Is the Universe Expanding from a Single Point and Can We Observe Its Center?

[cristo]

I'm not quite sure I'm understanding this question. Anyways, if a spaceshuttle is traveling in a given direction, and doesn't change this direction, it would soon "pass through" Earth's atmosphere (edge, or boundary) and continue on into space. Now if that spaceshuttle continues on and could travel at a speed greater than that of expansion (if the universe is expanding) and given an infinite amount of fuel and time, would this spaceshuttle not eventually pass through the edge, or boundary of our universe?

There is no such thing as the boundary of the universe (as I'm sure has been said several times in this thread alone. I've not read through it all for a while, so may be repeating things that have been said before, but I'll give it a go anyway.

The 'expanding balloon' analogy is meant to represent a finite yet unbound surface which is expanding. This analogy is two dimensional: that is, the 'people' on the surface of the balloon cannot move in the radial dimension of the sphere. They simply see a universe which is expanding. It is finite, since if they walk in one direction they will, at some point, end up back where they started, yet it is unbound: in walking around the sphere they do not encounter any sort of wall or boundary, or edge, or anything of the sort.
I agree with Russ that we could do with a sticky on this: if anyone wants to volunteer to write one, let me know (or write it an hit report).

 

[Sundance]

G'day from the land of ozzzzzzz

The balloon analogy is theoretical and not a fact.

The expanding universe is also a theory to support the Big Bang theory.

Observations of galaxies from near and far indicate a universe where the parts want to collect and eject matter and reforming as units and not as a total universe.

 

[cristo]

The balloon analogy is theoretical and not a fact.

The balloon analogy is simply that: an analogy used to show that it does make sense to talk about a finite yet unbound universe.

The expanding universe is also a theory to support the Big Bang theory.

The fact that the universe is expanding is a fundamental part of the 'big bang theory,' or, to use non-popular science terms the standard model of cosmology. To imply that these are 'theories' in that sense of the word is confusing: there is a huge amount of evidence supporting the standard cosmological model.

Observations of galaxies from near and far indicate a universe where the parts want to collect and eject matter and reforming as units and not as a total universe.

Sorry, I don't understand the point you are trying to make here.

 

[Sundance]

G'day from the land of ozzzzzzzzzzz

Cristo said

The fact that the universe is expanding is a fundamental part of the 'big bang theory,' or, to use non-popular science terms the standard model of cosmology. To imply that these are 'theories' in that sense of the word is confusing: there is a huge amount of evidence supporting the standard cosmological model.

If there is a huge amount of evidence supporting the SCM or the BBT. Please submit this evidence. Each theory has to stand on their own foundations and not to be dictated by emotional hangups of the past few decades.

Observations of galaxies from near and far indicate a universe where the parts want to collect and eject matter and reforming as units and not as a total universe.

Sorry, I don't understand the point you are trying to make here.

The point I'm trying to make is that observations tell the story. We have a good understanding of star formation and galaxy evolution, compact matter and jet formation and so on, explaining a cyclic process that we can physically observe and study. Observations do not indicate and expansion of the universe.

 

[cristo]

If there is a huge amount of evidence supporting the SCM or the BBT. Please submit this evidence. Each theory has to stand on their own foundations and not to be dictated by emotional hangups of the past few decades.

I'm not sure that one should have to 'defend' the standard cosmological model here since the evidence has been published and discussed in the community for years now. As a start, see here and the links therein.

 

[mysearch]

Response to #34:

Hi Jon,
Apologises for the length of this post, but wanted to try and table some equations for the dust ball based on your description. I don’t necessarily expect a response, but would welcome any correction of my assumptions as I find the model an extremely useful way of trying to understand some of the wider, albeit speculative, ideas within cosmology.

Thanks for the pointer to Newton’s Shells, this does indeed suggest the standard Newtonian equation is not applicable within the overall perspective of the dust ball model and should be replaced by the one you presented, i.e.

[1] $$F = \frac{4\pi Gm\rho r}{3}$$

In a spatially flat model, each galaxy's speed away from the center is equal to the Newtonian escape velocity of the total matter within the spherical ball defined by that galaxy's radius from the center.

