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Where did the Universe come from?





Bikerman
Something that repeatedly comes up in these forums is the origin of the universe and how we can reconcile the Big Bang theory with the conservation of energy. Surely, the argument goes, you cannot just 'create' all the matter in the universe out of nothing? Surely, the argument continues, the conservation of energy means that there must have been energy (or mass, remember that they can be treated as equivalent!) 'before' the Big Bang?
I normally cop-out of such debates because the explanation is difficult to phrase and involves a lot of maths. I've recently come across a nice illustration, however, by a colleague in the Science Forums which doesn't require a PhD in theoretical physics (which she actually has).

Imagine a hollow sphere of matter/mass. Now, we know that, outside the sphere, there is a gravitational attraction to the centre of the sphere. That is basic Newtonian physics. In Relativistic terms we say that spacetime outside the sphere is 'bent'. We also know that inside the sphere that gravity cancels to zero.* In Relativistic terms we can say that within the sphere spacetime is flat.
Now, imagine that we winch the shell outwards to make the sphere bigger. We are obviously putting energy into this operation. The gravitational field outside the sphere remains the same (ie the region of 'bent' spacetime is the same) but the volume of flat spacetime inside the sphere has increased.
Another way of saying this is that flat spacetime has more energy than bent spacetime. We could, therefore, say that gravity (bent spacetime) is actually negative energy, since the energy content of a volume of space without a gravitational field is more than the energy of a volume of space with a gravitational field.

Now, expand this concept to the whole universe. We have energy and mass (call these +ve energy) and we also have regions of bent spacetime, or gravity (call this -ve energy). It is quite easy to see that the two can cancel to zero, meaning that the net energy of the entire universe is zero.

The entire universe, therefore, could result from a single quantum fluctuation (1 plus -1 = 0).


* This can be proved quite easily using Gauss's law. There is a very simple demonstration HERE
ocalhoun
Bikerman wrote:


The entire universe, therefore, could result from a single quantum fluctuation (1 plus -1 = 0).

I had come across that the quantum fluctuation idea before, and I rather like it... It would be unbelievably unlikely, but then again, we couldn't be here to speculate about it until after it happened, and (I suppose) there would be an infinite amount of time to wait for that unlikely chain of events to happen.

I do like the gravity touch though... That seems more likely than there being equal amounts of matter and antimatter in the universe. (Though perhaps... if a distant galaxy were composed entirely of antimatter, how would we know the difference?)
DoctorBeaver
Chris - It is possible that the laws of physics as we understand them came into existence with our universe. As such, maybe conservation of energy is unique to our universe and doesn't actually apply to the Big Bang.

As for the argument you put forward in your post I was under the impression that +ive & -ive energy in the universe does indeed cancel so that the overall level is 0.
Bikerman
DoctorBeaver wrote:
Chris - It is possible that the laws of physics as we understand them came into existence with our universe. As such, maybe conservation of energy is unique to our universe and doesn't actually apply to the Big Bang.
Absolutely true...no argument from me on that one. The only proviso I would make is that it appears that we can extrapolate the laws back to a minute fraction of time after the BB (currently about 10^-35 seconds with some confidence and 10^-11 seconds with much greater confidence). I take the point, though, that the BB event itself is still entirely speculative and different physical laws could well have applied.
Quote:
As for the argument you put forward in your post I was under the impression that +ive & -ive energy in the universe does indeed cancel so that the overall level is 0.
Well, that seems to be a consensus view, but there are others (as always in physics Smile )
metalfreek
As far as I know laws of physics as we know is valid in our universe. During big Bang these laws were inalid. So, how can we use this law to predict a thing that is before the Big Bang. so this is a big problem in explaning the origin of universe.
Indi
metalfreek wrote:
As far as I know laws of physics as we know is valid in our universe. During big Bang these laws were inalid. So, how can we use this law to predict a thing that is before the Big Bang. so this is a big problem in explaning the origin of universe.

Yes, you understand where we're at correctly.

