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I heard Carl Sagan talking about the Universe 15 Billion years ago, and the Big Bang. He made the statement that it was the biggest explosion of all time (at first I thought this a subtle pun). This leads me to my question. What would time have been like at +1 "moment" after the big bang? What I'm trying to ask is, and I hate to say it because I'm afraid I'll sound foolish, did time flow at the same rate? Wouldn't all that mass in one place have distorted space/time (and why didn't it "rip" it)?

If I were inside of that mass with a stop watch, I'm guessing I wouldn't have been able to measure a difference because time would have effected me the same way as it effected all the other space/mass in the area. I'm guessing I would have to have something inside (the initial Big Bang mass) and something outside measuring time and see if there was a difference (intuitively this feels weird to consider, could I actually place something "outside" the "Big Bang mass").

Maybe I've said too much, or made the question too complicated. I apologize if this is the case.


A black hole is a lot of Mass collapsed into a small space. I believe that as mass increases time dilation increases. I remember hearing that if you fell into a black hole, you'd never experience the last second of your life...

If this is true of black holes, how did time pass in the mass/energy that is responsible for the big bang? As the big bang occurred, did time speed up with the expansion of the universe? I'm trying to explain what I'm getting at by asking more questions related to what I was asking. I'm trying to understand what time itself looked like. As I understand it time prevents everything from happening at once. If time was a line, were the ends smashed together into a point before the Big Bang ? Maybe as mass/energy expanded the "time line" expanded too?

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While you've got the seed of a interesting idea (distortions of spacetime in the immediate aftermath of the big bang), I don't think this is really a well-defined question. You've kind of identified the reason why: there's no way to place something outside the universe to compare it with what's in the universe, as you're asking about. –  David Z Nov 22 '10 at 0:04
I have to agree with David on his comment here. –  Noldorin Nov 22 '10 at 0:37
Thanks for keeping your question active, Everett - perhaps we can improve it into something we can answer. Maybe this will help: there's a lot of philosophical speculation about what time really is, but we don't deal with that here. In physics, time is just a coordinate, like distance, height, or latitude or longitude. Just as different observers can measure distance differently, they can also measure time differently. That's what time dilation is. So if you want to ask a meaningful question about time dilation, you generally need to identify two specific observers and two specific events. –  David Z Nov 22 '10 at 4:00
I think I may have used the wrong term. Thank you for your patience with me, I really want to understand this. Let me go back to the sentence with the word dilation in it, and restate it. As mass increases a change occurs in the flow of time. That was what I meant to say. More to follow, saving edit... –  Everett Nov 22 '10 at 4:12
That's certainly not the normal way of thinking of it. Generally, you write down the four-metric as $ds^{2}=-dt^{2} + a(t)^{2}\left(\gamma_{ij}dx^{i}dx^{j}\right)$, where the $\gamma_{ij}$ is a time-independent spatial metric. Thus, the easy way of thinking of the problem is to imagine the spatial dimensions getting really small and the time 'flowing' at the same rate. But that is just a special choice of time coordinate, and you can choose others to give the metric a different form. It's more relevant to think of a problem like this in terms of what I"m trying to measure THEN interpreting. –  Jerry Schirmer Nov 22 '10 at 4:33

2 Answers 2

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Here is a sense in which this can be answered a bit unambiguously--it is a known effect that gravitational fields both dilate time, by a factor $\sqrt{1-\frac{2\,G\,M}{c^{2}\,r}}$ and redshift light waves by that same factor.

It is also known that cosmological effects redshift gravitational waves. This time, it is done by a factor of $a(t)$, the so-called 'radius of the universe'. For example, the cosmic microwave background radiation was believed to have been radiated from a surface whose temperature (and therefore, emitted wavelength) is roughly equivalent to the surface of a hot star. It is a matter of simple algebra to find a value of $\frac{M}{r}$ for which the two effects are roughly equivalent, and, if you wish, you can think of this as describing a "different rate of time flow."

To my knowledge, there really isn't a useful reason to do this, though.

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I usually hear $a(t)$ called the "scale factor" which I think is a bit more accurate than "radius." –  David Z Nov 22 '10 at 4:38
Both terms are used---I see 'radius of the universe' more often in popular literature. In a closed cosmological model, the spatial section is a 3-sphere, and the radius is literally $a(t)$. Older researchers loathed boundary conditions in GR,and so there was a collective hope that the universe was closed. We now know that the closed model is almost certainly not our universe, but some people still say 'radius of the universe'. The question seemed non-technical enough that I stuck with the most common non-technical term in my answer. –  Jerry Schirmer Nov 22 '10 at 12:56

In the context of FRW cosmology, there is no difference in the rate of time between the epochs of the evolution of the universe. You can see that from the form of the line element


That is a result of the symmetries that you assume for the matter distribution (homogeneous, isotropic) and the choice of observers that you make. So the observers that follow the expansion of the Universe, which are the galaxies more or less, perceive the same time wherever and whenever they are. The cosmological time is the proper time of all the comoving observers, as it is evident from the line element.

In the case of a Schwarzschild metric and static observers


it is the factor in front of dt that makes the difference and you have different time rates for observers at different positions.

There is one more point. Someone mentions the redshift and the perceived difference of the rate of time for faraway objects. That would appear to contradict what I am saying, but it isn't. The redshift effect is an observer symmetric effect. Like in the case of SR where you have two inertial observers with different velocities and each of them thinks that the others time runs slower, when both of them actually experience proper time. That is very different from the case of the static observers near a gravitating object, where there is no such symmetry. The clock of the observer that is at bigger r runs faster than the clock of the one that is at smaller r.

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But, I could define a set $\{T(t,r),R(t,r)\}$ such that $g_{TT}=1, \,G_{TR}=0$. Sure, it would make the rest of the metric time-dependent and messy, but there's no abstract reason why I can't do it (It's two PDE's for two variables, after all, and would be equivalent to setting $\alpha=1,\,\beta_{r}=0$ in an ADM IVBP of the problem). –  Jerry Schirmer Nov 22 '10 at 13:00
Are you talking about the Schw. metric? If you do that, then you are making a different choice of observers and you are measuring time with respect to their proper time (these observers are probably the freefalling ones). The Schw. metric in its usual form is written with respect to the static observers at infinity. In any case, the physical picture wouldn't change. –  Vagelford Nov 22 '10 at 13:31
By the way, these observers (freefalling) wouldn't perceive any difference in the rate of time as they fell towards the r=0 of the Schw. spacetime. –  Vagelford Nov 22 '10 at 13:37
And there is one more point. In Schw. you can't define a global class of such observers. You would have to distinguish in different classes the observers that begin their freefall from different distances. On the other hand, you don't have a restriction like that in the case of the FRW comoving observers. –  Vagelford Nov 22 '10 at 13:46
An other way to put it, and I hope that this will make it clear, is that if the universe were closed, at the moment of the Big Crunch, all the observers would be on time, they would all shake hands and agree on the age of the universe. –  Vagelford Nov 23 '10 at 0:13

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