mass filled universe
penguinista
the gravitational effects be? As far as I can figure, in a Newtonian
universe gravity from mass on one side would cancel gravity from the
other side. Thus you couldn't detect the mass gravitationally. Does
this hold in GR? How would it vary with density?
thoma...@gmail.com
This is exactly the mistake (or rather one of the mistakes)
cosmologists are making: they assume that the universe self-gravitates
despite the assumption of an isotropic and homogeneous universe. I have
discussed this point on my website under
http://www.physicsmyths.org.uk/discussions/cosmology2.htm (see the
reply to Juan Casado (starting under the small print)).
Thomas
macro...@internetcds.com
singularity. For if it was its gravity would be infinite
making it a black hole without possibility of
expanding.
It must have started with an energy build up that
becomes matter where the energy is spread out
so the gravity remains finite. Only in this way can
the Big Bang proceed. Only in this way could
the universe expand/inflate.
Its as simple as that.
Tom Roberts
> Suppose an unbounded universe was uniformly filled with mass, what would
> the gravitational effects be? As far as I can figure, in a Newtonian
> universe gravity from mass on one side would cancel gravity from the
> other side. Thus you couldn't detect the mass gravitationally.
Yes.
> Does
> this hold in GR?
No, with the proviso that in GR "gravitationally" means "geometrically".
There are three classes of FRW manifolds which meet your criteria and
are the basis for modern cosmological models:
A a space with negative curvature and topology R^3 and unbounded time
extending from a big bang origin; the spatial submanifold orthogonal
to the cosmic time coordinate expands without limit and the mass
density decreases to zero.
B a flat space with topology R^3 and unbounded time extending from a
big bang origin; the spatial submanifold orthogonal to the cosmic
time coordinate expands without limit and the mass density decreases
to zero.
C a space with positive curvature and topology S^3 which expands
from a big bang origin, reaches a maximum size, and then
contracts to a big crunch ending; time is bounded between big bang
and big crunch. Space is unbounded in the sense that it has no
boundary, but there is at all times a largest possible distance
between any pair of points (when measured on a 3-surface of
constant cosmic time).
[Apparently you have not considered the possibility that
space can be "unbounded" topologically, just as the
surface of a sphere is finite but unbounded; that is
class C.]
In principle it is possible to distinguish these three classes via
geometrical measurements of the cosmos. Current measurements indicate
that the world we inhabit is consistent with class B. But there are
recent measurements that imply that the Cosmological constant is small
but nonzero, and that can have a profound effect on the possible
cosmological models (A, B, and C all assume it is exactly zero). I am
not up-to-date on the details.
> How would it vary with density?
There is a critical density below which class A applies, at which class
B applies, and above which class C applies.
Tom Roberts tjro...@lucent.com
Spoonfed
If I understand correctly, the idea here is that gravitation acts as an
attractive force in an unseen fourth spatial dimension. This means
that even though, as penguinista points out, a homogeneous, isotropic
matter distribution cannot cause the imbalance of force necessary to
pull an object in any direction, given only three dimensions, it is
possible for there to be a non-isotropic, non-homogeneous matter
distribution in four dimensions that WOULD result in a uniform pushing
or pulling between the matter.
I can see this is an interesting possibility. Is there any evidence to
suggest that there is such a fourth dimension?
> In principle it is possible to distinguish these three classes via
> geometrical measurements of the cosmos. Current measurements indicate
> that the world we inhabit is consistent with class B. But there are
> recent measurements that imply that the Cosmological constant is small
> but nonzero, and that can have a profound effect on the possible
> cosmological models (A, B, and C all assume it is exactly zero). I am
> not up-to-date on the details.
Hmmmm, so, it would seem that all evidence points to the flat universe,
in which case there is no extra dimension through which gravitational
forces can act. That's too bad. It would have been a much more
interesting universe if we had this fourth dimension.
Interesting, you say that we may have a small but nonzero cosmological
constant. If I read this article correctly
http://www.phys.washington.edu/users/dbkaplan/555/lecture_03.pdf the
parameter k can take on three different values: for k = 0, the above
line element describes ordinary at space in spherical coordinates; k =
1 yields the metric for S3, with constant positive curvature, while k =
-1 is AdS3 and has constant negative curvature.
So it is not possible for k to be slightly above 0, it is either 0, -1,
or 1.
>
>
> > How would it vary with density?
>
> There is a critical density below which class A applies, at which class
> B applies, and above which class C applies.
