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0

If you are working with the Newman Penrose null tetrad, i.e. with respect to which the metric looks like $g=\left(\begin{array}{cccc}0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{array}\right)$ then the standard notation is $\lbrace l,n,m,\overline{m}\rbrace$, where $l,n$ distinguish the ...


2

The issue here is that the Schwarzschild coordinates are divided into two disconnected patches by the coordinate singularity at $r=2m$. There is no physical connection and the two pieces can be viewed as separate solutions. The Eddington-Finkelstein coordinates take the outer solution and extend it beyond the horizon, but inside the horizon it is different ...


3

Relativists tend to use the proper time, $d\tau$, and the proper distance, $ds$, interchangably. If you're working with proper time you'd expect the equation for it to look like: $$ d\tau^2 = dt^2 + \text{other terms} $$ while if you're working with proper distance you expect: $$ ds^2 = dx^2 + dy^2 + dz^2 + \text{other terms} $$ The sign problem comes ...


0

Could you provide a simple reason for these two conventions? The reason behind the (-,+,+,+) convention (the "mostly plus metric") is that a positive length in 3 dimensional space (e.g., the distance from my head to my toes) should still be a positive length in 4 dimensional space-time. Why should the distance from my head to my toes all of a sudden ...


0

related: http://math.stackexchange.com/q/160882/224026 see also: http://en.wikipedia.org/wiki/Metric_tensor: " From the coordinate-independent point of view, a metric tensor is defined to be a nondegenerate symmetric bilinear form on each tangent space that varies smoothly from point to point. "


1

1- A degenerate matrix is a matrix whose rank is smaller thank its dimension. 2- A singular (non-invertible) matrix is one who has a vanishing determinant. Equivalence of the two : A matrix whose rank is smaller than it's dimension when diagonalized will have at least one zero eigenvalue, and consequently a vanishing determinant.


3

If $\text{det } g = 0$, then $\text{ker } g \neq \{\vec{0}\}$, ie there is some vector $X \in \text{ker } g$, such that $g(X,\ast) $ gives zero 1-form, so $g(X,Y)=0 $ for any $Y$.


0

the quantity $ds^2 = g_{\mu,\nu} \ dx^\mu \ dx^\nu $ is a measure for distance on a manifold. It is indeed invariant under coordinate transformations, since $g_{\mu,\nu}$ transforms like $$ \tilde{g}_{\mu,\nu} = \frac{\partial x^\alpha}{\partial \tilde{x}^\mu} \frac{\partial x^\beta}{\partial \tilde{x}^\nu} \ g_{\alpha,\beta} $$ and $dx^\mu$ transforms like ...


0

Based on the comments from @ACuriousMind, the question itself is poorly formed. To calculate an interval between two events on a general manifold, one needs both a metric and a connection. Using the process of parallel transport, one moves one of the points next to the other. Then one can subtract the two points. The metric at that point is then used to ...


5

Although the Lorentz boosts have "tensor indices", they are not defined as tensors on the space, but as elements of the isometry group. Let $M,M'$ be two (pseudo-)Riemannian manifolds with metric tensors $g,g'$. Given a diffeomorphism $\phi : M\to M'$, you could let the isometry $\Lambda : M\to M$ (which, on $\mathbb{R}^{1,3}$ would be given by ...


0

Gödel’s stationary rotating solution is both locally rotationally symmetric and everywhere rotating. The material source of the Gödel metric is a pressure-free perfect fluid: the definition from wikipedia is "a homogeneous distribution of swirling dust particles". Each of these dust motes can be thought of as carrying a clock that measures proper ...


6

I'm unsure if you're specifically asking about the Gödel spacetime or if your question is a more general one of whether time travel can exist. So let me try to give a general answer that addresses both. Einstein's equation tells us how the geometry of spacetime is related to the distribution of matter and energy. Leaving aside the vexed issue of quantum ...


12

Gödel's rotating universe solution does allow for time travel (closed timelike curves), but it has nothing to do with wormholes--in such a universe one could travel into one's own past just by taking a rocket on a long-term looping path through space, from any starting location. This page discussing Gödel's solution includes a spacetime diagram showing how ...


-2

The idea that space is expanding is a relation between it's size and time. Time expanding would be a relation between time and time, so it can't???


4

It's really an either/or proposition, i.e. either space is expanding or the time experienced by distant objects is dilated, depending on how you view the situation. We choose the former description because it is better. To expand (excuse the pun!) on what I mean, the measurable result of time dilation is red-shift and indeed distant cosmological objects ...


0

You ask Is there more time, longer time or is the time part not affected at all by expansion? First off, Is there more time? I don't know because I dont know exactly what time is? Do you? Did Einstein? No, I think he said he didn't in one of his books? Does anybody? Probably not. Is there longer time is an easier question because I don't think that ...


