22

There is a pure geometric definition of the Einstein tensor $G_{\mu\nu}$ in terms of derivatives of the metric. Independent of any physics. Likewise, given a field theory, you could in principle calculate the stress energy tensor. GR is a physical theory which couples the geometry, through the Einstein tensor, to the matter content, via the stress ...

20

The canonical energy-momentum tensor is exactly zero, due to the Einstein equation. The same holds for any diffeomorphism invariant theory. By saying ''it doesnt exist'' one just means that it doesn't contain any useful information.

20

The energy-momentum tensor is defined locally, and it's a tensor. In electromagnetism, or in Newtonian gravity, the way we form a local energy density is basically by squaring the field. The problem with applying this to GR is that the gravitational field $\mathbf{g}$ is zero, locally, in an inertial (i.e., free-falling) frame of reference, so any energy ...

19

For any matter/energy distribution we can in principle assemble it from point particles. So the stress-energy tensor of the whole system can be expressed as a sum of the stress-energy tensors of the point particles. The reason this helps is that the stress-energy of a point particle is very simple. It is: $$T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\... 19 You can certainly do this, and indeed it is regularly done. For example Alcubierre designed his FTL drive by starting with the metric he wanted and calculating the required stress-energy tensor. It is a straightforward calculation - it is somewhat tedious to do by hand but Mathematica would do the calculation in a few seconds. The problem is that the ... 14 When you ask "Why is gravity such a unique force?" then you should know that in the framework of General Relativity gravity is not a force at all. In General Relativity energy (for example the mass of an object) cause curvature. The movement of other objects is influenced by this curvature - they travel along the path of shortest distance between two points (... 14 You should think at the way the Noether current is obtained. When an infinitesimal symmetry transformation is made spacetime dependent, that is the parameters \omega^a that control the symmetry are taken as functions of the spacetime point \omega^a=\omega^a(x), the action is not longer left invariant$$ \delta S=-\int d^D x\, J^{a\,\mu}(x)\partial_\mu \...

14

TL;DR: It may be helpful to think of the above-diagonal orange components as "energy flux" rather than "momentum density"; if you do this, the interpretations in terms of shears and pressures become more natural. Here's another way to think of the stress-energy tensor. First, you're hopefully familiar with the notion of the energy-momentum four-vector: $... 13 The trick is given in equation 4.4 of the attached article: First couple the theory to gravity, (by introducing a metric tensor in the integration measure and for each index raising) obtaining the action:$S = \int_M d^4x \sqrt{-g} \mathcal{L}$Then vary the action with respect to the metric tensor:$T_{\alpha\beta} = \frac{1}{\sqrt{-g}} \frac{\delta S}{\...

13

Yes, everything generates a gravitational field, whether it is massive or massless like a photon. The source of the gravitational field is an object called the stress-energy tensor. This is normally written as a 4 x 4 symmetric matrix, and the top left entry is the energy density. Note that mass does not appear at all. We convert mass to energy by ...

12

$\require{cancel}$I) OP is considering Dirac fermions in a curved spacetime. OP's action has various shortcomings. The correct action reads$^1$ $$S~=~\int\!d^nx~ {\cal L}, \qquad {\cal L} ~=~e L, \qquad L~=~T-V,\qquad e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|},$$ $$T~=~\frac{i}{2} \bar{\psi} \stackrel{\leftrightarrow}{\cancel{\nabla}} \psi, \qquad V~=~ \... 12 The answer to your question is yes, the metric is influenced by the electromagnetic field, and a neutral particle will follow a geodesic of that metric. Thus, this implies that the neutral particle will indeed "feel" the electromagnetic field, but only in a very indirect way (from the geometry of spacetime around the particle). The best example of this is ... 11 Good question! From a physical perspective, the stress-energy tensor is the source term for Einstein's equation, kind of like the electric charge and current is the source term for Maxwell's equations. It represents the amounts of energy, momentum, pressure, and stress in the space. Roughly:$$T = \begin{pmatrix}u & p_x & p_y & p_z \\ p_x & ...

11

Surely regardless of the mathematics, in empty space it should still be forced zero by the fact that nothing is there? The moon orbits the Earth despite the fact that it is, for all intents and purposes, in a vacuum (i.e. the local matter/energy density is zero). The point is that matter and energy in one localized region produce spacetime curvature ...

10

The two derivations are indeed different, but the resulting object should be the same: it should be symmetric and conserved on-shell. In fact, perhaps the cleanest way to derive it is to couple the theory to gravity and then vary the resulting action with respect to the metric. If we let $$S = \int_M d^4x \sqrt{-g} \mathcal{L}$$ denote the action of the ...

10

No, spacetime curvature is not the same as matter. First of all, measuring two things and stating that the measures are equal doesn't mean that the two things are the same concept. For example, if a polygon has $E$ edges and $V$ vertices, then $E=V$, but that doesn't mean an edge is the same thing as a vertex. Second, there are many different ways of ...

10

First, there is no mechanical algorithm to solve a general differential equation. Einstein's equations are obviously no exception – in fact, they belong among the more complicated and less "solvable" equations among those one may learn about. Analytically writable solutions only exist in very special, simple, and/or symmetric cases (simple enough equations ...

10

As @Holographer has mentioned in a comment, the correct formula for the stress-tensor that enters the EFE is $$T_{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta S_{\text{matter}}}{ \delta g^{\mu\nu} }$$ whereas what you are computing is the canonical stress energy tensor. However, there is a subtle relation between the two, which I will elaborate upon here. ...

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. ...

9

The two derivations are actually identical, except for the fact that Weinberg didn't have the general form of the Noether theorem for symmetries acting on the space-time coordinates as well as on the fields (Equation 2.141 in Di Francesco, Mathieu and Sénéchal's book). As a consquence, Weinberg had to compute the variation of the action with respect to the ...

9

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 $$g_{\mu\nu}~=~e^a{}_{\mu} ~\... 9 Disclaimer: Let us here avoid the discussion of how to assign a stress-energy-momentum (SEM) pseudo-tensor t^{\mu\nu} to the gravitational field. The word pseudo here refers to the fact that t^{\mu\nu} is not a tensor wrt. general coordinate transformations; only a rigid subgroup thereof. In other words, the pseudo-tensor t^{\mu\nu}\frac{\partial}{\... 9 Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ... 8 Consider some basic unit analysis. Pressure is defined as force/area which is the same as momentum/area/time since F=dp/dt. Momentum flow would be the momentum passing through a unit area per unit time so it's the same units. More physically, think of a gas at constant pressure in a box. If you popped a little hole of unit area in the side of the box, the ... 8 Just like any observable in Quantum Mechanics, stress-energy tensor is an operator acting on the Hilbert space of the quantum field. For example, consider the Scalar field with Lagrangian$$ \mathcal{L} = \frac{1}{2} \partial_{\mu} \phi \, \partial^{\mu} \phi - \frac{m}{2} \phi^2. $$Its classical stress-energy tensor is given by$$ T_{\mu \nu} = \partial_{...

8

In most situations the distinction “matter first, geometry second” or “geometry first, matter second” is not that clear cut. Often assumptions are made that constrain both the geometry and stress energy tensor. Take for example the Schwarzschild metric. We derive it by writing down the most general metric compatible with restrictions imposed by physical ...

Only top voted, non community-wiki answers of a minimum length are eligible