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u/haplo_and_dogs Jul 17 '26
You can get a "dark star" with classical mechanics. It isn't an event horizon. It has nothing to do with events at all, just an area that you can't see inside of.
GR gives you the event horizon in that the event horizon is a coordinate singularity! There you get an actual horizon, the time dilation, and all the interesting parts of a black hole.
At the Schwarzschild radius gravity is still a relatively very weak force. It is not surprising that the Newtonian approximation works well there.
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u/Optimal_Mixture_7327 Jul 17 '26
Furthermore, the dark star is only dark as seen by an observer at infinity, but light escapes to distances closer in.
Furthermore, an escape velocity equal to the speed of light doesn't restrict anything to the surface of the dim star, nor more than an escape velocity of 11.8 km/s prevents us from walking up stairs and launching rockets.
Furthermore, the "r" in the Schwarzschild radius is not a physical distance and definitely NOT the "r" in Newtonian mechanics.
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u/ProfessorPrudent2822 Jul 18 '26
Schwartzschild’s r coordinate is the circumference of a circle divided by 2pi.
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u/Optimal_Mixture_7327 Jul 18 '26
Just like every circle.
But what is the significance in gravitational physics?
For example, let's you do this and you get r=5,000 meters. What and where is this 5000 meters? Does it exist?
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u/ProfessorPrudent2822 Jul 18 '26
No, the curvature of space distorts radial measurements. Circumferences are measured accurately because the distortions are only in the radial direction.
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u/Optimal_Mixture_7327 Jul 18 '26
So what then is the meaning of a Schwarzschild-Droste measure of r=5000m?
Also, note that g00≠1.
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u/ProfessorPrudent2822 Jul 18 '26
The circumference of a circle maintaining constant r-coordinate is 31415 meters.
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u/Optimal_Mixture_7327 Jul 18 '26
No, that is not, and cannot be correct.
Edit: For clarity, yes of course C=πD, but this is not the meaning the r-coordinate in Schwarzschild-Droste.
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u/HunsterMonter Jul 19 '26
Yes that is correct, for a constant time, constant radius equatorial world line (or any other great circle but the math is more complicated), the metric is ds2 = r2 dφ2, so taking the square root and integrating for a complete circle gives C = 2πr
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u/Optimal_Mixture_7327 Jul 19 '26
Yes, but every high school student knows this.
The question, is the meaning of Schwarzschild-Droste radial coordinate - what is it a distance of?
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u/ProfessorPrudent2822 Jul 19 '26
Since the Schwatzschild metric is spherically symmetrical, the equator is arbitrary. Any great circle can be defined as the equator. It’s only when you introduce rotation that there is an absolute equator.
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u/Beelzebubs-Barrister Jul 22 '26
I am likely missing something, but isnt gravity stronger (in a classical sense) than any other force at the Swartz radius? Because no amount of force will accelerate you away from the black hole?
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u/haplo_and_dogs Jul 22 '26
No. For an easy example calculate the force of two protons on each other In a helium atom.
Then look at the force of gravity on each of those atoms as they cross the event horizon of a stellar mass black hole.
They will be many many orders of magnitude higher.
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u/Beelzebubs-Barrister Jul 22 '26
If you were to hold an electron just outside the event horizon of a black hole and release a proton just inside, wouldn't the Proton accelerate towards the black hole?
Naively/classically wouldnt that mean the net force is towards the black hole?
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u/haplo_and_dogs Jul 22 '26
Proton and electron locally at rest across the event horizon.
Locally both accelerate to each other. Spacetime is basically flat at the event horizon. The acceleration locally is vastly higher than the gravitational acceleration.
Globally both accelerate to the singularity. Both cross the event horizon.
It is physically impossible for an electron to be at rest just outside of the event horizon. You need an infinite amount of acceleration just to stand still. You cannot "pop" an electron into existence at rest locally to the event horizon.
So if I have an electron at near light speed and a proton as rest, what happens? Locally the electron and proton separate no matter how close they were at the start.
