Black holes

[Farsight]

I have to go, and will catch up later, but very briefly, this one is hugely important:

But I can't see how he can possibly be right re the second (the unit of time), even allowing for sloppy language.

We define the second using the hyperfine transition and microwaves. These microwaves only have a defined frequency after we've defined the second. To do that we count 9 192 631 770 microwaves into the detector and then say a second has elapsed. When the microwaves are moving slower, the second is bigger. Then we use it and the motion of light to define the metre. (This doesn't change because the slower light and the bigger second cancel each other out). Then we use the speed of light to measure the speed of those microwaves.

I'm not kidding about this. There is no "time flowing" in the NIST caesium fountain clock. Just electromagnetic hyperfine spin flips occurring and microwaves moving. I would urge you to look into this, it's probably about the most important thing there is. That and the parallel-mirror light clock that runs slower at the lower elevation. Again there's no time flowing in it, just light moving.

 

[sol invictus]

My answer is: I don't know.

Well, that's at least an honest answer. Thanks for admitting that.

However, it leaves me at a loss as to how you can claim to have identified a mistake in all general relativity textbooks, while at the same time acknowledging that you cannot answer a simple question about a trivial metric (which in fact is nothing more than flat spacetime). Indeed, it's clear from your response that you do not know how to manipulate metrics or extract physical predictions from them.

 

[Ziggurat]

Not for that reason. Your reference frame does not actually exist.

Then no reference frame exists, relativity is nonsense from top to bottom, and we may as well all go home.

That light beam does not curve instantly the moment you accelerate towards it.

Yes, it does.

You know that instantaneous action-at-distance does not occur.

Nothing which transmits information can move faster than light. But no information was transmitted, so no problem exists.

You know that the way you see the world is relative to your motion, not the way the world is.

That's wrong even in special relativity. Hell, that's wrong in Galilean relativity.

It doesn't. Light exists. Space exists. You exist. But you are not surrounded by a real object called a reference frame.

Yes, it's an abstraction. But it's an abstraction that we can use to predict the results of measurements. Unless the theory is wrong, then the abstraction will work. Whenever you claim that there's no such thing as a reference frame, you're really saying that general relativity is nonsense.

 

[sol invictus]

If I give you the metric in terms of some variables the geometry is determined in an abstract sense but not in a concrete way. Unless the geometry has a concrete meaning, I do not see how anyone can say they understand what such a geometry is really like.

Then you don't understand anything about this. Given that metric, I have all the information I need to compute all of its invariant curvatures, its geodesics, etc. Those are concrete facts about the geometry.

The claim that one can understand what the geometry of a space is like when the metric is only given abstractly is false on its face.

More assertions without evidence. I can tell you exactly what geometry the metric you posted describes.

The first part of the quote is still wrong, the metric does not determine the coordinate system and the geometry.

No, it is quite correct.

How are those coordinates measured? Unless one is told how coordinates are measured for specific cases in a given metric, I can guarantee you will not know what the geometry is really like, as I can introduce all sorts of screwy ways to measure an x, or a y, or a whatever. I guess that was not considered though by you perhaps?

I don't think you've grasped what coordinates are, or what a metric is. Coordinates aren't "measured", they're simply a set of labels for spacetime points. For example, your metric describes a 2D space with points labelled by the pairs (x,y), and it tells me what the distance is between such points, what its geodesics are, etc. I can tell you exactly what the geometry of the space is, including all curvatures, singularities, sum of interior angles of a geodesic triangle, etc. (there is only one piece of information lacking, namely whether or not there are identifications on x or y).

 

[DeiRenDopa]

I have to go, and will catch up later, but very briefly, this one is hugely important:

But I can't see how he can possibly be right re the second (the unit of time), even allowing for sloppy language.

We define the second using the hyperfine transition and microwaves. These microwaves only have a defined frequency after we've defined the second. To do that we count 9 192 631 770 microwaves into the detector and then say a second has elapsed. When the microwaves are moving slower, the second is bigger.

Thanks for this.

Ya know Farsight, having read many a post by MM, I'm getting déjà vu all over again, reading your post. To be sure, you don't make extensive use of "double quotes", nor CAPITALS, nor ...

"To do that we count 9 192 631 770 microwaves into the detector and then say a second has elapsed" - ?!?!

