Does an event horizon ever exist?

[Nugatory]

[Edit] Found it here
http://www.astronomycafe.net/qadir/ask/a11339.html

Well, that just shows that you shouldn't ask an astronomer when you can ask a physicist

Seriously, kidding aside, that link is basically just plain wrong, or at least worded so imprecisely that it might as well be wrong (it doesn't mention which coordinates it's talking about - that's a huge red flag right there). See also #114 in this thread.

 

[ExecNight]

I guess if there wasn't an event horizon or singularity. As outside observers we wouldn't be able to observe a black hole getting larger in the universe?

So i guess again even if the objects just redshift at the event horzion, they are gravitationally adding up to it. And at a finite time.

I am guessing yet again, those objects don't create a new gravitational point at the event horizon. As they move infinitly slow at that point. Since we don't observe event horizons expanding asymetrically around a singularity. We can assume the object hits the singularity at a finite time inside for us outside observers.

What is the speed an object's redshifting at the event horizon?

If i sent a telescope moving at 0.99999999999999999999999999c towards the event horizon, would i see a sunlike object instead of a black hole with the combined light of all the objects that's falling into it?

Posting a thought process, be calm. Take it easy.

 

[Nugatory]

Posting a thought process, be calm. Take it easy.

No problem at all, but you might want to consider starting another thread instead of sending this one down a tanget.

 

[PAllen]

[Edit] Found it here
http://www.astronomycafe.net/qadir/ask/a11339.html

So you did get it from somewhere. Unfortunately it is still completely false. Sorry, not all sources on the internet can be trusted.

 

[PAllen]

Not necessarily; this also depends on the coordinate system you pick. There are coordinate charts in which all four coordinates are spacelike inside the horizon--for example, Painleve coordinates. (AFAIK it's not possible to have *more* than one timelike coordinate, but you can have less than one.)

This just makes it even more important to understand that coordinates don't tell you about the physics; you have to compute invariants to understand the physics.

On further thought, I think what you propose is also not correct.

First, it is possible to have two time like coordinates. Consider in Minkowski space u and v defined by:

x = .5 (u+v)
t= (v-u)/c
y and z unchanged

Then the metric in timelike signature is:

c^2d\tau^2 = .75 dv^2 + .75 du^2 - 1.5 dudv - dy^2 - dz^2

Clearly u and v are timelike and there is nothing wrong with such coordinates.

The more interesting question is possibilities for an orthonormal basis at an event. Here, I don't see how it could be true that you could have 4 orthonormal spacelike vectors. Please show me how this could be if you think so. For orthonormal basis, I can demonstrate two possibilities:

- 3 spacelike, 1 timelike
- 2 spacelike, 2 lightlike

I am not sure if there is any other possibility, but I certainly don't see how to have 4 orthonormal spacelike vectors.

 

[Dale]

The more interesting question is possibilities for an orthonormal basis at an event. Here, I don't see how it could be true that you could have 4 orthonormal spacelike vectors. Please show me how this could be if you think so. For orthonormal basis, I can demonstrate two possibilities:

- 3 spacelike, 1 timelike
- 2 spacelike, 2 lightlike

I could be wrong, but I don't think that you can consider lightlike vectors to be normalized. So I don't think that an orthonormal basis can have any lightlike vectors.

 

[PAllen]

I could be wrong, but I don't think that you can consider lightlike vectors to be normalized. So I don't think that an orthonormal basis can have any lightlike vectors.

They can't be normalized, but they can be orthogonal or not to another vector via dot product. You can definitely have coordinate systems with light like coordinates, so what do you call the basis vectors of such a coordinate system?

[edit: maybe need to distinguish orthogonal from orthonormal? ]

 

[George Jones]

Peter didn't specify that the coordinates are orthogonal. It is not possible to have four orthogonal spacelike vectors, but it is possible to have a basis of four spacelike vectors, i.e., to have four linearly independent spacelike vectors (e.g., Painleve-Gullstrand inside the event horizon). It is also possible to have four independent timelike vectors, four linearly independent lightlike vectors, etc.

The metric components with respect to Painleve-Gullstrand coordinates contain non-zero cross-terms, and thus Painleve-Gullstrand is not an orthogonal coordinate system.

 

[PAllen]

Peter didn't specify that the coordinates are orthogonal. It is not possible to have four orthogonal spacelike vectors, but it is possible to have a basis of four spacelike vectors, i.e., to have four linearly independent spacelike vectors (e.g., Painleve-Gullstrand inside the event horizon). It is also possible to have four independent timelike vectors, four linearly independent lightlike vectors, etc.

