Scientists readying to create first image of a black hole

February 20, 2017 by Bob Yirka report
Sagittarius A*. This image was taken with NASA's Chandra X-Ray Observatory. Credit: Public domain

(Phys.org)—A team of researchers from around the world is getting ready to create what might be the first image of a black hole. The project is the result of collaboration between teams manning radio receivers around the world and a team at MIT that will assemble the data from the other teams and hopefully create an image.

The project has been ongoing for approximately 20 years as project members have sought to piece together what has now become known as the Event Horizon Telescope (EHT). Each of the 12 participating receiving teams will use equipment that has been installed for the project to record data received at a frequency of 230GHz during April 5 through the 14th. The data will be recorded onto hard drives which will all be sent to MIT Haystack Observatory in Massachusetts, where a team will stitch the data together using a technique called very long baseline array interferometry—in effect, creating the illusion of a single radio telescope as large as the Earth. The black hole they will all focus on is the one believed to be at the center of the Milky Way galaxy—Sagittarius A*.

A black hole cannot be photographed, of course, light cannot reflect or escape from it, thus, there would be none to capture. What the team is hoping to capture is the light that surrounds the black hole at its , just before it disappears.

Sagittarius A* is approximately 26,000 light-years from Earth and is believed to have a mass approximately four million times greater than the sun—it is also believed that its event horizon is approximately 12.4 million miles across. Despite its huge size, it would still be smaller than a pin prick against our night sky, hence the need for the array of radio telescopes.

The researchers believe the image that will be created will be based on a ring around a black blob, but because of the Doppler effect, it should look to us like a crescent. Processing at Haystack is expected to take many months, which means we should not expect to see an image released to the press until sometime in 2018.

Explore further: Astronomers poised to capture image of supermassive black hole

More information: www.eventhorizontelescope.org/

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wduckss
1.7 / 5 (11) Feb 20, 2017
"As the cloud approached the black hole, Dr. Daryl Haggard said" It's exciting to have something that feels more like an experiment, "and hoped that the interaction would produce effects that would provide new information and insights.
Nothing was observed during and after the closest approach of the cloud to the black hole, which was described as a lack of "fireworks" and a "flop". .." Wikipedia

Persistently society, they want a to prove that there is fiction notwithstanding on evidence. Simply must be black holes, fail "minds" a few generations!
 
http://www.svemir...ack-hole
antialias_physorg
4.5 / 5 (15) Feb 20, 2017
Persistently society, they want a to prove that there is fiction notwithstanding on evidence.

Since they're trying to get the evidence - I wonder what you'll say if they actually produce an image?
Whydening Gyre
4.4 / 5 (13) Feb 20, 2017
Persistently society, they want a to prove that there is fiction notwithstanding on evidence.

Since they're trying to get the evidence - I wonder what you'll say if they actually produce an image?

Benni might have to shut-up with his request for pictures of them...:-)

Wducks,
Whats your explanation for the quite large gravitational anomaly at the center of most (if not all) galaxies, then?
andyf
4.3 / 5 (6) Feb 20, 2017
Tsk, tsk...

"received at a wavelength of 230GHz"

230 GHz is a frequency not a wavelength - as enny fule nose!
Da Schneib
3.7 / 5 (6) Feb 20, 2017
Another quibble: they won't be "seeing" the shape of the event horizon in visible light, but in radio. For an object of this size, though, the resolution should easily be high enough.
Chris_Reeve
1.7 / 5 (6) Feb 20, 2017
Quite confident that they'll see no well-defined "dark" spot -- either radio or optical.
andyf
3 / 5 (2) Feb 20, 2017
Are the two ovals surrounding the bright sources in the upper left quadrant of the main image due to gravitational lensing by intervening massive objects?

It might be worth having a close look at them too.
Benni
1 / 5 (9) Feb 20, 2017
Another quibble: they won't be "seeing" the shape of the event horizon in visible light, but in radio. For an object of this size, though, the resolution should easily be high enough.


Hey, hey there schneibo, you could just send your pics to them, it would save them the bother of going through all this extra expense of modeling things that don't exist, except of course as seen through your super secret telescope through which your pics were taken.
baudrunner
1 / 5 (4) Feb 20, 2017
There is danger in inspiration derived from an artist's depiction. I'm glad to see that some effort to see the 'halo' is being done. According to accepted theory of the singularity that is a black hole, light cannot reflect or escape from it because at the event horizon an atom's electrons are stripped from its nuclei, and stable atomic configurations are required for the exchange of information between the particles in a medium which participate in light propagation (Helium is an exception). During this process, photons are decoupled from atoms and very briefly retain their information. This we would see as a halo. What appear to be black holes in random places in the cosmos, and not in the centers of galaxies, are probably objects traveling faster than light.
Da Schneib
4.1 / 5 (9) Feb 20, 2017
I wonder what you'll say if they actually produce an image?
He'll just claim they're lying. Not all that imaginative, our Lenni.
Benni
1.4 / 5 (9) Feb 20, 2017
I wonder what you'll say if they actually produce an image?
........of what?
Benni
1.7 / 5 (6) Feb 20, 2017
I wonder what you'll say if they actually produce an image?
He'll just claim they're lying. Not all that imaginative, our Lenni.
............No schneibo, I won't ......at least I won't if I can get equal time on your telescope.
Da Schneib
3.4 / 5 (5) Feb 20, 2017
I'll hold you to that, Lenni. When they're done and produce an image, I'll be reminding you of it on a regular basis.

Remember:
-m'' + m'n' - m'² - 2m'/r = 0
m'' + m'² - m'n' - 2m'/r = 0
e⁻²ⁿ (1 + m'r - n'r) - 1 = 0
R₂₂ sin² ϕ = 0
cgsperling
2.3 / 5 (3) Feb 20, 2017
Ooh ! Ooh ! Let me guess ...... it will be Round and Black. Did I get it right ?
Benni
1.4 / 5 (9) Feb 20, 2017
I'll hold you to that, Lenni. When they're done and produce an image, I'll be reminding you of it on a regular basis.

Remember:
-m'' + m'n' - m'² - 2m'/r = 0
m'' + m'² - m'n' - 2m'/r = 0
e⁻²ⁿ (1 + m'r - n'r) - 1 = 0
R₂₂ sin² ÄŽď�� = 0

Let's see those pics of BHs you told us you've seen. Remember? It won't be an "image" they come up with Schneibo, it'll just be another model based only on how they base a case for interpreting the radio waves they come up with.

