Talk:Schwarzschild radius

message: The formatting of the article after the Parameters table is broken. Text of the remaining article is aligned to the right column of the table? Checked in three different web browsers.
Anon  — Preceding unsigned comment added by 2A0C:5BC0:40:1028:694E:373B:C8BD:777 (talk) 10:49, 14 December 2020 (UTC)
thread_title:  Formatting is broken? 

message: http://scienceworld.wolfram.com/physics/SchwarzschildRadius.html
claims the derivation here is incorrect. that is it is incorrect to set the escape velocity to c the speed of light to solve for r.  On consideration, it is OBVIOUS that there is something wrong with an equation (for a finite mass) that claims v = c, isn't there? (since this is impossible.)  —Preceding unsigned comment added by 69.40.245.3 (talk) 22:12, 23 August 2008 (UTC)
thread_title:  Incorrect derivation? 

message: Shouldn't the radius =  square root ( 2GM/c^2) ???
:No, see , , , etc. -- Tim Starling 04:25, Nov 18, 2004 (UTC)
::Or, more simply, using dimensional analysis:  r\, is in meters (m\,), so 2Gm/c^2\, must be in meters also.  Let's see:
::G \to N\cdot m^2/kg^2\,, where N\, is force = mass * acceleration = kg\cdot m/s^2\,, so G\, is in units of (kg \cdot m/s^2) \cdot (m^2/kg^2)\,.
::m/c^2\, is in units of kg/(m/s)^2 \to kg \cdot s^2/m^2.
::So when combined Gm/c^2\, is in units of (kg \cdot m/s^2) \cdot (m^2/kg^2) \cdot (kg \cdot s^2/m^2).
::You can see directly that units of kg\, and s\, are equal in numerator and denominator, leaving m^3/m^2 \to m\, (meters).
::QED.  Slowmover 20:23, 22 March 2006 (UTC)
thread_title:  Needs square-root? 

message: What is the physical meaning of an object's Schwarzschild radius when it is bigger than the radius? -- Myria 05:35, 29 December 2005 (UTC)
replies: text: Giving this some thought. -- Slowmover 19:34, 30 June 2006 (UTC)
replies: text:  I think that one useful way to look at it is this: If you take an object with a particular mass, it has a certain amount of gravity, all of which has to exit through its surface. If you then squish the object into a smaller volume, the intensity of the gravitational field at the surface increases, and when you increase this surface gravity, you increase the escape velocity, the speed at which somethng has to be thrown with at the surface to have a chance of completely escaping. For a given mass M, the r=2M radius tells us how compact the material would have to be in order for this surface escape velocity to reach the critical situation of being equal to the nominal speed of light. John Michell did the calculations for this back in the Eighteenth Century, but when Isaac Newton's original ideas about light took a hammering, the idea became unfashionable and dropped out of sight until Karl Schwarzchild brought it back, more unambigiuously, in the context of Einstein's general theory of relativity, in the early Twentieth Century. ErkDemon 15:59, 29 December 2006 (UTC)
replies: text:  That is no answer to Snaxalotl's question: Was the universe a black hole in its early stages? 145.97.222.84 (talk) 17:39, 15 January 2012 (UTC)
thread_title:  Meaning for non-black holes? 

message: I added a link to the hyperphysics sight because it distinguishes between a gravitational radius and a Scwarzschild radius. The distinction is an important one even though it is quite common to confuse the two.Lucretius 07:53, 15 January 2006 (UTC)
thread_title:  Gravitational versus Schwarzchild radius 