>
>
> Tom Roberts tjro...@lucent.com
Hmmm, so if I understand this analogy correctly, it is much like
scribbling on a page with a charcoal pencil for such a long time that
eventually, the page begins to bend upward. When you add too much
stuff to the universe, of course, it cannot remain flat, it must bend
into the fourth dimension.
I think sometimes Einstein is given credit for discovering the fourth
dimension, but I think Madeline L'Engle's "A Wrinkle In Time" really
demonstrates the best available physical evidence. Seriously, how else
could those kids have folded space? Also, how would you explain the
wormholes in StarTrek and Babylon 5? How would you explain Disney's
"The Black Hole?" Carl Sagan's "Contact?" We've seen all of this good
evidence of the folding of space.
I jest, but for me, it was kind of painful to realize the possibility
that all of the science fiction I've ever read is not even plausible.
I would much rather have this possibility of zipping around the galaxy,
and coming back home again. This positive curvature model seems to
offer the slightest glimmer of hope that such is possible. And I think
that is the main reason that it stays alive.
Jonathan Doolin
Ben Rudiak-Gould
> Suppose an unbounded universe was uniformly filled with mass, what would
> the gravitational effects be? As far as I can figure, in a Newtonian
> universe gravity from mass on one side would cancel gravity from the
> other side. Thus you couldn't detect the mass gravitationally.
No, in a Newtonian universe it's ill-defined. One can argue that there's no
net force, as you did above, but one can also argue that there's a net force
with any magnitude and direction you please. Recall that the gravitational
field inside a spherical shell of matter is zero, while the gravitational
field outside is the same as if the mass were concentrated at the center;
thus, partitioning the mass of the universe into concentric spherical shells
about some point P, you can show that the net acceleration of a test
particle at some point Q is equal to 4/3 pi G rho (P-Q), for arbitrary P. If
you take P=Q, you get your argument as a special case.
> Does this hold in GR? How would it vary with density?
In GR the problem doesn't arise because of locality: the geometry is only
affected by mass within the visible universe, which is finite. So it's easy
to talk about spatially infinite homogeneous universes in GR. Because GR is
local, and local (finite) arrangements of matter gravitate together, it must
be the case that an infinite homogeneous arrangement of matter will also
gravitate together in GR. In the infinite case, there are no empty regions
left behind (homogeneity is preserved), so we can say that the universe as a
whole is shrinking.
(Not that the real universe is shrinking, but that's because the initial
conditions of the real universe are different from what I assumed above.)
-- Ben
Androcles
"Ben Rudiak-Gould" <br276d...@cam.ac.uk> wrote in message
news:de7m27$cnj$1...@gemini.csx.cam.ac.uk...
Very nice analysis.
Androcles
Spoonfed
Ben Rudiak-Gould wrote:
> penguinista wrote:
> > Suppose an unbounded universe was uniformly filled with mass, what would
> > the gravitational effects be? As far as I can figure, in a Newtonian
> > universe gravity from mass on one side would cancel gravity from the
> > other side. Thus you couldn't detect the mass gravitationally.
>
> No, in a Newtonian universe it's ill-defined. One can argue that there's no
> net force, as you did above, but one can also argue that there's a net force
> with any magnitude and direction you please. Recall that the gravitational
> field inside a spherical shell of matter is zero, while the gravitational
> field outside is the same as if the mass were concentrated at the center;
> thus, partitioning the mass of the universe into concentric spherical shells
> about some point P, you can show that the net acceleration of a test
> particle at some point Q is equal to 4/3 pi G rho (P-Q), for arbitrary P. If
> you take P=Q, you get your argument as a special case.
>
So you're saying, if you have a test mass, and you partition the
universe into an infinite number of concentric spherical shells, the
test mass should be attracted to the center of those shells?
This makes it so that the force of gravity from the infinite universe
can act in any direction you want, simply by choosing the origin from
which you create these spheres.
As a check, you could re-do the problem with cartesian instead of
spherical coordinates, subdividing the universe into an infinite number
of cubes instead of an infinite number of concentric spheres. In this
case, I'm sure you would find that no matter where you placed the
origin, the calculated force would equal zero.
> > Does this hold in GR? How would it vary with density?
>
> In GR the problem doesn't arise because of locality: the geometry is only
> affected by mass within the visible universe, which is finite.
> So it's easy
> to talk about spatially infinite homogeneous universes in GR. Because GR is
> local, and local (finite) arrangements of matter gravitate together, it must
> be the case that an infinite homogeneous arrangement of matter will also
> gravitate together in GR. In the infinite case, there are no empty regions
> left behind (homogeneity is preserved), so we can say that the universe as a
> whole is shrinking.