2

Suppose that the vector field $X$ is tangent to the hypersurface. Then $X\cdot v$ must vanish, since $v$ is a constant on the surface. But $X \cdot v$ is precisely what is meant by $dv(X)$. So the tangent space is precisely the null space of $dv$. Now since we have a metric, we can "raise the index", $$dv \mapsto g^\sharp dv = g^{\mu\nu} (dv)_\nu = V^\mu$$ ...


0

Curvature isn't really a problem -- find the normal vector at an arbitrary point, since then this is just a linear algebra problem. Lorentzian signature is also not a problem -- as long as we have an inner product, we're good. The procedure is analogous to the "3+1 split" of spacetime used, for instance, in the ADM formalism. Often we foliate spacetime ...


2

I don't think your confusion is related to curvature, and I think your question would persist even in the following flat space example. Consider a surface in Minkowski space $$\{(t,x,y,z): t,x,y,z, \in \mathbb{R}\},$$ with metric determined by $$\begin{bmatrix} -c^2 & 0 & 0 & 0 \\[0.3em] 0 & 1 & 0 & 0 \\[0.3em] 0 ...


0

If you have coordinates $(v,r,\theta,\phi)$ on $M$, and your hypersurfaces $\Sigma_v$ are characterized by $v=const$, then $$ \frac{\partial}{\partial r},\ \frac{\partial}{\partial \theta}\ \mathrm{and}\ \frac{\partial}{\partial \phi} $$ are coordinate basis fields on $\Sigma_v,$ and $\left.\partial/\partial v\right|_v$ is a vector field that pointwise ...


0

You can try using this graph for reference: This is from the equation (in this, units are used such that the speed of light is 1) $$ E=\sqrt(1-2GM/r)/r $$ Where the x axis is the distance from the black hole and the y axis is energy. Here G is the gravitational constant, r is the radius from the black hole and M is the mass of the black hole Honestly I ...


0

When you have a rank (1,1) tensor (one raised and one lower index and no other indices) you can treat the nxn coordinates in a basis as a matrix, and then as a matrix you can compute the inverse, then interpret those numbers in the matrix as the coordinates in the same basis for a (possibly different) rank (1,1) tensor. That's literally all that is going ...


0

You can think of the metric $g$ (with components $g_{ij}$) as being a (0,2) tensor (0 upper indices, 2 lower indices) that eats two vectors $v$ and $w$ (components $v^{i}$ and $w^{i}$, each vector being a (1,0) tensor; the contraction of the $i$ and $j$ indices is the 'eating') and spits out a number. Similarly, the Weyl tensor eats 4 vectors to spit out a ...


2

Well they are real numbers... if you like you can express to another physicist your relation just as well like so: Suppose $\eta_{\mu \nu}$ is a metric in $N$ dimensions with components who vary contravariantly and $\bar{\eta}_{\mu \nu}$ the corresponding covariant components. Then $\eta_{\mu\nu} = ...


0

spacelike, timelike and lightlike [...] What, however, is the physical intuition behind? Using "physical" terminology means (foremost, and even exclusively) to refer to distinguishable "participants" (a.k.a. "principal identifiable points", or "material points") where each is (thought as being) capable, at least in principle, of determining with whom ...


2

Spacelike separation means that there exists a reference frame where the two events occur simultaneously, but in different places. Timelike separation means that there exists a reference frame where the two events occur at the same place, but at different times. Lightlike means that, well, light could travel between those points.


0

Expand your line element and obtain the metric $g_{ij}$. It is of the form $$g_{ij}=\delta_{ij}-n_in_j$$ where $n=\langle \frac3{13}, \frac4{13}, \frac{12}{13}\rangle$ and so $n_in^i=1$ What you have now is a projection operator (because $g_{ij}g_{jk} = \delta_{ik}$, check it symbolically) which does this: It takes any 3D vector $v$ and gives you its ...


1

It is not a space-time because it is not Lorentzian. It is actually Riemannian. This exercise may be from a general relativity book, but is in fact a geometry question. So I take it that the question is to show that it represents a two dimensional space. But since it is in the general relativity tag one can be smart and guess the following. Consider the ...


0

Having a zero column in a diagonalization is bad (since the metric would be degenerate), but also bad would be if somehow it looked like $$dl^2=dx^2+dy^2+dz^2$$ or $$dl^2=-dx^2-dy^2-dz^2.$$ S you also want to avoid the metric being positive definite or negative definite. For more dimensions you'd also want to worry about having two spatial directions and ...


1

As I answered to the linked question, it is almost never the case that we write out the field equations in full then work through some procedure to solve them. I think you will struggle to find any reference that does this. Solutions are almost invariably found by exploiting symmetry or using approximations such as the weak field. Have you Googled for the ...


4

It's the Eddington-Finkelstein metric in the case where $d\phi = d\theta = 0$ i.e. a radial trajectory. The $t$ coordinate is actually the Eddington-Finkelstein $u$ coordinate.


