einstein tensor वाक्य
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- In the mid 1980s, a decade after Lovelock proposed his generalization of the Einstein tensor, physicists began to discuss the quadratic Gauss Bonnet term within the context of string theory, with particular attention to its property of being ghost-free in Minkowski space.
- However, due to the second Bianchi identity of the Riemann curvature tensor, the divergence of the Einstein tensor is zero which means that four of the ten equations are redundant, leaving four degrees of freedom which can be associated with the choice of the four coordinates.
- The coefficients of the characteristic will often appear very complicated, and the traces are not much better; when looking for solutions it is almost always better to compute components of the Einstein tensor with respect to a suitably adapted frame and then to kill appropriate combinations of components directly.
- Where \ scriptstyle G _ { ab } is the Einstein tensor, \ scriptstyle \ Lambda is the cosmological constant, \ scriptstyle c is the speed of light in a vacuum and \ scriptstyle G is the gravitational constant, which comes from Newton's law of universal gravitation.
- Einstein formulated this relation by using the Riemann curvature tensor and the metric to define another geometrical quantity "'G "', now called the Einstein tensor, which describes some aspects of the way spacetime is curved . " Einstein's equation " then states that
- In his theory, Yilmaz wishes to retain the left hand side of the Einstein field equation ( namely the Einstein tensor, which is well-defined for any Lorentzian manifold, independent of general relativity ) but to modify the right hand side, the stress energy tensor, by adding a kind of gravitational contribution.
- I have heard that the energy density is negative in the ergosphere of Kerr, although perhaps that's not immediately relevant to your intuition regarding negative mass . ( Of course Kerr, being vacuum, has vanishing Einstein tensor, so " energy " is possibly used in a different sense . ) BiaBy 22 : 08, 31 January 2016 ( UTC)
- The main reason for this is that " gravitational field energy " is not a part of the energy momentum tensor; instead, what might be identified as the contribution of the gravitational field to a total energy is part of the Einstein tensor on the other side of Einstein's equation ( and, as such, a consequence of these equations'non-linearity ).
- Where G _ { \ alpha \ beta } = R _ { \ alpha \ beta }-{ 1 \ over 2 } g _ { \ alpha \ beta } R is the Einstein tensor, which produces the correct left-hand side to the Einstein field equations, without the cosmological term, which however is trivial to include by replacing S _ { EH } with
- Variation of the first term of the action with respect to the tetrad e ^ { \ alpha } _ { \ I } gives the ( mixed index ) Einstein tensor and variation of the second term with respect to the tetrad gives a quantity that vanishes by symmetries of the Riemann tensor ( specifically the first Bianchi identity ), together these imply Einstein's vacuum field equations hold.
- These employ a great deal of differential geometry, and quite a few things that I have not worked much with preveously, such as affine connections, the torsion tensor, parallel transport, affine geodetics, affine flatness, tangent / fibre bundle, sections, Lie derivative, covariant derivative, absolute derivative, Riemann tensor, Ricci tensor, Einstein tensor, Christoffel symbols, Levi-Civita connection, etc.
- In the case of a Lorentzian manifold, n = 4, the Einstein tensor G _ { ab } = R _ { ab }-1 / 2 \, g _ { ab } R has, by design, a trace which is just the negative of the Ricci scalar, and one may check that the traceless part of the Einstein tensor agrees with the traceless part of the Ricci tensor.
- In the case of a Lorentzian manifold, n = 4, the Einstein tensor G _ { ab } = R _ { ab }-1 / 2 \, g _ { ab } R has, by design, a trace which is just the negative of the Ricci scalar, and one may check that the traceless part of the Einstein tensor agrees with the traceless part of the Ricci tensor.
- Here, the Einstein tensor G _ { \ mu \ nu } describes the curvature of space-time, whilst the energy-momentum tensor T _ { \ mu \ nu } describes the local distribution of matter . ( \ kappa is a constant . ) The Einstein equations express " local " relationships between the quantities involved specifically, this is a system of coupled non-linear second order partial differential equations.
- The GR field equation ( ignoring some uninteresting constants ) looks like G + ?g = T, where G is the " Einstein tensor " describing spacetime curvature, ? is the " cosmological constant ", g is the " metric tensor " describing . . . well . . . spacetime curvature again, I suppose, and T is the " stress-energy tensor " describing everything else ( the other forces and the fermions ).
- Regardless of whether you use an inertial frame of reference or an accelerating frame of reference in this problem, the Riemann curvature tensor will be zero everywhere, so the Einstein tensor will be zero everywhere, and the Einstein field equations simplify down to 0 = 0 . ( The mass of the objects involved here has a negligible effect on the curvature, and so is taken to be zero . ) In other words, in this problem, even if you use an accelerating frame of reference, the Einstein field equations that describe gravity say nothing at all about the observed acceleration.
- Where G _ { \ mu \ nu } = R _ { \ mu \ nu }-{ R \ over 2 } g _ { \ mu \ nu } is the Einstein tensor, which combines the Ricci tensor, the scalar curvature and the metric tensor, \ Lambda is the cosmological constant, 0 T _ { \ mu \ nu } energy-momentum tensor of matter, \ pi is the irrational number originally introduced as the ratio of the circumference of a circle to its diameter, c is the speed of light, G Newton's gravitational constant.
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