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I am a mathematically minded individual. I do not intuitively comprehend physics, and as a sophomore in high school who has only taken Intro to Physics in his freshman year, I may very well have a completely erroneous view of many concepts within physics.

Recently, Einstein's model of gravity has been on my mind. I was watching the following lecture here wherein Professor Sera Cremonini explains that gravity is due to the curvature of 4 dimensional spacetime, and that this curvature is caused by the mass of objects within the universe. Furthermore, the higher the curvature (equivalently, the higher the mass of an object) the higher the force of gravity near that object. To illustrate this, she uses a stretched rubber disk and puts a large ball on it whilst smaller balls are on the disk. The large ball causes the rubber to stretch, and this curvature causes the smaller balls to be pulled toward the large ball just as gravity causes orbits of the solar system. This made me think.


Theoretically, is it possible for the curvature of the universe at one particular point to be so strong that the gravitational force is infinite? If infinite is ridiculous in this context, consider the question reformulated: Is there, or can there be, a point in the universe where the curvature is greater than any other point in the universe? If so, why doesn't the strength of gravity at this point cause the entire universe to be contracted to this point? That is, isn't this just like the Big Crunch?


Once again, I don't know if I've conveyed this using the proper terminology and I'm aware it's a very bizarre idea. I hope you all can entertain my thoughts and attempt to answer my question as best as possible, and correct my understanding and terminology as warranted. Thank you.

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Basically, there's a lot of momentum left over from the Big Bang. (Tells me my intuition) –  Deiwin Nov 30 '12 at 2:07
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2 Answers

In physics, places where physically meaningful properties take on infinite values are known as singularities, and certainly they can exist. An ideal Schwarzschild black hole has a singularity at its center (a point in space, or a one-dimensional curve in spacetime, if you want to think about it that way). A Kerr black hole, which is Schwarzschild + rotation, has a ring-shaped singularity inside it.

Two important points, though. First, our current understanding of physics is not complete. You can take any physical theory out there, and if you push it hard enough, you will find ways to break it. What better way to push something than by letting quantities (like the curvature of spacetime) go to infinity? At some point, reality will stop being described well by your theory, and you will need a better one (theory that is - better realities are hard to come by). Thus the presence of singularities often correlate with the breakdown of our understanding.

Second, and more directly answering your question, a theory can be self-consistent even with some things going to infinity. Just because the curvature is infinite does not mean it has an infinitely powerful effect on the rest of the universe.

Here's a typical example, without any GR even. Suppose the Sun is a uniform-density sphere of radius $R_\odot$, mass $M_\odot$, and surface gravity (aka acceleration at the surface) $g_\odot$, used here as a proxy for curvature. Consider sending $R_\odot$ to $0$, compressing the Sun but keeping the mass constant. The surface gravity is $$ g_\odot = \frac{GM_\odot}{R_\odot^2}, $$ which will clearly go to infinity as $R_\odot$ goes to $0$.

But how much is the Earth influenced by the surface gravity? We can write the acceleration $a$ we feel from the Sun as a function of $g_\odot$, $R_\odot$, and our distance $d$ to the Sun. Applying the inverse-square nature of gravity, we simply have $$ a = \left(\frac{R_\odot}{d}\right)^2 g_\odot. $$ Sure, $g_\odot$ may be going to infinity, but $R_\odot^2$ is going to zero at the same rate. In fact, $$ a = \frac{GM_\odot}{d^2} $$ independent of $R_\odot$. Thus the Sun could become a one solar mass black hole and we would still go around in the same orbit.

GR is no different in that while some quantities are free to go to infinity, others may be going to $0$, and these things often cancel.

Addendum: I should also add that it has been conjectured by some that singularities in GR are always hidden behind an event horizon, which if nothing else relieves some of the quasi-philosophical qualms about what might happen if you try to probe a singularity directly.

Also, if you are familiar with the delta function, that provides another line of intuition. Even though the function is "infinite" at a point (mathematicians reading this, I'm talking colloquially), all quantities of interest in theories that use it will somehow integrate over it, producing appropriately finite results.

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"Thus the Sun could become a one solar mass black hole and we would still go around in the same orbit." THIS IS COOL. –  000 Nov 30 '12 at 23:51
Okay. Putting my childish amazement aside, this is probably the most understandable answer thus far. I like that you used equations to address my point in a manner more mathematical than physical. (I translate your remarks into simple statements of calculus, particularly involving limits of $f(x)=1/x$. Also, your final statement is simply algebraic substitution.) It would appear that you addressed my question very well. However, I would like to understand the other answers before I stated that sincerely. Regardless, $+1$. –  000 Nov 30 '12 at 23:54
Indeed - I almost wrote an answer in terms of limits, but I refrained since not all high school sophomores know about them. And take your time studying the other answers - no need to rush physics. –  Chris White Dec 1 '12 at 0:49
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A side note: gravitation is nothing but free fall in General Relativity, so the force is zero.

Now, such an infinite curvature is basically a singularity. This IS possible within finite time in General Relativity.

The curvature of a point is a local property. It propagates through the universe with the speed of light. This explains why 'Big Crunch' doesn't immediately happen. And as with Newtonian gravitation, if the objects affected are of high enough velocity, then they can escape from the gravitational well, so it is not inevitable to happen.

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I have a few questions. I am sorry if they are dense or obvious: Why is gravitation simply free fall in General Relativity? Is this because all gravitation is perceived as the "falling" of objects into other objects? (Or, better yet, is your use of 'falling' different from my perception of the word?) How does the surrounding curvature of a plane which a point lies in (I don't think it's correct to say "curvature of a point", although I did :/) propagate through the universe with the speed of light? I don't understand this. (continued below) –  000 Nov 30 '12 at 23:43
Are you saying that the curvature created by an object is almost instantaneously caused? That is, whenever a new object accumulates into a lot of compact mass, its effect on the universe is nearly instantaneous? I understand escape velocity and I follow you until that part, but what confuses me is this: "[. . .] then they can escape from the gravitational well, so it is not inevitable to happen." Are you simply saying that it is not possible for gravity to be so immense that the escape velocity is greater than the speed of light (since that is the 'speed limit' of the universe [I think])? –  000 Nov 30 '12 at 23:47
@Limitless Firstly: in General Relativity, gravitation exerts no 'force' on objects. Instead, gravitation works by changing the properties of spacetime and the effect of gravitation is manifested in the paths of 'straight lines'(geodesics)(aka free fall) in which objects travel are changed. For the second question, see en.wikipedia.org/wiki/Speed_of_gravity . Thirdly, in "[. . .] then they can escape from the gravitational well, so it is not inevitable to happen." it refers to the Big Crunch. I'm saying as long as objects have a velocity >= escape velocity, they can escape!(continued) –  namehere Dec 1 '12 at 3:48
Of course some objects(e.g. black holes) have 'escape velocities' larger than the speed of light in some regions, but that's not what I'm referring in this context. –  namehere Dec 1 '12 at 3:49
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