Hopefully this antennae won't be aimed in your direction, what with 80-95% of you being missing who knows what they'd come up with & mis-interpret as an EH.
Whydening Gyre
3.7 / 5 (6) Feb 20, 2017
Another quibble: they won't be "seeing" the shape of the event horizon in visible light, but in radio. For an object of this size, though, the resolution should easily be high enough.

ANd yet another quibble - They won't be "seeing" the Black Hole. Just the event horizon.
That'll be Benni's next go to...
All the while NOT denying it's an horizon as a result of the gravitational anomaly INSIDE it...
Whydening Gyre
4 / 5 (8) Feb 20, 2017
I'll hold you to that, Lenni. When they're done and produce an image, I'll be reminding you of it on a regular basis.
Remember:
-m'' + m'n' - m'² - 2m'/r = 0
m'' + m'² - m'n' - 2m'/r = 0
e⁻²ⁿ (1 + m'r - n'r) - 1 = 0
R₂₂ sin² ÄĹ�Ä�ď��ď�� = 0

Good one, DS. That one always cracks me up...
Let's see those pics of BHs you told us you've seen.

I just wanna see a couple pics of those "Ski trails". Or maybe Benni in a Nuclear Power plant...
It won't be an "image" they come up, it'll be another model based only on how they ... interpret the radio waves they {see}

Kinda like your spectroscopy...?
Hopefully this antennae won't be aimed in your direction, what with 80-95% of you being missing who knows what they'd come up with & mis-interpret as an EH.

Wouldn't a Nuclear Engineer already know about the space between atoms?
Benni seems to be misinterpreting it as solid flesh...
tblakely1357
5 / 5 (1) Feb 20, 2017
Given that space-time is greatly distorted close to a black hole I've no idea what light escaping from superheated gas close to the event horizon would look like many lightyears away.
Da Schneib
3 / 5 (4) Feb 21, 2017
The article kind of describes it. It should look like a sphere with a donut around its plane of rotation.

From our viewpoint, we'll probably see a crescent.
Azrael
3.5 / 5 (8) Feb 21, 2017
I'll hold you to that, Lenni. When they're done and produce an image, I'll be reminding you of it on a regular basis.

Remember:
-m'' + m'n' - m'² - 2m'/r = 0
m'' + m'² - m'n' - 2m'/r = 0
e⁻²ⁿ (1 + m'r - n'r) - 1 = 0
R₂₂ sin² ÄĹ�Ä�ď��ď�� = 0

Let's see those pics of BHs you told us you've seen. Remember? It won't be an "image" they come up with Schneibo, it'll just be another model based only on how they base a case for interpreting the radio waves they come up with.

Hopefully this antennae won't be aimed in your direction, what with 80-95% of you being missing who knows what they'd come up with & mis-interpret as an EH.


I would say "I can't wait for them to produce a picture so you'll finally shut up", but you already dismiss the existence of black holes despite tremendous evidence that they exist, so you'll probably claim the scientists doctored the telescope or something...
antialias_physorg
4.5 / 5 (8) Feb 21, 2017
it'll just be another model based only on how they base a case for interpreting the radio waves

Photons is photons. They're really looking at this with a telescope this time. There's no "interpretation", here.
Benni
1 / 5 (9) Feb 21, 2017
I would say "I can't wait for them to produce a picture so you'll finally shut up", but you already dismiss the existence of black holes despite tremendous evidence that they exist, so you'll probably claim the scientists doctored the telescope or something


Well Azzie, here's the problem.....you simply don't understand why a pic of a BH can never be made, it's the so-called Event Horizon which according to all the artistic renderings completely hide the actual surface of a BH.

If you can never peer through the depths of an EH due to features of infinite gravity extending above the surface of a BH, how can you expect to take a picture? You think some radio waves being emitted by particle decay from the top of a so-called EH will provide information for what's below that point?
antialias_physorg
4.5 / 5 (8) Feb 21, 2017
how can you expect to take a picture?

Much like you take a picture of a solar eclipse. Just because you see nothing in the middle doesn't mean the sun isn't there. You can see it by what you get around the rim. And the rim of a black hole should give off a very characteristic image.
Benni
1 / 5 (5) Feb 21, 2017
how can you expect to take a picture?

Much like you take a picture of a solar eclipse. Just because you see nothing in the middle doesn't mean the sun isn't there. You can see it by what you get around the rim. And the rim of a black hole should give off a very characteristic image.
.........you don't know what you're talking about, not the same thing as what I'm talking about., but to you it sounds good so you post it simply because it makes print as an opposition opinion.

Da Schneib
4.5 / 5 (8) Feb 21, 2017
Lenni hasn't figured out there're stars behind it. You know, like the rest of the galaxy. And stuff.
Azrael
3.7 / 5 (6) Feb 21, 2017
Well Azzie, here's the problem.....you simply don't understand why a pic of a BH can never be made, it's the so-called Event Horizon which according to .....


Benni, I assure you that I'm not misunderstanding that a black hole cannot be directly photographed.

Also, you don't have to "see past the event horizon" to detect and photograph the effects that a 12.4 million mile wide, 400-million solar mass black hole has on it's immediate environment including some very specific optical effects that we expect to see, and none of which can be reasonably explained by any other phenomenon that has been observed to exist in our universe.

Merrit
not rated yet Feb 21, 2017
It just impossible with our current technology to observe a BH directly. It may be possible with two black holes spiraling together to have a spaceship traveling near lightspeed to shoot between them and make a direct observation.

Also, there is the question whether black holes have a charge. Do charged particles add their charge to the total charge of the black hole? If so, you could theoretically add only positive or negatively charged matter to a black hole until the electromagnetic force is greater than the gravity causing the black hole to explode since gravity is order of magnitudes less powerful.
Benni
1 / 5 (6) Feb 21, 2017
Lenni hasn't figured out there're stars behind it. You know, like the rest of the galaxy. And stuff.


......if only I had access to your super secret telescope through which you made those pictures of BHs you claim to have seen.
Benni
1 / 5 (5) Feb 21, 2017
you don't have to "see past the event horizon" to detect and photograph the effects


Oh yes, here we go.....the usual duck dodge & weave routine using the same argument as used for evidence of DM......INFERRED GRAVITY.

All these narratives used by DM & BH Enthusiasts have one thing in common: UNOBSERVABILITY

DM - 80-90% of the Universe is MISSING, but we can't OBSERVE it.

BH - backwardation application of the Inverse Square Law creating infinite gravity & infinite density on the surface of a finite stellar mass, but we can't OBSERVE it.

........talk about building houses on sand.

Da Schneib
4.4 / 5 (7) Feb 22, 2017
See, tolja Lenni would make up a story about how it's all a lie. It's already laying the groundwork for it:

Oh yes, here we go.....the usual duck dodge & weave routine


All these narratives


And notably, using Derridista memology, too.

