Does an event horizon ever exist?

[Dale]

The argument about a Rindler horizon seems unhelpful, as that is a theoretical concept which also involves infinities.

The Rindler horizon involves exactly the same infinities as the Schwarzschild horizon. What is your justification for dismissing the analogy, besides the obvious fact that it is damaging to your argument? You have asserted the physical significance of Schwarzschild time through the use of radar coordinates, just like those used for Rindler time.

 

[PeterDonis]

The argument about a Rindler horizon seems unhelpful, as that is a theoretical concept which also involves infinities.

If the Rindler horizon is a "theoretical concept", then so is the black hole event horizon. Rindler coordinates are the natural coordinates for observers with constant proper acceleration in Minkowski spacetime in the same way as Schwarzschild coordinates are the natural coordinates for observers with constant proper acceleration in Schwarzschild spacetime. So Rindler coordinates, and their associated horizon, are only "theoretical" to the same extent that Schwarzschild coordinates, and their associated horizon, are.

Yes, Rindler coordinates become singular (which is how I would rephrase your "involves infinities") at the Rindler horizon, in the same way as Schwarzschild coordinates become singular at the event horizon. That's the point: any argument that events can't happen "after" an infinite Schwarzschild time coordinate would also prove, if it were valid, that events can't happen "after" an infinite Rindler time coordinate. Which would prove, if valid, that a whole region of Minkowski spacetime can't exist. Which is why the argument isn't valid to begin with, for Rindler coordinates or for Schwarzschild coordinates.

 

[Jonathan Scott]

Do you really think it is a natural view to assume it takes light just as much time to go down into a gravitational well as it takes to come back out? Remember the speed of light is always c locally, once you look at the speed of light over an extended curved region you can show just about anything with the appropriate coordinate chart.

I would consider it very unnatural to assume anything else, as outside the event horizon of a black holes, classical mechanics is entirely reversible with respect to time.

In this case, I'm specifically referring to a static metric for a spherically symmetrical situation. That means that the coordinate speed of light inwards at a given point in the metric is the same as the speed outwards.

 

[Passionflower]

In this case, I'm specifically referring to a static metric for a spherically symmetrical situation. That means that the coordinate speed of light inwards at a given point in the metric is the same as the speed outwards.

No it does not.

We have the metric, the one you are talking about is the external Schwarzschild metric and we have a coordinate chart for a particular metric. Just because a particular coordinate chart of a given metric shows that the coordinate speed of light is isotropic does not mean it is a physical statement about the speed of light over a larger distance.

By analogy, on a Mercator projection it shows that Greenland is larger than Australia, however in reality it is much smaller than Australia.

 

[photonkid]

Does Montgomery, Alabama "exist"? How do you know it exists?

I'm sure one could write buckets of philosophical prose about this question :-(

However, it's perfectly possible to go visit Montgomery Alabama in a finite amount of time. The major differences between Montomery Alabama and a black hole event horizon , according to current theory and experiment, are as follows:

1) The nearest event horizon, at the center of our galaxy, is a lot further away than Montgomery. So the "finite" amount of time is a lot longer, and the vehicles one would need to use aren't technologically feasible.

But the time needed to get there is still finite according to theory.

2) The more troubling question is that if you do reach the event horizon, you won't be able to report your findings back to the people on Earth. But you'll still reach there in a finite time - according to current theory.

But the person who goes to Alabama can tell when he has arrived at Alabama. The person who goes to visit an event horizon has no way to know when he has reached it. Is there some way to know when you've reached an event horizon? Does the absence of radiation tell you? Can you calculate size versus mass reliably if you're close enough?

 

[PAllen]

But the person who goes to Alabama can tell when he has arrived at Alabama. The person who goes to visit an event horizon has no way to know when he has reached it. Is there some way to know when you've reached an event horizon? Does the absence of radiation tell you? Can you calculate size versus mass reliably if you're close enough?

These are good questions without good answers. Nobody has observed either Hawking radiation or Unruh radiation. There is strong and increasing observational evidence of horizons, which used to be entirely theoretical (like Hawking and Unruh radiation). If the black hole firewall proposal is true, then a well established horizon would be very detectible as you passed through it. However, classically, there is no way to detect it locally as you pass through it. Indirectly, as you observe from a distance, it is detectible, thus you can compute when you've passed it. (The firewall proposal is just that: some big names associated with it, but also prominent QG researchers who believe it is refuted).