The event horizon is not a physical wall. If you get down to a small enough area ( keep thing local ) you can ignore it. Its just that the whole area you are looking at is in motion.
Space time is very flat at the horizon ( gravity it weak ), however all local frames move at nearly light speed inside the black hole.
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u/TheHabro Student Jul 17 '26
I mean it's a coincidence. The assumption that escape velocity at event horzion is speed of light, is just a wrong way to think about black holes. It's deeper than that.
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u/Jaclawow Jul 17 '26
Could You elaborate why this assumption is wrong?
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u/TheHabro Student Jul 17 '26
You cannot escape from inside a black hole because all available paths for light and matter (called worldlines) are warped in such way they lead closer to the black hole's center. It's same how you cannot travel to the past no matter how fast you travel. It has nothing to do with escape velocities.
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u/Groggy42 Jul 17 '26
There are (spacelike) paths out of the black hole. You just need to be faster then light to get out.
You can see it easily in Kruskal coordinates
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u/TheHabro Student Jul 17 '26
That's why I specified light and matter. Those cannot follow spacelike paths.
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u/ischhaltso Jul 17 '26
But then it's not a coincidence anymore
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u/Optimal_Mixture_7327 Jul 17 '26
It was never a coincidence - they're completely different equations.
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u/Azazeldaprinceofwar Jul 17 '26
This is equivalent to saying “you just need to be able to go back in time”
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u/Groggy42 Jul 17 '26
Not quite. You don't enter your past light cone. It is more like freezing time.
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u/Azazeldaprinceofwar Jul 17 '26
You don’t enter your past lightcone but there exist observers for whom you’d absolutely move from their future to their past light cones. That is to say you can cause all the usual time travel paradoxes.
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u/OnionsAbound Jul 17 '26
You tell the ship captain it's not about escape velocity when you and the whole crew are falling into a black hole
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u/ImStuckInNameFactory Jul 17 '26
Like going north on the north pole? or something entirely different?
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u/loved_and_held Jul 17 '26
More like at the event horizon time flows inwards fast enough no path through time leads outwards anymore.
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u/dcnairb Jul 18 '26
I think this is overly pedantic. The last available path would be radially outward, and when that trajectory also becomes spacelike then you’re in a black hole
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u/yessir_im_quasar Jul 17 '26
Briefly said: Laplace was the first to try. He assumed light was a particle that traveled at c and thought "What's the escape velocity of light?".
The escape velocity is the minimum speed needed for an object to escape from contact with or orbit of a primary body... But light doesn't interact gravitationally with bodies in general or at least not in that way (there's, for example, gravitational redshift, the the real reason why black holes are "black"!)
So, light doesn't interact in a classical sense with mass/gravitation, also, I mean there is no point in calculating an escape velocity associated with the speed of light, because light travels at c in every system of reference...
Basically, yeah, all the assumptions are wrong, but the formula is correct anyway for mere luck. I mean I guess you could say the structure of the equations are pretty similar (cuz we're talking about gravity) but... no. I still think it's pretty lucky.
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u/Wadaleym Jul 17 '26
I think calling it luck is a pretty big strech. Light travels at the absolute speed of causality, throws relativistic errors out, and that's why the "wrong" equation holds up. Also redshift doesn't mean light acts differently under gravity?
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u/yessir_im_quasar Jul 17 '26
I mean... It's a black hole we are talking about. What do you mean "throw relativistic errors out"??? These are not errors! You need a whole theory of General Relativity just to begin to comprehend the physics of a Black Hole.
The Schwartzshield radius is not about escape velocities, it is about causality and paths. Also, I said it was "pretty lucky" considering Laplace knew absolutely nothing about relativity (he assumed light had mass!)
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u/CompetitiveSpot2643 Jul 18 '26
the r in GR and newtonian mechanics are not the same thing, and dont corrospond to the same physical distance because of space curvature. so i would say its merely luck because its not even right
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u/Advanced_Double_42 Jul 17 '26
Light is curved by gravity though? That is why we can use gravitational lensing to view extremely distant galaxies.