Tsk tsk, Lenni.
Benni
1 / 5 (6) Feb 22, 2017
See, tolja Lenni would make up a story about how it's all a lie. It's already laying the groundwork for it:


Prove it.........
antialias_physorg
4.6 / 5 (9) Feb 23, 2017
It just impossible with our current technology to observe a BH directly.

Really depends on what you think of 'directly'. Taking the insanely stupid Benni-definition and limiting ourselves to optical wavelengths then we can't. We can see the event horizon and all the cool emissions and distortions that go on there, however.

But we actually do have sensors that can tell what goes on within black holes (even inside the event horizon): LIGO. gravitational waves are not prevented from emanation from a black hole. E.g. when binary black holes merge we get a 'ringdown' signal after the two black holes' event horizons have merged. With enough LIGO like stations spread at great distances one could envision a 'gravity telescope' that could produce an actual picture of mass distributions inside the event horizon during such a phase.
jonesdave
4 / 5 (8) Feb 23, 2017
All these narratives used by DM & BH Enthusiasts have one thing in common: UNOBSERVABILITY


Rubbish.
http://www.galact...ons.html

As I keep saying, all that Benni has to do is sit down and work out what is causing the above. Shouldn't be rocket science to come up with a mass estimate for the unseen object that these stars are orbiting. And then come up with an explanation for what is causing it. You know, with the relevant maths thrown in. Get to it. And I still haven't seen a picture of the electron that I asked for. Do electrons exist?
Benni
1 / 5 (5) Feb 23, 2017
As I keep saying, all that Benni has to do is sit down and work out what is causing the above. Shouldn't be rocket science to come up with a mass estimate for the unseen object that these stars are orbiting


OK Jonesy cricky mate, you want me to do what you won't (can't) do? If you believe in this so stridently, then YOU come up with the easier than "rocket science" math. Write up your PAPER, get it peer reviewed, then tell us where to go to find it.

After you've written your PAPER, include some kind of rendition of what a REAL EH should look like, but don't make the inane mistake of making it look like the eye of a hurricane that we so often see in artist's renderings of BHs.

But before you get into your PAPER too far, you should read what Hawking said in his April 2016 News Conference about APPARENT EVENT HORIZONS, Google it.

jonesdave
3.9 / 5 (7) Feb 23, 2017
OK Jonesy cricky mate, you want me to do what you won't (can't) do? If you believe in this so stridently, then YOU come up with the easier than "rocket science" math. Write up your PAPER, get it peer reviewed, then tell us where to go to find it.


Don't need to. Already been done: https://arxiv.org...2870.pdf
Secondly, I'm not the one suggesting BHs don't exist. Therefore it is incumbent on those who do claim such things to actually offer up their own analysis of the data, and to suggest a plausible alternative. Until such time, it is all just hot air on the comments section of a sci-news website. i.e. irrelevant.

Ojorf
4.2 / 5 (5) Feb 23, 2017
... it is incumbent on those who do claim such things to actually offer up their own analysis of the data, and to suggest a plausible alternative.


I also can't wait for his analysis.
Benni
1 / 5 (6) Feb 23, 2017
I'm not the one suggesting BHs don't exist. Therefore it is incumbent on those who do claim such things to actually offer up their own analysis of the data
......No, cricky mate, it isn't up to me to prove BH Perpetual Motion math is falsifiable, it's up to those of you who created the hypothesis that INFINITE DENSITY & INFINITE GRAVITY can exist on the surface of a given FINITE MASS.

and to suggest a plausible alternative


Why? It isn't up to me to suggest any such thing.......it's up to you to come up with Laws of Physics whereby you can PROVE infinite density of mass can create infinite gravity that can exist on the surface of a given finite stellar mass, to date all you have is your untenable Schwarzschild Black Hole Math to fall back on.......yeah you need pics all right & if you need to you'll FAKE them as well.

jonesdave
4.1 / 5 (9) Feb 23, 2017
^^^^Oh dear, Benni's losing it. So show us the mathematical proofs that BHs can't exist. Because there are plenty of well qualified people that say they can. And it's nothing to do with what I can or cannot prove. YOU are the one making a claim that goes against established science, and yet you have nothing to offer, other than continual whines that BHs don't exist! SO PROVE IT. Otherwise shut up, because in that case you're just another fanboy of some physics crank or other, with nothing better to do than spamming your evidence-free nonsense on an irrelevant news website.
antialias_physorg
4.5 / 5 (8) Feb 23, 2017
SO PROVE IT.

Don't hold your breath. In all the years he's been on here he hasn't yet once even tried. Weasel. Duck. Cop-out. Rinse. Repeat.
It's really quite predictable.
Whydening Gyre
4.4 / 5 (7) Feb 23, 2017
Oh yes, here we go.....the usual duck dodge & weave routine using the same argument as used for evidence of DM......INFERRED GRAVITY.
All these narratives used by DM & BH Enthusiasts have one thing in common: UNOBSERVABILITY
DM - 80-90% of the Universe is MISSING, but we can't OBSERVE it.

Guess it's sorta like the "inferred" strong and weak force...
Don't Nuclear Engineers utilize that??

BH - backwardation application of the Inverse Square Law creating infinite gravity & infinite density on the surface of a finite stellar mass, but we can't OBSERVE it.

........talk about building houses on sand.

More like - building houses on diamond...
How many times do you have to be reminded that it is not "infinite"?
YOU are the only one inferring that.
Vewy Inteles - .. Naaaagh. Just Ssshtupid. (Thanks, Arty J.)

Benni
1 / 5 (4) Feb 23, 2017
So show us the mathematical proofs that BHs can't exist
.......simple the Inverse Square Law.

Below from WkiiPedia, Black Hole:

"At the center of a black hole lies a gravitational singularity, a region where the spacetime curvature becomes infinite.[61] For a non-rotating black hole, this region takes the shape of a single point and for a rotating black hole, it is smeared out to form a ring singularity that lies in the plane of rotation.[62] In both cases, the singular region has zero volume. It can also be shown that the singular region contains all the mass of the black hole solution.[63] The singular region can thus be thought of as having infinite density."

Jonesy, it is because you are totally ignorant that in GR Einstein applied the Inverse Square Law where maximum gravitational attraction of a gravitating body exists at the surface of a stellar mass, not it's center. Obviously you don't know this. Yet the above definition for BH clearly states the opposite.
Da Schneib
4 / 5 (4) Feb 23, 2017
New acronym: ALTANBHT

Another Lenni There Aint No Black Holes Thread
Benni
1 / 5 (4) Feb 23, 2017
New acronym: ALTANBHT

Another Lenni There Aint No Black Holes Thread


Nah, that ain't it Schneibo........just having lots of fun pointing out the alternative universe guys like you & Jonesy like to pretend exists.

