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## Hot answers tagged stress-energy-tensor

19

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 ...

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 ...

9

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 & ... 7 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 ... 7 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 ... 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 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 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 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

Actually, in the context of general relativity, $c$ has no (physical) unit. More precisely, $c$ is meter per second. Meter is a measure of length. Second is a measure of time. In GR we unified space and time, and hence a meter and a second are different units of measurement for the "same thing". The number $c$ is a pure scalar that is just a conversion ...

4

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 ...

4

Every term contains one $\lambda$ in the superscript and one in the subscript, so you sum over those. The only indices which don't appear in both superscript and subscript in the same term are $\mu$ and $\nu$. Example: $$\Gamma_{\lambda\sigma}^\lambda\Gamma_{\mu\nu}^\sigma = \Gamma_{00}^0\Gamma_{\mu\nu}^0 + \Gamma_{01}^0\Gamma_{\mu\nu}^1 + \cdots + ... 4 I'll try and answer in an intuitive way as best I can (as you asked for on the crosslink). The relation between the true, or physical, surface area of a sphere with radius r_{meas} and the surface area one expects from standard Euclidean space is a measure (as you say) of the average curvature (more precisely it is a measure of the scalar curvature R: ... 4 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 ... 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 Your question is interesting from an historical perspective: this is exactly what William Kingdon Clifford conjectured. But this is not what modern physics thinks is going on. "Stuff" - matter and energy - it's all the same as far as the relevant physics is concerned - does influence the curvature of spacetime. The amount and distribution of "stuff" is ... 3 1) I gather you mean gravitation potential energy of the test particle. Well, any such thing is only useful in so far as it is related to a constant of motion throughout the geodesic--in the case of gravitational potential, being part of the conserved mechanical energy, kinetic + potential. (Another example could be angular momentum.) In GTR, these ... 3 Start with (iii)  T^\mu{}_\mu = g_{\mu\nu}T^{\mu\nu} I don't think this can be correct because both indices appear twice. What's wrong with  g_{\mu\nu}T^{\mu\nu}? Both indices are contracted. Explicitly it means$$ \sum_{\mu=0}^3\sum_{\nu=0}^3 g_{\mu\nu}T^{\mu\nu}$$which is a perfectly good scalar. g^\mu{}_\mu =2 here I summed all the ... 3 Keep in mind that the stress-energy tensor is described by the matter and energy fields. Each model of matter or field will have different expressions for this tensor the first one is for a scalar field \phi under the usual E^2 - c^2 P^2 = m^2 c^4 relativistic identity defined by the Klein-Gordon equation, the second is something entirely different, it ... 3 This object is related to the Schouten tensor,$$S_{ab} = \frac{1}{n-2}\left(R_{ab} - \frac{R}{2(n-1)} g_{ab}\right).$$We find$${C_{ab}}^{cd} - {R_{ab}}^{cd} = -4 S_{[a}^{[c} \delta_{b]}^{d]}.$$As @Luboš Motl mentions, this tensor depends only on the Ricci tensor and scalar curvature. That is, it doesn't "know" any more than Ricci---on the geometry ... 3 No, it is surely not possible to write every general tensor in terms of 7 parameters. Clearly, the space of possible stress-energy tensors is 10-dimensional in d=4 (10 parameters), so you can't make it 7-dimensional (7 parameters). The special choices of the tensor that you mentioned are isotropic in a frame – treating x,y,z on equal footing (they are ... 3 This is a clever method used to derive Noether's current for any global symmetry; for the translational symmetry, it produces the stress-energy tensor. We have to consider a local transformation because the variation of the action, \delta S, vanishes for the global transformation because the global transformation is by definition a symmetry:$$\delta S = ...

3

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 ...

3

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 $T^{\alpha\beta} = \mathrm{diag}(e, p, ... 3 The gravitational field can indeed be assigned an energy. Unfortunately though whereas for, say, the EM field you can define an energy density at a point ($\bf{E}^2+\bf{B}^2$), for the gravitational field you can't do this. - Whichever way you define the energy in terms of the Christoffel symbols, you run into the problem that you can make them, and hence ... 3 Speaking about Cauchy stress tensor in classical mechanics, the answer to your first question is that it does not matter, as you have metric in arbitrary coordinates induced by dot product of underlying Euclidian space. You can exploit symmetry of Cauchy stress tensor from balance of angular momentum assuming no couple-stresses, i.e. sources of angular ... 3 Here is my own answer to the first part of the question. I don't know the answer to the second part. Let's pick a local set of Minkowski coordinates$(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 ...

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