message: below are the calculations for the mass and radius of a black hole with the density of water. The resulting mass is 135 million times the mass of the Sun. The radius is 400 million kilometres.
\begin{align}
&\rho=\frac{m}{\frac{4}{3}*\pi*r^3}\\ &\frac{m}{\left(\frac{2Gm}{c^2}\right)^3*\frac{4}{3}*\pi}=1000Kg/M^3\\ &\frac{m}{\frac{8G^3m^3}{c^6}*\frac{4\pi}{3} }=1000Kg/M^3\\ &\frac{3mc^6}{32*\pi*m^3G^3}=1000Kg/M^3\\ &3c^6=32000*\pi*m^2G^3*Kg/M^3\\ &m=\sqrt{\frac{3c^6}{32000*\pi*G^3*Kg/M^3} }=\frac{c^3/100}{\sqrt{3.2/3*\pi*G^3*Kg/M^3} }\\ &m\approx\frac{2.7*10^{23}M^3/\text{sec}^3}{\sqrt{10/3*3*10^{-31}M^9*\text{sec}^{-6}*Kg^{-3}*Kg/M^3} } = \frac{2.7*10^{27}M^3/\text{sec}^3}{\sqrt{10*10^{-31}M^6/\text{sec}^6*Kg^{-2} } }\\ &m\approx\frac{2.7*10^{23} }{10^{-15} }Kg=2.7*10^{38}Kg\\ &r=\frac{2Gm}{c^2}=4*10^8M \end{align}
:That is the density of an object that perfectly fits its Schwarzschild radius. Pity that, once the object fits the object it will collapse into a singularity of infinite density. From the article: The Schwarzschild radius is the radius for a given mass where, if that mass could be compressed to fit within that radius, no known force or degeneracy pressure could stop it from continuing to collapse into a gravitational singularity -- which according to General Relativity means it will reach infinite density.
: I'm not sure about what you wanted to prove. --badpazzword (talk) 08:25, 29 March 2008 (UTC)
replies: text: I think the point here is only that the density claimed is the average density for everything inside the Schwarzschild radius. A nerd fact for people who like trivia.  —Preceding unsigned comment added by Snaxalotl (talk • contribs) 18:25, 7 June 2009 (UTC)
thread_title:  Low density black hole 

message: I just wanted to let the editors of this article know this article's far better than it was before; I didn't have a good idea of the Schwarzchild radius earlier, but it's clear now. This seems to be applicable to a lot of the physics articles in particular. Just wanted to point it out- thanks a lot for your contributions! Robinson0120 01:47, 17 January 2007 (UTC)
thread_title: Great Job on the Article

message: I removed this sentence:
:The Schwarzschild radius of a sphere with a uniform density equal to the critical density is equal to the radius of the visible universe.
because it's not true--the Schwarzschild radius in this case is about 250 Gly, and the radius of the visible universe is about 50 Gly. There's certainly a meaningful relationship here, but it's not quite so simple. (And I'm not sure exactly what it is.) -- BenRG 16:11, 26 March 2007 (UTC)
thread_title:  Removed incorrect claim 

message: From the article:
"Note that although the result is correct, general relativity must be used to properly derive the Schwarzschild radius. Some consider it to be only a coincidence that Newtonian physics produces the same result, yet this may be an indication of a deeper underlying symmetry in nature."
The first sentence and the first half of the second sentence are my own contribution from a while back.  However, someone seems to have mangled the message by using the much weaker phrase, "Some consider it..." and the highly questionable, "yet this may be an indication of a deeper underlying symmetry in nature."  I won't blindly revert it to, "It is only a coincidence that Newtonian physics produces the same result," but I'm quite skeptical about this "underlying symmetry" they're referring to.  Newtonian physics is an approximation, plain and simple.  Saying there's a symmetry between general relativity and Newtonian physics is like saying there is a symmetry between cars and engines.  Cars encompass engines-- they're not on equal footing.  You can compare different models of cars, just as you can compare general relativity to less reputable theories.  If someone knows what this underlying symmetry is, I'd love to hear about it.  Otherwise, I'll go ahead and revert that sentence to its much clearer form.--134.173.200.14 20:13, 3 April 2007 (UTC)
thread_title:  Underlying symmetry? 