 

[pervect]

But the person who goes to Alabama can tell when he has arrived at Alabama. The person who goes to visit an event horizon has no way to know when he has reached it. Is there some way to know when you've reached an event horizon? Does the absence of radiation tell you? Can you calculate size versus mass reliably if you're close enough?

You certainly won't be able to tell when you've passed the absolute horizon by any local measurement. It would also require an infinite amount of time to determine even a global measurement, as it's defined by the fact that light rays emitted never reach infinity.

This may be an issue, it becomes most important when a black hole is evaporating.

I *think* you can detect when you've passed an apparent horizon. This is defined by having the property that the expansion parameter \theta is negative for both ingoing and outgoing null geodesics (light rays).

I don't know specificially what apparatus you'd need to measure it, but I think the expansion parameter should be locally measurable. It might be good to get independent confirmation of this if it's really an important point.

 

[photonkid]

There is a theorem (originally due to Einstein) that says that an object in a static equilibrium, like a neutron star (or a white dwarf or anything else that holds itself up statically against its own gravity) can't have a radius smaller than 9/8 of the Schwarzschild radius for its mass. The time dilation factor at that radius is only 3 (i.e., time at the surface of such an object "flows" 1/3 as fast as it does at infinity); it certainly isn't approaching infinity.

One of the key things about black hole candidates is that at least some of them appear to be confined within a radius smaller than the above limit. (At least, that's my understanding.) So whatever is in there, it can't be something in static equilibrium like a neutron star.

So I'm guessing that the time dilation factor approaches infinity as you approach the Schwarzschild radius - because on the relativity faq website it says
<quote On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon.</>
and
<quote> In any case, this doesn't indicate that I take forever to fall in, since the proper time involved is actually finite.</>

I have to confess I have no understanding whatsoever of why a "finite proper time" means it won't take an infinite amount of "earth time" for something to reach an event horizon - even if the collapsing star was the only thing in the universe. However, it's also difficult to imagine matter pancaking indefinitely at the edge of the event horizon. The whole situation is very bizarre.

 

[Jonathan Scott]

No it does not.

We have the metric, the one you are talking about is the external Schwarzschild metric and we have a coordinate chart for a particular metric. Just because a particular coordinate chart of a given metric shows that the coordinate speed of light is isotropic does not mean it is a physical statement about the speed of light over a larger distance.

By analogy, on a Mercator projection it shows that Greenland is larger than Australia, however in reality it is much smaller than Australia.

I still really don't know what you're going on about here. For any spherically symmetrical static metric, isn't it obvious that the coordinate radial speed of light is the same both ways, and the coordinate time taken for light to travel between two points in a radial direction is the same regardless of direction?

 

[PAllen]

Would you consider the simultaneity of a thrown baseball compared to your own unphysical, or implausible? Then consider a static observer choosing to consider, moment to moment, the simultaneity of fastball thrown toward the BH. Whether you use Fermi-Normal simultaneity for the baseball at that moment, or computed radar (assuming extension of inertial trajectory of baseball the moment it leaved your hand, extended to the distant past and future) simultaneity, you would find by this slightly different point of view:

- at every point of the static world line, there are simultaneous events at and inside the horizon
- in the case of SC geometry, even the pure vacuum case, there is an earliest moment on your world line where thrown baseball simultaneity reaches the singularity (before this, it (Normal simultaneity) goes inside the horizon to some minimum r>0 then r grows (on the way to alternate KS sheet we are not considering here). Thus you have a sense of 'when' the singularity formed.
- all later history of static world line, your baseball 'now', includes events all the way to the singularity.

Whether one believes such predictions of classical GR apply to our universe, the above is unambiguously true of classical GR.

A correction and addition to above:

- Using radar simultaneity for the baseball thrown at the BH at the moment it leaves your (static observer) hand (with its then inertial trajectory extended to the distant past), there will be events at and inside the horizon simultaneous with this event (of leaving your hand). However, the singularity will never be radar simultaneous to any event on a free fall trajectory before it reaches the singularity.

