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I am the first in the world

Freedom for transylvania

FREEDOM FOR TRANSYLVANIA
I HAD 0 INCOME LAST YEAR IN GARBAGE COUNTRY ROMANIA
THE GARBAGE ROMANIAN SECRET POLICE IS AFTER ME EACH DAY
My blog: http://adrianferent.blogspot.com

THE POLICE CALLED ME IN THE COURT AGAIN FRIDAY 27 JAN, 2012, SINCE 2009 BECAUSE I DID NOT LET THEM TO STEAL FLOWERS FROM A WOMAN. THEY PUSHED ME, THEY BROUGHT ME WITH FORCE AT THEIR LOCATION, THEY INTERROGATE ME..
THE CORRUPT JUDGES SAID IS OK FOR POLICE TO STEAL ANYTHING IN ROMANIA.
IN CLUJ NAPOCA THE POLICE AND THE JUDGES ARE THE MOST CORRUPT IN EUROPE.
http://www.spete.info/99-dosar-proces-209472112009-cluj-672670/
 
Alas, however, they left you with your caps lock button intact.

On a serious note, however, do you have a cogent argument or point to make? Is this about democracy and the rule of law in Romania? Is it about perceived discrimination against minorities? Really, what are you saying?
 
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If the secret police are after you, then run off to Jupitier.

After all, you can get to Jupiter in one hour and the secret police will not be able to follow you there.
 
alfa1 said:
FREEDOM FOR TRANSYLVANIA
I HAD 0 INCOME LAST YEAR IN GARBAGE COUNTRY ROMANIA
THE GARBAGE ROMANIAN SECRET POLICE IS AFTER ME EACH DAY
My blog: http://adrianferent.blogspot.com

THE POLICE CALLED ME IN THE COURT AGAIN FRIDAY 27 JAN, 2012, SINCE 2009 BECAUSE I DID NOT LET THEM TO STEAL FLOWERS FROM A WOMAN. THEY PUSHED ME, THEY BROUGHT ME WITH FORCE AT THEIR LOCATION, THEY INTERROGATE ME..
THE CORRUPT JUDGES SAID IS OK FOR POLICE TO STEAL ANYTHING IN ROMANIA.
IN CLUJ NAPOCA THE POLICE AND THE JUDGES ARE THE MOST CORRUPT IN EUROPE.
http://www.spete.info/99-dosar-proces-209472112009-cluj-672670/
Click to expand...

Shame on you, stealing flowers. Buy your own.
 
alfa1 said:
FREEDOM FOR TRANSYLVANIA
I HAD 0 INCOME LAST YEAR IN GARBAGE COUNTRY ROMANIA
THE GARBAGE ROMANIAN SECRET POLICE IS AFTER ME EACH DAY
My blog: http://adrianferent.blogspot.com

THE POLICE CALLED ME IN THE COURT AGAIN FRIDAY 27 JAN, 2012, SINCE 2009 BECAUSE I DID NOT LET THEM TO STEAL FLOWERS FROM A WOMAN. THEY PUSHED ME, THEY BROUGHT ME WITH FORCE AT THEIR LOCATION, THEY INTERROGATE ME..
THE CORRUPT JUDGES SAID IS OK FOR POLICE TO STEAL ANYTHING IN ROMANIA.
IN CLUJ NAPOCA THE POLICE AND THE JUDGES ARE THE MOST CORRUPT IN EUROPE.
http://www.spete.info/99-dosar-proces-209472112009-cluj-672670/
Click to expand...

Cool_story_bro_1359.jpeg
 
alfa1 said:
FREEDOM FOR TRANSYLVANIA
I HAD 0 INCOME LAST YEAR IN GARBAGE COUNTRY ROMANIA
THE GARBAGE ROMANIAN SECRET POLICE IS AFTER ME EACH DAY
My blog: http://adrianferent.blogspot.com

THE POLICE CALLED ME IN THE COURT AGAIN FRIDAY 27 JAN, 2012, SINCE 2009 BECAUSE I DID NOT LET THEM TO STEAL FLOWERS FROM A WOMAN. THEY PUSHED ME, THEY BROUGHT ME WITH FORCE AT THEIR LOCATION, THEY INTERROGATE ME..
THE CORRUPT JUDGES SAID IS OK FOR POLICE TO STEAL ANYTHING IN ROMANIA.
IN CLUJ NAPOCA THE POLICE AND THE JUDGES ARE THE MOST CORRUPT IN EUROPE.
http://www.spete.info/99-dosar-proces-209472112009-cluj-672670/
Click to expand...