You like to be entitled to your own pseudo-science, you just have no tolerance for the actual Laws of Physics governing the dynamics of gravitating bodies, this because you do not understand the Equivalence Principle in GR or the Mass/Energy Equivalence Principle in SR.

OK, you & Jonesy think the maximum gravitational of a certain kind of finite stellar mass can exist at it's center & that it's INFINITE........OK, prove it.
jonesdave
3.7 / 5 (6) Feb 23, 2017
^^^Give it up Benni. You're clueless. I've told you, tell us what all those stars are orbiting at the galactic centre. What is its mass? What is your hypothesis? You aren't capable of coming up with one of your own, are you? So who do you follow? Crothers? Lol.
You are just a very sad individual, with no clue about the science you criticise, and no alternative to offer. You are just an anti-science troll, with a boring, repetetive posting history. Same old, same old. Get a life.
Benni
1 / 5 (4) Feb 23, 2017
You are just a very sad individual, with no clue about the science you criticise
.........not criticizing "science" jonesy, I'm criticizing pseudo-science, the stuff you purvey in..........like your blatant ignorance of the existence of the Inverse Square Law........Yeah, you'd like to have that law repealed wouldn't you? You're not alone, so would Schneibo.

You challenge me to put up the math, I do just that & you go dumb & silent. Have you ever even seen the formula for the Inverse Square Law, or is it just something you've heard about ? Maybe if I get ambitious i'll put it up for you so you won't need to be bothered with the grueling arduous task of googling it. Nah, I guess I won't do it after all, it's more fun watching you trip, stumble, cough & gag all over yourself trying to figure it out......after all it is high school physics.
Whydening Gyre
4 / 5 (4) Feb 24, 2017
... like your blatant ignorance of the existence of the Inverse Square Law........Yeah, you'd like to have that law repealed wouldn't you? You're not alone, so would Schneibo.

You challenge me to put up the math, I do just that

No ya didn't. You put up a wiki definition.
Have you ever even seen the formula for the Inverse Square Law,....after all it is high school physics.

One problem here, Benni. The ISL is NOT infinite. One end is 0 (in the case of BH's, this means "space to apply it to"). You can only reach maximum (contiguous) density. Which is only an accretive addition of the contributing mass. IOW, the surface (from which it is applied) describes the center.
Your own understanding of it's limitations needs some work.
Maybe some remedial HS Physics?
Da Schneib
3.7 / 5 (3) Feb 24, 2017
Lenni doesn't know enough math to understand why the event horizon of a black hole isn't a singularity. That's because he doesn't know about Rindler coordinates.
Whydening Gyre
3.7 / 5 (3) Feb 24, 2017
Lenni doesn't know enough math to understand why the event horizon of a black hole isn't a singularity. That's because he doesn't know about Rindler coordinates.

Neither do I. So why do I get it?
SiaoX
2.5 / 5 (2) Feb 24, 2017
why the event horizon of a black hole isn't a singularity
But it looks like so at the Schwarzschild spacetime diagram - the space-time inversion occurs there.
antialias_physorg
4 / 5 (4) Feb 24, 2017
why the event horizon of a black hole isn't a singularity

Because at a singularity curvature would be infinite - and at the event horizon it isn't.
Benni
1 / 5 (4) Feb 24, 2017
why the event horizon of a black hole isn't a singularity
CENTER, not EH

The ISL is NOT infinite.


Which is the reason it is foolishness applying it to Schwarzschild's BH Math, but that has never stopped BH Enthusiasts from doing it anyway, and when they do it they always reverse the application of the force of gravity whereby gravity is maximum (INFINITY) at the center of the gravitating body, whereas by application of the Inverse Square Law it is exactly ZERO and not INFINITY.

Math is tough, right WhyGuy?

The quintessential definition of a BH is:

"At the center of a black hole lies a gravitational singularity, a region where the spacetime curvature becomes infinite"

......and this is exactly a reverse application of the ISL & is the reason a FINITE stellar body defined as having INFINITE GRAVITY & DENSITY at it's center cannot exist.

SiaoX
3 / 5 (1) Feb 24, 2017
Because at a singularity curvature would be infinite
I'm indeed aware, that the textbook physics places the actual singularity at the center of black hole instead of event horizon - but I don't understand why it does so, once the event horizon is already singular for light spreading. At the event horizon the time is said to stop. The horizon forms at t=infinity in terms of Schwarzschild time - which already looks like quite singular situation for me. And it would also imply, that the space-time curvature is already infinite there, because the speed spreading is indirectly proportional the space-time curvature and the speed of time cannot get lower than zero. To know something doesn't mean, you can explain it - and so far I did not met anybody, who could explain me this conceptual controversy.
Whydening Gyre
3.7 / 5 (3) Feb 24, 2017
The quintessential definition of a BH is:

"At the center of a black hole lies a gravitational singularity, a region where the spacetime curvature becomes infinite"

......and this is exactly a reverse application of the ISL & is the reason a FINITE stellar body defined as having INFINITE GRAVITY & DENSITY at it's center cannot exist.

You're the only one attempting to apply it in that manner.
Everyone else understands it as it is - a useful APPROXIMATION.
A curious math artifact - not reality.
Benni
1 / 5 (1) Feb 24, 2017
textbook physics places the actual singularity at the center of black hole instead of event horizon- but I don't understand why it does so, once the event horizon is already singular for light spreading. At the event horizon the time is said to stop. The horizon forms at t=infinity in terms of Schwarzschild time - which already looks like quite singular


You put your finger dead on the conundrum that is the crux of the problem with Black Hole Math.

The point of "no return" supposedly occurs somewhere inside the bounds of the EH, long before you even arrive at the surface of the BH where infinite gravity already exists. The conundrum being that the SINGULARITY is somehow separate from the surface, I guess BH Enthusiasts imagine the SINGULARITY has a greater magnitude of infinite gravity than the infinite gravity at the surface, I'd like to see jonesy's calculation for that.
SiaoX
4 / 5 (1) Feb 24, 2017
The point of "no return" is actually the photon sphere - way above the event horizon. Which is also what I don't understand: if some volume allows not light to escape, it should also look black - or not?
Benni
1 / 5 (1) Feb 24, 2017
The point of "no return" is actually https://i.imgur.com/BOFCnDxm.png - way above the event horizon. Which is also what I don't understand: if some volume allows not light to escape, it should also look black - or not?


