# Is the Schwarzchild Radius Relativistic or Newtonian?

• I
I noticed that the Schwarzchild Radius Formula and the Escape Velocity Formula are actually identical. The Schwarzchild Radius is supposed to be one of the great equations generated from Relativity, while the Escape Velocity is something that was generated just using Newtonian gravity. All you need to do is plug in the speed of light as your escape velocity and you get the Schwarzchild Formula. My question is, historically speaking was Schwarzchild actually derived from Relativity or not?

r = (2 G M)/c^2 |
M | mass
G | Newtonian gravitational constant (≈ 6.674×10^-11 m^3/(kg s^2))
c | speed of light (≈ 2.998×10^8 m/s)

Escape Velocity:
v = sqrt((2 G m)/r) |
v | speed
m | mass
G | Newtonian gravitational constant (≈ 6.674×10^-11 m^3/(kg s^2))

Dale
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historically speaking was Schwarzchild actually derived from Relativity or not?
Yes. It was actually derived from relativity.

The Schwarzchild Radius is supposed to be one of the great equations generated from Relativity, while the Escape Velocity is something that was generated just using Newtonian gravity.
A Newtonian surface with escape velocity c does not behave like a Schwarzschild event horizon.

• Ibix
Yes. It was actually derived from relativity.

A Newtonian surface with escape velocity c does not behave like a Schwarzschild event horizon.
Then why is the standard escape velocity formula identical to it? Just replace the v with c in the escape velocity equation, and you got the Scwarzchild Radius equation.

Orodruin
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Then why is the standard escape velocity formula identical to it? Just replace the v with c in the escape velocity equation, and you got the Scwarzchild Radius equation.
Numerical coincidence. It clearly cannot depend on the variables in any other way due to dimensionality (there is no other combination that leads to a length), so the only question is about the prefactor, which happens to be the same.

• Ibix
Numerical coincidence. It clearly cannot depend on the variables in any other way due to dimensionality (there is no other combination that leads to a length), so the only question is about the prefactor, which happens to be the same.
That would be quite the coincidence, considering that the two equations are meant for basically the same purpose and come from a related field in physics!

I'm wondering how Karl Schwarzchild originally derived his equation? And who originally derived the Escape Velocity equation, and how did they derive it? The Schwarzchild Radius at best might make use of Special Relativity (i.e. the speed of light limit), but it doesn't look like it requires anything from General Relativity at all.

Ibix
All you need to do is plug in the speed of light as your escape velocity and you get the Schwarzchild Formula. My question is, historically speaking was Schwarzchild actually derived from Relativity or not?
It's called the Schwarzschild radius because it falls out of Karl Schwarzschild's first solution to Einstein's equations.

According to Wikipedia, "dark stars" with escape velocity too high for light to escape were discussed as early as 1783 (https://en.m.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics)). They are very different from black holes. Light does escape their surface but it doesn't make it to infinity, whereas light cannot cross the event horizon of a black hole. I don't think the idea is completely coherent, since we don't have a proper non-relativistic theory of light. So, as Orodruin says, the fact that the numbers match is just a (not terribly huge) coincidence.

Interestingly, they seem to have immediately twigged that if such dark stars existed some of them would be in binary systems and proposed looking for unexplained wobbles in the paths of visible stars. That's how we look for non-luminous companions now.

• rrogers and Dale
According to Wikipedia, "dark stars" with escape velocity too high for light to escape were discussed as early as 1783 (https://en.m.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics)). They are very different from black holes. Light does escape their surface but it doesn't make it to infinity, whereas light cannot cross the event horizon of a black hole. I don't think the idea is completely coherent, since we don't have a proper non-relativistic theory of light. So, as Orodruin says, the fact that the numbers match is just a (not terribly huge) coincidence.
But the maximum distance the light could make it out of such "Dark Stars" would actually be exactly the same distance as the radius of an event horizon of a black hole. So again, the exact same numbers are arrived at from the two equations of separate origins.

Ibix
That would be quite the coincidence, considering that the two equations are meant for basically the same purpose and come from a related field in physics!
It's not a huge coincidence. As Orodruin says, both have to be a constant times ##GM/c^2## or the units don't work out.
I'm wondering how Karl Schwarzchild originally derived his equation?
It's derived from assuming a spherically symmetric vacuum solution to Einstein's field equations.
And who originally derived the Escape Velocity equation, and how did they derive it?
Newton, by conservation of energy so far as I am aware.
The Schwarzchild Radius at best might make use of Special Relativity (i.e. the speed of light limit), but it doesn't look like it requires anything from General Relativity at all.
Special relativity works only when there is no gravity. So - no, you can't get any gravity-related results from special relativity. Attempts were made to patch Newtonian gravity into SR between 1905 and 1916, but were never completely successful. And Eddington's eclipse measurements of gravitational lensing in 1921 nailed the coffin lid shut.

