Hot answers tagged

24

The shape of a black hole's event horizon depends on who is asking. Observers who are moving quickly towards a hole, for example, will see a different shape compared to those who are not. Per @benrg the event horizon of a static black hole is not observer dependent, similarly to how the shape of an expanding flash of light is not. In the coordinates ...


21

This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to ...


18

Yes, they are perfect spheres. But let's understand what is the sphere and why. The spherical surfaces are the horizons. They are surfaces in space that have perfect spherical geometry. Now, that only holds true for various conditions: 1) they are static black holes, and if they arose from matter/energy collapsing they are in their equilibrium states. ...


9

Actually the result is even stronger: Given a timelike geodesic $\gamma$ and a point $p \in \gamma$, there is a neighborhood $U \ni p$ equipped with coordinates, $x^0,x^1,x^2,x^3$ such that in the portion of $\gamma$ included in $U$, exactly along $\gamma$, the derivatives of the metric vanish in the said coordinates. Equivalently the Christoffel symbols $\...


3

from your profile you seem to be an amateur (gifted?, teenager?, still in school?) self-studying GR. Great! So drop the Old Man's Book (the Meaning of Relativity) and get yourself a modern intro to GR or - my suggestion - the wonderful MTW's Gravitation. I did the same when I was 16. It was published in 1973 and Kip Thorne himself told me two weeks ago that ...


3

Distance measurements in $n$ dimensional flat space follows the same pattern for $n$ equal 1,2,3, or higher values. I'm going to assume a straight line, change in position to simplify the math (that is we're measuring what a introductory book would call the "displacement" $s$ rather than distance. But then distance is just an accumulation of many magnitudes ...


2

No. When they merge their horizons will change shape, and eventually become the static or stationary shape of a BH horizon. Nothing inside either horizon while this is happening can escape. At all times the timelike curves stay inside, and the deformed horizons are where the lightlike curves end up. In each, and after they merge. The area of each horizon ...


2

One meter is a unit defined in the "real world" around us – places we can actually visit. Or it is used for the lengths and dimensions of objects we can touch. It only makes sense to use the same "meter" for other worlds if we can actually get to those worlds. If two worlds are completely separated from each other, it makes no sense to apply the units of ...


2

Whatever unit you're using for distance in 1D is still good in any number of dimensions. Kilometers in manifold of dimension n is fine (assuming non-compactified dimensions).


2

The answer is simply that not every space-time has a corresponding effective potential in the sense that we have a coordinate $x$ such that $\dot{x}=\sqrt{2(E-V_{eff})}$. But this is true even in Newtonian mechanics, consider a problem with a Lagrangian $$L = \frac{m}{2}(\dot{r}^2 + r^2 \dot{\varphi}^2) - V(\varphi)$$ Obviously, $p_r\equiv m \dot{r}$ is ...


2

That $\Delta g_{ij} = 0$ as you define it is equivalent to saying that the gradient of all metric components have vanishing divergence $$ g_{ij;k}{}^k \equiv g^{k\ell}g_{ij;k\ell} = 0. $$ Here it is important to remember that the indices $i,j$ denote functions. To clarify this we will let $g_k$ represent the gradient of an arbitrary component function $g_{ij}...


2

This definition doesn't depend on the metric signature convention. Note that in definition of $\gamma$ the metric doesn't appear anywhere. It is defined purely in terms of "3-vectors" and "3-scalars" measured by particular observer. So it is impossible for metric to appear here explicitly.


2

As commentators have indicated Hilbert space is a vector space. A manifold is a space with an atlas-chart construction with maps on overlapping regions that define connection coefficients and ultimately curvature. It is certainly possible to think of a finite dimensional complex vector space that is a locally flat region in an otherwise curved space. This ...


2

Hilbert spaces are vectorspaces by definition. If you interpret a vector space as a manifold (which you can do) then it's a flat manifold.


2

I have just now finished an article, "Geometry of the 3-sphere", in which at the end of the paper I give a simple derivation of the Riemann curvature bivector for the unit 3-sphere, using (Clifford) geometric algebra. I also discuss the Lie group $SU(2)$ and Lie algebra $SU(2)$ on the unit 3-sphere, using the powerful, but still rather unknown geometric ...



Only top voted, non community-wiki answers of a minimum length are eligible