# I The Singularity and the Schwarzschild radius

#### stg213

Summary
if the singularity appears when space-time reaches ifninity how can this 'grow further' ?
A singularity would be:
a location in spacetime where the gravitational field of a celestial body is predicted to become infinite by general relativity in a way that does not depend on the coordinate system. (wiki)

If the threshold to get a singularity is reached then space-time curvature becomes infinite ->

2GM/c^2

=> the more mass the singularity has the bigger the black hole

Now my question may be naive but: how can more mass curve space-time beyond infinity ?

Let's suppose there is a minimal stable singularity (doesn't matter how small) => the particular curvature of a singularity appears (according to relativity this is already infinite)

Given that we know black holes vary in size and gravitational strength this would imply:

a) variable 'infinite' curvature
b) that curvature is not infinite within the singularity
c) that the relationship between mass and space-time geometry is flawed

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#### Dale

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Consider the Dirac delta function $\delta(x)$. At x=0 it is infinite. At x=0 the function $2\delta(x)$ is also infinite.

In what sense is $2\delta(x)>\delta(x)$?

#### Nugatory

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(It’s not completely clear from your post, but I’ll assume that you understand that the “singularity” at the Schwarzschild radius is just an artifact of the coordinates and you’re talking about the real singularity at $r=0$)

That definition of a singularity as a point where the curvature is infinite is misleading. This may be one of the places that Wikipedia is repeating a common oversimplification, or you may have misled yourself by combining four different Wikipedia articles written by different people who never expected the articles to form a single coherent presentation.

Strictly speaking, the singularity isn’t a point in spacetime at all, sort of like how the hole in the bottom of a funnel isn’t part of the surface of the funnel. Thus, the infinities we find when we try calculating the curvature at $r=0$ aren’t telling us that the curvature is infinite there, they’re telling us that we’re using a formula where it doesn’t apply. We can calculate the curvature for non-zero values of $r$ arbitrarily close the singularity, and we will find that the curvature increases without limit as $r$ get smaller - but it’s always finite.

#### Dale

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we will find that the curvature increases without limit as r get smaller - but it’s always finite
Hmm. I don’t agree with this. Increases without limit is what infinite means. It means that there is no finite number which is the maximum. If it is larger still than any finite number, then it is infinite. Yes, the value at any finite r>0 is finite, but that value is not the biggest value.

#### Orodruin

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Hmm. I don’t agree with this. Increases without limit is what infinite means. It means that there is no finite number which is the maximum. If it is larger still than any finite number, then it is infinite. Yes, the value at any finite r>0 is finite, but that value is not the biggest value.
He is not talking about the limit, he is talking about the value at any finite $r$. For all $r$, the curvature is finite. $r = 0$ is not part of the Schwarzschild spacetime.

#### stg213

First, seeing all the disagreement, i'm glad that my question isn't totally dumb :D

I'll try to respond to all points raised:

The dirac delta function example is in a way just a fancy way of saying:

2∞ > 1∞ (which mathematically may make sense but in this case isn't exactly useful as we are talking about finite black holes with finite mass and finite event horizons and finite gravitational effects)

I know that wasn't the point of the example and I had to do a bit of digging to get to the bottom of it but the step function is meant to describe a spike... you have nothing.... nothing... nothing... then suddenly a singularity (something that tends to infinity) then goes back to nothing.

Secondly:
What i meant was indeed r=0.

From all my knowledge (and my knowledge may indeed be very faulty) singularities are are all points of r=0. I.e. a point of infinite density, infinite space-time curvature but somehow of finite mass.

I will again use wiki as it is a convenient way to quote:
While in a non-rotating black hole the singularity occurs at a single point in the model coordinates, called a "point singularity", in a rotating black hole, also known as a Kerr black hole, the singularity occurs on a ring (a circular line), known as a "ring singularity".

so the singularity is a 1-dimensional-point OR a 2 dimensional circle. Onward oversimplifying -> That means that in that 1 dimensional point or 2 dimensional circle 'all mass is contained' beyond it there is 'nothing' up to the event horizon. i.e. a point 'where all the stuff is'... then the space-time curvature bends outward to create a sphere around that point. But in my understanding mass isn't spread within the event horizon but all concentrated in that one point/ring.

What bugs me about this description is the 'space-time geometry' singularity caused by finite mass that in turn causes variable size black holes (size of the event horizon, not of the point singularity) with variable gravitational effect. Something like a variable infinity causing finite variance... something here doesn't work (or i'm too dumb to get it).

PS: by the above understanding i would expect all black holes to be of the exact same size/mass/gravitational effect regardless of what went into making them -> one infinity causing one effect (this is obviously not the case)

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#### Nugatory

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I will again use wiki as it is a convenient way to quote.....
"While in a non-rotating black hole the singularity occurs at a single point in the model coordinates"
As I suggested in my previous post, you do want to approach wikipedia with some caution. That statement is not quite wrong, but without context including a thorough discussion of the relationship between coordinates and points in a manifold it is easily misunderstood. And of course there's no reason to expect to find that context in articles trying to explain the difference between Kerr and Schwarzschild spacetimes.

