New answers tagged metric-tensor
6
For $a<0$, it's a possible metric for spacetime. For $a>0$, it's a possible metric for a 4-dimensional Euclidean space. For $a=0$, it's degenerate, and in many cases it's not possible to work with a degenerate metric, e.g., the machinery of general relativity requires that the metric be nondegenerate. It doesn't matter whether $a$ has a particular ...
1
The FLRW metric can be static, this is the solution that Einstein concocted before Hubble observed the expansion of the universe. The only way that Einstein could make his equations static was by introducing the infamous cosmological constant $\Lambda$. The general FLRW metric has the form
$$
\text{d}s^2 = -c^2\text{d}t^2 + a(t)\left[\frac{\text{d}r^2}{1 - ...
1
I'm not sure what OP exactly is requesting, but OP's equation follows e.g. from the general fact that for an arbitrary 2D surface, the Ricci tensor
$$ R_{\mu\nu} ~\propto~g_{\mu\nu} $$
is always proportional to the metric tensor $g_{\mu\nu}$. This is basically a consequence of that in 2D the Riemann curvature tensor is complete determined by the scalar ...
2
Let's recap: upper indices are vectors ($x^\mu$), the inner product on Minkowski space is given by $g_{\mu \nu}$ so "dual vectors" have lower indices $x_\nu = g_{\nu \rho} x^\rho.$
Then you see that a matrix (in the sense of linear map between vectors) has one upper and one lower index, because it maps a vector to another vector:
$$x^\mu \mapsto ...
1
The Schwarzschild metric is, in $-+++$ sign convention and units of $c = 1$ is
$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{1-\frac{2M}{r}} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\text{.}$$
We can index the coordinates arbitrarily, but let's take them in the typical order: $(U^0,U^1,U^2,U^3) ...
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, ...
4
Since the metric and inverse metric are related by
$$
g^{\mu\lambda}g_{\lambda\nu} = \delta^\mu_\nu
$$
taking the variation of both sides gives
$$
\delta g^{\mu\lambda}g_{\lambda\nu} + g^{\mu\lambda}\delta g_{\lambda\nu} =0
$$
or in other words
$$
\delta g_{\mu\nu} = -g_{\mu\rho}g_{\nu\sigma}\delta g^{\rho\sigma}
$$
It follows that there is a ...
0
Obviously, you are talking about the derivation in Landau-Lifshitz book. I admit that this is not the clearest explanation of what is going on. (On the other hand that is not the first and not the last place like that in Landau-Lifshitz...)
Anyway, I don't really see what is your problem -- just use Stokes' theorem from (6.19):
$$\Delta A_i = \frac12\int ...
1
First of all, we don't usually talk about the direction of propagation of a plane wave in QFT. Plane waves are said to exist at all spacetime coordinates with a certain internal momentum, k. And, in reference to some of the comments, in QFT, we don't normally operate with wavefunctions. We promote wavefunctions to operators and act on states. But in this ...
0
I think there is a method that I believe is rather simple. Take a look:
There is a thing called 'normal Riemann coordinates'. In this coordinates the metric is expanded around the origin, and the coefficients of expansion are expressed in terms of the Riemann tensor. I suggest that you read about them and check whether the coordinates described below are ...
0
Given that your metric is diagonal, it simplifies a lot these calculations. However, the Riemman tensor is such an object...
First, start with the Christoffel Symbols
$$ \Gamma^i{}_{k\ell}= {1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m})$$
Note that $g_{im}=0$ for $i \neq m$ so it simplifies to
$$ \Gamma^i{}_{k\ell}= {1 \over 2} g^{ii} ...
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 ...
4
You are incorrect to suppose that this spacetime is curved. In fact, up to some conditions on the coordinate ranges, this is simply a piece of Minkowski spacetime. Let me put it in this form:
$$ds^2 = dt^2 - t^2(d\psi^2 + \sinh^2\psi\,d\Omega^2)\text{,}$$
where $d\Omega^2 = d\theta^2 + \sin^2\theta\,d\phi^2$ is the metric for a unit $2$-sphere, and we can go ...
0
The point is that saying "spacetime is curved" does not necessarily mean anything specific about the metric or any of the curvature tensors. I think for physics (meaning, for general relativity), "spacetime is curved" should mean that the scalar curvature is not zero, since that's what shows up in the action. Of course, what matters is the variation of the ...
2
Firs of all, the metric depends on the specific coordinate system. If metric is not diagonal, it does not mean that the space is curved -- just take some crazy coordinates in the flat space and you will have a really messy metric. Futhermore, only vanishing Riemann tensor ensures that there are local coordinates in which metric is diag(1,-1,-1,-1), that is, ...
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