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Ibix
But the maximum distance the light could make it out of such "Dark Stars" would actually be exactly the same distance as the radius of an event horizon of a black hole.
That is not what an escape velocity means. Something with escape velocity can escape to infinity. Something with less velocity goes up but returns, like a ball thrown straight up. Light was hypothesised to leave "dark stars". It just wouldn't get all the way away, so you'd see them glowing if you were close enough. Light cannot cross the event horizon of a black hole full stop (edit: not outward, anyway).

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• Dale and m4r35n357
Orodruin
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It's not a huge coincidence. As Orodruin says, it has to be a constant times ##GM/c^2## or the units don't work out.

Let me expand on this a bit. Apart from the functional form, there could be a numerical overall constant. This would typically be expected to not be too far from one or a fraction of numbers not too far from one. At worst it would have some geometrical factor involving pi or a square root. There really are not that many options for these expectations.

As it happens, the factor comes out to be two both in the case of the event horizon and the escape velocity, which may be the most common factor appearing in this type of consideratiobs apart from one.

PeroK
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That would be quite the coincidence, considering that the two equations are meant for basically the same purpose and come from a related field in physics!

Well, both theories are trying to explain the same physical reality. You are bound to have the same approximate equations in a number of cases. It's something of a coincidence that you get exactly the same equation for escape velocity, but it's not that big a deal.

Dale
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Then why is the standard escape velocity formula identical to it?
Coincidence.

Actually they are not really identical because the r coordinates have different meanings in curved spacetime and in flat spacetime. So they have the same form but refer to different things. This happens all the time in physics.

the two equations are meant for basically the same purpose
This is incorrect. As I said in post 2 a Newtonian surface with escape velocity c is not at all the same as a Schwarzschild event horizon

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• Ibix and PeroK
stevendaryl
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The relationship between the General Relativistic derivation of the Schwarzschild radius and the Newtonian derivation is kind of interesting.

Here is an interesting fact: If you have a point-mass of mass ##m## whose velocity is directed radially away from a large spherically symmetric mass ##M##, then you have the following equation of motion for the object's position ##r##: (in the limit as ##m \ll M##):

EOM: ##E = \frac{1}{2} m (\dot{r})^2 - \frac{GmM}{r}##

where ##E## is a constant of the motion. (Note: the presence of ##m## on the right-hand side is irrelevant, but is just thrown in so that it looks like a Newtonian energy equation. The equations of motion are independent of ##m##, if you fix ##E/m##).

This equation is exact, not an approximation, in two very different cases:
1. Newtonian physics, where ##E## is interpreted as the total Newtonian total energy ##E_N = \frac{1}{2} mv^2 + U##, and ##\dot{r}## means the derivative with respect to coordinate time ##t##.
2. General Relativity, where ##\dot{r}## means the derivative with respect to proper time ##\tau##, and ##r## is the Schwarzschild radial coordinate.
In the General Relativity case, the interpretation of ##E## is a little complicated. It's actually:

##E_{rel} = \frac{c^2}{2m} ((p_0)^2 - m^2)##

where ##p_0 = \frac{dt}{d\tau}## the ##t## component of the 4-momentum ##p^\mu = m \frac{dx^\mu}{d\tau}##. ##p_0## is conserved in any spacetime in which the metric is independent of ##t##.

Without knowing any of the details about solutions to Einstein's field equations, one should guess that EOM is approximately true, when ##\dot{r} \ll c## and when ##r## is large. That's because we know that Newtonian gravity is approximately true (in the appropriate coordinate system) when gravitational fields are weak and when objects are moving slowly compared to the speed of light (so that ##\frac{dr}{d\tau} \approx \frac{dr}{dt}##).

What's complicated to show is that EOM is exact when using Schwarzschild coordinates, even when ##\dot{r}## is comparable to the speed of light, and when gravitational fields are strong.

But given EOM, the Schwarzschild radius is easily derivable. To escape to infinity, the object must have ##E \geq 0##. So the limiting case is when ##E = 0##, which implies that: ##\dot{r} = \sqrt{\frac{2GM}{r}}##. Since ##\dot{r} < c##, it follows that escape is only possible if ##r > 2GM/c^2##.

So the odd and wonderful fact about the Schwarzschild solution is that a very careful, rigorous derivation gives exactly the same answer as an approximate, heuristic argument. It's a little like how Bohr's heuristic quantization procedure gives the same answer for hydrogen's energy levels as Schrodinger's equation.

• DanielMB, Sorcerer, bbbl67 and 2 others
martinbn
May be it is good to keep in mind that ##r## may be the same letter, but what it stands for are two different things in GR and in Newtonian gravity.

