# 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 ...

14

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} ... 13 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 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 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}{\... 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} $$denote the action of the ... 10 Good question! From a physical perspective, the stress-energy tensor is the source term for Einstein's equation, kind of like the electric charge and current is the source term for Maxwell's equations. It represents the amounts of energy, momentum, pressure, and stress in the space. Roughly:$$T = \begin{pmatrix}u & p_x & p_y & p_z \\ p_x & ...

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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 ...

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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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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. ...

7

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^\... 7 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 ... 6 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 ... 6 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 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 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 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 ... 6 If classical mechanics were valid at cosmological levels, the answer would be yes. But general relativity is what describes the dynamics at this larger scale, and it is not generically possible in GR (in an arbitrary spacetime) to define conservation of momentum or conservation of mass-energy. 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 Einstein's equation is$$G_{ab} = 8 \pi T_{ab}.$$The left-hand side of the equation, G_{ab}, is called the Einstein tensor. It is an expression involving second derivatives of the metric, g_{ab}. The right-hand side of the equation, T_{ab}, is called the Stress-Energy tensor. Each stress-energy tensor is associated with a unique matter ... 5 I don't know much about supersymmetry but in absent of any other answers, maybe you will benefit infinitesimally from my guesses. Lets think in terms of a non-interacting SUSY theory with a bosonic and a fermionic field S \sim \int\text d^2z\left(\partial X\bar{\partial}X - \left(\psi\bar{\partial}\psi + \bar{\psi}\partial\bar{\psi}\right)\right). Then ... 5 This is a clever method used to derive Noether's current for any global symmetry; for the translational symmetry, it produces the stress-energy tensor. We have to consider a local transformation because the variation of the action, \delta S, vanishes for the global transformation because the global transformation is by definition a symmetry:$$\delta S = ...

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A black hole won't form. The reason why is that the boosted particle is equivalent by a boost to a reference frame where there is no black hole, and the presence/abscence of a black hole is coordinate-independent. While the energy of, say, an object with Earth's density profile can be made arbitrarily large through a boost, the boosted Earth will still ...

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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 ...

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A perfect fluid is defined by the property that, in the local rest frame, it allows no energy fluxes and no anisotropic stresses. Thus, at a given space-time point, in the local rest frame [in which the components of the 4-velocity are $u^{\alpha} = (1, 0, 0, 0)^{\mathsf{T}}$], the energy momentum tensor components are $T^{\alpha\beta} = \mathrm{diag}(e, p, ... 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 L}}{\... 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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