Later you make the statement above, which led me to wonder if we also need to revise the standard definition of the escape velocity for the dust ball universe. Normally, the ‘classical’ escape velocity is derived from the balance of kinetic energy (outward expansion) against potential energy (inward gravitation). However, on the basis that gravitational potential is the integral of force, given in [1], would the minimum dust ball escape velocity be given by:

[2] $$E_k = \frac {mv^2}{2} = \frac{4\pi Gm\rho}{3} * \frac {r^2}{2}= E_p$$

[3] $$v = r \sqrt {\frac{4\pi G\rho}{3}}$$

I recognise that the premise of this derivation is weak, as the expression for kinetic energy is only valid for non-relativistic velocity, but suggests that:

[4] $$v = r*X$$ but similar in form to Hubble $$v = Hr$$

However, if we set [k=0] and $$[\Lambda=0]$$ in Friedmann’s equation we appear to get

[5] $$H^2 =\frac {v^2}{r^2} = \frac{8\pi G\rho}{3}$$

[6] $$v = r \sqrt {\frac{8\pi G\rho}{3}}$$

Is the inference that recession velocity of the dust ball is twice the required escape velocity, i.e. comparing [3] to [6]? On the assumption that we can still equate equation [1] to [ma] to derive an expression for [a=-g]:

[7] $$a=\frac{4}{3}\pi G\rho r=-g$$

Which suggests that not only does gravity not fall off via the inverse square law, but that it actually increases with radius [r]. However, this also seems to have some obvious similarities to Friedmann’s Acceleration equation,

[8] $$\left(\frac {\ddot a}{a}\right )= - \frac{4}{3}\pi G (\rho+3P/c^2)$$

However, on the assumption that the expansion of the dust ball is actually accelerating, does this suggest that $$[3P/c^2 > -2 \rho]$$, i.e. that $$\ddot a > g$$

Finally, if possible, I wanted to check a couple of issues associated with expansion of the dust ball:

- Given that the dust ball expands under the pressure of dark energy, there was an implication that the dark energy per unit volume does not get ‘diluted’, but rather remains constant. If so, wouldn’t the effective mass associated with this energy increase and therefore cause the mass-density $$[\rho]$$ of the dust ball to increase with the expansion over time? If so, does the increase of [g] with radius [r] remain linear?

- Overall, the logic of your thought experiment seems sound. However, I have a attached a diagram showing 3 points in space (2-D only), which defines a triangle. Each point of the triangle is subject to different levels of expansive acceleration and gravitational deceleration, which is proportional to [r], subject to the caveat above. Is it the assumption of a spatially flat, homogeneous model that this triangle remains geometrically similar under expansion?

Again, apologises for the length, but wanted to table any equations and issues for future reference. Anyway, thanks again for helping me with this model.

 

[jonmtkisco]

Hi mysearch,

Later you make the statement above, which led me to wonder if we also need to revise the standard definition of the escape velocity for the dust ball universe. Normally, the ‘classical’ escape velocity is derived from the balance of kinetic energy (outward expansion) against potential energy (inward gravitation). However, on the basis that gravitational potential is the integral of force, given in [1], would the minimum dust ball escape velocity be given by:

[2] $$E_k = \frac {mv^2}{2} = \frac{4\pi Gm\rho}{3} * \frac {r^2}{2}= E_p$$

[3] $$v = r \sqrt {\frac{4\pi G\rho}{3}}$$

I too had wondered whether there is an inconsistency between the normal "external" Newtonian escape velocity equation and the use of the Shell Theorem equation for the interior of a solid ball. Although there may seem to be a superficial inconsistency, I am comfortable that the normal external escape velocity equation continues to apply in this context.

Remember that only the spherical surface of the dust ball (or any large radius within the dust ball centered at the center of gravity) moves outward at the escape velocity of the mass contents of the ball defined by that radius. The interior portions of that ball move outward at slower speeds as one approaches the center. In fact, the surface of each smaller radial ball (nested inside the larger ball) moves outward at the escape velocity of its own smaller mass contents.

I think it makes sense to think of the spherical surface of any such ball as being external to the body of the ball itself. Therefore, the external escape velocity equation applies.

That moving spherical surface does not need to worry about "running into" the next incremental shell of dust located outward from it, because that outward shell is itself moving outward fast enough to stay out of the way. The Shell Theorem (and the relativistic version of it, Birkhoff's Theorem) assures us that any mass shells located outside our spherical surface have NO gravitational effect on the ball of mass inside our spherical surface. Our spherical surface experiences only the gravity of the mass inside of it, which reinforces that its own movement is external to the gravitating mass.

I would also point out that the normal escape velocity formula is what is baked into the Friedmann equations, and I see no reason to question the accuracy of those equations.