The metric components with respect to Painleve-Gullstrand coordinates contain non-zero cross-terms, and thus Painleve-Gullstrand is not an orthogonal coordinate system.

I knew that and never disputed that. I disagreed with the claim of at most one timelike, and separately raised the question of orthogonal coordinates to have something more restrictive to discuss. I saw the possible basis of Peter's claim of at most one timelike coordinate as that he was thinking of orthogonal coordinates - then I believe it is true.

 

[George Jones]

I knew that and never disputed that. I disagreed with the claim of at most one timelike, and separately raised the question of orthogonal coordinates to have something more restrictive to discuss. I saw the possible basis of Peter's claim of at most one timelike coordinate as that he was thinking of orthogonal coordinates - then I believe it is true.

Sorry, I misinterpreted. Even though you specified "orthonormal" and Peter didn't, from

The more interesting question is possibilities for an orthonormal basis at an event. Here, I don't see how it could be true that you could have 4 orthonormal spacelike vectors. Please show me how this could be if you think so.

I thought you were taking issue with the first sentence in the quote below, and that you wanted Peter to justify his claim.

There are coordinate charts in which all four coordinates are spacelike inside the horizon--for example, Painleve coordinates. (AFAIK it's not possible to have *more* than one timelike coordinate, but you can have less than one.)

This just makes it even more important to understand that coordinates don't tell you about the physics; you have to compute invariants to understand the physics.

 

[Dale]

They can't be normalized, but they can be orthogonal or not to another vector via dot product. You can definitely have coordinate systems with light like coordinates, so what do you call the basis vectors of such a coordinate system?

[edit: maybe need to distinguish orthogonal from orthonormal? ]

Yes, exactly. Orthogonal, but not orthonormal.

I don't know if there is a term besides orthogonal for a basis which contains null vectors and for which the non-null vectors are orthonormal.

 

[PeterDonis]

Peter didn't specify that the coordinates are orthogonal. It is not possible to have four orthogonal spacelike vectors, but it is possible to have a basis of four spacelike vectors, i.e., to have four linearly independent spacelike vectors (e.g., Painleve-Gullstrand inside the event horizon). It is also possible to have four independent timelike vectors, four linearly independent lightlike vectors, etc.

Yes, I didn't intend to restrict discussion to orthogonal coordinates, though I agree with PAllen that that is an interesting separate question, and when I made the statement about more than one timelike coordinate being possible I think I was relying on intuition that only applies to orthogonal coordinates. I think it's ultimately due to an intuitive tug to conflate "linearly independent" with "orthogonal".

(I remember bcrowell giving an example a while back of a chart with four null coordinates, but I have never seen an example of a chart with more than one timelike coordinate.)

 

[George Jones]

Yes, it is impossible to have four orthonormal spacelike vectors. To see this, let \left\{\bf{e}_\mu \right\} be a standard orthonormal basis that consists of one timelike and three spacelike vectors. There is nothing special about the orthonormal spacelike vectors \left\{\bf{e}_i \right\}, so try to use them as elements of an orthonormal basis of spacelike vectors.

Let \bf{v} be orthogonal to all three elements of \left\{\bf{e}_i \right\}. This gives that \bf{v} must be proportional to \bf{e}_0, i.e., \bf{v} must be timelike.

Geroch inverts this to give a basis-independent definition of Minkowski spacetime that I like.

 

[WannabeNewton]

What are orthogonal coordinates? Coordinates on an open subset $U$ of a real differentiable manifold are just $C^{\infty}$ functions $x^{\mu}: U\rightarrow \mathbb{R}$. It doesn't make any sense to talk about orthogonal coordinates using the metric tensor on $M$ (you can try and find an inner product on $C^{\infty}(M)$ but as far as I know there is no natural one and even if you did this wouldn't even be physically relevant). I don't get why there is such an interchange of the term coordinates and the term coordinate vector fields; they are quite different objects.

You can of course always choose the coordinate vector fields evaluated at some point in the open subset as the basis for the tangent space there but this basis will not be orthogonal in general let alone orthonormal with respect to the metric tensor evaluated at that point (of course you can always find some orthonormal basis using Gram Schmidt). Also, the coordinate vector field case is a very special one in that the coordinate basis has a direct relation to the coordinates but there are a plethora of bases I can choose for my tangent space at some point, consisting of some permissible combination of time-like and space-like vectors, and they have no a priori relation to the coordinates at that point at all (you can of course relate them to the coordinate basis via the change of basis transformation but that is extra).

Separately, null tetrads show up all the time e.g. in the context of spinors.

 

[George Jones]

What are orthogonal coordinates?