- I had an error in my analysis of the Normal simultaneity for this event. Corrected, it becomes too complex for me to be willing to solve in detail. All that I can reliably show is that (as for the radar case) there exist events at and inside the horizon simultaneous to the moment it leaves your hand. My guess is that the qualitative features would be the same as radar, meaning that the singularity would never be Normal simultaneous to the thrown baseball.

 

[PeterDonis]

I have no understanding whatsoever of why a "finite proper time" means it won't take an infinite amount of "earth time" for something to reach an event horizon

Because they are two different kinds of questions: one is about an invariant, the other is about a coordinate-dependent simultaneity convention.

The first question, how much proper time does it take, is simple to answer because it's an invariant: just integrate the proper time along the infalling object's worldline. This will give a finite answer that is independent of the coordinates you choose.

The second question, how much "earth time" does it take, doesn't have a well-defined answer because it depends on which coordinates you adopt and the simultaneity convention that goes with them. That's true of *any* question about how to assign a "time" for some observer to events that are not on that observer's worldline. The infalling object is not on the Earth's worldline once it starts to fall, so any assignment of "earth time" to events on it is coordinate-dependent.

If you adopt the most natural simultaneity convention for "earth time", the one that goes with Schwarzschild coordinates, then you can't assign *any* finite "earth time" to events on or inside the horizon. Saying that "it takes infinite Earth time" for an infalling object to reach the horizon is a sloppy way of putting it because it assumes that assigning "earth time" to events on the horizon is even well-defined in the first place, and it isn't.

Also, the Schwarzschild coordinate simultaneity convention is not the only one you can adopt for assigning "earth time" to events; there are others, as PAllen has noted, which allow a finite "earth time" to be assigned to events on or inside the horizon.

 

[photonkid]

ok, yeah, I pretty much gathered that that's what the rationale was but...

Because they are two different kinds of questions: one is about an invariant, the other is about a coordinate-dependent simultaneity convention.

The first question, how much proper time does it take, is simple to answer because it's an invariant: just integrate the proper time along the infalling object's worldline. This will give a finite answer that is independent of the coordinates you choose.

...there seems to be a problem with this. If I move towards a gravitational field, time runs progressively more slowly (relative to where I was) and I cannot tell that it's doing so. If time actually stopped or ran infinitely slowly I would never know. If I'm one millimeter from the event horizon and my clock tells me it takes one microsecond to move one micron, how many billion years would pass on a clock one light year away while I move one micron?
It's too strange for me to grasp. Equally strange is that there are supposedly three time coordinates inside the event horizon. Why can't we just say that there is nothing happening inside an event horizon. If something is happening inside the event horizon (events are occurring) then time does exist inside the event horizon.

The second question, how much "earth time" does it take, doesn't have a well-defined answer because it depends on which coordinates you adopt and the simultaneity convention that goes with them.

If we had a way of determining whether an event horizon is present (e.g. size / mass) and that it is growing, we could say that the time taken on our clock for at least some of the matter to cross the event horizon is between zero and 13 billion years.

 

[PAllen]

...there seems to be a problem with this. If I move towards a gravitational field, time runs progressively more slowly (relative to where I was) and I cannot tell that it's doing so. If time actually stopped or ran infinitely slowly I would never know. If I'm one millimeter from the event horizon and my clock tells me it takes one microsecond to move one micron, how many billion years would pass on a clock one light year away while I move one micron?

For what observer a light year away? For someone a light year away moving sufficiently rapidly toward the BH, your clock would appear to be going at the same rate as theirs.

It's too strange for me to grasp. Equally strange is that there are supposedly three time coordinates inside the event horizon. Why can't we just say that there is nothing happening inside an event horizon. If something is happening inside the event horizon (events are occurring) then time does exist inside the event horizon.

Where did you get the idea there are 3 time coordinates inside the EH? There is one, just like outside. What really happens inside an EH we don't know (and many believe GR breaks down somewhere inside the EH; some believe it breaks in some ways at the EH). However, what GR predicts about the interior is unambiguous - sufficiently local physics is the same as anywhere else.

 

[Nugatory]

Equally strange is that there are supposedly three time coordinates inside the event horizon

I think that you may be misunderstanding the common (and correct) statement that inside the event horizon the Schwarzschild t coordinate becomes spacelike and the Schwarzschild r coordinate becomes timelike. This does indeed happen, but the other two spatial coordinates (θ and ψ) remain spacelike so we still have three space and one time coordinate.