message: Is it possibly so that if an objects rotates fast you could puch some more mass within the S-radius without the object collapsing? (As Schwarzschild is, as I understand, a solution for the non-rotating case.)If so the first paragraph is not entirely correct. -- Agge1000 (talk) 18:44, 16 November 2007 (UTC)
thread_title:  Rotating objects 

message: Quote from the article:
:while the Earth's is only about 9 mm, the size of a walnut.
Most walnuts I have known are quite a bit larger than 9 mm. Perhaps peanut would be more appropriate. --66.92.218.124 (talk) 19:28, 22 May 2008 (UTC)
replies: text: As of 12/26/2014 it reads "9 mm, the size of a peanut." This is wrong.  First, a peanut contains between 2-4 seeds, and is quite a bit larger than 1.8 cm. Second, the article isn't clear if it is referring to a peanut or a peanut seed (my guess is it means peanut seed). Third, "size" can be interpreted as volume, and the peanut seed is NOT round, so the volume of a seed with a given semi-major axis compared to a ball with that diameter will be quite a bit smaller. It is a really bad comparison, imho. I don't understand why ANY comparison is necessary, simply say a its the size of a small ball of diameter about 1.8 cm or 0.7 inches. Or are we attempting to inform children who don't know what a centimeter (or inch) is (or don't understand fractions)??Abitslow (talk) 16:56, 26 December 2014 (UTC)
replies: text: |--------------|
That's the size of an Earth-mass black hole.
Now, your computer screen might be smaller or bigger than mine, but I've noticed that as long as they're modern screens, bigger screens tend to show more things rather than bigger things. But if you don't trust that size, find a penny. US pennies have a diameter of 19 mm, which is only about 6% bigger than the Earth hole's 18 mm event horizon. And as you can tell from where I said "19 mm," I actually measured the penny, rather than just guessing. 50.5.96.209 (talk) 22:07, 16 May 2017 (UTC)
thread_title:  9 mm Walnut? 

message: r_s
Sun    	2953.2500765 m
Mercury	0.49028 mm
Venus  	7.22909 mm
Earth  	8.870056078 mm
Mars   	0.95305 mm
Jupiter	2.81915558 m
Saturn 	0.84408275 m
Uranus 	0.1289325 m
Neptune	0.1521333 m
The International System of Units (SI) has a unique unit for inertial mass (the kilogram) and a unique unit for energy (the Joule), but it does not have a unique unit for gravitational mass.  Therefore, gravitational masses are commonly listed in terms of inertial mass equivalents: for example, the mass of the earth as 5974.2 yottagrams.  However, this tradition is unfortunate, and even bizarre, when one considers that there is no known way to measure the inertial mass of a planet sized object.
The masses of planets and stars are currently estimated by first measuring their gravity, and then using G in Newton's theory of gravity to estimate the inertial mass.  In fact, the inertial masses of the sun and planets have never been directly measured.  The values listed in the literature are just estimates using Newton's theory, and there is no known way to prove that these estimates are correct.
See the following article:
"http://curious.astro.cornell.edu/question.php?number=452"
As an alternative, I would propose using the “standard gravitational parameter” to directly calculate the Schwarzschild radius.  The fact that c is defined exactly allows the Schwarzschild radius of a gravitating object to be determined to the same accuracy to which one knows the gravity of the object (See the table to the right: the “standard gravitational parameters” of the sun and earth are known to over 9 digits).  And using the “standard gravitational parameter” to directly calculate the Schwarzschild radius removes the unnecessary step of estimating the Newtonian inertial mass.  Furthermore, the Schwarzschild radius can be indirectly measured by measuring time dilation near the object.  So the Schwarzschild radius has the advantage of being both accurate and measurable, whereas the inertial mass is neither measurable nor accurately determinable.
Current alternatives for listing precise gravitational masses are to list mass ratios as is done in this NASA web site, or to list large masses in terms of their standard gravitational parameter.
70.176.112.179 (talk) 04:54, 19 April 2009 (UTC)
thread_title:  A metric for gravitational mass 

message: Images toward the center of the Milky Way as seen by Hubble show that there is so much galactic matter, both stars and dust, between us and the center that it is impossible to see what is at the center.  No black hole has ever been seen at the center of our galaxy, nor any galaxy.  The luminosity at the center of galaxies is so great that detail cannot be discerned.  That black holes exist at the center of galaxies is pure speculation.  My Flatley (talk) 18:27, 1 October 2009 (UTC)
replies: text: If you look at Supermassive black hole they discuss the evidence for such objects there, stating that there is a consensus amongst astronomers that such things do exist. Read about the evidence, and take your discussion to their talk page if not convinced.Puzl bustr (talk) 15:12, 6 December 2009 (UTC)
thread_title:  Supermassive Black Hole at the center of the Milky Way 