That's rather different than what it says on the couple of country's time standard webpages. For example, NIST says this:

This process is repeated many times while the microwave signal in the cavity is tuned to different frequencies. Eventually, a microwave frequency is found that alters the states of most of the cesium atoms and maximizes their fluorescence. This frequency is the natural resonance frequency of the cesium atom (9,192,631,770 Hz), or the frequency used to define the second.

How did you derive "we count 9 192 631 770 microwaves into the detector" from "the microwave signal in the cavity is tuned to ... 9,192,631,770 Hz"?

I'm not kidding about this. There is no "time flowing" in the NIST caesium fountain clock. Just electromagnetic hyperfine spin flips occurring and microwaves moving.

"microwaves moving"?

Sorry, I just cannot follow what you're trying to say here.

I would urge you to look into this, it's probably about the most important thing there is. That and the parallel-mirror light clock that runs slower at the lower elevation. Again there's no time flowing in it, just light moving.

So, when you have time, what - in your own words - is "the parallel-mirror light clock"? (Later I may ask about "that runs slower at the lower elevation").

"there's no time flowing in it, just light moving"? Who cares? And why?

 

[tensordyne]

(there is only one piece of information lacking, namely whether or not there are identifications on x or y).

What, pray tell, do you mean by identifications on x or y?

As per the rest, seriously sol, try and think like an experimentalist and perhaps then you will see what I am referring to. In case you do not, since you claim you can tell what the geometry is just given the form of a metric, here is one.

ds² = u² du² + v² dv²
So what is the geometry like in this case?

 

[Ziggurat]

What, pray tell, do you mean by identifications on x or y?

Here are two seemingly identical metrics:

ds² = dr² + r²dθ²
ds² = dn² + n²dm²
If r ranges from 0 to infinity, and θ varies from 0 to 2pi, then the first geometry is a Euclidean plane with polar coordinates. If n ranges from 0 to infinity and m varies from 0 to pi, then the second geometry is a cone. The identification of my coordinates includes the range they span, and I need that to determine my geometry.

As per the rest, seriously sol, try and think like an experimentalist and perhaps then you will see what I am referring to. In case you do not, since you claim you can tell what the geometry is just given the form of a metric, here is one.

ds² = u² du² + v² dv²
So what is the geometry like in this case?

Sol's probably better at this than I am, but that looks to me like an ordinary Euclidean plane with funny coordinates.

 

[tensordyne]

Any other takers? I would hate to say what it is before everyone has their say in the matter of predicting what the geometry should be in this case.

 

[W.D.Clinger]

What, pray tell, do you mean by identifications on x or y?

As per the rest, seriously sol, try and think like an experimentalist and perhaps then you will see what I am referring to. In case you do not, since you claim you can tell what the geometry is just given the form of a metric, here is one.

ds² = u² du² + v² dv²
So what is the geometry like in this case?

Sol's probably better at this than I am, but that looks to me like an ordinary Euclidean plane with funny coordinates.

Any other takers? I would hate to say what it is before everyone has their say in the matter of predicting what the geometry should be in this case.

Working independently, I got the same answer Ziggurat got via the coordinate transformation

[latex]
\[
\begin{align*}
x = \frac{u^2}{2} \\
y = \frac{v^2}{2} \\
\end{align*}
\]
[/latex]​

ETA:

What, pray tell, do you mean by identifications on x or y?

sol invictus answered that by giving an example of two sets of coordinates mapping to the same point.

Speaking as a mathematician, I'd quibble with sol invictus on this. With the mathematical definition of a chart (aka coordinate patch), you can't have two distinct sets of coordinates for the same point. Physicists get sloppy about that.

Physicists also get sloppy about describing the range of coordinate values, and I made that same mistake in my coordinate transformation above. As I have defined x and y, they can't be negative. That means I've described a proper submanifold of the manifold on which your u-v metric is defined.

That shouldn't be hard to fix, though. I look forward to your explanation of this example.
​

 

[Ziggurat]

Working independently, I got the same answer Ziggurat got via the coordinate transformation

[latex]
\[
\begin{align*}
x = \frac{u^2}{2} \\
y = \frac{v^2}{2} \\
\end{align*}
\]
[/latex]​

That mapping only works for one quadrant. If we want it to cover all quadrants, we need something like

[latex]
\[
\begin{align*}
x = \frac{u^3}{2|u|} \\
y = \frac{v^3}{2|v|} \\
\end{align*}
\]
[/latex]​

 

[tensordyne]

Waiting on sol invictus for now. Got to sleep. Thanks go to Ziggurat and W.D.Clinger for responding. I will say what the geometry is like when I wake up.

