Is time, inside an event horizon, time-like or space-like?

[DrGreg]

You say "the photons are going in the opposite direction to the particles" … which presumably is in inertial coordinates … but then surely $$c_- < v < c_+ < 0$$ is wrong?

I presume the < 0 part was only meant to be inside the horizon, in a coordinate system like Einstein-Finkelstein coordinates. In an inertial coordinate system, the velocity of one light beam is always -c < 0, while the velocity of the other light beam is always +c > 0.

Sorry, I got this wrong. c_- and c₊ must always be opposite signs, unless one of them is zero or infinite. My mental picture was switching between different coordinate systems and got confused. In fact, in Schwarzschild coords, c_- = -c₊.

The condition $c_- < v < c_+$ applies when all three velocities are coordinate distance divided by coordinate time, where "coordinate distance" is a space-like coordinate, and "coordinate time" is a timelike or null coordinate. ("Null 'time'" allows for the possibility c₊ = 0.) So in Schwarzschild coords, inside the event horizon, it applies to dt/dr but not dr/dt. It's not clear to me whether this helps at all in Eddington-Finkelstein coords, where (in the radial 1D case) there is no spacelike coordinate.

(Now I'm feeling confused!)

The above observations arise from considering that, in 2D spacetime, any metric is a quadratic form that can be factorised as

$$ds^2 = A(c_- dt - dx)(c_+ dt - dx)$$

$$= A c_- c_+ dt^2 - A (c_- + c_+)dt dx + A dx^2$$​

for some A. If x is spacelike, A must be negative (with a +--- metric signature - put dt=0). For a massive particle, $dx = v dt$, ds² has to be positive.

 

[JesseM]

Sorry, I got this wrong. c_- and c₊ must always be opposite signs, unless one of them is zero or infinite. My mental picture was switching between different coordinate systems and got confused. In fact, in Schwarzschild coords, c_- = -c₊.

But they aren't opposite signs in Eddington-Finkelstein coordinates inside the horizon, are they? Both the infalling and outfalling worldlines are going from lower right to upper left (if the singularity is to the left of the event horizon).

 

[tiny-tim]

… a stitch-up … ?

Hi DrGreg!

Thanks for the clarification, and for the previous help on light-cones.

I've done a lot of thinking, and found a succinct way of describing my worries.

Imagine an observer O freefalling down a radius inside a horizon, and a particle P freefalling faster down an adjacent radius.

In Schwarzschild coordinates, O has a forward and backward light-cone which move with him. Let P meet the surface of these cones at events A and B, respectively. Then a photon from event A goes down the surface of the backward cone (infinitely fast in Schwarzschild coordinates) to meet O at C, say. At a later event D, a photon goes down the surface of the forward cone (infinitely fast in Schwarzschild coordinates) to meet P at B.

Before O reaches C, O can receive photons from P, from positions before A. But O cannot send photons to P (wherever P is).

After D, O can send photons to P, at positions after B, but cannot receive photons from P (wherever P is).

In inertial coordinates, with time and radial distance "interchanged", events A and C have the same r coordinate (because in Schwarzschild coordinates they are at the same time), and so do events B and D.

So O sees the "y" coordinate of P stay constant, but the r coordinate decrease until it is 0 (so P is on the y-axis).

Then O sees P no more … but O can wait until event D, and then send photons to P, at B and beyond; and at D, P is again on the y-axis.

And O can neither send photons to P nor receive them from P, when P is between A and B

So, if O uses inertial coordinates, O can tell he is inside a horizon because everything disappears (or appears) on the y-axis … and this is in any neighbourhood of O, however small.

Perhaps the paradox comes from correctly saying that we can interchange dt² and dx², but not noticing that that leaves an ambiguity between ±t?

The inertial coordinates are literally a stitch-up, since they open out the two Schwarzschild half-cones into hemispheres on the y-z-plane, and join them without regard either to the missing space between or to the sign of t.

Where have I gone wrong? ​

 

[DrGreg]

But they aren't opposite signs in Eddington-Finkelstein coordinates inside the horizon, are they? Both the infalling and outfalling worldlines are going from lower right to upper left (if the singularity is to the left of the event horizon).

In my previous post #31, I said that the lightspeeds had opposite signs when speed was (spacelike coord) divided by (timelike or null coord). Due to my confusion over what is a null coord (see this new thread), I must withdraw that and say I think it applies only if speed is spacelike divided by timelike. One of the Eddington-Finkelstein coordinates is null, so it doesn't apply in that case. (In fact one of the lightspeeds is infinite in that case -- don't forget EF diagrams are often "skewed" to show the the surfaces of constant v as cones when really they should be horizontal planes).

 

[DrGreg]

tiny-tim

I'm having great difficulty understanding post #33. I think your idea of a light cone still differs significantly from mine. It's not clear whether each of the various letters you use refers to an event (location at a specific time, a zero-dimensional "point" in spacetime), or to a particle/photon (lots of locations spread over time, a one-dimensional worldline (curve or line) in spacetime).

Don't forget, nothing moves is spacetime, as one of the dimensions in the diagram is time. A point that moves in space corresponds to a static line in spacetime.

It would really help if you could work out a way to enclose a diagram.

E.g. if you possesses a copy of Microsoft Word, you can draw diagrams in that, then use Alt+PrintScreen to copy a snaphsot to the clipboard, paste that into Microsoft Photo Editor, crop to size, save as a PNG and then attach that (via "manage attachments") to your post. If you don't have Microsoft Word, you may have something else you can draw with.