You can see this if you look carefully at the Schwarzschild metric:
<br /> ds^2 = -(1-\frac{R}{r})dt^2+(1-\frac{R}{r})^{-1}dr^2+r^2(d\theta^2+sin^2{\theta}cos^2\phi)<br />
For r>R, outside the horizon, the coefficient of dt^2 is negative and the coefficient of dr^2 is positive. For r<R these signs flip; but the coefficients of the d\theta^2 and d\phi^2 terms remain positive.

 

[PeterDonis]

If I move towards a gravitational field, time runs progressively more slowly (relative to where I was)

You keep talking as if this statement has a physical meaning, when the whole point is that it doesn't; it's only a statement about the coordinate convention you pick. There is *no* way of assigning a physical meaning to the statement "time runs more slowly relative to where I was". You have to pick a simultaneity convention to assign a time "relative to where I was", and a simultaneity convention is just a convention; it doesn't tell you anything about the physics.

If I'm one millimeter from the event horizon and my clock tells me it takes one microsecond to move one micron, how many billion years would pass on a clock one light year away while I move one micron?

Same problem here. To answer this question you have to pick a simultaneity convention, and that's just a convention; it doesn't tell you anything about the physics.

If we had a way of determining whether an event horizon is present (e.g. size / mass) and that it is growing, we could say that the time taken on our clock for at least some of the matter to cross the event horizon is between zero and 13 billion years.

Once again, to do this you would need to pick a simultaneity convention, and that wouldn't tell you anything about the physics.

The problem you appear to be having is that you keep wanting to ask a question that doesn't have a well-defined answer. I know it seems to you like it should, but it doesn't. Repeating the question in different ways won't change that.

 

[PeterDonis]

Where did you get the idea there are 3 time coordinates inside the EH? There is one, just like outside.

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.

 

[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.

True, as I often stress. You could also have 2 spacelike and two light like coordinates. What I should have said is that inside the EH you can set up normal coordinates at any event the look locally like Minkowski coordinates.

 

[photonkid]

You keep talking as if this statement has a physical meaning, when the whole point is that it doesn't; it's only a statement about the coordinate convention you pick. There is *no* way of assigning a physical meaning to the statement "time runs more slowly relative to where I was".

Yes there is. If I have two identical clocks on top of a mountain and take one down to sea level for a while, then back to the top of the mountain, it will be behind the clock that stayed on top of the mountain.

 

[photonkid]

For what observer a light year away? For someone a light year away moving sufficiently rapidly toward the BH, your clock would appear to be going at the same rate as theirs.

For an observer not moving towards the black hole.

Where did you get the idea there are 3 time coordinates inside the EH?

I read it somewhere on the internet but I can't find where just at the moment.

There's a kind of allusion to it here but this is not where I read it.
http://www.einstein-online.info/spotlights/changing_places

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

 

[Nugatory]

Yes there is. If I have two identical clocks on top of a mountain and take one down to sea level for a while, then back to the top of the mountain, it will be behind the clock that stayed on top of the mountain.

That only works when the two clocks begin and end at the same location so there is no simultaneity problem at the beginning and end of the roundtrip. Take a look at my long answer reply to opti_mus back in post #81 of this thread - you're describing case #1 where we're measuring the proper time between two events (clocks separate, clocks rejoin). Unless and until they've rejoined, we're in case #2 of that post - no universal notion of "at the same time".

Here's a simpler example from special relativity:
Two spaceships traveling in opposite directions both set their clocks to the same time as they fly past one another, heading off to infinity never to rejoin. Because of time dilation and relativity of simultaneity, both will say that the other one has the slow clock, and they'll both be right. In this situation, the statement "this clock is really the slow one, that one is running normally" has no absolute physical significance - it can't because they're both right and they can't both be the slow one.
But if instead of flying off to infinity, one of the ships turns around so that eventually the two meet again, then we have an unambiguous and observer-indepedent "at the same time" at the ends of the journey, so we can compare the time elapsed. That's analogous to your example of the clock going down the mountain and coming back up and turns the mutual time dilation into the classic twin paradox.

(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)

 

[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.