message: I noticed someone questioned the relationship between core black hole mass and the mass of the galaxy - and this is only an average relationship with much scatter, as evidenced by their example comparing Andromeda and Milky Way. I found that there is a much tighter relationship and an article M-sigma relation discussing it, and the previous work in this area, so added a ref and the wikilink. Puzl bustr (talk) 15:12, 6 December 2009 (UTC)
thread_title:  M-sigma relation 

message: I found this interesting quote and would like to start a section in the article for similar size comparisons: "The Schwarzschild radius for the Sun is about two miles, 1/200,000th of its current width; for the Earth to become a black hole, it would have to be squeezed into a ball with a radius of one centimetre." --Are we living in a designer universe? --Thorwald (talk) 03:24, 22 November 2010 (UTC)
thread_title:  Schwarzschild radius for the Sun 

message: I made a few edits in the intro when I noticed that TWO distinct definitions were given.  I ultimately saw that these two definitions were intricately muddled throughout the article.
Unless I have some big misunderstandings, a singularity is not the same as a black hole.  
Just because the Schwarzschild radius (an event horizon-like-thing) is at the surface doesn't mean the ability to resist gravitational pressure is all lost and an automatic compression to a singularity ensues.  A "singularity" is defined to have a radius of ZERO, black holes generally don't have a surface radius of zero, they just have a lot of mass and are small enough that the Schwarzschild radius is above the surface.
In so many places in the article, the SR is referred to as a function of MASS ONLY, which implies it is NOT related to the ability or inability of the matter to resist the gravity pressure, yet in many places the ability to resist gravity pressure and singularities are muddled in with it all.
So, what's the answer?  It's a pretty serious problem and, if you'll forgive me, I'm raising the EXTREMELY DUBIOUS flag for the whole article because of it.
I can't fix it.  I don't have enough detailed background, nor enough time, nor enough inclination for that matter.  But you people who watch this article CAN fix it, and you need to get on it!  It's incredibly bad!
Okay, if I'm out-to-lunch it's no big deal, just explain to me why.  For that matter, explain it in the article itself.  It is unclear in the article just how the ability to resist gravitational pressure is related to the SR being at the surface, etc.  If they are indeed related, please try to make that clear in the article.  I'm a really smart guy, and it's unclear to me.  :-).
All the best to you all,
Dave (today on IP 108.7.171.191 (talk) 05:31, 27 May 2011 (UTC)
thread_title:  A singularity is not the same as a black hole 

message: On somewhat closer inspection, it looks like the only remaining of these muddlements are a sentence remaining in the intro and a bunch woven into the HISTORY section.  I am going to remove the History section because that is the quickest and easiest way to correct the problem.  I know it's a big hairy deal to delete a section, but if you want to put it back, you are certainly welcome, just please do your best to correct the problem as you do (by removing the muddlements or adding explanation why they aren't muddlements)
Yours,
Dave (today on IP 108.7.171.191 (talk) 05:49, 27 May 2011 (UTC)
thread_title: Taking a shot at fixing it

message: I made those changes I described above.  But, they took a long time to show up in the article, AND they haven't yet appeared in the edit history page.  It's been 5 or 6 days so far.  There were other strangenesses while I was doing the edits too (too much to go in to).  So if you want to see the edits, click on "My" IP address above, you'll see they were actually made, they just don't show up in the edit history for some reason.  All other articles I've played with since then are behaving normally.  I am an experienced editor by the way, so I do know when WP is behaving strangely.
Dave (today on another IP: 108.7.162.218 (talk) 18:11, 2 June 2011 (UTC)
thread_title: Wikipedia bugs in evidence here

message: the Sun has a Schwarzschild radius of approximately 3.0 km (1.86 miles). - Brad Watson, Miami 71.196.121.70 (talk) 19:44, 11 July 2011 (UTC)
thread_title:  The Sun has a Schwarzschild radius of app. 3 km (1.86 miles) 

message: The top image on the page should be a diagram showing at least one radius. The Schwarzschild radius might be a good one to include. I don't know what the diagram shown there is but it doesn't seem to have anything to do with gravity or spacetime or light paths.
Michael McGinnis (talk) 06:33, 22 July 2011 (UTC)
thread_title:  The top image on the page should be a diagram showing at least one radius 