However. ^_^

Some theories are attempting to go beyond our universe (as we know it). These are technically not scientific theories yet, they're just hypotheses (we tend to use the term theory for things that will probably be theories, but technically aren't yet - for example, M-theory - which is kinda bad, but eh), but they are extremely promising. For example: in M-theory, spacetime "tears" all the time, and when this tear happens, a "flop transition" occurs and a bubble forms around the tear - so the tear doesn't affect the universe around it (you can think of it as the universe scabbing over). The interesting notion some people raise is what happens inside this tear... it may be another universe unto itself. Neat, eh?
DoctorBeaver
I've just read Chris's original post and a couple of things occured to me.

Quote:

We also know that inside the sphere that gravity cancels to zero


Do you mean that the total gravity is zero? Or that it is zero at every point within the shell? I can see that at the centre it would be perfectly balanced; but if you moved away from the centre towards one part of the interior of the shell, wouldn't the gravitational attraction from that side of the sphere increase?

Quote:

Now, imagine that we winch the shell outwards to make the sphere bigger. We are obviously putting energy into this operation. The gravitational field outside the sphere remains the same (ie the region of 'bent' spacetime is the same)


Does the gravitational field outside the sphere remain the same? If you expand the shell then it gets further from the centre. The inverse square law says that the gravitational field outside the shell must consequently get weaker.
Bikerman
DoctorBeaver wrote:
I've just read Chris's original post and a couple of things occured to me.

Quote:

We also know that inside the sphere that gravity cancels to zero


Do you mean that the total gravity is zero? Or that it is zero at every point within the shell? I can see that at the centre it would be perfectly balanced; but if you moved away from the centre towards one part of the interior of the shell, wouldn't the gravitational attraction from that side of the sphere increase?

No. Gauss clearly shows that the gravity at every point inside the sphere is 0 (assuming no outside influence).
Here's an explanation:
http://www.merlyn.demon.co.uk/gravity1.htm#FiSSh
and here's some 'meat on the bones'
http://en.wikipedia.org/wiki/Gauss%27s_law_for_gravitational_fields
Quote:

Quote:

Now, imagine that we winch the shell outwards to make the sphere bigger. We are obviously putting energy into this operation. The gravitational field outside the sphere remains the same (ie the region of 'bent' spacetime is the same)
Does the gravitational field outside the sphere remain the same? If you expand the shell then it gets further from the centre. The inverse square law says that the gravitational field outside the shell must consequently get weaker.
No. Mass produces a distortion in spacetime. The mass stays constant so the distortion is the same - just spread over a greater volume.
DoctorBeaver
Quote:

The mass stays constant so the distortion is the same - just spread over a greater volume.


How can that be? A black hole becomes such when the mass is concentrated in a small enough area. Spacetime in the region becomes distorted to the extent that light cannot escape. Spread the same mass over a larger area and the effect is diminished allowing light to escape. If the mass is spread over a large enough area there is hardly any distortion.
Bikerman
DoctorBeaver wrote:
Quote:

The mass stays constant so the distortion is the same - just spread over a greater volume.


How can that be? A black hole becomes such when the mass is concentrated in a small enough area. Spacetime in the region becomes distorted to the extent that light cannot escape. Spread the same mass over a larger area and the effect is diminished allowing light to escape. If the mass is spread over a large enough area there is hardly any distortion.
A black hole is a special case.
Consider a large sphere of mass m. Expand the sphere so that the mass is the same. The gravitational 'pull' at the surface is reduced according to the inverse square law (m/r^2) but there is more surface. In total the amount of distortion is constant for m.
DoctorBeaver
Ah, the total amount is the same. Now I get you. Thank you for the clarification.
Bikerman
DoctorBeaver wrote:
Ah, the total amount is the same. Now I get you. Thank you for the clarification.

Yep, that's it. Now, given that, then it is inherent in the logic that matter/energy is 'cancelled' by gravity/distorted space time. That gives a zero sum for the entire universe.
It's an interesting way of looking at it isn't it?
Smile
Indi
Bikerman wrote:
A black hole is a special case.
Consider a large sphere of mass m. Expand the sphere so that the mass is the same. The gravitational 'pull' at the surface is reduced according to the inverse square law (m/r^2) but there is more surface. In total the amount of distortion is constant for m.

i don't know if you need to write off black holes as a special case for this. All you need to do is refine the claim.