Well you've certainly done the right thing bringing this to the attention of the JREF Forum. I can think of no organisation on Earth better equipped to deal with this crisis.
 
In Romania all the stealing is done with secret police, if you like to be free you have to share the money with them.
They stolen money from banks, pyramidal networks, gangs…because they have the INFORMATION!

In any country where the economy is going down secret police is the number one mafia organization, and of course they love the country.

For example the CEO from CFR Mihai Necolaiciuc stolen 100 million $ and Hilary Clinton signed expulsion papers because he was hiding in USA. After 9 months in jail 2 days ago he was released and now is free because he shared the money with secret police and politicians.
 
So the entire OP deserves upper case, but not the first letter in the name of your proposed country?
 
Who in their right mind would go to war with Transylvania? Armies the world over would be too terrified to set foot in the place!
 
Stellafane said:
Who in their right mind would go to war with Transylvania? Armies the world over would be too terrified to set foot in the place!
Click to expand...

It would actually be quite inexpensive. All you need is a few pitchforks, burning torches and a crowd of angy peasants to carry them.
 
alfa1 said:
For example the CEO from CFR Mihai Necolaiciuc stolen 100 million $ and Hilary Clinton signed expulsion papers because he was hiding in USA. After 9 months in jail 2 days ago he was released and now is free because he shared the money with secret police and politicians.
Click to expand...

He shared it with me too.
 
Back in the 1700s most of what is now Kentucky was still part of Virginia, and it was known as Transylvania. But it wasn't named for the Romanian province, it's just Latin for "beyond the forest".

/useless trivia
 
WildCat said:
Back in the 1700s most of what is now Kentucky was still part of Virginia, and it was known as Transylvania. But it wasn't named for the Romanian province, it's just Latin for "beyond the forest".

/useless trivia
Click to expand...

I have wondered for some time why there is a Transylvania University in Lexington. Now I know.
 
Sorry to have been away for a while guys. It was the wife's birthday, then we had a lot of work done on the house, and I've had scant free time.

dlorde said:
Not everybody ;)

If, as you say, everything arriving at the event horizon gets 'frozen' in time, the event horizon must be a sphere of temporal stasis where nothing ever happens, so how does the singularity ever grow? It sounds as if a huge event horizon (formed how?) would be a vast frozen shell of no-longer infalling junk.
Click to expand...
Like a hailstone. The infalling junk accretes and the frozen region gets "bigger". The bigger is in quotes because you can't measure distance when light doesn't move, and the radial length contraction makes things even trickier.

dlorde said:
How would there be Hawking radiation?
Click to expand...
There wouldn't be any Hawking radiation. That's the whole point. Look at the thread title. This discussion between me and the guys brewed up because Susskind and Hawking are buddies, staging a mock "war" about a load of old tosh to promote one another. Take a look at wiki and see the bit that says this:

"A slightly more precise, but still much simplified, view of the process is that vacuum fluctuations cause a particle-antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole whilst the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole)".

Those vacuum fluctuations blithely ignore the immense gravitational time dilation, and there are no negative energy particles. Sheesh, talk about Emperor's new clothes.
 
Mashuna said:
Farsight said:
Your mistake was not to link to them, and not to list them. Now everybody reading this thread knows you haven't got any. The abuse proves it. And everybody can see that RC is in horror-struck denial, and that sol is disingenuous because of this:

"Actually, photons passing through a gravitational field do slow down, in the sense that passing through a region containing a non-zero gravity field takes longer than passing through a region (of the same size) that doesn't. The extreme case is a black hole, where the photon never comes out the other side".

