# Can time equal 0 in a Singularity?

1. Apr 25, 2015

First off, I want to say hello. I have been interested in gravity for most of my life, I've just taken forever to do anything about it. Better late than never. That's my story and I'm sticking to it. :)

By no means do I consider myself a professional by any means, but I have been putting out an concerted effort, but I do dabble a bit.

I was wondering. If one was to enter a singularity, does the passage of time reach zero or always approaching zero going towards infinity?

2. Apr 25, 2015

### mathman

How do you approach zero going towards infinity?

As an object approaches a black hole, outside observers see time slowing down for the object. However in the reference frame of the object, time proceeds normally.

3. Apr 25, 2015

I'm sorry, I messed that question up. I realize that now. I'm just nervous and now slightly embarrassed. Your answer made me realize that. Let me pose a more proper one.

I understand the "Bob and Alice" scenario and should have remembered that.

As an object gains mass, gravity increases. As the latter increases, passage of time slows to someone with an external reference as the person near the mass observes the time of person further away as passing faster, unless I have mistaken something.

What I guess I should have asked is, According to the math from an external reference, does the passage of time slow to just an extremely slow rate, or can it reach zero?

4. Apr 25, 2015

### Staff: Mentor

A couple of clarifications are in order here. First of all, where is the observer for whom the passage of time is supposed to slow? I'm going to assume you intend for the observer to be on the surface of the object. (Note that that will create issues once the object gains mass or reduces its radius past a certain point--see below.)

Second, what is important for determining how much time slows at the surface of the object is not just the object's mass or "gravity"; it's the ratio of the object's mass to its radius. In conventional units, this is the quantity $GM / c^2 R$, where $G$ is Newton's gravitational constant, $M$ is the object's mass, $c$ is the speed of light, and $R$ is the object's radius. In GR, we often use "geometrized units", where the "mass" $M$ is actually in units of length, i.e., it's what we would write as $GM / c^2$ in conventional units. So in these units the ratio is simply $M / R$, and the factor by which the passage of time slows for an observer at rest on the object's surface, as compared to an observer very far away, is $\sqrt{1 - 2M / R}$. (Note that this assumes the object is not rotating.)

If we require that the object is stable and has a surface that someone can stand on, then there is a limit to how much time can slow on its surface. That is because, for an object to be stable and have a surface, the ratio $M / R$ can't be any larger than 4/9 (in geometrized units), which means the factor $\sqrt{1 - 2M / r}$ can't be any smaller than $\sqrt{1 - 8/9} = 1/3$.

Any object with a radius smaller than 9/4 of its mass $M$ in geometrized units cannot be stable; it must be collapsing into a black hole. A black hole does not have a surface that someone can stand on; to get close to its horizon, you have to be in something like a rocket that can accelerate you so you can maintain altitude against the hole's gravity. Assuming you have something like that, you could in principle get as close to the horizon as you wanted, i.e., you could make your radius $R$ as close to $2M$ as you wanted; and that means you could, in principle, make the factor $\sqrt{1 - 2M / R}$ as close to zero as you wanted. But the acceleration you felt would increase without bound as you got closer to the horizon.

5. Apr 26, 2015

Thank you. That does clarify it for me. Since I have had no classes or anyone to ask until now, or at least had the nerve to ask. I realized just how bad my question forming is. I must work on that.
After reading what you posted, I think I know what I should have asked. It should have been worded something like, can gravitational time dilation cause time to stop passing in any system.
Let me make sure I understand. One could get close to doing so, but time can't stop for any reason within our universe.

6. Apr 26, 2015

### harrylin

As far as we know, indeed: even for an outside observer at rest with the black hole, the physical processes of a body that is falling towards the black hole will never totally stop. Note that nevertheless a digital time counter may stop according to such an observer; there are nice simulation programs for download that illustrate that effect.

7. Apr 26, 2015

### nitsuj

The stage that all physics "plays out" on is spacetime, having no time dimension spoils the metric. "inside" a black hole is a region that cannot be measured. One perspective could be because there is no time dimension in that region.

8. Apr 26, 2015

### DrGreg

That depends what you mean by "inside".

Inside the event horizon, but outside the singularity, there is still one time dimension and 3 space dimensions just like anywhere else in spacetime. Someone in that region (who hasn't yet been spaghettified by high tidal gravity) can make local measurements without problem, although they can't send the results to the rest of us outside the event horizon. Things appear to go wrong with time only if you try to use the wrong coordinates for the region you are studying.

Inside the singularity, we cannot say anything. According to classical GR theory, spacetime comes to an end just outside the singularity. But when you get very close to the singularity, quantum effects become significant and GR takes no account of that. So until someone comes up with a viable theory of quantum gravity, we don't know what happens close to the singularity.

9. Apr 26, 2015

### nitsuj

I meant the part on the "inside" of the surface that has everything "on" it.

10. Apr 26, 2015

Are you meaning "firewall?"

11. Apr 26, 2015

### nitsuj

I don't know what a firewall is in this context. Though it would prohibit hacking for sure!

12. Apr 26, 2015

Some think that just "inside" an event horizon, there is a place where all the matter consumed resides. They call it a firewall. Stephen Hawking then said in rebuttal (paraphrased) "If that is so, then the event horizon does not exist", or something to that extent. An "apparent" horizon I think it was called.

Dealing with stellar mass BH that is.

13. Apr 26, 2015

### Staff: Mentor

If this means the horizon, then you are not correct; as DrGreg pointed out, spacetime inside the horizon still has a time dimension. (This is true of both types of horizon that I refer to below.)

What Hawking means is this: if a "firewall" is present--which really means, if quantum effects become strong enough to affect the spacetime geometry when an event horizon is close to being formed--then an event horizon never actually forms; there is never a region of spacetime that can't send light signals out to infinity. An apparent horizon can still form--this is a surface where, locally, outgoing light does not move outward but stays in the same place. But the light that is trapped inside the apparent horizon, in this model, does not stay trapped forever; eventually the apparent horizon goes away and the light escapes out again. This is in contrast to the standard black hole model, in which an apparent horizon forms, and at least some of the light that is locally trapped inside it never escapes--i.e., the apparent horizon is associated with an event horizon.

If this "firewall" model turns out to be true (I personally think it's unlikely, but it's way too early to tell for sure), then there would never be a singularity; quantum effects would prevent a singularity from forming, because they become strong enough to change the spacetime geometry well before a singularity is even close to forming.

14. Apr 27, 2015

### nitsuj

^^ I don't know black holes in near the detail you and Greg have, or most on pf lol. Thinking of them I've come up with some simple questions, is there an area between the event horizon and the surface of the object; what is the surface area (geometric) of an event horizon as measured by an outside observer I imagine it would be nill, and not some huge surface.

15. Apr 27, 2015

### Staff: Mentor

The surface of what object?

It is $4 \pi$ times the horizon radius squared, i.e., $4 \pi$ times $2M$ squared, or $16 \pi M^2$. Here $M$ is in geometric units; in conventional units the area would be $16 \pi G^2 M^2 / c^4$.

Note that this does not mean that the "radius" of the horizon is an actual distance from the "center" to the horizon; it isn't. The radial coordinate $r$ is simply defined so that the area of a 2-sphere at $r$ is $4 \pi r^2$. Under that definition, the radial coordinate of the horizon of a black hole is $r = 2M$, which gives the area I gave above.