New answers tagged tensor-calculus
2
Let's recap: upper indices are vectors ($x^\mu$), the inner product on Minkowski space is given by $g_{\mu \nu}$ so "dual vectors" have lower indices $x_\nu = g_{\nu \rho} x^\rho.$
Then you see that a matrix (in the sense of linear map between vectors) has one upper and one lower index, because it maps a vector to another vector:
$$x^\mu \mapsto ...
2
I) Pragmatically speaking, the most important property of $\sqrt{-g}$ for model building purposes, is not per se the fact that $\sqrt{-g}d^{4}x$ measures the volume element of a 4-dimensional Parallelepiped with infinitesimal edges $dx^0, \ldots, dx^3$.
II) A more important property is that $\sqrt{-g}d^{4}x$ transforms as a scalar (i.e. is invariant) under ...
4
A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise.
What you probably find surprising is that $\delta g_{\mu\nu}$ and $\delta g^{\rho\sigma}$ are not related to each other by simply raising the indices $\mu,\nu$ or lowering the indices $\rho,\sigma$. Indeed, they're not related in this way. In this case, ...
3
A perfect fluid is defined by the property that, in the local rest frame, it allows no energy fluxes and no anisotropic stresses. Thus, at a given space-time point, in the local rest frame [in which the components of the 4-velocity are $u^{\alpha} = (1, 0, 0, 0)^{\mathsf{T}}$], the energy momentum tensor components are $T^{\alpha\beta} = \mathrm{diag}(e, p, ...
3
First off, please don't use units with $c\ne 1$ in GR. It makes everything horribly messy.
What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like $\Delta x^\mu$. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that $u^\mu=(1,0,0,0)$, not $(-1,0,0,0)$. This is ...
1
As soon as you get something like $\delta_{bd}$, alarm bells should ring, as this is not a tensor.
The inverse metric $g^{ac}$ is defined by the identity
$$
g^{ac}g_{cb} = \delta^a_b
$$
If you plug this into your expression (and use the fact that $g$ is symmetric), you will obtain the correct equation.
0
It can be show easily by the next reasoning.
$$
DA_{i} = g_{ik}DA^{k},
$$
because $DA_{i}$ is a vector (according to the definition of covariant derivative).
On the other hand,
$$
DA_{i} = D(g_{ik}A^{k}) = g_{ik}DA^{k} + A^{k}Dg_{ik}.
$$
So,
$$
g_{ik}DA^{k} + A^{k}Dg_{ik} = g_{ik}DA^{k} \Rightarrow Dg_{ik} = 0.
$$
So, it isn't a condition, it is a ...
1
The elements of this particular metric tensor do not depend on time. $dx^j$ and $dt$ are treated as constants when you take the time derivative.
You should also have $\Gamma^{\lambda}_{00} = -\frac{1}{2} g^{\nu \lambda} \frac{\partial g_{00}}{\partial x^\nu}$.
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