Still no response from sol on that. Not looking too good is it? Ah, I do love our little chats. Time for bed.
Click to expand...
As part of "everybody reading this thread", I'd like to inform you that your post is in error.
Click to expand...
No it isn't. Ben hasn't got any counterargument, check back through the thread. All he did was bang on about the distinction between a falling observer and an observer at the event horizon, which is academic when light clocks don't tick because light doesn't move any more. RC is definitely in denial, and sol has gone quiet because he's been caught bang to rights. Oh, and Clinger is trying to blind you with erudition, using long dull posts that you don't understand followed by and therefore Farsight is wrong. Don't fall for it. This black hole stuff is simple. Just think it through for yourself instead of letting other people tell you what to think.
 
Farsight said:
No it isn't. ....
Click to expand...
You are lying - the "counterargument" is the basic GR that you remain ignorant of.
I "went silent" because you have been merely parroting this ignorance (and it was a public holiday here!).

Basic GR, Farsight:
An observer falling into a black hole passes through an event horizon with no effects.
Another observer observing that observer falling into the black hole can never see them reach the event horizon. They will see that clock running slower and slower but they can never see the clock stop. That observer will state that the clock can never stop according to GR because it takes an infinite time for it to reach the event horizon, i.e. it never gets there.

Your "light clocks don't click" fantasy never happens in this scenario.

FYI: infinite time means that whatever time you think of, there is more time to go.
 
Farsight said:
No it isn't. Ben hasn't got any counterargument, check back through the thread. All he did was bang on about the distinction between a falling observer and an observer at the event horizon, which is academic when light clocks don't tick because light doesn't move any more.
Click to expand...

I "bang on" that distinction because it's your mistake. If you meet someone who thinks 1+1=3, you will find yourself "banging on" 1+1=2 a lot. If I meet someone who says "time stops at R=2M", I don't have much to say except "that's a coordinate-dependent statement".

There are some observers who think the falling clock (light-based or otherwise, it doesn't matter) "stops" approaching R = 2M. There are other observers who think the falling clock "stops" approaching R = 3M. There are other observers who don't think the falling clock ever stops at all.
 
some homework problems for Farsight, part 1

Farsight said:
Oh, and Clinger is trying to blind you with erudition, using long dull posts that you don't understand followed by and therefore Farsight is wrong.
Click to expand...
...whereas Farsight is defending his arguments by accusing me of erudition and by accusing Mashuna and others of not understanding my posts.
:D

Farsight said:
Don't fall for it. This black hole stuff is simple. Just think it through for yourself instead of letting other people tell you what to think.
Click to expand...
It's pretty simple, all right. It's mostly just high school algebra and first year college calculus (although it might involve second year calculus if Farsight went to a weak school or didn't take the version of calculus that math and physics majors take).

I've got some homework problems here for Farsight.

I'm not asking Farsight to do anything I haven't done myself, but I'm not so foolish as to assume he'll take his own advice and try to think these problems through for himself. These exercises are mostly for those who really are willing to think for themselves, and have enough background in high school algebra and calculus to do the calculations.

I'm going to start out with some exercises that involve the 2-sphere, because two of Farsight's beloved Schwarzschild coordinates are exactly the same as spherical coordinates on a unit 2-sphere. In part 2, I'll move on to some exercises that involve all four Schwarzschild coordinates and the Schwarzschild metric. In part 3, I'll introduce Lemaître coordinates and assign some homework involving those coordinates.

By the time we're done with all of the homework exercises, we'll have talked about three spacetime manifolds: M, M', and M*. M is the spacetime manifold that can be covered by as few as two charts using Schwarzschild coordinates. M is a submanifold of the larger manifold M' that can be covered using Lemaître coordinates and the Lemaître metric. (One of the homework exercises in part 3 will ask Farsight to prove the Lemaître metric coincides with the Schwarzschild metric on the submanifold M that Schwarzschild coordinates can reach.) M* is the largest spacetime manifold that extends both M and M'.