Note the qualifying "point in the model coordinates". That's a pretty strong hint that it's not describing a point in spacetime, but rather a particular set of $(t,r, \theta, \phi)$ coordinate four-tuples, namely the ones for which $r=0$ and which map to no point in spacetime.

#### Orodruin

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2∞ > 1∞ (which mathematically may make sense but in this case isn't exactly useful as we are talking about finite black holes with finite mass and finite event ho
On the contrary, it is very applicable here. There is no finiteness to the singularity. By your same argumentation, you could not have point charges with different charge in classical electrodynamics. The delta function is at the heart if this, since it is useful in describing the charge density of a point charge.

What i meant was indeed r=0.
This is not part of the Schwarzschild spacetime. It is also a space-like singularity, more related to a moment in time than to a place in space.

What bugs me about this description is the 'space-time geometry' singularity caused by finite mass that in turn causes variable size black holes (size of the event horizon, not of the point singularity) with variable gravitational effect. Something like a variable infinity causing finite variance... something here doesn't work (or i'm too dumb to get it).
Do you understand that different point charges can have different electric fields even though the field goes to infinity as you approach the charge. This is principally no different.

I strongly suggest not trying to draw inferences from Wikipedia articles at this level.

#### stg213

On the contrary, it is very applicable here. There is no finiteness to the singularity. By your same argumentation, you could not have point charges with different charge in classical electrodynamics. The delta function is at the heart if this, since it is useful in describing the charge density of a point charge.
Well luckily they don't and classical electrodynamics has been superseeded by QED.

But i do take it that the same principle applies in the description of singularities if i get that right ? I.e. that's the principle that was used to describe them.

#### PeterDonis

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how can more mass curve space-time beyond infinity
The "mass" of a Schwarzschild black hole is not a property of the singularity. It's a global property of the spacetime geometry. Holes with different masses have different curvatures in their spacetime geometries. There is no need to attribute any different properties to their singularities.

#### Orodruin

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Well luckily they don't and classical electrodynamics has been superseeded by QED.
Do you seriously believe that just because there is a newer and more accurate theory, you can just get out of studying classical electromagnetism? Otherwise I do not see any point in bringing up QED at all. Not to be overly blunt, but based on what we have discussed here, I doubt you have a working understanding of quantum field theory.

#### Dale

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The dirac delta function example is in a way just a fancy way of saying:

2∞ > 1∞ (which mathematically may make sense but in this case isn't exactly useful as we are talking about finite black holes with finite mass and finite event horizons and finite gravitational effects)
Actually, that is not the way that $2\delta(x)>\delta(x)$. What happens is that $\int \delta(x)\; dx = H(x)$ which is well defined and finite everywhere. So $\int 2\;\delta(x) \;dx = 2\;H(x)$.

Similarly for a singularity. None of the quantities are well defined “at” the singularity. But the curvature invariants near the singularity are related somewhat like the integral. So a bigger singularity is one where those nearby integrals are larger. This even though the singularity itself is not well defined, the integral is and so the nearby quantities are well defined.

#### Dale

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He is not talking about the limit, he is talking about the value at any finite $r$. For all $r$, the curvature is finite. $r = 0$ is not part of the Schwarzschild spacetime.
For any finite $r$ the curvature is finite, but for any finite $r$ there is a real number that is smaller and whose curvature is correspondingly larger. So the values of the curvature in the manifold itself (not the singularity) are not finite. You can pick any finite curvature and there is a point in the manifold with a larger curvature. Something that is larger than any finite number is infinite. Again, this is talking about values in the manifold, not the singularity.

#### Orodruin

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For any finite $r$ the curvature is finite, but for any finite $r$ there is a real number that is smaller and whose curvature is correspondingly larger. So the values of the curvature in the manifold itself (not the singularity) are not finite. You can pick any finite curvature and there is a point in the manifold with a larger curvature. Something that is larger than any finite number is infinite. Again, this is talking about values in the manifold, not the singularity.
And again I think it was pretty clear that @Nugatory was talking about single points. It is the limit that is infinite, not any value inside the manifold, all values inside the manifold are finite because, as you have pointed out, there are larger numbers. What you are describing is an unbounded function (which the curvature is), but it is not how you would go about checking that the value of the function is finite everywhere.

#### Dale

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I think it was pretty clear that @Nugatory was talking about single points
It certainly wasn’t clear to me. He said that “the curvature increases without limit as r get smaller - but it’s always finite”, which certainly seems to be talking about more than single points. In my understanding the word “infinite” is usually shorthand for “increases without limit”. So saying that something “increases without limit” is the same as saying it is “infinite” and thus it is contradictory to also say that it is always finite. I just don’t think that the two statements are compatible.

#### Orodruin

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It certainly wasn’t clear to me. He said that “the curvature increases without limit as r get smaller - but it’s always finite”, which certainly seems to be talking about more than single points. In my understanding the word “infinite” is usually shorthand for “increases without limit”. So saying that something “increases without limit” is the same as saying it is “infinite” and thus it is contradictory to also say that it is always finite. I just don’t think that the two statements are compatible.
I would say the correct nomenclature if you want to say that it grows without bound in some limit is "tends to infinity", not that it is infinite. It is always finite whatever value of $r$ you choose so indeed it is finite for all $r$.