message: The formula given was 
: \frac{t_r}{t} = \sqrt{1 - \frac{r_s}{r}} = \frac{1}{2}
I removed the 1/2 as it clearly doesn't make sense here, (otherwise tr = 1/2 t regardless of the size of the mass) and doesn't correspond to the formula given in the reference article (http://en.wikipedia.org/wiki/Gravitational_time_dilation) where it says
:t_0 =  t_f \sqrt{1 - \frac{2GM}{rc^2}} = t_f \sqrt{1 - \frac{r_0}{r}} 
Neill Jones
87.113.10.142 (talk) 15:09, 16 February 2012 (UTC)
thread_title:  The Schwarzschild radius in gravitational time dilation 

message: "Schwarzschild" translates literally as "black shield" or liberally as "shield of blackness".
To a German newcomer to astrophysics this would be an obvious name for the phenomenon, before being told that this is the name of the discoverer/hypothesizer.  Not unlikely enticed by his own name, the astonomer Karl Schwarzschild thought harder than others of the possibility of a capsule or shield of blackness around a very dense mass.  Half a century had to pass after his death until evidence appeared that this was not just a pipedream.
Puddington (talk) 22:57, 17 June 2012 (UTC)
thread_title:  "Shield of blackness" 

message: The milky way's diameter as mentioned in the parameter table is incorrect. It is mentioned at ~0.2ly whereas wikipedia on the 'Milky way' page states it as 100-120kly. Please make the necessary change.  — Preceding unsigned comment added by 115.242.197.202 (talk) 16:27, 9 December 2012 (UTC)
replies: text: The number quoted in the table is not the Milky Way's diameter, it's its Schwarzschild radius. I didn't check if the number given is correct but it should certainly be much smaller than the diameter. If it was the same we wouldn't be here :-) — HHHIPPO 21:41, 9 December 2012 (UTC)
replies: text: I fixed the Milky Way's actual radius which was being quoted as  but it is more like . However the mass of the Observable Universe is much more problematic Mmom (talk) 14:36, 21 April 2021 (UTC)
thread_title:  Error in the Milky Way's diameter 

message: I think the table would be a lot more helpful if instead of showing only the Schwarzschild radius, it also showed the actual radius of the objects. If both of these are next to each other, the reader can better appreciate how far those objects are from black holes. Βαll (talk) 07:19, 19 December 2012 (UTC)
thread_title:  Table of Schwarzchild radii 

message: There's a nice explanation at: . I think it should be paraphrased for this article, though I don't have the time to do it myself right now. I may return later, but if someone who is more knowledgeable about relativity could do this, I'd be very appreciative.
Insurrectionist (talk) 05:58, 20 August 2013 (UTC)
replies: text: This is true, however the mass of the Observable Universe quoted here is off by a factor of 10. This is a real problem as there is no general consensus on this mass - but having different values across different pages should be addressed: Observable universe. The exact value is very important as multiplying the mass by 10 will in fact bring the Schwarzschild to the Hubble radius, thus giving credibility to the Black hole cosmology theory. If no professional astrophysicists are willing to update this, I eventually will.   Mmom (talk) 14:42, 21 April 2021 (UTC)
replies: text:  Where did this value of 8.8E52 kg even come from? Observable Universe calculates more than this just for the mass of ordinary matter! I'm flagging it. So this value corresponding with the age of the universe is going to have to be corrected as well. I've seen it disputed on Physics Stack Exchange (+) but more obviously it implies a value for mass of the observable universe that's far too low. DAVilla (talk) 15:43, 4 January 2023 (UTC)
thread_title:  A possible reference for the Schwartzschild Radius of the observable universe 