• Dale
Ibix
To pick nits, the relativistic definition ##r=\sqrt{A/4\pi}## actually does give you the Newtonian ##r## when applied to a Euclidean space. It isn't the distance to the centre in general in GR, of course, even in cases where that's defined.

• bbbl67
The relationship between the General Relativistic derivation of the Schwarzschild radius and the Newtonian derivation is kind of interesting.
To add a tiny bit to what you have said for the OP's benefit, the radial free-fall times (from EOM) are identical if the GR time is regarded as proper time (as it would be for an EOM!). The coordinate time relationship is of course very different.

The relationship between the General Relativistic derivation of the Schwarzschild radius and the Newtonian derivation is kind of interesting.
Now, that's a good explanation! I'll have to go through the details to understand it a bit more later, but I understood the gist of it.

Has anyone else noticed the extreme similarity between the Newtonian and Relativistic equations before? I mean historically, not among the denizens of this forum of course. LOL!

Dale
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Has anyone else noticed the extreme similarity between the Newtonian and Relativistic equations before?
Sure. It is well known. I am not aware of any professional scientific reference that thinks it is more than an interesting coincidence.

Ibix
Has anyone else noticed the extreme similarity between the Newtonian and Relativistic equations before?
Chapter 7 of Sean Carroll's GR lecture notes has a comparison of massive and massless particles in GR and Newton. Note that the status of massless objects in Newtonian gravity is rather dubious. In Carroll's analysis of this approach they turn out to be unaffected by it.

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stevendaryl
Staff Emeritus
May be it is good to keep in mind that ##r## may be the same letter, but what it stands for are two different things in GR and in Newtonian gravity.

Well, in the nonrelativistic, weak field limit, they are the same.

martinbn
Well, in the nonrelativistic, weak field limit, they are the same.
Well, in that limit the two theories are the same. But we are talking about a black hole.

• Dale
stevendaryl
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Well, in that limit the two theories are the same. But we are talking about a black hole.

Well, I just don't think that it is actually meaningful to say that ##r## means something different in the two theories. In the regions where the two theories are similar, they mean the same thing. In the regions where the theories are completely different, I don't know what it would mean to say that it's the same coordinate or not.

martinbn
Well, I just don't think that it is actually meaningful to say that ##r## means something different in the two theories. In the regions where the two theories are similar, they mean the same thing. In the regions where the theories are completely different, I don't know what it would mean to say that it's the same coordinate or not.
Well, the same later is used in different contexts, there is a possibility (I think that is often the case) that misleading intuition is used.

PeterDonis
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I just don't think that it is actually meaningful to say that ##r## means something different in the two theories.

Sure it does. In Newtonian gravity, ##r## is the physical distance from the center. In GR in Schwarzschild coordinates, ##r## is the areal radius. Both of these are physically meaningful quantities, and conceptually it is certainly meaningful to consider the question of whether or not they are the same. In Newtonian gravity the two are the same; in GR they are not (and for a black hole "physical distance from the center" isn't even well-defined).

• martinbn and Ibix
Dale
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Well, in the nonrelativistic, weak field limit, they are the same.
Right, but that limit requires an object much larger than its event horizon, which defeats the whole point of the thread.

In the regions where the theories are completely different, I don't know what it would mean to say that it's the same coordinate or not.
This is related to the point I was trying to get across to the OP. A Newtonian surface with escape velocity c is not comparable to a Schwarzschild event horizon. The latter is a region that simply has no analogy in Newtonian gravity.

• martinbn
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Well, I just don't think that it is actually meaningful to say that ##r## means something different in the two theories.
I'd say it is just a rephrasing of the statement that "space is flat" in Newtonian gravity (in the Newtonian limit the spatial velocity decouples from the spatial curvature), while in GR it is (in general) not. The comparison can be made even more by restating Newtonian gravity in a general-covariant way.

stevendaryl
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This is related to the point I was trying to get across to the OP. A Newtonian surface with escape velocity c is not comparable to a Schwarzschild event horizon. The latter is a region that simply has no analogy in Newtonian gravity.

Well, the issue was whether it's a "coincidence" that you can use Newtonian physics to calculate the radius ##r## such that not even light can escape if it starts at a smaller radius than that. There is one part of it that is definitely not a coincidence, and there is another part that is, sort of a miracle that it works.

In both cases (Newtonian gravity and GR with a spherically symmetric spacetime) the equation of motion for a test mass can be written as a potential problem:

##E = \frac{1}{2} m^2 (\frac{dr}{ds})^2 - \frac{GMm}{r}##

where ##E## is a constant of the motion, and ##r## is a radial coordinate, and ##s## is a path parameter.

Given the fact that the equations are the same, it's no coincidence that the Newtonian calculation gives the right answer.

But the fact that that equations are exactly the same in the two cases is something of a miracle.