\left\{ x^\mu \right\} is an orthogonal coordinate if at every p in the coordinate chart, \left\{ \partial_\mu \right\} evaluated at p is an orthonormal basis for the tangent space at p.

For nice spacetimes, there always exist (orthonormal) tetrad fields \left\{ \bf{e}_\mu \right\}, but there don't always exist orthogonal coordinate systems.

 

[WannabeNewton]

\left\{ x^\mu \right\} is an orthogonal coordinate if at every p in the coordinate chart, \left\{ \partial_\mu \right\} evaluated at p is an orthonormal basis for the tangent space at p.

Ah ok so it isn't a literal thing, it is based off of the coordinate vector fields. Regardless, given any orthonormal basis for $T_{p}M$, if $M$ is to be a space-time with a lorentzian metric then there can only be one time-like basis vector in the orthonormal set otherwise we won't have a lorentzian signature. The signature is independent of orthonormal basis that's for sure. Now the finite dimensional spectral theorem says that any symmetric matrix with real entries can be diagonalized and $g_{ab}$ is symmetric and contains real entries, regardless of what basis it is represented in. When diagonalized we can still read off the signature since the eigenvalues will just be the diagonal entries and looking at their signs will give us the signature. However this will in general change the basis it is represented in so: if $(t,x^1,x^2,x^3)$ are coordinates on some open subset of space-time then do we only define $t$ to be time-like if $g(\partial_t,\partial_t) < 0$? If so then I would off the cuff find it hard to see why there can only be one time coordinate.

 

[George Jones]

but I have never seen a result that says it must be independent of all possible bases for $T_{p}M$ e.g. the coordinate basis; it is possible such a result exists but I have never seen it in a text.

I am not sure what you are saying here.

 

[WannabeNewton]

I am not sure what you are saying here.

Would the signature of the metric be readable off of any basis (and independent of any basis), in particular the coordinate basis? I don't suppose so, considering possible cross terms in the metric that could come out of the coordinate basis such as for the kerr metric. So I do agree with PAllen.

 

[George Jones]

Would the signature of the metric be readable off of any basis (and independent of any basis), in particular the coordinate basis? I don't suppose so, considering possible cross terms in the metric that could come out of the coordinate basis such as for the kerr metric.

Right. I wrote

The metric components with respect to Painleve-Gullstrand coordinates contain non-zero cross-terms, and thus Painleve-Gullstrand is not an orthogonal coordinate system.

If the metric components with respect to a particular coordinate system contain non-zero cross-terms, then that particular coordinate system is not orthogonal. It doesn't necessarily mean that an orthogonal coordinate system doesn't exist, but it might be case that an orthogonal coordinate doesn't exist. I think that general results are a bit tricky to obtain.

 

[photonkid]

(I seem to remember saying somewhere back in this thread that you (or someone?) are are making things harder for yourself by jumping into a fairly subtle problem of GR before you've solidly nailed down the basics of special relativity and classical mechanics. If I didn't say it already, I'm saying it now)

Yep, that was Optimus you said it to but I did take notice. I've had a goal of understanding special relativity for a long time but nothing makes any sense. Like, how did Einstein figure out the speed of light is the same for all observers - do you need a full understanding of Maxwell's equations to understand how Einstein got there. And where on Earth did E = MCsquared come from. I bought the spacetime physics book a few years ago but the first section "parable of the surveyors" didn't make any sense at all and I gave up. I'm frustrated because I can't tell if special relativity is hard or if the stuff I read is superficial or if I'm just blind or if there's nothing much to understand.

Every couple of years I come back to trying to understand special relativity and my next plan is to read the special relativity wikibook thing on Wikipedia.

 

[PAllen]

Right. I wrote


If the metric components with respect to a particular coordinate system contain non-zero cross-terms, then that particular coordinate system is not orthogonal. It doesn't necessarily mean that an orthogonal coordinate system doesn't exist, but it might be case that an orthogonal coordinate doesn't exist. I think that general results are a bit tricky to obtain.

To sum up this side discussion:

- for general coordinates, you can have any combination light like, space like, and time like coordinates; further their character can change from place to place.

- For orthogonal coordinates, the only possibility is 3 spacelike, 1 timelike.

----

My earlier claims about orthogonal coordinates involving null coordinates were incorrect. This follows from two facts:

- two linearly independent null vectors cannot be orthogonal (funny, I helped someone prove this for homework exercise some time ago, but forgot about it for this discussion).

- you can choose 4 vectors at an event such that 3 are orthonormal spacelike, one is null and orthogonal to two of the spacelike vectors; but then it won't be orthogonal to the last one.

These facts rule out any orthogonal coordinates involving any light like coordinate.