 

[W.D.Clinger]

Waiting on sol invictus for now. Got to sleep. Thanks go to Ziggurat and W.D.Clinger for responding. I will say what the geometry is like when I wake up.

While you're sleeping, I'll give another example:

[latex]
\[
\begin{align*}
ds^2 &= dx^2 + dy^2 \\
- \infty &< x < \infty \\
- 5 &< y < + 5
\end{align*}
\]
[/latex]​

That's for one chart. There's a second chart with exactly the same metric and the same inequalities on the coordinates.

Without knowing how those charts compose, you can't tell whether the metric above is for a Euclidean plane or for a cylinder.

Without the inequalities shown, you wouldn't even suspect the metric describes anything other than a Euclidean plane.

I suspect tensordyne's going to tell us his metric was for a 2-dimensional locally Euclidean manifold that isn't globally homeomorphic to Euclidean 2-space.

 

[Reality Check]

...

Originally Posted by sol invictus

Hence when sol asks:

"So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?"

My answer is: I don't know.

How do you know what happens with the Schwarzschild metric

at r= r_s?
Let see: The dt component vanishes and the dr component is undefined (1/0). We call that a coordinate singularity.
You think that means that the coordinate speed of light goes to zero at this event horizon.

And in sol invictus's metric what happens?
Let see: The dt component vanishes and the dr component is undefined (1/0). We call that a coordinate singularity.
Why do you not also think that means that the coordinate speed of light goes to zero at this "event horizon"?

And what about Kruskal–Szekeres coordinates?

...
Note that the metric is perfectly well defined and non-singular at the event horizon.

What happens to the coordinate speed of light at this event horizon.


If you cannot answer these questions then you have to rely on the answers of others. What the textbooks state is:

• In Schwarzschild coordinates the coordinate speed of light is zero at the event horizon.
• In Kruskal–Szekeres coordinates the coordinate speed of light is c at the event horizon.

In other words: The coordinate speed of light at the event horizon depends on the coordinates that you use !

 

[Farsight]

I'm sorry, but many speakers of English use "no" when affirming a negative question, and your "don't exist as locations per se" suggests you may be doing that here.

Please clarify what you mean by your "No" that I highlighted. Here are three things you might have meant by that word...

This is old news now Clinger. I thought he was talking about a black hole. You've seen my later post addressing his expression and questions.

 

[Farsight]

And one more time - you are wrong.
Alice is not a popsicle - she is an observer.
Bob is not a popsicle - he is an observer.
I will make this post even clearer:
Let there be an observer Alice who is using Schwarzschild coordinates.
Let there be an observer Bob who is using Kruskal–Szekeres coordinates
Neither observer is falling into the black hole. They are both observing the black hole from outside.
Alice measures that there is a singularity at the event horizon.
Bob measures that nothing special happens at the event horizon.

This is embarrassing, RC. Here's the deal: they measure what they measure. They chuck in a few rocks: gone. They shine a light on it: black. They look through their telescope: black. Bob doesn't say to Alice: hey, take a gander through my Kruskal-Szekeres telescope. You can see right past that event horizon right into the infinite future! Wow, look at that point singularity in the middle!

All: back me up or you lose credibility.

If Alice or Bob were falling into the black hole then GR states that they measure nothing special at the event horizon.

No it doesn't. The good book says that, not GR.

 

[Ziggurat]

All: back me up or you lose credibility.

I think you're confused about whose credibility is at risk here.

No it doesn't. The good book says that, not GR.

And... there goes your credibility.

 

[Complexity]

All: back me up or you lose credibility.

You lost all credibility a long time ago.

Do your homework before posting again.

 

[Farsight]

Farsight, are you really stating that you are the only observer in the universe with a reference frame.

No. I'm saying a reference frame is an artefact of measurement. I can take you outside and point up to the clear night sky and say that's the moon, those are the stars, that's a satellite, and the black between them is space. I can show you these things, and the light you use to observe. But I cannot show you some grid in space with some rectangle around it. I cannot show you a reference frame.

