Hot answers tagged metric-tensor
15
The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not ...
13
A homogeneous cosmology is one in which there are no "special" places in the universe: at a given instant in time, the universe appears the same at every location (on large enough spatial scales).
An isotropic cosmology is one in which there are no "special" directions: at a given instant in time, the universe appears the same in every direction (again, on ...
12
It means that the laws of physics are the same everywhere and the same in every direction. It is of fundamental importance as these symmetries give rise to conservation laws. The isotropy of the universe means that angular momentum is conserved; its homogeneity means that momentum is conserved. A similar symmetry, that the laws of physics are the same for ...
8
An interesting question indeed :-) Yes, you can flip the overall sign of the Minkowski metric, and in fact a lot of physicists do this! The sign choice $\operatorname{diag}(-1, 1, 1, 1)$ is conventional in fundamental quantum field theory and in quantum gravity, if I remember correctly, whereas $\operatorname{diag}(1, -1, -1, -1)$ is conventional in particle ...
8
You tell if a space (or spacetime) is curved or not by calculating its curvature tensor. Or more unambiguously one of the curvature scalars (e.g. Ricci, or Kretschmann) since these don't depend on the coordinate system, but all of the information in the scalars is also contained in the Riemann tensor.
It is not necessarily obvious whether a given metric is ...
6
Expanded on Michael Brown's comments, you seem to have confused $dt$ with $d\tau$. They are not the same. It is true that $ds^2 = dt^2$, but also true that $ds^2 = g_{tt} dt^2 + \ldots$ (other terms omitted).
What this means physically is that, even for two events that take place at the same location but different coordinate times (according to a given ...
6
Each of the indices in a tensor have a particular left-right ordering. This ordering cannot be changed unless the tensor has some particular symmetry that permits it (or rather, that equates different components on interchange).
The up-down positions of indices tells us about whether the index is associated with using a basis vector (up) or a basis ...
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 ...
5
The expression $A^{\mu}B_{\mu}$ simply means that
$$A^{\mu}B_{\mu}=A^{0}B_{0}+A^{1}B_{1}+A^{2}B_{2}+A^{3}B_{3}$$
Using the Minkowski metric with signature $(+---)$ you write this as
$$A^{\mu}B_{\mu}=A^{\mu}\eta_{\mu\nu}B^{\nu}=A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}$$
The metric simply tells you have how the components of a vector and its dual vector ...
4
Edit edit: as has been pointed out, I was incorrect to say $\partial_t = \partial_{t'}$ and so on. Serves me right for trying to look at it by inspection instead of being rigorous.
Nevertheless, I do think cylindrical coordinates simplifies the problem somewhat. Recall the cylindrical line element:
$$ds^2 = -dt^2 + dr^2 + r^2 \, d\phi^2 + dz^2$$
Now, ...
4
1) OP is asking about the use of the word flat metric. It means a pseudo-Riemannian metric (of arbitrary signature) whose corresponding Levi-Civita Riemann curvature tensor vanishes.
2) However, the word Euclidean space may potentially cause confusion among mathematicians and physicists. For a mathematician an Euclidean space is always an affine space, ...
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
The problem arose when you wrote $ds^2 = g_{00} d\tau^2$. Generally one of your coordinates $x^\mu$ will be timelike, and the others spacelike, but the timelike one is not in general the proper time of someone whose spatial coordinates are not changing. That is, $t \neq \tau$. Using your sign convention,1 $d\tau^2 = ds^2$, so the (arbitrarily large) lapse in ...
4
When you write the five dimensional Kaluza-Klein metric tensor as
$$ g_{mn} = \left(
\begin{array}{cc}
g_{\mu\nu} & g_{\mu 5} \\
g_{5\nu} & g_{55}\\
\end{array}
\right) $$
where $g_{\mu\nu}$ corresponds to the ordinary four dimensional metric and $ g_{\mu 5}$ is the ordinary four dimensional vector potetial, $g_{55}$ appears as an ...
4
An easy way to see that they are distinct is to consider what happens upon raising (or lowering) all indices.
For example, upon lowering,
$$
T_{ab}{}^{cde}
$$
becomes $T_{abcde}$, whereas
$$
T_{a}{}^{cd}{}_{b}{}^{e}
$$
becomes $T_{acdbe}$, and similarly
$$
T_{a}{}^{cde}{}_{b}
$$
becomes
$$
T_{acdeb}.
$$
You need to "slant" the indices so as to keep track ...
