# Tag Info

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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 ...

12

Euler density is simply the integrand in $2n$ dimensions of the integral that is equal to the Euler characteristic. The Euler characteristic may be written as the integral of the following Euler density in $2n$ dimensions: $$E_{2n} = \frac{1}{2^n} R_{i_1 j_1 k_1 l_1} \dots R_{i_n j_n k_n l_n} \epsilon^{i_1 j_1 \dots i_n j_n} \epsilon^{k_1 l_1 \dots k_n l_n} ... 12 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 ... 12 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 ... 11 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 ... 11 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. 11 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 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} $$denote the action of the ... 10 No, spacetime curvature is not the same as matter. First of all, measuring two things and stating that the measures are equal doesn't mean that the two things are the same concept. For example, if a polygon has E edges and V vertices, then E=V, but that doesn't mean an edge is the same thing as a vertex. Second, there are many different ways of ... 9 Good question! From a physical perspective, the stress-energy tensor is the source term for Einstein's equation, kind of like the electric charge and current is the source term for Maxwell's equations. It represents the amounts of energy, momentum, pressure, and stress in the space. Roughly:$$T = \begin{pmatrix}u & p_x & p_y & p_z \\ p_x & ...

6

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 ...

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I recently re-derived these equations with all the dimensionful constants in place. Your last statement in the "Edit" is correct: $T_{00} = \rho_{E}\,c^{2} = \rho\,c^{4}$. It's easy to lose track of factors of $c$ in calculations like this; the usual culprit is mixing up $t$ and $x^{0} = c\,t$, and $\partial_t$ and $\partial_0 = c^{-1}\,\partial_{t}$. For ...

6

Here's part of my answer to the derivvation of the EM tensor for the ghost action. It does not match the expression you gave, but I may have made a mistake. CAn you check my work? We start with the action \begin{split} S_{gh} &= - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} \nabla_\mu c^\beta \\ \end{split} ...

6

Here is my own answer to the first part of the question. I don't know the answer to the second part. Let's pick a local set of Minkowski coordinates $(t,x,y,z)$. Then $T_{\mu\nu}$ represents a flux of the $\mu$ component of energy-momentum through a hypersurface perpendicular to the $\nu$ axis. For example, say we have a bunch of particles at rest in a ...

6

Apparently the behavior of tachyons in general relativity has been analyzed, though I don't have information on your second question about couplings to tardyons. P. 127 of J. Richard Gott III's book Time Travel in Einstein's Universe, available on google books here, says: A tachyon would have to be accompanied by gravitational waves, just as an ...

6

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: $$T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha ... 5 No, that transformation is not legit. Your final statement (A) says that the covariant derivative of the stress-energy tensor vanishes for every component. It has three free indices which means it represents 4\times4\times4=64 real-valued equations. Your initial statement (B) states only that four-at-a-time linear combinations of those derivatives ... 5 This is an answer to the question as qualified in a comment. The stress energy tensor is a tensor field so it is a function of position in spacetime. In the Schwarzschild coordinates the geometry is time independent so the local value of the stress-energy tensor is just a function of the position in space. Everywhere outside the spherical object it is zero ... 5 The energy momentum tensor is found by varying the metric and holding all other fields constant. Since clearly$$\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}$$and comparison with ... 5 Yes. The tear is initiated at stress concentrations around the holes, where stress is highest. After initiation, the tear continues to propagate along the line of highest stress. Stress is a function of force and geometry (\sigma_{n} = \frac {F}{A_{n}}). In a piece of paper without holes, the stress is uniform, and the paper tears when stress exceeds ... 5 Isotropy and homogeneity are different. The former is a consequence of invariance under rotations while the latter comes from invariance under translations. The stress tensor of an isotropic fluid then must be invariant under any orthogonal transformation, and this implies that it is a multiple of the "identity" tensor. More precisely, assume matrix notation ... 5 As @Holographer has mentioned in a comment, the correct formula for the stress-tensor that enters the EFE is$$ T_{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta S_{\text{matter}}}{ \delta g^{\mu\nu} } $$whereas what you are computing is the canonical stress energy tensor. However, there is a subtle relation between the two, which I will elaborate upon here. ... 5 The tensor itself is coordinate independent, however its components with respect to a basis in the tensor product space are. You can switch back and forth between tensor components of the same type (such as 2 times covariant T_{\mu\nu}) using the general transformation law for tensor components that you can find in any introductory diff. geometry or ... 5 Recall that the Chern-Simons action in terms of differential forms is given by,$$S= \frac{k}{4\pi}\int_M \mathrm{Tr} \left[ A\wedge \mathrm{d}A + \frac{2}{3}A \wedge A \wedge A\right]$$where A is our gauge connection. We now employ a definition of the stress-energy tensor which we would normally also apply when varying the Einstein-Hilbert action back ... 5 This is really a comment, but it got a bit long for the comment field. I'd guess that, like me, your experience in physics is from an area where solving differential equations is a routine part of the job. We're used to analysing a problem, writing down a differential equation that encapsulates the physics and solving it, analytically if we're lucky or in ... 5 See Edit below, the original answer is not completely correct. There is no gauge freedom in F. F is gauge invariant. In fact, F is completely measurable. It's components are the Electric and Magnetic fields, so you just go out with a set of test charges and measure E and B and you've got F. One hint that T and F do not contain the same ... 5 First off, please don't use units with c\ne 1 in GR. It makes everything horribly messy. What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like \Delta x^\mu. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that u^\mu=(1,0,0,0), not (-1,0,0,0). This is ... 5 Symmetry of the canonical energy-momentum tensor can be related to the spin of the object(s) that contribute to it (in other words, the representation of the Lorentz group under the fields transform). Note that the canonical EM tensor is obtained by using the Noether's procedure for translational symmetry$$ T_{\mu\nu} = \sum\limits_r \frac{\delta {\cal ...

5

The two quantities don't correspond because they are conserved quantities corresponding to different symmetries. One is a symmetry from shifting your field, the other from shifting space-time itself. Here is what is going on precisely: Let us do a simpler case first: In a particle mechanics system, let's say a free particle with $L = \frac{1}{2}m\dot{x}^2$, ...

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