I have a reference frame.

Then show it to me. When you can't, you will have perhaps learned something about the important distinction between reality and mathematical abstraction.

I am in a gravitational field and I have chosen to use Kruskal–Szekeres coordinates. I observe that "the metric is perfectly well defined and non-singular at the event horizon".

No you don't. You observe refracted starlight around a black round thing that in itself you cannot see. You can measure the gravitational time dilation and the local "force" of gravity at various locations around it, taking care not to get too close. You can map out these measurements, on a map. Then you can observe that map. But when you do, you're looking at a map, and the map is not the territory. Your well-defined non-singular metric is not something you can actually observe.

 

[Vorpal]

While you're sleeping, I'll give another example:

[latex]\[
\begin{align*}
ds^2 &= dx^2 + dy^2 \\
- \infty &< x < \infty \\
- 5 &< y < + 5
\end{align*}
\][/latex]​

...
Without knowing how those charts compose, you can't tell whether the metric above is for a Euclidean plane or for a cylinder.

If that covers the entire manifold, that should be an infinite strip rather than a cylinder. Though a cylinder would only take a minor variation.[/nits]

A more physical example is an isotropic universe of positive spatial curvature:
[latex]$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\left[\mathrm{d}\chi^2 + \sin^2\chi(\mathrm{d}\theta^2 + \sin^2\theta\mathrm{d}\phi^2)\right]$[/latex]
Is that spatially a 3-sphere? We don't know without specifying more information about the coordinates!

 

[W.D.Clinger]

While you're sleeping, I'll give another example:

[latex]
\[
\begin{align*}
ds^2 &= dx^2 + dy^2 \\
- \infty &< x < \infty \\
- 5 &< y < + 5
\end{align*}
\]
[/latex]​

That's for one chart. There's a second chart with exactly the same metric and the same inequalities on the coordinates.

Without knowing how those charts compose, you can't tell whether the metric above is for a Euclidean plane or for a cylinder.

If that covers the entire manifold, that should be an infinite strip rather than a cylinder. Though a cylinder would only take a minor variation.[/nits]

As I said, that was for one chart. There's a second chart with exactly the same metric form and exactly the same inequalities on the coordinates.

Neither of those charts covers the entire manifold.

Until I tell you more, you can't possibly know whether the manifold is an infinite strip or a cylinder.

Let's call the two charts f and g. Suppose both f and g do have the entire manifold as their domain, and for every point p: if f(p)=<x,y> then g(p)=<-x,-y>. Then you know the manifold is an infinite strip.

Suppose, however, that the points of the manifold are the elements of a set X defined by

[latex]
\[
X = \left\{ \langle a, b \rangle \; : \; - \infty < a < \infty \hbox{ and }
0 \leq b < 1 \right\}
\]
[/latex]​

Let the topology on X be generated by the basis consisting of all open neighborhoods of the form

[latex]
\[
\begin{align*}
N( \langle a, b \rangle, \epsilon ) &=
\left\{ \langle a^\prime, b^\prime \rangle \; : \;
| a - a^\prime | < \epsilon \hbox{ and }
| b - b^\prime | < \epsilon \hbox{ and }
0 \leq b^\prime < 1 \right\} \\
& \cup
\left\{ \langle a^\prime, b^\prime \rangle \; : \;
| a - a^\prime | < \epsilon \hbox{ and }
| b - b^\prime - 1 | < \epsilon \hbox{ and }
0 \leq b^\prime < 1 \right\}
\end{align*}
\]
[/latex]​

where ε > 0.

Suppose further that chart f is defined on the open subset of X consisting of points whose second (b) label (not coordinate!) is nonzero, and

f < a, b > = < a, 10 b - 5 >​

Suppose chart g is defined on the open subset of X consisting of points whose second (b) label is not 1/2, and

g < a, b > = < a, 10 b > (if 0 <= b < 1/2), and
g < a, b > = < a, 10 b - 10 > otherwise​

Then (unless I've made some mistake) the manifold is a cylinder.

Physicists tend to forget that coordinates are defined by charts, and are not attached to the points themselves.

Is that spatially a 3-sphere? We don't know without specifying more information about the coordinates!

We appear to be in complete agreement.