This is Hawkings "APPARENT HORIZON" that he discusses in his April 2016 News Conference. I'd suggest you google that news conference & read what he says about the EH being "apparent", it was a really big deal last year when he used that word "apparent", such descriptive terminology has sent the world of cosmology into a tailspin because if the EH does not exist, neither do BHs.
Whydening Gyre
3.7 / 5 (3) Feb 25, 2017
The point of "no return" is actually https://i.imgur.com/BOFCnDxm.png - way above the event horizon.

At which point, due to momentum, light begins moving in a generally tangential manner around the gravitational event.
Which is also what I don't understand: if some volume allows not light to escape, it should also look black - or not?
Not if it's allowed to move sideways...
Technically, the singular mass affecting the light is the only place it will be "black". Other than that, it's varying shades of grey. Trajectories, if you will.
Benni intentionally neglects prior angular momentum of light (AND the "singularity") in his description, indicating his misleading application of the inverse/square Law (Not to mention, the terms EH and BH).
BTW, it's not infinite, it's MAXIMUM additive value of collected mass.

Da Schneib
3.7 / 5 (3) Feb 25, 2017
why the event horizon of a black hole isn't a singularity
But it looks like so at the http://casa.color.../st0.gif - the space-time inversion occurs there.
This particular singularity isn't an actual feature of real black holes; it appears due to the peculiar coordinates chosen for the representation. It's an artifact of the math, not of the physical situation. And the same appears to be true for the supposed "singularity" at the centers of black holes; in fact, we know for certain that without a theory of quantum gravity, which we do not have, our representation of anything beyond the event horizon is lacking, and we can only observe by gravity radiation which we have only begun to be able to even detect.
Da Schneib
3.7 / 5 (3) Feb 25, 2017
why the event horizon of a black hole isn't a singularity

Because at a singularity curvature would be infinite - and at the event horizon it isn't.
And you can tell that when you choose the correct coordinates, i.e. Rindler coordinates.
Da Schneib
4.2 / 5 (5) Feb 25, 2017
I usually don't actually read anything Lenni says, but I had to check the context of the quote from Whyde. In this case, the context is exactly as expected.
The quintessential definition of a BH is:

"At the center of a black hole lies a gravitational singularity, a region where the spacetime curvature becomes infinite"
Whyde's right; Lenni's lying again.

The quintessential definition of a black hole is enough mass gathered within a small enough radius that the escape velocity is greater than the speed of light. Period. There is no speculation about what's inside the radius where that escape velocity crosses the value of c.
Da Schneib
4.2 / 5 (5) Feb 25, 2017
The point of "no return" is actually https://i.imgur.com/BOFCnDxm.png - way above the event horizon.
Your "photon sphere" is actually the event horizon according to the diagram you showed.

Which is also what I don't understand: if some volume allows not light to escape, it should also look black - or not?
That's, you know, why they call it, ummm, a "black hole."

Just sayin'.
Da Schneib
4 / 5 (4) Feb 25, 2017
To be more pedantically accurate, there is of course lots of *speculation* about what's inside the event horizon, but there aren't any real testable hypotheses much less theories.
RealityCheck
2 / 5 (4) Feb 25, 2017
Hi Da Schneib, SiaoX (and PS to Whyde). :)

From SiaoX:
The point of "no return" is actually https://i.imgur.com/BOFCnDxm.png - way above the event horizon. Which is also what I don't understand: if some volume allows not light to escape, it should also look black - or not?


Response from Da Shcnein:
The point of "no return" is actually https://i.imgur.com/BOFCnDxm.png - way above the event horizon.
Your "photon sphere" is actually the event horizon according to the diagram you showed.
@ Da Schneib, from looking at the referenced diagram I see the black mark as the black hole extent including black EH; and I see the DOTTED CIRCLE denoting the photon spehere which SiaoX alluded to.

So Da Schneib, perhaps you misread the diagram? Please clarify. Thanks. :)

PS: @ Whyde, photons 'trapped' tangentially, moving around/above the BH feature's actual EH, will not be coming OUT of there for us to 'see' that sphere by, hence it should be 'black' there too? :)
RealityCheck
2 / 5 (4) Feb 25, 2017
Hi XiaoX (and Whyde). :)

The only photons which could leave that sphere region would be the photons emitted by particles being disintegrated/accelerated to produce photons while IN that spherical region and sufficiently directed radially outwards so that they aren't 'trapped' into tangential motion around/within that region. So perhaps the only way we could 'see' that photon sphere would be via photons emitted by matter while traversing/colliding in that region and being strongly 'tidally' disrupted/accelerated to emit photons outwards direction. That is probably where the most 'flaring' emissions of gamma-rays and X-rays occurs (I can't immediately what the current hypotheses say about that bit; can you/Da Schneib (or anyone) remind me about that? Thanks in advance). :)
Da Schneib
3.7 / 5 (3) Feb 25, 2017
@ Da Schneib, from looking at the referenced diagram I see the black mark as the black hole extent including black EH; and I see the DOTTED CIRCLE denoting the photon spehere which SiaoX alluded to.

So Da Schneib, perhaps you misread the diagram? Please clarify. Thanks. :)
Or perhaps the diagram isn't very accurate. Technically the "photon spehere" [sic] is the event horizon.

Maybe you should brush up on black hole physics. Or, considering your web site, maybe you should discard everything you think you know and start over from scratch.

On Earth. The RealityCheck just bounced again.
Benni
1 / 5 (1) Feb 25, 2017
"At the center of a black hole lies a gravitational singularity, a region where the spacetime curvature becomes infinite"
Lenni's lying again.
......shame on you for calling WikiPedia experts liars, I copied their quote under Black Hole

a black hole is enough mass gathered within a small enough radius that the escape velocity is greater than the speed of light.
.....and this requires a force of INFINITE GRAVITY, and attaining infinite gravity requires infinite mass.

However I understand Einstein's General Relativity is so far beyond the grasp of your high school physics that it isn't any wonder you're unable comprehend that GRAVITY IS MASS DEPENDENT & not dependent on whatever figment of your imagination that you want to conjure up.

Da Schneib
4 / 5 (4) Feb 25, 2017
shame on you for calling WikiPedia experts liars, I copied their quote under Black Hole
You claimed this was "the quintessential definition of a black hole." That was a direct lie. And now you're lying about what Wikipedia said.

a black hole is enough mass gathered within a small enough radius that the escape velocity is greater than the speed of light.
.....and this requires a force of INFINITE GRAVITY, and attaining infinite gravity requires infinite mass.
You're lying again, Lenni.

