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1 vote
3 answers
86 views

Does (covariant) divergence-freeness of the stress-energy tensor ${T^{\mu\nu}}_{;\nu}=0$ follow from the Bianchi identity?

I'm working through Chap. $30$ of Dirac's "GTR" where he develops the "comprehensive action principle". He makes a very slick and mathematically elegant argument to show that the ...
Khun Chang's user avatar
1 vote
0 answers
45 views

A covariant derivative computation in General Relativity [duplicate]

I am trying to compute $\nabla^\mu\nabla^\nu R_{\mu\nu}$. I proceed as follows: \begin{align} \nabla^\mu\nabla^\nu R_{\mu\nu}&=g^{\mu\rho}g^{\nu\lambda}\nabla_\rho\nabla_\lambda R_{\mu\nu} \\ &...
vyali's user avatar
  • 392
1 vote
4 answers
230 views

How to find the double covariant derivative of a general vector?

I have been reading through Carroll's GR textbook and there is a line in the derivation of the Riemann tensor that I do not understand. $$\nabla_\mu \nabla_\nu V^\rho=\partial_\mu(\nabla_\nu V^\rho) - ...
Chris G's user avatar
  • 51
8 votes
2 answers
825 views

How does the covariant derivative satisfy the Leibniz rule?

In Carroll's "Spacetime and Geometry", he states on page 95 (section 3.2) that the covariant derivative, $\nabla$, is a map from $\left(k, l\right)$ tensor fields to $\left(k, l+1\right)$ ...
MBar2269's user avatar
  • 103
1 vote
0 answers
79 views

Scalar curvature from Riemannian metric

I want to compute the scalar curvature for points on an empirical manifold (sampled data). I have already an algorithm that learns the Riemannian metric and computes geodesics, so from the metric I ...
can't stop me now's user avatar
0 votes
1 answer
122 views

Einstein field equations from covariant derivative of a general linear gauge transformation

A general linear transformation is given by \begin{align} \psi'(x) \to g \psi(x) g^{-1}, \end{align} The gauge-covariant derivative associated with this transformation is \begin{align} D_\mu \psi=\...
Anon21's user avatar
  • 1,548
1 vote
0 answers
160 views

What is the meaning of $\nabla _{\mu}\nabla _{\nu}\phi(r)$ in general relativity?

I know the covariant derivative of a tensor is $$\nabla_{\mu} V_{\nu}=\partial_\mu V_\nu-\Gamma_{\mu\nu}^{\lambda}V_{\lambda}$$ Now I want to obtain $\nabla_{\mu}\nabla_{\nu}\Phi(x)$ where $\Phi(x)$...
Alice's user avatar
  • 67
0 votes
1 answer
1k views

Covariant derivative, directional derivative, and curvature tensor

I'm confused about how to connect those three things together, so hopefully the question doesn't end up vague. The main problem is understanding how the curvature tensor is a commutator of covariant ...
EM_1's user avatar
  • 860
1 vote
1 answer
250 views

Are Riemann curvature tensors defined up to a total derivative?

Consider a case where the Riemann Tensor is given by $$R^{\mu}_{~~~\nu\rho\sigma} = P^{\mu}_{~~~\nu\rho\sigma} +\nabla_{\rho}A^{\mu}_{~~~\nu\sigma}-\nabla_{\sigma}A^{\mu}_{~~~\nu\rho}$$ It seems to me ...
Dr. user44690's user avatar
0 votes
0 answers
99 views

Derivatives of the metric in the local Lorentz frame

In the local Lorentz frame (local flatness at point P) we have: $$ g_{\alpha \beta} (P) = \eta_{\alpha \beta}, \quad \Gamma^{\rho}_{\alpha \beta}(P) = 0. $$ In the reference that I am following (The ...
Edison Santos's user avatar
1 vote
1 answer
350 views

Covariant derivative of spherical harmonics

Given is the metric $\gamma_{jk}$ for the surface of a Sphere $S^2$ with $\gamma_{22}=1,\gamma_{23}=\gamma_{32}=0$ and $\gamma_{33}=\sin^2(\theta)$. The coordinates are $x=$($t,r,\theta,\phi$) and $j$ ...
Trafalgar's user avatar
4 votes
1 answer
620 views

Derivation of Gauss-Codazzi type equation (Ricci relation)

I am following Padmanabhan's book Gravitation for the particular derivation. The derivation goes as follows, \begin{align} R_{abst}n^t&=\nabla_a\nabla_b n_s-\nabla_b\nabla_a n_s=\nabla_a(-K_{...
Faber Bosch's user avatar
1 vote
0 answers
98 views

Question on differentiation, Sakharov equation (related to a previous post) [closed]

I'll keep this as short and sweet as possible, the other day I was going through some old physics notes I had wrote. I'm quick to admit, my knowledge on physics is good but I'm in no way a pure ...
MaskedMagician's user avatar
3 votes
1 answer
2k views

Physical meaning of the vector Laplace operator

I have seen here a question asking for the physical interpretation of the Laplace operator for a scalar field. However, there is also a vectorial version of this operator, the vector laplace operator, ...
Invenietis's user avatar
3 votes
2 answers
171 views

Does covariant derivative include magnitude change of a vector as well as direction change of the same vector?

Does covariant derivative include magnitude change of a vector as well as direction change of the vector? In some explanations I followed I have not noticed mentioning of magnitude change along with ...
Janko Bradvica's user avatar

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