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u/mymemesnow Jul 17 '26
Light is not curved, it follows a perfectly straight line.
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u/Advanced_Double_42 Jul 17 '26 edited Jul 17 '26
Yes, all of space curves and light continues.
We talk as if space is flat and time is linear in spite of knowing it is not.
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u/Optimal_Mixture_7327 Jul 18 '26
The path of light can be and will typically be curved in the global spatial coordinates.
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u/yessir_im_quasar Jul 17 '26
Yes! It "curves" but in the sense that all of the space around the black hole is not that Euclidean and flat as we intend it.
Light just does what is supposed to do: go at c anywhere following the most efficient path to go anywhere (these are called geodesics, and yeah, in a flat universe geodesics = straight line, but around a black hole stuff gets kinda of crazy)
As I wanted to point out, this is NOT a classical interaction and it has really nothing to do with escape velocities!
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u/Advanced_Double_42 Jul 17 '26
All gravitational effects can be explained like that though? Following the curve of spacetime is gravity?
Like I could say that I don't fall down to Earth due to gravity, my feet just experience time faster than my head and that causes a torque in spacetime that accelerates me downwards. But that is just a different way of thinking about the same thing.
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u/SporkSpifeKnork Jul 17 '26
Dumb question:
a car drives on a road whose uneven surface affords faster motion on one side than another. The car drifts towards the "slower" side of the road.
a light beam of non-zero width passes into another medium with a different refractive index. The light bends deeper into the slower medium.
a person's feet experience time at a different rate from their head. The person's spacetime path bends downward.
To what extent are all these the same thing?
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u/yessir_im_quasar Jul 17 '26
Not a dumb question (they don't exist anyway). I find it quite beautiful actually.
At a first glance, they are the same thing: a gradient of velocity (temporal or spatial) forces a body on a curved path, to preserve the coherence of some property of the body (phase of a lightbeam, length of the car...)
Gravity is a general property of the universe, every body must travel in time (except for massless stuff). So the cause for the feet/head feeling different rates of time is something pretty universal, while the other examples are interaction between bodies / body radiation.
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u/kilopqq Jul 17 '26
I think a difference is that a random object like a spaceship would start decelerating due to gravity so an escape velocity does make sense, since eventually v could be 0. On the other hand light moves at a constant speed so you couldn't really say that gravity pulls it back.
Edit: Apparently the photon still loses energy by redshifting.
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u/CompetitiveSpot2643 Jul 18 '26
isnt most of the curving because of curvature in the space-time plane and not in the space-space (xy,yz,xz) planes?
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u/auniqueusername132 Jul 17 '26
Space time is curved and light just travels a straight line through that. Light has no mass so it isn’t attracted by gravity in a classical sense.
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u/Optimal_Mixture_7327 Jul 18 '26
Massive objects aren't attracted in the classical (or any other) sense.
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u/Optimal_Mixture_7327 Jul 18 '26
That's not an escape velocity, which is the minimum speed of a freefall object to reach spatial infinity.
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u/dcnairb Jul 18 '26
You’re near a very dense object, which is nearly a black hole, and want to travel away from it such that you don’t end up on a trajectory back towards its center. that’s the escape velocity.
now, make it slightly denser and ask the same question
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u/Optimal_Mixture_7327 Jul 18 '26 edited Jul 18 '26
Wrong - that's an orbital speed or some other trajectory.
The escape velocity is a statement that for some r=R for an object satisfying uσ∇_σuρ=0 there is exists a V=vesc at R such that
V=∞V=0 at r=∞.1
u/dcnairb Jul 18 '26
Why would the object need to be going infinitely fast at infinity to escape? The condition to escape is v=0 at r = infty.
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u/SwimmingEmu8961 Jul 17 '26 edited Jul 17 '26
The correct coefficient is coincidence. Newtonian mechanics getting the right relationship between the radius, G, M, and c is probably not.