• bbbl67
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But the fact that that equations are exactly the same in the two cases is something of a miracle.
Or a consequence of using coordinate systems in which the equations are the same....

What happens to the similarity if the central mass has significant angular momentum?

A.T.
In Newtonian gravity, ##r## is the physical distance from the center.
You could just as well state this as:

In Newtonian gravity, ##r## is the areal radius.

because as you say:

In Newtonian gravity the two are the same

PeterDonis
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You could just as well state this as

No, because the fact that the physical distance from the center happens to be the same in Newtonian mechanics as the areal radius does not mean that ##r## is defined as the areal radius. ##r## is defined as the physical distance from the center; the fact that this happens to be equal to the areal radius is a theorem that has to be proved.

A.T.
No, because the fact that the physical distance from the center happens to be the same in Newtonian mechanics as the areal radius does not mean that r is defined as the areal radius. r is defined as the physical distance from the center;
Newtonian mechanics assumes flat space, so in its context the two definitions are equivalent and interchangeable. GR doesn't assume flat space, so it has to be more specific and differentiate between the two.

Dale
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Well, the issue was whether it's a "coincidence" that you can use Newtonian physics to calculate the radius rrr such that not even light can escape if it starts at a smaller radius than that.
But that is not what a surface with escape velocity c is. Light can easily escape that surface, as can material objects.

Escape velocity is not the velocity required to escape that radius, it is the velocity required to ballistically escape to infinity from that radius. Light could escape the Newtonian radius as could rockets and even bullets. Light and bullets wouldn’t make it to infinity, they would orbit, but a rocket could even make it to infinity.

The Newtonian surface has none of the interesting properties of an event horizon. The form of the equation is coincidentally the same. That’s it.

stevendaryl
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But that is not what a surface with escape velocity c is. Light can easily escape that surface, as can material objects.

Escape velocity is not the velocity required to escape that radius, it is the velocity required to ballistically escape to infinity from that radius. Light could escape the Newtonian radius as could rockets and even bullets. Light and bullets wouldn’t make it to infinity, they would orbit, but a rocket could even make it to infinity.

The Newtonian surface has none of the interesting properties of an event horizon. The form of the equation is coincidentally the same. That’s it.

I disagree completely. If we have the equations of motion:

##E = \frac{1}{2} m (\dot{r})^2 - \frac{GMm}{r}##

we can easily sketch the motion, qualitatively: If ##E < 0## and ##\dot{r}## is initially positive, then ##r## will increase up until the point where

##r_{max} = -\frac{GMm}{E}## (that's a positive number, since ##E < 0##.

Then it will turn around, and ##\dot{r}## will be negative until it falls into singularity at ##r = 0##. Note that ##r_{max}## will typically be greater than the Schwarzschild radius, ##\frac{2GM}{c^2}##. So it's not true, at least not based on the Schwarzschild geometry alone, that nothing can rise above the event horizon.

Now, if you study this "orbit" from the point of view of the Schwarzschild geometry, what you'll see is that, in terms of the Schwarzschild coordinates ##r, t##, the initial rise above the event horizon occurs at a time that is not covered by the Schwarzchild coordinates, as does the fall into the event horizon. But speaking informally, the rise happens at an event that has Schwarzschild time ##t = - \infty##, while the fall happens at an even that has Schwarzschild time ##t = + \infty##. From the point of view of an observer stationary from the black hole, the test particle takes forever to reach its maximum height, then turns around and takes forever to dip below the event horizon again (although both events happen at finite proper time).

Typically in studying black holes, the portion of the Schwarzschild geometry with ##t < 0## is considered an unphysical "white hole", and the real black hole geometry only includes the section ##t > 0##. That makes sense, because black holes aren't actually eternal, but are formed from the collapse of massive stars. So there is no actual black hole prior to that, and so no white hole geometry in the limit ##t \rightarrow -\infty##. But as far as the Schwarzschild metric is concerned, you can include the white hole region for the sake of studying solutions of the field equations and what the paths of test particles look like. In the complete spacetime that includes both the ##t >0## and ##t < 0## regions, the meaning of escape velocity for a test particle is very much analogous to what it is in Newtonian physics.

• bbbl67
PeterDonis
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Newtonian mechanics assumes flat space, so in its context the two definitions are equivalent and interchangeable. GR doesn't assume flat space, so it has to be more specific and differentiate between the two.

Yes, which means that if we are discussing a comparison between Newtonian mechanics and GR, which we are in this thread, we have to differentiate between the two. ##r## is actually defined as the physical distance from the center in Newtonian mechanics, so that's the relevant definition for this discussion.

Dale
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If E<0E<0E < 0 and ˙rr˙\dot{r} is initially positive
This is another difference. In GR ##\dot r## cannot be initially positive below the horizon. In Newtonian gravity it can.