#### Dale

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Is “finite for all r” the same as “finite for any r” to your understanding?

I think those are different statements and only the second is true. If you pick any single point the curvature at that point is finite. But if you look at the set of all points that set has no finite upper bound to the curvature. So the curvature is not finite for all r.

I am no mathematician, so I could very well be getting the terminology wrong, but that is my understanding of how this should be expressed.

#### PeterDonis

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Is “finite for all r” the same as “finite for any r” to your understanding?
First we have to know what the allowed range of values for $r$ is. For Schwarzschild spacetime, it is $r > 0$. And the curvature is finite for all values in that range. Or, equivalently, it is finite for any value of $r$ in that range.

if you look at the set of all points that set has no finite upper bound to the curvature
That means the limit of the curvature as $r \to 0$ is infinite. But that's a different statement from saying the curvature is not finite for all $r$. The value $r = 0$ is not in the allowed range of $r$ for the spacetime.

#### Orodruin

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But if you look at the set of all points that set has no finite upper bound to the curvature. So the curvature is not finite for all r.
Yes it is. It is finite for all allowed $r$. Again, what you are describing is an unbounded function, not a function that has an infinite value.

#### Dale

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The value r=0r=0r = 0 is not in the allowed range of rrr for the spacetime.
I am not taking about r=0. That is not in the manifold. I am only talking about the set of all points actually in the manifold. The curvature for that set of points has no finite upper bound.

what you are describing is an unbounded function, not a function that has an infinite value.
Infinity is not a value that a function has. Infinite simply means that it is unbounded. I understand those two terms to be synonyms. Hence my original objection.

At least when my calculus teacher first taught me this concept she made a point of teaching that infinity was not a value but was just shorthand way of saying that the function was unbounded.

#### Orodruin

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The curvature for that set of points has no finite upper bound.
Again, the nomenclature for that is "unbounded". It is still finite for all $r$ because whatever $r$ you take it will have a finite value.

#### stg213

Do you seriously believe that just because there is a newer and more accurate theory, you can just get out of studying classical electromagnetism? Otherwise I do not see any point in bringing up QED at all. Not to be overly blunt, but based on what we have discussed here, I doubt you have a working understanding of quantum field theory.
I fail to see how 'getting out of studying' something is any part of this.

Point charges are an idealized point particle that appears in classical electromagnetism because it deals with electrical fields and the infinities that appear are rather a matemathical artefact not a description of reality (as far as we understand it today).

QED deals with the exchange of photons between charged particles... thus point charges wouldn't appear at all in this model nor their infinities. Columb's law might be a convenient and 'good enough' way to do practical calculations but it isn't an accurate model to use to describe reality, thus to infer it's consequences as a descriptive model for other phenomenae is imo flawed unless specifying why and how it applies.

Similarly for a singularity. None of the quantities are well defined “at” the singularity. But the curvature invariants near the singularity are related somewhat like the integral. So a bigger singularity is one where those nearby integrals are larger. This even though the singularity itself is not well defined, the integral is and so the nearby quantities are well defined.
Thank you! this is actually helpful in understanding the current model.

#### martinbn

I don't think that the Dirac delta is appropriate here. A better analogy would be $\frac1x$ for $x>0$. The function is unbounded but well defined and finite valued for all positive $x$. The point $x=0$ is not part of the domain of the function.

#### jbriggs444

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Is “finite for all r” the same as “finite for any r” to your understanding?
It depends on context. In this context they are synonymous. They are both equivalent to the statement $\forall r: isfinite(curvature(r))$

In some context one might want to ask whether there exists at least one r for which the curvature is finite. i.e. $\exists r: isfinite(curvature(r))$. But clearly, that cannot be an intended meaning here. We already know that there are lots of r values for which curvature is finite.

You seem to want to consider the set of curvatures that can be found over the range of strictly positive r values. And you want to express that this set includes values larger than any chosen finite bound. As has been pointed out, the proper terminology for this is not that the set of values "is infinite" but that the set of values "is unbounded", "is not bounded above" or "has no upper bound".

[More completely, we'd need to speak of an order relation and a set from which putative upper bounds could be drawn, but those can be implied]

When I took real analysis many years ago, our instructor trained us not to use the word "infinity" but to instead use circumlocutions for which precise mathematical definitions exist.

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#### Orodruin

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QED deals with the exchange of photons between charged particles... thus point charges wouldn't appear at all in this model nor their infinities.
If you think QED gets you out of dealing with infinities you are, sadly, mistaken. The problems involving infinities are even worse in QED and in QFT in general.

I don't think that the Dirac delta is appropriate here. A better analogy would be $\frac1x$ for $x>0$. The function is unbounded but well defined and finite valued for all positive $x$. The point $x=0$ is not part of the domain of the function.
This depends on whether you are trying to make an analogue for the mass distribution or for the potential.

"The Singularity and the Schwarzschild radius"

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