- Given that the dust ball expands under the pressure of dark energy, there was an implication that the dark energy per unit volume does not get ‘diluted’, but rather remains constant. If so, wouldn’t the effective mass associated with this energy increase and therefore cause the mass-density $$[\rho]$$ of the dust ball to increase with the expansion over time?

Well, the total mass of the dust ball increases a lot due to the added cosmological constant, but the density decreases. Keep in mind that the matter density is constantly decreasing, while the density of the cosmological constant, in kg/cubic meter, does not increase or decrease as the number of cubic meters increases. That's why it's called a constant.

- Overall, the logic of your thought experiment seems sound. However, I have a attached a diagram showing 3 points in space (2-D only), which defines a triangle. Each point of the triangle is subject to different levels of expansive acceleration and gravitational deceleration, which is proportional to [r], subject to the caveat above. Is it the assumption of a spatially flat, homogeneous model that this triangle remains geometrically similar under expansion?

Yes, I believe that an expanding dust ball model which is spatially flat and homogeneous will remain homogeneous over time. The triangle's angles will remain constant, and the lengths of each side will retain their same proportionality. If you find any flaw in the geometry I'm applying, please let me know. But it appears to work in a very straightforward way.

Jon

 

[Sundance]

G'day from the land of ozzzzzzz

Cristo said

I'm not sure that one should have to 'defend' the standard cosmological model here since the evidence has been published and discussed in the community for years now. As a start, see here and the links therein.

Just because you place the word "Standard" in front of model does not mean it is correct beyond question.
As a matter of fact the the so called evidence that supports the BBT is just ad hoc theories with the aim to support the BBT.

Refering to the link that you post and that I'm fully awear of is just an opinion and not evidence.

Frequently Asked Questions in Cosmology
http://www.astro.ucla.edu/~wright/cosmology_faq.html#BBevidence

What is the evidence for the Big Bang?
The evidence for the Big Bang comes from many pieces of observational data that are consistent with the Big Bang. None of these prove the Big Bang, since scientific theories are not proven. Many of these facts are consistent with the Big Bang and some other cosmological models, but taken together these observations show that the Big Bang is the best current model for the Universe. These observations include:

Does this make it correct.

No it allows people to think that "I KNOW" or this is "CORRECT" and stops further investication and research. Most papers assume first that the Standard model is correct than proceed to fit the data to that model.


======================================================

So again, what evidence supports the BBT?

 

[cristo]

Refering to the link that you post and that I'm fully awear of is just an opinion and not evidence.

Sorry, but that is nonsense. The evidence in that link was precisely that: evidence.

Most papers assume first that the Standard model is correct than proceed to fit the data to that model.

This is how science works: one cannot start over from scratch everytime one writes a paper. The fact of the matter is that the evidence strongly favours the standard model, thus it is upon the doubters to refute every single piece of evidence.


Finally, note that, as Spacetiger told you last year, PF is not the place to be debating the big bang theory. If you wish to do that, then write a paper and present it to the community by way of submitting to a peer-reviewed journal.

 

[Eraniamayomii]

Assuming we are not at the center of the universe:

If V is the vector from the center of the universe to us, then -V would be the vector from the center of the universe in the direction away from us. Wouldn't a star (or whatever) on the -V vector be accelerating away from us at a larger rate than other objects, and especially objects on the +V vector that are further away from the center of the universe than we are? This is obviously assuming that everything in the universe is expanding from a single point.

Can we observe (through redshift or some other means) where the center of the universe is and where we are in relation to it?

Hi. Nice to meet you.

Umm, sorry to not read all the other posts. It's pretty late, but isn't cosmic background radiation a fairly simple answer to this question?

Scientist think that we are moving at about 1.3 million miles per hour relative to a single point in space "like your center."

It does make me wonder why so slow? If all mass is the result of a catastrophic explosion, shouldn't we be moving a lot faster then a measly 1/500 the speed of light?

I read that we seem to be moving toward a dense clump of galaxies also. This makes me wonder why we are moving toward other galaxies. If we were blown out of an explosion, we should be all moving at the same acceleration, at a fairly constant pace as we gradually, and equally expand further apart. The fact that we are steering toward another galaxy makes me wonder how much momentum the big bang caused.

Please correct me if I'm off.

Great subject by the way!

Love

Eraniamayomii

 

[mysearch]

Response to #42, #45

I wanted to see whether there was any consensus (or correction of my assumptions) that the dust ball model, see link in #22, might address the original question in #1, and also the questions raised by Eraniamayomii in #45, albeit in a speculative way, i.e.