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 ...
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
The covariant derivative of a scalar is just its gradient because scalars don't depend on your basis vectors:
$$\nabla_j f=\partial_jf$$
Now it's a dual vector, so the next covariant derivative will depend on the connection. Assuming the Levi-Civita connection, i.e. the Christoffel symbols, the covariant derivative will be:
$$\nabla_i \nabla_j f=\nabla_i ...
3
There are two manifolds that are involved in string propagation.
The spacetime in which the string propagates.
The worldsheet of the string itself.
The fields $X^\mu$ are embedding coordinates of the worldsheet in the spacetime manifold. This means that for each point $(\sigma^1, \sigma^1)$ on the worldsheet, $X^\mu(\sigma^1, \sigma^2)$ gives the ...
3
Solving the Einstein equation for a system as complex as the Solar System could only be done numerically, and in any case it's not terribly useful. Nothing in the Solar System, is relativistic enough to need more than a linearised treatment (this is how Einstein calculated the precession of Mercury).
Actually even solving Newton's equation for a system as ...
3
Since the metric and inverse metric are related by
$$
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}
$$
It follows that there is a ...
2
The motivation for this construction is explained here:
He first gives the example of a space of constant positive curvature - the 3-sphere, given by taking a flat Euclidean space of one higher dimension (4) and restricting to the subspace $(x_1, x_2, x_3, x_4)$ s.t. $$x_1^2+x_2^2+x_3^2+x_4^2=a^2$$ for some $a$. The metric on the sphere is just the ...
2
I recently re-derived these equations with all the dimensionful constants in place. Your last statement in the "Edit" is correct: $T_{00} = \rho_{E}\,c^{2} = \rho\,c^{4}$. It's easy to lose track of factors of $c$ in calculations like this; the usual culprit is mixing up $t$ and $x^{0} = c\,t$, and $\partial_t$ and $\partial_0 = c^{-1}\,\partial_{t}$. For ...
2
Because your solution is mathematically valid but unacceptable physically because $a(t)$ is required to be real, you have proved that the equation has no physical solutions. In other words, a negative-cosmological-constant space can't be sliced into flat spatial slices (you have assumed that the spatial curvature term $k$ is absent as well: with a negative ...
2
If you use different units on different coordinates and if you use a different unit on the distance metric then the diagonal elements of the metric matrix can differ from $1$. For example if you want to measure distances in meters, but you use inches for the x dimension and centimeters for the y dimension, then the 2D metric would be:
$$\begin{bmatrix}
...
2
According to the generally accepted theory of the Big Bang, the universe originated between 10,000 and 20,000 million ago years ago and has been expanding ever since.
The uncertain future of the universe: the expansion could be limited (closed universe), shrinking the universe upon itself, or it could be infinite (open universe), in which case the universe ...
2
If $\lim_{x\rightarrow 0}\frac{\alpha(x)}{\beta(x)}=A$ ($A$ is a number different from zero), then the functions $\alpha(x)$ and $\beta(x)$ are called infinitesimals of the same order [1].
The proportionality at $x\rightarrow 0$ should be obvious from this.
[1] http://www.math24.net/infinitesimals.html
2
It will be better if you were mentioned the sources.
However, as I remember, that in some old works on this theory, they used to assume that $\phi=const$ because the main purpose of the theory was looking for a geometrical unification of Gravity and electromagnetism, and no physical scalar fields was known at that time
Also it will be not very accurate to ...
2
In my opinion it is better to work in an explicit covariant form. In my answer I will use two different definitions, the Greek indexes always run from $0$ to $3$ and Latin indexes from $1$ to $3$ and the metric $g_{\mu\nu}$ has signature $(-1,1,1,1)$.
To translate the expressions to a explicit covariant form we define some timelike vector field $v^\mu$. We ...
2
Consider the $4\times 4$ matrix $g_{\mu\nu}$ with zeroth row $g_{0\nu}$.
Now for $i=1,2,3$, add to the $i$'th row the zeroth row times $-g_{i0}/g_{00}$.
This produces the following matrix
$$\begin{bmatrix}
g_{00} & g_{01} & g_{02}& g_{03} \\
0 & -\gamma_{11} & -\gamma_{12}& -\gamma_{13} \\
0 & -\gamma_{21} & ...
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