If you take that same mass m, whatever it is, and contract the sphere until its radius is smaller than its Schwarzschild radius, you have a black hole. The only difference between this and the "normal" case is that once you've done this, you can't talk meaningfully about what is happening at the surface of the sphere - you have to talk about what's happening at the Schwarzschild radius/event horizon (or further away).

So all you have to do is say "the net distortion caused by any mass is the same regardless of how that mass is distributed once you are outside of the Schwarzschild radius".

Now... that being said ^_^... i wonder whether it is correct.

It sounds like you are saying that gravity curves according to mass, and the total volume integral of that distortion is the same for any given (uniformly distributed) mass, regardless of how it is distributed. i'm... not so sure. The curvature (first derivative) at any given point a distance r from the centre of a given (uniformly distributed) mass is constant, regardless of the density. But i don't know that that means the integral between r and the centre of mass is always the same. Is it?
Bikerman
Hmm...I'm going to have to get my slide rule out here and get back to you Smile

Having pondered...
The surface area of the sphere is given by 4.pi.r^2. This surely means that we have a constant 'total gravity' given the inverse square law? Do we need then to consider curvature ? Surely we can just equate total gravitational force with total curvature and cut out the need to integrate?
Indi
i'm not so sure. ^_^;

If i look at the problem in one dimension, with a mass of some size and distribution, i get the following graphs for the gravitational acceleration due to that mass with respect to distance from the centre (0):

R is the outer surface of the body in both cases - the first case is a less dense mass and the second is a more dense mass.

Each graph has two sections: within the body, and outside of the body (beyond the surface, of which R is the boundary). Now the first section can be anything (with certain constraints, and the actual shape depends on the distribution of mass within the body, i just made it a straight line)... but once you cross R, the second section is just Gm/rē, which is the curvature of space.

Note that regardless of what is going on within the body (r < R), once you are outside of the body, the curve is identical. That means g, the acceleration due to gravity due to the mass, the curvature of spacetime, the first derivative of spacetime (all of which are the same thing, more or less), are all the same for both cases.

But the total distortion - the integral of the curve between 0 and r, the area under the curve - ... is that the same?

i don't think so. i don't see the area under those two curves being the same. And the situation will just get more pronounced with additional dimensions.

(And as for a black hole, either or both of those curves could describe a black hole, if at any point on the curve 2·g·r is greater than cē.)

So we do have a constant gravity - which is the same as curvature... but not a constant total gravity (or total curvature).
Bikerman
Hang on, let's break this down.
Firstly do you agree that within the sphere there is no acceleration due to gravity (ie r<R)?
Indi
Bikerman wrote:
Hang on, let's break this down.
Firstly do you agree that within the sphere there is no acceleration due to gravity (ie r<R)?

No.

Even within the body there is still gravity/acceleration-due-to-gravity/spacetime-curvature (all the same thing, more or less). If you could pass through the mass's body unrestricted, (and there were no other forces at play) then you would come to rest at the centre of mass/gravity (same thing). If you left that point - even if you were still within the body - you would be pulled back to it by the unbalanced gravitational forces of the body's parts.
Bikerman
Indi wrote:
Bikerman wrote:
Hang on, let's break this down.
Firstly do you agree that within the sphere there is no acceleration due to gravity (ie r<R)?

No.

Even within the body there is still gravity/acceleration-due-to-gravity/spacetime-curvature (all the same thing, more or less). If you could pass through the mass's body unrestricted, (and there were no other forces at play) then you would come to rest at the centre of mass/gravity (same thing). If you left that point - even if you were still within the body - you would be pulled back to it by the unbalanced gravitational forces of the body's parts.

But what about Gauss' law? Surely the Gauss integral shows us that the net gravitational attraction at every point within the sphere is 0?
Indi
Bikerman wrote:
Indi wrote:
Bikerman wrote:
Hang on, let's break this down.
Firstly do you agree that within the sphere there is no acceleration due to gravity (ie r<R)?

No.