8

Let's separate out some definitions: metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ such that the following axioms hold for all $x,y,z \in X$: $d(x,y) \geq 0$, $d(x,y) = 0 \Leftrightarrow x = y$, $d(x,y) = d(y,x)$, and $d(x,z) \leq d(x,y) + d(y,z)$. pseudo-metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ ...


3

Yeah, you've not yet adapted. That's OK. Let me take you through it. In this conventional world of classical physics we have separate notions of distance and time, with the idea that either two events happen at the same time and therefore have an objective distance between them, or two events happen at different times and therefore have an objective time ...


9

I think it might help to think about the spacetime interval $\text{d}s^2$ as a measure of movement in spacetime relative to the speed of light. Let's say that you want to move from a point $p=(0,0,0,0)$ to another point $p'=(t,x,0,0)$. The quantity $\text{d}s^2 = c^2\text{d}t^2-\text{d}x^2$ is then: Positive if $x<ct$, which means that you traversed the ...


0

These equations are often done in units where c=1 to make things easier, in this case: $$ c^2=1 $$


0

If you have a metric, then you also have the manifold itself and all the points in it, and you can use the metric to compute the Einstein tensor at each point, and then multiply by a scalar constant to get the stress-energy tensor (assuming you know the value of your cosmological constant). But the reverse direction is very different. For instance, if you ...


1

As you described, if you have a metric, then you also have the manifold itself and all the points in it, and can use the metric to compute the Einstein tensor at each point, and then multiply by a scalar constant to get the stress-energy tensor (assuming no cosmological constant). But the reverse direction is very different. For instance, if you started ...


1

You can also do the following, which may not be as general as you want it, but the idea might be usefull for other problems. You already know that the given metric is Minkowski metric in different coordinates, so look at the null geodesics. In the usual coordinates $(t',x')$ they are given by $x'\pm t'=const$. Then find the null geodesics in the given ...


3

The problem with general relativity is it is too general and allows solutions that it would be very difficult to believe are physical. The first problem is that the metric is defined upon a topology. For example the physically extremely prohibitive condition that a spacetime is empty and with zero cosmological constant is not enough to uniquely define a ...


0

In general the answer is no, but if $g_{12}=const, g_{11}=const$ and there is a cosmological constant then yes. To see this recall the Einstein field equations $$ R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}+\Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}.$$ Now if the metric is constant then the Ricci tensor and Ricci curvature is trivial and we get $$ \Lambda ...


3

In the general case you want the Cartan-Karlhede algorithm. It is an algorithm for producing a complete set of classifying invariants for a metric, expressed as functions of the coordinates. Given the components of the metric $g$ in the coordinates $x_1, x_2, \ldots$, the algorithm produces a list \begin{align} \Lambda & = \Lambda(x_i) \\ \Psi_k & = ...


2

If you were to Wick rotate $t \rightarrow i \theta$, the metric would be $ds^2 = dr^2 + r^2 d\theta^2$, which is just flat space in polar coordinates. The standard cartesian coordinates can be obtained by $x=r\cos\theta$, $y=r\sin\theta$. The same procedure works in the original Lorentzian signature metric, but with hyperbolic trig functions instead of sines ...


2

$$g_{\mu\nu}=\begin{pmatrix}g_{00}&g_{01}&g_{02}\\g_{10}&g_{11}&g_{12}\\g_{20}&g_{21}&g_{22}\end{pmatrix}$$ $\mu,\nu=0,\ldots,N$ are the matrix indices of the metric (and of tensors in general) in $N+1$ dimensions.


7

Those Greek letters are indices indexing the components of $g$. Generally if one expresses a rank-2 tensor like $g$ as a matrix, the first index indexes the rows, the second the columns. In your example, we have $g_{rr} \equiv g_{11} = 1$, $g_{\theta\theta} \equiv g_{22} = r^2$, $g_{r\theta} \equiv g_{12} = 0$, etc. As you can see, we sometimes use numbers ...


7

Why is the scalar product of four-velocity with itself -1 The scalar product is invariant In the coordinate system in which the object is (momentarily) at rest, the only non-zero component is the temporal component. See that, in the rest frame, $\gamma = 1$ thus $d\tau = dt$. Then, (setting $c = 1$) we have $$\frac{dx^0}{dt} = 1,\,\frac{dx^i}{dt}=0 ...


4

The derivatives with respect to $\tau$ very much are numbers, but they are not all $1$. Consider your worldline as a curve $\gamma$ parameterized by $\lambda$. We have \begin{align} \gamma : \mathbb{R} & \to \mathbb{R}^4 \\ \lambda & \mapsto (x^0, x^1, x^2, x^3). \end{align} At any point in your worldline you have a position $(x^0, x^1, x^2, x^3)$, ...


10

First I just want to point out that saying that the four velocity $u_\mu$ satisfies $u_\mu u^\mu=-1$ is a convention, it is not a requirement. It amounts to a choice of the parameterization $\tau$. However, it is a very useful parameterization, it's not common to use other choices. In this parameterization, the four velocity takes the form \begin{equation} ...



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