To motivate the exercises in this part 1, I'll state the Schwarzschild metric and then focus on the unit 2-sphere part of that metric before returning to the full Schwarzschild metric in part 2. I'll use the (-,+,+,+) signature convention with natural units in which c=G=1. The Schwarzschild coordinates are (t, r, θ, φ), and the Schwarzschild metric is

[latex]
\[
ds^2 = - (1 - 2M/r)dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 d\Omega^2
\]
[/latex]​

where

[latex]
\[
d \Omega^2 = d \theta^2 + \sin^2 \theta d \phi^2
\]
[/latex]​

is the usual metric on a 2-sphere in spherical coordinates and

[latex]
\[
\begin{align*}
r &> 2M \\
- \pi &< \theta < \pi \\
- \pi &< \phi < \pi
\end{align*}
\]
[/latex]​

Exercise 1. Prove that the portion of the 2-sphere that's covered by spherical coordinates (θ, φ) with the above range restrictions is an open subset of the 2-sphere.

I'll give hints in blue. To prove that a subset A of the 2-sphere is open, all you have to do is to prove that to each point p of A there exists a positive real number δ such that every point q on the 2-sphere whose distance from p is strictly less than δ lies within A. Take δ to be the smallest distance from p to the four boundaries implied by the two inequalities that restrict θ and φ.​

Exercise 2. Consider the map from the 2-sphere to 2-dimensional Euclidean space that maps (θ, φ) to the point with x=φ and y=θ, with (θ, φ) restricted to the ranges shown above. From exercise 1, we already know the domain of this map is an open subset of the 2-sphere. Prove that its image (or range) is an open subset of Euclidean 2-space.

Hint: Figure out the boundaries for x and y, and use the same method as in the hint for exercise 1.​

Exercise 3. Prove that the map defined in exercise 2 is a bijection (aka one-to-one correspondence) between the open subset of the 2-sphere mentioned in exercise 1 and the open subset of the 2-dimensional Euclidean plane that's the image (or range) of the map.

Hint: All you have to do is to prove that if (θ1, φ1) and (θ2, φ2) both map to the same (x, y), then θ1 = θ2 and φ1 = φ2.​

Exercise 4. Prove that the map defined in exercise 2 is continuous.

Hint: You can do this by considering limits of Cauchy sequences, or the Weierstrass (epsilon-delta) definition of continuity, or you can simply prove that every open disk that lies within the image of the map is the image of an open subset of the map's domain.​

Exercise 5. Prove that the inverse of the map defined in exercise 2 is continuous.

Combining the results of exercises 1 through 5 tells us the map defined in exercise 2 is a homeomorphism between an open subset of the 2-sphere and an open subset of 2-dimensional Euclidean space. That means it's one of the infinitely many charts that provide a flat view of some open subset of the 2-sphere. (This particular chart resembles a Mercator projection, but it's less sophisticated than a Mercator projection because it doesn't stretch the distances between latitudes as you get further away from the equator.)

Exercise 6. Prove that the chart defined in exercise 2 cannot be extended to a larger chart that covers the entire 2-sphere by changing one all of the strict (less than) inequalities to a less than or equal inequalities.

Hint: Show that the extended map would no longer be a bijection, or even a well-defined function: some point on the 2-sphere would map to two distinct pairs of (x, y) coordinates. (ETA: If only some of the inequalities were changed, the map might still be a well-defined function but its domain or range might no longer be open, which would disqualify the map from being a chart.)​

The usual metric on a 2-sphere, when expressed in spherical coordinates restricted by the inequalities stated above, has coordinate singularities at (or opposite, depending on how you look at it) the prime meridian and at the poles. Exercise 6 tells us we can't remove those coordinate singularities by making the obvious extension. We can, however, choose a slightly different set of spherical coordinates that leaves out a different set of points.

Exercise 7. Prove that the entire 2-sphere can be covered by the open sets of just two distinct charts.

Hint: We already have one chart that covers most of the 2-sphere. Slide its prime meridian around 180 degrees, keeping the same poles, and then rotate that new prime meridian around its center by 90 degrees (which changes its poles).​

Exercise 8. Prove there is no single chart that covers the entire 2-sphere.

That's by far the most difficult of these exercises to prove, and the proofs I know involve some upper division math, so you might need to take exercise 8 on faith for the moment or ask some other participant for a proof. Exercise 8 will serve its purpose if you just try to come up with a single chart that covers the entire 2-sphere, and figure out why every map you come up with isn't really a chart. (Recall that a chart is a homeomorphism from an open subset of the 2-sphere to an open subset of the 2-dimensional plane; a homeomorphism is just a bijection that's continuous in both directions.)​
 
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