# Tag Info

29

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: $$E_{2n} = \frac{1}{2^n} R_{i_1 j_1 k_1 l_1} \dots R_{i_n j_n k_n l_n} \epsilon^{i_1 j_1 \dots i_n j_n} \epsilon^{k_1 l_1 \dots k_n l_n} ... 10 When you ask "Why is gravity such a unique force?" then you should know that in the framework of General Relativity gravity is not a force at all. In General Relativity energy (for example the mass of an object) cause curvature. The movement of other objects is influenced by this curvature - they travel along the path of shortest distance between two points ... 10 The trick is given in equation 4.4 of the attached article: First couple the theory to gravity, (by introducing a metric tensor in the integration measure and for each index raising) obtaining the action: S = \int_M d^4x \sqrt{-g} \mathcal{L} Then vary the action with respect to the metric tensor: T_{\alpha\beta} = \frac{1}{\sqrt{-g}} \frac{\delta ... 10 The two derivations are indeed different, but the resulting object should be the same: it should be symmetric and conserved on-shell. In fact, perhaps the cleanest way to derive it is to couple the theory to gravity and then vary the resulting action with respect to the metric. If we let$$ S = \int_M d^4x \sqrt{-g} \mathcal{L} $$denote the action of the ... 10 No, spacetime curvature is not the same as matter. First of all, measuring two things and stating that the measures are equal doesn't mean that the two things are the same concept. For example, if a polygon has E edges and V vertices, then E=V, but that doesn't mean an edge is the same thing as a vertex. Second, there are many different ways of ... 8 Good question! From a physical perspective, the stress-energy tensor is the source term for Einstein's equation, kind of like the electric charge and current is the source term for Maxwell's equations. It represents the amounts of energy, momentum, pressure, and stress in the space. Roughly:$$T = \begin{pmatrix}u & p_x & p_y & p_z \\ p_x & ...

6

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

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 ...

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 ...

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

General relativity is a classical theory. I will restate your dilemma as follows, since this is how Einstein stated it: We have an abstract manifold consisting of points, vectors that link nearby points, and a metric tensor that tells you the distance between nearby points. What makes these points physical? How can we tell point A apart from point B? Since ...

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 $L = \frac{1}{2}m\dot{x}^2$, ...

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

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

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

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

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

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 ...

4

The electromagnetic stress tensor can be defined in all spacetimes: $$\frac{\delta \mathscr{L}}{\delta g^{ab}} = F_{a}{}^{c}F_{bc} - \frac{1}{2}g_{ab}F^{cd}F_{cd}$$ Which reduces to the expression in the Wikipedia article for the case of flat spacetime. Note that it is still fine to define this in flat spacetime becase: 1) Electromagnetism is perfectly ...

4

I henceforth assume $c=1$. Consider a 3-volume $\Delta \Sigma_0$ at rest with the dust. For the rest observer (this is its definition) $u^\mu$ has only (unit) temporal component. The energy (i.e the mass) associated with that portion of system is $\Delta \Sigma_0 \rho_0$. The 4-momentum of that portion is therefore $\Delta p^\mu := \Delta \Sigma_0 \rho_0 ... 4 No, there is no contribution to the pressure from the gravitational attraction between the particles. To see this you need to appreciate that the pressure is an ensemble property, and look at the stress-energy tensor for a single point particle. This is: $$T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\beta \delta\left( x - x_p(t) \right)$$ where$v$... 4 OP's question (v6) asks: Is it possible to obtain the symmetric energy-momentum tensor directly from variational principles by adding a total derivative term to the Lagrangian? In other words, by shifting$\Delta{\cal L}=d_\mu X^\mu$, and choosing$X^\mu$appropriately, can we exactly get the shift in the energy-momentum tensor required, in order to ... 4 The gravitational field can indeed be assigned an energy. Unfortunately though whereas for, say, the EM field you can define an energy density at a point ($\bf{E}^2+\bf{B}^2$), for the gravitational field you can't do this. - Whichever way you define the energy in terms of the Christoffel symbols, you run into the problem that you can make them, and hence ... 4 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 ... 4 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 ... 4 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 ... 4 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} ... 4 @Prahar is right, the variation of the Christoffel symbol is a tensor, even if the Christoffel itself is not. We have$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}\delta\bigg(g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})\bigg)=\frac{1}{2}\delta g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})+ ...

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