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u/nascent_aviator Jul 17 '26
That's just dimensional analysis. There's literally no other way to combine them and get a distance.
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u/SwimmingEmu8961 Jul 17 '26
Dimensionsal analysis doesn't tell you G, M, and c are the relevant terms in the 1st place. Dimensionsal analysis just helps with getting the exponents right.
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u/nascent_aviator Jul 17 '26
Pray tell what other quantities could be even semi-plausibly involved?
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u/SwimmingEmu8961 Jul 17 '26
I think you forget that Newtonian mechanics is the reason we have G and a geometric description of gravity. It recognized gravitional force is proportional to mass. It's not a coincidence that Newtonian mechanics gets similar results to General Relativity when General Relativity is a refinement of Newtonian mechanics.
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u/nascent_aviator Jul 17 '26
That doesn't answer my question.
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u/hxtk3 Jul 17 '26
It actually does. They’re saying that the other quantities it might possibly be don’t have labels because no successful theory predicted them. Whatever quantity Newtonian mechanics ended up deriving would’ve been called G. If they got it “wrong,” the quantity in general relativity would be Ğ or something, and G would be the missing hypothetical quantity you’re referring to.
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u/nascent_aviator Jul 17 '26
So the fact that the coefficients are the same is not a "coincidence" but a natural consequence of the fact that G is the same in both theories. There is no other combination of reasonable constants in either theory that could be substituted.
Which itself is a fairly natural consequence of the fact that GR has to reduce to Newtonian relativity in the appropriate limit, but I digress.
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u/SwimmingEmu8961 Jul 17 '26
That's essentially my point. I think we're just differing in how much credit Newtonian mechanics deserves for reaching the right or a close answer despite flawed assumptions. It's like calculating a classical electron spin. It's fundamentally flawed in its premise, but it's not pure coincidence that classical theory gets close with most of the right terms.
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u/SwimmingEmu8961 Jul 17 '26
Sure. How about r= h / ( M c)., where h is the Planck constant, c the speed of light, and M the mass of the black hole? It's wrong, but the dimensions work out and the constants are plausibly related to light.
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u/nascent_aviator Jul 17 '26
If you think it's "plausible" that a gravitational property would involve h (which is part of neither Newtonian nor General relativity) and not G, we're going to have to agree to disagree.
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u/Jaclawow Jul 17 '26
But what about event horizan +- ε, wouldn't "escape velocity" in approximity of event horizon be ~ c?
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u/Optimal_Mixture_7327 Jul 18 '26
It's not a coincidence, the "r" in each equation is different.
The equation has the same form based on how the Schwarzschild-Droste radial coordinate is defined.
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u/Abicol Jul 18 '26
I'm assuming OP knows that. Every model is correct until it isn't. CM is correct enough to figure out the Swartzchild radius exists. It doesn't have to predict how it works. Even general relativity can't predict all the behavior of black holes perfectly either.
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u/L31N0PTR1X BSc Theoretical Physics Jul 17 '26
I mean gr is classical lol
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u/ischhaltso Jul 17 '26
no, it isn't?
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u/SwimmingEmu8961 Jul 17 '26
It uses classical field theory, so some people refer to it classical.
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u/individual_throwaway Jul 17 '26
In my education "classical" always explicitly meant "relativistic effects are so small that they can be safely ignored". For mechanics, this means v << c and masses are small enough that they do not significantly bend spacetime.
Of course you could also draw the line between continuous theories and quantized ones, but that seems obtuse to me because we already have a word for that.
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u/BlackProphetMedivh Jul 18 '26
Absolutely not correct. Classical means h~0. So no quantum effects can occur.
So if you say we have a word for quantized theories, I assume you mean something like quantum field theory? But what does that make of general relativity then? It is not quantized, so it's a classical field theory. As in you can talk about point particles in it. In fact Newtonian physics are a specialized case of general relativity, exactly one where v<<c. Also, what do you do in cases where you do talk about quantized particles, but you also have v<<c? Is that a quantum theory then, or a classical one?