For the Earth's mass, the Schwarzchild radius is about the size of a pea.

There isn't any need for gravity to be infinite to give a finite value of the escape velocity. The speed of light isn't infinite.

On Earth.
Da Schneib
3.7 / 5 (3) Feb 25, 2017
You can vote it down all you like @Lenni, but the truth will out.
Benni
1 / 5 (2) Feb 25, 2017
For the Earth's mass, the Schwarzchild radius is about the size of a pea.
This has nothing to do with the measurement of gravity.

There isn't any need for gravity to be infinite to give a finite value of the escape velocity. The speed of light isn't infinite
............and there you go again schneibo, removing all doubt about how little you know about Einstein's thesis of General Relativity. What's the problem here, a few Differential Equations too tough for you?

Benni
1 / 5 (2) Feb 25, 2017
You can vote it down all you like @Lenni, but the truth will out.
.....then post some "truth" & take some of the load off my back.
Da Schneib
4 / 5 (4) Feb 25, 2017
For the Earth's mass, the Schwarzchild radius is about the size of a pea.
This has nothing to do with the measurement of gravity.
It has to do with mass and radius. Mass has gravity. You're lying again, @Lenni.

There isn't any need for gravity to be infinite to give a finite value of the escape velocity. The speed of light isn't infinite
............and there you go again schneibo, removing all doubt about how little you know about Einstein's thesis of General Relativity. What's the problem here, a few Differential Equations too tough for you?
And there you go again, @Lenni, lying. Escape velocity is dependent upon mass and radius. No differential equations required.

On Earth.

You can vote it down all you like @Lenni, but the truth will out.
.....then post some "truth" & take some of the load off my back.
I did. It's not my problem if you choose to lie about it, but I'll certainly point it out. Get over it.
Benni
1 / 5 (2) Feb 25, 2017
And there you go again, @Lenni, lying. Escape velocity is dependent upon mass and radius. No differential equations required


Yeah schneibo, Escape Velocity is dependent on mass & radius as a function of PARTICULATE behavior as governed by the equations for KINETIC ENERGY: KE= 1/2 mv²

Your problem is that you don't know photons are not subject to the laws of Kinetic Energy. Photons are subject to the laws Einstein laid out in his Mass/Energy Equivalence Principle in Special Relativity, We also call these photons ELECTRO-MAGNETISM: E=mc².

A photon can be only be reduced below the speed of light by reducing its frequency to ZERO in which case it becomes non-existent because photons do not exist at any speed but that of light. No one has ever measured a photon traveling at half the speed of light ,but you don't know these things, which is the reason you're so caught up in the swill of your usual name calling routines & getting mad at me because I can figure it out.
Da Schneib
4 / 5 (4) Feb 25, 2017
Escape Velocity is dependent on mass & radius as a function of PARTICULATE behavior
No, it's not a function of PARTICULATE [sic] behavior. It's simply how much mass/energy inside what radius; the composition of the mass is immaterial to the escape velocity. In fact, there is a type of conjectured black hole made of energy, called a "kugelblitz." No matter in it at all. None required; just the requisite amount of mass/energy within the requisite radius.

You're lying again, @Lenni. It's pretty obvious. Nobody who knows any actual physics could ever believe your claim. I'd suggest you stop claiming to be a nuclear engineer and stop claiming to be able to solve differential equations when you obviously aren't a nuclear engineer and obviously can't solve differential equations, not to mention making spurious claims about relativity when you don't know anything about relativity.
RealityCheck
1.5 / 5 (2) Feb 26, 2017
Hi Da Schneib. :)
@ Da Schneib, from looking at the referenced diagram I see the black mark as the black hole extent including black EH; and I see the DOTTED CIRCLE denoting the photon spehere which SiaoX alluded to.

So Da Schneib, perhaps you misread the diagram? Please clarify. Thanks. :)
Or perhaps the diagram isn't very accurate. Technically the "photon spehere" [sic] is the event horizon.
The issue was your 'reading' of that diagram. If you didn't actually mean to refer to that diagram, but another diagram, then you should have said so, in the interests of clarity.
Maybe you should brush up on black hole physics. Or, considering your web site, maybe you should discard everything you think you know and start over from scratch.
Since it was obvious from the exchanges in another thread that you are laboring under a misapprehension (due to you trusting to bot-voting ignoramuses for your 'facts' about me/my site etc), I'll let that one slide, mate :)
Da Schneib
3.7 / 5 (3) Feb 26, 2017
Hi Da Schneib. :)
@ Da Schneib, from looking at the referenced diagram I see the black mark as the black hole extent including black EH; and I see the DOTTED CIRCLE denoting the photon spehere which SiaoX alluded to.

So Da Schneib, perhaps you misread the diagram? Please clarify. Thanks. :)
Or perhaps the diagram isn't very accurate. Technically the "photon spehere" [sic] is the event horizon.
The issue was your 'reading' of that diagram. If you didn't actually mean to refer to that diagram, but another diagram, then you should have said so, in the interests of clarity.
The diagram in my judgment isn't very accurate. That's not my problem; if you knew enough relativity to spot its inaccuracy, you wouldn't have asked the question in the first place.

If you choose to comment on subjects you don't understand you're not going to get any slack from me. I suggest you understand relativity physics before you comment on it.
Da Schneib
3.7 / 5 (3) Feb 26, 2017
As far as my judgments on your web site, you're not going to like them much. It looks pretty wacky. Very much about a lot of personal and idiosyncratic conjectures, and not much connection to real experimental and observational data. Never mind connection to well-established physical theory, which it also doesn't exhibit. It doesn't have anything to do with other peoples' judgments, it's my own and I approach such things without prejudice. This may be difficult for you to believe, but it's how I roll.
RealityCheck
1.5 / 5 (2) Feb 26, 2017
Hi Da Schneib. :)
@ Da Schneib, from looking at the referenced diagram I see the black mark as the black hole extent including black EH; and I see the DOTTED CIRCLE denoting the photon sphere which SiaoX alluded to. So Da Schneib, perhaps you misread the diagram? Please clarify. Thanks. :)
Or perhaps the diagram isn't very accurate. Technically the "photon spehere" [sic] is the event horizon.
The issue was your 'reading' of that diagram. If you didn't actually mean to refer to that diagram, but another diagram, then you should have said so, in the interests of clarity.
The diagram in my judgment isn't very accurate. That's not my problem;...
I was merely asking for clarification, because YOU said to SiaoX:
Your "photon sphere" is actually the event horizon ***according to the diagram you showed***[my asterisks].
Wherein you specifically but incorrectly said HIS diagram showed them as being the same, despite it showing them as different. OK? :)
Da Schneib
3.7 / 5 (3) Feb 26, 2017
The diagram in my judgment isn't very accurate. That's not my problem;...
I was merely asking for clarification, because YOU said to SiaoX:
Your "photon sphere" is actually the event horizon ***according to the diagram you showed***[my asterisks].
Wherein you specifically but incorrectly said HIS diagram showed them as being the same, despite it showing them as different. OK? :)
Again, the fact the diagram wasn't very accurate isn't my problem, and if you knew enough relativity physics to ask the question in the first place you would have looked at it and known that.