message: In the original 1916 paper, alpha='Schwarzschild Radius' arises as a constant of integration and must be computed empirically using the scalar radius r and angular velocity. Nowhere in the 1916 paper does he derive a value alpha=2GM/c^2. Furthermore, his metric differs from that in the modern treatment. His metric does not have a one-way event horizon at alpha where space and time reverse roles as in the modern treatment. See http://de.wikisource.org/wiki/%C3%9Cber_das_Gravitationsfeld_eines_Massenpunktes_nach_der_Einsteinschen_Theorie . And here is an English translation: http://arxiv.org/pdf/physics/9905030v1.pdf — Preceding unsigned comment added by Mathdoktor (talk • contribs) 15:10, 3 September 2013 (UTC)
thread_title:  Schwarzschild Radius (1916 version) 

message: The radius of the observable universe is given in the table as 4.4 x 10^25 m (~4.4 billion light years)
, which is an order of magnitude too small.  I did not correct it (yet) because the number is correctly flagged as needing a reference, which I don't have at the moment.
Sprite82 (talk) 23:12, 4 September 2014 (UTC)
thread_title: Incorrect radius for the Universe 

message: |-
| Universe     ||align="right" | 4.46 (~4.7 Gly) || align="right" | 8 (9.9WMAP- Content of the Universe)
This information in correct and the paper is flawed.  The paper asserts that because the universe is flat that it is near the critical density, and that's wrong.  The flattness of the universe comes from dark energy, and the mass of the universe is lower than the critical mass.
Roadrunner (talk) 16:51, 19 December 2014 (UTC)
thread_title:  Removed incorrect information. 

message: In the See Also section, the “classification of black holes by Schwarzschild radius” includes the visible universe “if its density is the critical density.” The idea that the visible universe could be a black hole (regardless of its density) is famously refuted; I'm not an expert, but certainly the claim isn't supported by either the visible universe link or the critical density link, much less a citation.  — Preceding unsigned comment added by Luke Maurer (talk • contribs) 01:25, 10 November 2015 (UTC)
replies: text: I wouldn't call someone's opinion, in a non-reviewed medium no less, refutation of a very obvious idea, that we, nor light, nor anything else, will ever exit the universe. How exactly that should be phrased, and how strong of a claim can be made to this end, is another matter, but certainly a white hole as postulated holds some analogy to the inside of a black hole. DAVilla (talk) 16:22, 4 January 2023 (UTC)
replies: text:  Even if an idea might seem "obvious", it does not belong in WP if we cannot find anything reputable that presents the idea.  No refutation whatsoever is needed.  The mass of the observable universe is not even clearly defined here – for example, it includes the mass density due to the cosmological constant, but this is not mentioned.  The radius of the observable universe is also ill-defined (my guess it is the comoving distance to what emitted observable radiation that we are receiving now, and this has long ago passed the horizon of what we can still receive radiation from that is emitted today.  In all, this is convoluted.  I've just removed it.  —Quondum 00:36, 17 April 2023 (UTC)
thread_title:  Visible universe as black hole 

message: I don't know where to start. The lede, as I read it, says that Rs is a property of a sphere. Wrong. At least on a meta-physics level, Rs is a property of a MASS. And the lede utterly fails to make that clear. It also goes on to talk about compression of a mass, as if that were generally possible. It fails to distinguish spherical mass distributions from alternatives, for instance a long thin wire. It talks about escape velocity (as if that explains anything to someone who doesn't have a good grasp of Freshman physics). It talks about light not able to "escape" from an star's surface, yet BHs emit light (called Hawking Radiation (although I realize that it is moot whether HR is "escaping", what is meant by surface, etc.)) It talks about a "sphere". Does it mean a ball (3 dimensional) or a surface (2 dimensional)? In other words, the lede makes all sorts of unwarranted assumptions about the reader's interpretation of its quite poor choice of words. Rs is a property of any mass which is defined to be the distance from a point of that given mass at which the escape velocity is the speed of light. Earth's Rs is a bit more than one-third of an inch (0.89 cm) (see table). I visited this article previously and it was much better then. The Formula section is virtually incomprehensible and is an extremely LOUSY attempt to DERIVE the formula. Here are my suggestions. Define the terms of the equation. For instance, no where do I see a clear explanation that M is rest mass...or is it? Rs clearly depends NOT just on mass of an object, but on its total energy. You'd think that would be made clear. Anyway, why not clean up all the garbage? 
Formula
Rs is directly proportional to Mass(energy). Rs ∝ M.
The proportionality constant is 2G/c² where G is the UNIVERSAL Gravitational Constant and c is the absolute Speed of Light.
Rs = 2GM/c²
all the rest is superfluous.(and the lame attempt at a derivation is risible...limiting maximum radius of a massive object indeed.)216.96.79.179 (talk) 16:01, 29 January 2016 (UTC)
thread_title:  This article is garbage. 