Even within the body there is still gravity/acceleration-due-to-gravity/spacetime-curvature (all the same thing, more or less). If you could pass through the mass's body unrestricted, (and there were no other forces at play) then you would come to rest at the centre of mass/gravity (same thing). If you left that point - even if you were still within the body - you would be pulled back to it by the unbalanced gravitational forces of the body's parts.

But what about Gauss' law? Surely the Gauss integral shows us that the net gravitational attraction at every point within the sphere is 0?

Gauss's law is referring to a Gaussian surface, like a shell, not a complete solid body. The way to apply it here is to imagine a solid spherical mass as a bunch of layered spherical shells - like an onion. For a point within the body, there are some shell-layers that are outside of that point (the point is inside of those shells) and some that are inside of that point (the point is outside of the shell). As per Gauss's law, the shells outside of that point contribute nothing - the net curvature due to the surrounding shells is 0. But the shells inside of that point do contribute... they cause a gravitational pull toward the centre of mass of those shells.

Hm, this is hard to explain in words and trivially easy to show in pictures. ^_^; Tomorrow, if i have time, i will try to illustrate this.
Bikerman
Ahh...I think I see the confusion - I think we are talking at cross purposes here Indi.
The thought experiment is to take a hollow shell of matter and winch it outwards, not a solid body.
Obviously in a solid sphere then there is gravity within the boundary of the sphere.* Here I am talking about a notional shell of material surrounding space (think of a Dyson sphere). We then winch the shell outwards (using energy) and thus increase the 'flat' spacetime inside the shell.

* I did quite a bit of work on gravity inside 'solid' spheres when answering the question - does earth's gravity decrease or increase as you go downwards below the surface? (The answer, which surprised me when I calculated it, is it increases for most of the way. If you integrate with the different densities at the main 4 layers you find it rises to about 10.8 at the outer core).
Speaker_To_Animals
Bikerman wrote:
I've recently come across a nice illustration, however, by a colleague in the Science Forums which doesn't require a PhD in theoretical physics (which she actually has).


Hello, I'm the chap that Bikerman was referring to. I'd just like to point that this illustration was of course not thought up by me, but was borrowed from some popular science book I was looking at one day in a book shop -- which book and by whom, I can't remember. Just don't like to claim credit for someone else's neat explanation!

As regards matter, of course the energy due to anitmatter is positive, just as the energy due to matter is. Why there is a universe of matter, rather than antimatter, is thought to be due to a slight asymmetry in the laws of physics as regards the way matter behaves compared to antimatter.

So, when everything was rather hot, we had photons creating particle-antiparticle pairs, and pairs of particles and antiparticles annihilating to photons, which would give equal amounts of matter and antimatter. A slight symmetry in some other reactions meant that there was eventually just slightly more matter than antimatter. So, when things cooled down enough that the photons couldn't do the creating particle pairs trick anymore, in effect all the antimatter had annihilated, leaving just the unbalanced matter. The photons that were created from the matter-antimatter annihilation are the photons of the cosmic microwave background. The leftover scum after most stuff had annihilated to photons is the whole of the matter we see in the universe today.

If you count them, we have a lot more photons than leftover matter particles, so in terms of numbers, the universe is mostly photons!

We have already detected these asymmetries in the laws of physics, just not quite in the right channels yet to do the job as regards the Big Bang. But the fact that there is an asymmetry in thelaws of physics has already been established (called CP violation, if you want to go google!).
Bikerman
Thanks Speaks....
I should point out that I know a bit of which I speak, but speaker knows more... Smile
nikhilio
If you read up on string theory, and M theory, they have a description of the beginning of the "Universe" as collisons of Branes
Speaker_To_Animals
nikhilio wrote:
If you read up on string theory, and M theory, they have a description of the beginning of the "Universe" as collisons of Branes


True, but you have to believe string theory first.

Loop quantum gravity predicts a universe before ours, as the latest issue of New Scientist explains.
poppat
Although the universe seems vast to us humans, I believe that we are part of something much bigger and indescribable. Like the cliche grain of sand theory.

poppat
ocalhoun
poppat wrote:
Although the universe seems vast to us humans, I believe that we are part of something much bigger and indescribable. Like the cliche grain of sand theory.

poppat

Any reason at all for believing that, or are you trying to start a new religion?
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