Your sarcastic and demeaning style has betrayed you. Perhaps you'll consider this in the future before you start making nasty posts referencing mine.
SiaoX
3 / 5 (1) Feb 26, 2017
This is Hawkings "apparent horizon" that he discusses in his April 2016 News Conference
The apparent horizon was coined by Visser in 2001 as a surface at which radially outgoing light no longer moves outward locally, but stays at the same radius. A photon sphere is a surface around black hole containing all the possible closed orbits of a photon.
Your "photon sphere" is actually the event horizon according to the diagram you showed.
In Schwarzchild metric the photon sphere has by one and half bigger radius than the event horizon.
Da Schneib
3.7 / 5 (3) Feb 26, 2017
Your "photon sphere" is actually the event horizon according to the diagram you showed.
In Schwarzchild metric the photon sphere has by one and half http://www.gothos...tic.gif.
The photon sphere is the distance at which an orbiting (by definition tangent) photon will orbit forever (assuming the mass of the BH does not change). This is by definition the event horizon. Above this, the photon will escape; below it it will be absorbed.

If you argue otherwise please present the math to prove it.
Whydening Gyre
5 / 5 (3) Feb 26, 2017
The point of "no return" is actually https://i.imgur.com/BOFCnDxm.png - way above the event horizon. Which is also what I don't understand: if some volume allows not light to escape, it should also look black - or not?

My bad. Shoulda actually looked at the diagram...
It IS the EH.
Shucks, DS beat me to it.... Guess I should avoid plumbing projects when I'm trying to discuss a subject...
SiaoX
1 / 5 (1) Feb 26, 2017
@Da Schneib: You told, that "photon sphere" is actually the event horizon according to the diagram I showed. But my diagram shows photon sphere, not even horizon. The relation of boths. If you still argue otherwise please present the math to prove it.
The photon sphere is the distance at which an orbiting (by definition tangent) photon will orbit forever (assuming the mass of the BH does not change). This is by definition the event horizon
Nope, it isn't - event horizon is 1.5x smaller. But at least I'm not the only one, who cannot see any difference there... ;-)
RealityCheck
1 / 5 (2) Feb 26, 2017
Hi Da Schneib. :)
Again, the fact the diagram wasn't very accurate isn't my problem, and if you knew enough relativity physics to ask the question in the first place you would have looked at it and known that.

Your sarcastic and demeaning style has betrayed you. Perhaps you'll consider this in the future before you start making nasty posts referencing mine.
Again, please stick to the context and thrust of my relevant post asking you for clarification. I specifically asked for clarification re YOUR reading of HIS diagram. I asked NOT about correctness of diagram, but only asked for clarification of YOUR claim HIS diagram showed them as same when actually it showed them as different.

You have since clarified that you were actually commenting based on your own source/diagram, not based on his source/diagram. Hence all is clarified, and no need for further exchanges on this with me. OK? :)
Da Schneib
5 / 5 (1) Feb 26, 2017
Sorry, @SiaoX, not going to argue about diagrams. If you don't have the math, you don't have any evidence and I dismiss your funny diagrams with prejudice.
Da Schneib
3.7 / 5 (3) Feb 26, 2017
Again, please stick to the context
Sorry, not if you don't, @BouncedRealityCheck. If you don't like your own tactics applied back to you, consider using different ones, because I don't like you much because of them and will definitely stick them back up your nose until you choke on them.
SiaoX
1 / 5 (1) Feb 26, 2017
This argument is silly, but at least it applies in completely symmetric way: you cannot dismiss "my funny" diagrams without math: if you don't have the math, you don't have any evidence. BTW did you have a look, from where I linked "my funny" diagrams? You can find pile of math there.
Da Schneib
3 / 5 (2) Feb 26, 2017
Sorry, @SiaoX, but my perception of silliness is using inaccurate diagrams instead of the math that the diagrams are inaccurately derived from.

I can do the math and see the inaccuracies in the diagrams. If you cannot you are not qualified to have this discussion.
SiaoX
3 / 5 (2) Feb 26, 2017
@Da Schneib: These diagrams are exact - and this one is generated just by math, you're calling for. 'BTW You're the first BH "expert", who I know how he is trying to teach people here and who doesn't know about difference between event horizon and photon sphere. Try to think logically: if these two concepts would be equal, they wouldn't be named differently - don't you think?
RealityCheck
1 / 5 (1) Feb 26, 2017
Hi Da Schneib. :)
Again, please stick to the context
Sorry, not if you don't, @BouncedRealityCheck. If you don't like your own tactics applied back to you, consider using different ones, because I don't like you much because of them and will definitely stick them back up your nose until you choke on them.
So it was a 'tactic' from you, and not a genuine attempt at mutual understanding based on clarification of matter posted by you? Well, when you feel like discussing based on mutual respect/respect for genuine science discourse principles of objectivity (and lack of bias and tactics etc), then I will be glad to again point out that I was the one being on-science and informed of the subject matter while you keep kneejerking to ego-tripping tactics etc. Please look at your own side of the street before casting aspersions at my side of the street, hey?

PS: What was that you said you were drinking.... "Old B**stard" beer? Drinking and Internet-ting is dangerous! :)
SiaoX
1 / 5 (1) Feb 26, 2017
I can do the math and see the inaccuracies in the diagrams.
Well, prove it with math, after then. Show us your math at least once, if you have full mouth of it all the time.
Da Schneib
3.7 / 5 (3) Feb 26, 2017
These diagrams are exact
Knowing the math, I know they are not.

Nor is your clumsy attempt at shifting the burden of proof going to work either.

Sorry, but onto ignore you go. I only deal with cranks intermittently when I get bored enough.
SiaoX
5 / 5 (2) Feb 26, 2017
Without sorry, the burden of proof is up to you - my diagram is linked from page, which contains computer program, which generates it. The same diagram you can find at all sites, which deal with this topic. But even without it, every troll who is confusing photon sphere with event horizon is crackpot - no matter how much of math he is able to show or not. You're just an incompetent parasite of this forum, who is trolling and annoying others with your lack of knowledge. You should really deal more with yourself for not to behave like the sociopath.
Benni
1 / 5 (1) Feb 26, 2017
A photon sphere is a surface around black hole containing all the possible closed orbits of a photon
.