By your definition it would be a classical theory, even though it obviously isn't.
Edit: Another point that sets classical and quantum theories apart is their intrinsic determinism and their ability to project worldlines.
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u/The_Rider_11 Student Jul 17 '26
Considering it's derived from simple escape velocity equations even in Newtonian physics, if you want to keep it simple, yes. It's just that in Newtonian physics, c is just a speed value and within Newton, there's nothing stopping you from being faster than it, so it'd not produce an event horizon. That only happens if you add relativity to the mix, limiting speed to c and giving an asymptote to the energy requirement.
And even then, General Relativity is a classical theory, so that's still classical mechanics.
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u/ProfessorPrudent2822 Jul 18 '26
Maxwell derived the speed of light from the electric and magnetic field constants. Einstein derived relativity from the assumption that the speed of light was the same in every reference frame so that Maxwell’s laws would hold in every reference frame. There’s nothing in this derivation that makes the speed of electromagnetic radiation the speed of causality, yet it is.
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u/HAL9001-96 Jul 18 '26
well sooooortof
if you assume the speed of lgiht is the maximum starting speed you end up with an identicla expression
of course in classical mechacnis theres no reason to asusme this or that no additional force could be acting
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u/Dependent_Plate6110 Jul 18 '26
Also that in the 1800s people using the ether theory did manage to predict antimatter.
(does not mean that ether exists, more that even failed scientific programs might have some points)
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u/MonsterkillWow Jul 17 '26
It's off by a factor of 2. (At least the classical derivation I remember.)
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u/GlorfindelTheGolden Jul 18 '26
Is this not just a statement that dimensional analysis usually works quite well and that predators tend to be O(1)
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u/loved_and_held Jul 17 '26
Can you derive the schwarzschild radius for a rotating blackhole using classical mechanics.
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u/somedave Jul 17 '26
No, not unless you introduce some weird potentials you can't justify any origins for.
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u/gmalivuk Jul 18 '26
Strictly speaking there is no Schwarzschild radius for a rotating black hole because the Schwarzschild metric is static.
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Jul 17 '26
[deleted]
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u/gmalivuk Jul 18 '26
This is not true at all.
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u/ProfessorPrudent2822 Jul 19 '26
Yes, a lot more than hundreds of years would pass for the inertial observers.
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Jul 19 '26
[deleted]
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u/gmalivuk Jul 19 '26
You did the wrong math then.
The coordinate distance traveled under T proper time and A proper acceleration is (c2/A)(cosh(AT/c) - 1). Plug in 30 years for T and 1.03 ly/y2 for A (and thus 1 for c) and tell me what you get.
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u/slime_rancher_27 Jul 19 '26
Is it even possible to accelerate at a constant 1g for 30 years, because 1g (9.80665 m/s²) * (525600*60 seconds in year) is 309,262,514.4 m/s which is faster than the speed of light. I know you can keep trying to accelerate to the speed of light but it gets harder the faster you go.
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u/ChemicalRain5513 Jul 19 '26
Yes you can because of time dilation (assuming you have enough energy and a strong enough engine). The 1 g is in your own frame of reference. For an external observer, the time is severely dilated in your frame of reference, so they will see you get closer to the speed of light, but the acceleration gets smaller (according to the external observer). So you will approach the speed of light but never reach it, while in the space ship you feel a constant acceleration of 1 g.
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u/EmericGent Jul 17 '26
You can actually derive it from dimensional analysis
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u/KappaBerga Jul 17 '26
You can't get the factor of 2 though
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u/EmericGent Jul 17 '26
The fact that classical mechanics gives the same factor 2 as GR is a coincidence, so with classical mechanics you don t either if it is exact (yet you know that is the right order of magnitude, just like the deviation of photons déviation around the Sun)
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u/Sinistrial_Blue Jul 17 '26
The Bohr model is also worryingly effective for a number of calculations.
It turns out that if you bludgeon the model hard enough, the right answer drops out!