#1: Can we observe (through redshift or some other means) where the center of the universe is and where we are in relation to it?

I believe, based on Jon’s assessment of the dust ball model, the answer is no?

#45: Scientist think that we are moving at about 1.3 million miles per hour relative to a single point in space "like your center."

Again, based on dust ball model, there is a velocity with respect to a centre of gravity, but this would be proportional to [r], as per Hubble recession, plus the model is still homogeneous as per the standard model?

#45: This makes me wonder why we are moving toward other galaxies. If we were blown out of an explosion, we should be all moving at the same acceleration, at a fairly constant pace as we gradually, and equally expand further apart. The fact that we are steering toward another galaxy makes me wonder how much momentum the big bang caused.

My assumption about colliding galaxies is that this would be treated as a local aberration on the very large scale of the universe? The issue of momentum is interesting because within the context of the dust ball model as I am not sure how the perception of greater than [c] recessional velocity is addressed with respect to SR?

In #42, Jon summarised some of the arguments surrounding the concept of escape velocity within the dust ball model. What seems of interest is that dust ball model allows a centre of gravity, but Jon has explained why this centre might not be perceived based on Newton’s shell theorem, i.e. the gravitational effects within the dust ball are proportional to [r] not $$[1/r^2]$$, see #34 & #41.

One final point with respect to a comment in #42

Well, the total mass of the dust ball increases a lot due to the added cosmological constant, but the density decreases. Keep in mind that the matter density is constantly decreasing, while the density of the cosmological constant, in kg/cubic meter, does not increase or decrease as the number of cubic meters increases. That's why it's called a constant.

If the `cosmological constant` is constant for each unit volume, does it not raise the question as to what is the net effect given the volume of space is increasing? Now the relationship between mass, density and volume is normally given by the simple equation $$M=\rho*V$$. However, there is the implication of an equivalence between mass and energy, such that if dark energy/cosmological constant remains constant per unit volume, its effective mass per unit volume must also remain constant?

[1] $$\rho_\Lambda = \frac{M}{V} \equiv \frac {(E=\Lambda)*(4/3)\pi r^3}{c^2*(4/3)\pi r^3}$$

In the broader context of an evolving universe, is it correct to assume that matter and CDM do not remain constant per unit volume, e.g

[2] $$\rho_M = \frac{M}{V} \equiv \frac {M+CDM}{(4/3)\pi r^3}$$

Furthermore, an earlier radiation-dominated universe would be affected by an additional [1/r] factor linked to Doppler shift, e.g.

[3] $$\rho_\lambda = \frac{M}{V} \equiv \frac {E=hf}{c^2 (4/3)\pi r^4}$$

So does the dust ball model remain consistent with an evolution from a radiation to matter to dark energy dominated universe? Would a $$\Lambda$$-dominated universe simply expand forever, if its energy per unit volume is constant? What physics might account for this energy source?

 

[Eraniamayomii]

Dear mysearch

Can you please provide links about this dust ball model? I havn't any idea what it is, and I'd like to study up on it before I respond.

I'v searched google. It doesn't seem very common.

Thank you


Love

Eraniamayomii

 

[mysearch]

Dear mysearch
Can you please provide links about this dust ball model? I havn't any idea what it is, and I'd like to study up on it before I respond.

Hi Eraniamayomii,
I did reference the original link via post #22, but here's the link to click on:
http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-11778.pdf

 

[Eraniamayomii]

Dear mysearch

This link doesn't seem to work on my blackberry cell phone. Can you please provide me with the official name of this concept so I can more accurately search for it? I'm terribly sorry to be such an inconvenience, but I'm very interested. Thanks again for your patience.


Love

Eraniamayomii

 

[mysearch]

Hi Eraniamayomii
Not quite sure what you are up to, if you look at the properties of the link provided in #48 or #22 it will give you the full web reference as:

http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-11778.pdf

I don’t understand why you can’t just click on it, blackberry cell phone or otherwise, but failing this, here is the title information of the document:

Finite cosmology and a CMB cold spot
Ronald J. Adler, James D. Bjorken† and James M. Overduin
Gravity Probe B, Hansen Experimental Physics Laboratory, Stanford University,
Stanford, CA 94305, U.S.A.
SLAC-PUB-11778 gr-qc/0602102

Hopefully 3rd time lucky