Maybe you have a pic of a "photon sphere"? I'd be curious as to what its "surface" looks like?

I just don't see how this team will ever get a picture of the surface of a BH with the EH being composed of so man different layers of stuff. Maybe the team is hoping all these hypothetical layers of stuff really don't exist & they'll be able to get their picture.
Benni
1 / 5 (1) Feb 26, 2017
The photon sphere is the distance at which an orbiting (by definition tangent) photon will orbit forever (assuming the mass of the BH does not change). This is by definition the event horizon. Above this, the photon will escape; below it it will be absorbed.

If you argue otherwise please present the math to prove it.


Hey, schneibo........it's your statement, you be the one to put up the math to prove it.
Da Schneib
4.2 / 5 (5) Feb 26, 2017
@Lenni, sure:

-m'' + m'n' - m'² - 2m'/r = 0
m'' + m'² - m'n' - 2m'/r = 0
e⁻²ⁿ (1 + m'r - n'r) - 1 = 0
R₂₂ sin² ϕ = 0

No problem.

Solving these equations for a massless particle moving at c, we find that there is a single distance at which such a particle can orbit the BH defined by the mass within a given radius. This is because such a particle can move at only one speed. For objects or particles with mass, this represents an inner limit beyond which they cannot go without being trapped permanently within the BH; this is the EH.
cantdrive85
1 / 5 (1) Feb 26, 2017
Scientists readying to create first image of a black hole

This is an abomination, BH images are bountiful;
https://www.googl..._AUIBigB
And like the one above, they too are all "created". This is the value of the artistic types like Whyde who can employ their fanciful imaginings to create "science".
Benni
1 / 5 (3) Feb 26, 2017
it's not a function of PARTICULATE [sic] behavior. It's simply how much mass/energy inside what radius; the composition of the mass is immaterial to the escape velocity. In fact, there is a type of conjectured black hole made of energy, called a "kugelblitz." No matter in it at all. None required; just the requisite amount of mass/energy within the requisite radius.


......got a section from within General Relativity you can point to that backs up all this slop & swill Fake Science you continue regurgitating?
Da Schneib
5 / 5 (4) Feb 26, 2017
For exact pedantic accuracy I should note that these equations are only exactly true for a non-rotating, non-charged BH, a Schwarzchild BH; they are formidably more complex for the Kerr BH with rotation, or the Kerr-Newman BH with rotation and charge. These solutions have a region called the "ergosphere," within which the possible orbits are not defined only by proximity but also by angle of incidence,and within which frame dragging does not permit an object with zero kinetic energy to exist. (That's perhaps oversimplified, but if you know about this stuff you'll get it.)
Da Schneib
5 / 5 (5) Feb 26, 2017
......got a section from within General Relativity you can point to that backs up all this slop & swill Fake Science you continue regurgitating?
Lenni doubles down on shifting the burden of proof. It's up to you, @Lenni, to prove that there's anything in GRT regarding PARTICULATE [sic] matter. See, all you have to do is look at the entire right side of the EFE and note there's no indication in the stress-energy tensor of whether the mass/energy is mass or energy, and no indication of any needed calculations that are different for various kinds of matter. If you claim there are, show where they occur in the EFE.
Benni
1 / 5 (4) Feb 26, 2017
Solving these equations for a massless particle moving at c.....This is because such a particle can move at only one speed...


.....photons are not particles. There is no such thing as "a massless particle". All PARTICLES are governed within the limits of KE=mv², this by necessity of the limits imposed by the Mass/Energy Equivalence Principle of E=mc² excludes Electro-magnetic radiation (photons)

Lenni doubles down on shifting the burden of proof. It's up to you, @Lenni, to prove that there's anything in GRT
....you can't even invoke the proper thesis, it's SR not GR.

Da Schneib
5 / 5 (5) Feb 26, 2017
photons are not particles
Sorry, man, back on ignore you go. Either you're psychotic or you're just making stuff up.

https://en.wikipe...i/Photon
First sentence: "A photon is an elementary particle..."
Benni
1 / 5 (2) Feb 26, 2017
Schneibo.......your brain rivals the density of a BH.......infinite.
Da Schneib
5 / 5 (6) Feb 26, 2017
So, just to recap: @Lenni claims the event horizon of a black hole represents some infinite quantity. This is shown to be an artifact of the coordinate system traditionally used to define black holes, a coordinate system that is inappropriate. Use of the correct coordinate system removes the infinities, proving this.

@Lenni claims black holes are impossible, but ignores the fact that the metric can be recovered by observing the motions of stars, dust, and gas surrounding the site of the putative object, and cannot present math that shows there is no event horizon, in either set of coordinates mentioned above. To put this another way, @Lenni claims that black holes are impossible, but neglects (or ignores) the fact that whatever is making those stars move like that, it is compact and it has an event horizon. Maybe @Lenni wants to use some other terminology. But the physical facts are there in the motions of the stars, and there is only one explanation.
Da Schneib
5 / 5 (6) Feb 26, 2017
Continuing, @Lenni denies black holes because they are supposed to have a singularity at the center, totally ignoring the fact that without a quantum theory of gravity, all accounts of what happens inside the event horizon are mere speculation and there is no proof of a singularity within the black hole since we already know that all the math we have is invalid inside the EH.

Now, back to your regularly scheduled reality.
Da Schneib
5 / 5 (5) Feb 26, 2017
As for the project of creating an image of a black hole, it's obvious that any light source behind or even around the object will have some of its light curved so that it comes directly at an observer along the photon sphere. We already know this because Eddington showed that the positions of stars in the background show deviation from their normally observed locations when they can be observed close to the Sun during an eclipse.

So what we expect to see is a dark center with a rim of light where the light of surrounding and background objects is redirected toward the observer. The spectral characteristics of this rim of light will confirm its origins, showing that there is a photon sphere and that it is at the predicted distance from the center of the black hole.
Whydening Gyre
5 / 5 (1) Feb 26, 2017
Continuing, @Lenni denies black holes because they are supposed to have a singularity at the center, totally ignoring the fact that without a quantum theory of gravity, all accounts of what happens inside the event horizon are mere speculation and there is no proof of a singularity within the black hole since we already know that all the math we have is invalid inside the EH.

Now, back to your regularly scheduled reality.

He doesn't really deny BH's. He just thinks of it as clever way of getting attention from adults.
Only child syndrome...

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