message: In the section Formula, the formula for potential energy, GMm/r, cannot be used. This formula assumes that the masses are constant. Obviously, relativistic mass is not. At best, it's an upper bound on the energy but would be too high to calculate the event horizon. Shawn H Corey (talk) 14:05, 15 June 2016 (UTC)
thread_title:  Wrong formula for potential energy 

message: This article appears to have been written by people who don't know any general relativity. The section titled "Formula" is at best a pseudo-derivation using Newtonian gravity, but it isn't labeled as such. I'm going to add a tag to say that it needs attention from an expert. It might actually make more sense to delete it, since the relevant information is contained in the article Black hole.--76.169.116.244 (talk) 22:37, 10 August 2016 (UTC)
replies: text: Replaced with link to existing Derivation article. NE Ent 17:34, 7 October 2016 (UTC)
thread_title: Needs attention from an expert, or deletion

message: a black hole doesn't choose to rotate, it fundamentally rotates, nothing has to initiate that rotation, that's the only way space-time works at these degenerate concentrations of energy. Some people claim allow some myth in order we test rare cases, but the point is that fake physics doesn't leed to solutions but mistakes.  — Preceding unsigned comment added by 2A02:587:4110:2B00:F0D5:204:77E2:3637 (talk) 19:08, 23 October 2016 (UTC)
thread_title:  fictional theory 

message: In this article, we should pay attention to the fact that the expression for the Schwarzschild radius r_s =2\,G\,M/c^2=2\,(G/c^3)\,Mc in form coincides with the integral form of the Einstein equations for a small region of spacetime: R_i=2\,(G/c^3)P_i=2\,(G/c^3)\,Mc\,U_i=r_s\,U_i, where R_i is the i-component of the Schwarzschild radius r_s; P_i is the four-momentum of matter; U_i is the four-velocity. See details  here in the second proof. 178.120.43.16 (talk) 08:20, 3 April 2017 (UTC)
thread_title:  The Schwarzschild radius and the Einstein equations 

message: I've just been writing a R(s) to-from Mass tool for my "toolbox spreadsheet", and checking it against the data here.
The Earth, for example, for a mass of 5.98E+24kg gives a R(s) of 0.00887m, which is right (9mm) ; Sun, 2E+30kg, 2968m (3km) ; 70kg human, 1.03895420951896E-25m ... which is twice the size given.
I've corrected the table by doubling the "human" data line for a 35 kg child and a 70kg adult. The simple factor-of-2 difference may make the point of the simple proportionality between mass and R(s).
I don't have independent mass data for SgrA*, or the Andromeda SMBH.
AKarley (talk) 11:07, 5 September 2017 (UTC)
replies: text: It's very interesting that the larger the mass of a black hole (or its progenitor object), the lower the density inside the hole. Most people naturally assume that if matter is compressed to the point where not even light can escape it, then it will end up at roughly the same density no matter how much matter there was in the progenitor. But instead, if a notional solar-mass black hole had the same high density as its notional earth-mass counterpart, the radius of the former would be in the range of 70 cm, not 3 km! 188.150.64.57 (talk) 05:21, 25 February 2023 (UTC)
thread_title:  Error in sizes table 

message: According to the Encyclopædia Britannica, the notation specifically for the Schwarzschild radius of an object is "Rg" (R sub g), not (r sub s).
Or am I stale in my understanding? Be kind.. Rodent of Unusual Size (talk) 07:13, 1 July 2019 (UTC)
thread_title:  Wrong notation? Shouldn't it be 'R sub g' instead of 'r sub s'? 

message: Schwarzschild radius has an abbreviation: Sch. R. 2601:580:7:7AE8:457C:E737:3CD4:D165 (talk) 16:01, 18 April 2020 (UTC)
thread_title:  Sch. R. is the Abbreviation for Schwarzschild Radius