>
> (Not that the real universe is shrinking, but that's because the initial
> conditions of the real universe are different from what I assumed above.)
>
> -- Ben
Let me see if I understand:
Finite subsections of the infinite universe must pull together because
they are finite. Since the finite sections are not "visible" to each
other, they do not halt each other from shrinking. Moreover, the
infinite universe, as a whole must shrink because it is made of an
infinite number of these shrinking finite parts?
Tom Roberts
> Tom Roberts wrote:
>>There are three classes of FRW manifolds which meet your criteria and
>>are the basis for modern cosmological models:
>
> If I understand correctly, the idea here is that gravitation acts as an
> attractive force in an unseen fourth spatial dimension.
No. The idea is that gravitation is an aspect of geometry is a 4-d
spacetime.
>>In principle it is possible to distinguish these three classes via
>>geometrical measurements of the cosmos. Current measurements indicate
>>that the world we inhabit is consistent with class B. But there are
>>recent measurements that imply that the Cosmological constant is small
>>but nonzero, and that can have a profound effect on the possible
>>cosmological models (A, B, and C all assume it is exactly zero). I am
>>not up-to-date on the details.
>
> Hmmmm, so, it would seem that all evidence points to the flat universe,
No. Re-read what I wrote.
> Interesting, you say that we may have a small but nonzero cosmological
> constant. If I read this article correctly
> http://www.phys.washington.edu/users/dbkaplan/555/lecture_03.pdf the
> parameter k can take on three different values: for k = 0, the above
> line element describes ordinary at space in spherical coordinates; k =
> 1 yields the metric for S3, with constant positive curvature, while k =
> -1 is AdS3 and has constant negative curvature.
>
> So it is not possible for k to be slightly above 0, it is either 0, -1,
> or 1.
k is not the cosmological constant, it is a discrete topological
parameter for the FRW manifolds. k=-1 is my class A with a density less
than the critical density; k=0 is my class B with a density equal to the
critical density, and k=1 is my class C with a density greater than the
critical density.
> [... goes off into never-never land]
Tom Roberts tjro...@lucent.com
Tom Roberts
> penguinista wrote:
>> Suppose an unbounded universe was uniformly filled with mass, what
>> would the gravitational effects be? As far as I can figure, in a
>> Newtonian universe gravity from mass on one side would cancel gravity
>> from the other side. Thus you couldn't detect the mass gravitationally.
>
> No, in a Newtonian universe it's ill-defined. One can argue that there's
> no net force, as you did above, but one can also argue that there's a
Your argument is invalid because you assume that an infinite set can be
partitioned sensibly however you choose and that they are "equivalent".
Infinite sets have subtleties....
In Newtonian gravitation the gravitational force at any point for such a
universe is zero. This is clear and unambiguous both from the
translational symmetry and from solving the Poisson equation for the
potential.
[When solving Poisson's equation here be sure to use a Green's
function appropriate for the boundary conditions (the usual
one does not apply as it requires \rho=0 at r=infinity).]
> In GR the problem doesn't arise because of locality: the geometry is
> only affected by mass within the visible universe, which is finite.
Not true. In GR one must model the entire universe, not just the portion
visible to one specific observer located at one specific location in
space and time.
> So
> it's easy to talk about spatially infinite homogeneous universes in GR.
Yes, but not for the reason you cite. Rather it's because Friedmann,
Walker, and Robertson solved the Einstein equation for such conditions
(indeed for more general conditions than yours).
> Because GR is local, and local (finite) arrangements of matter gravitate
> together, it must be the case that an infinite homogeneous arrangement
> of matter will also gravitate together in GR.
No. See my other posts in this thread. The FRW solutions do not all do
this; only class C does so.
Tom Roberts tjro...@lucent.com
shevek
Tom Roberts wrote:
> Ben Rudiak-Gould wrote:[...]
>
> > In GR the problem doesn't arise because of locality: the geometry is
> > only affected by mass within the visible universe, which is finite.
>
> Not true. In GR one must model the entire universe, not just the portion
> visible to one specific observer located at one specific location in
> space and time.
>
Thanks Tom. Please forgive this speculative question, but what is
meant by the "entire universe" ? I have seen cosmologists claim -the
universe- is infinite, and still basically desrribed by the FW metric.
Others claim it is finite, and also basically decribed by the FW
metric. However, mathematicians will tell you that there is no such
thing as "the universe", as a universal set poses logical
inconsistencies.
Is there a common physics definition of "universe"? Is it the set of
all matter and energy, or only the matter and energy that could
possibly be observed or have an effect on us? Do you think these
questions are somehow relevant to cosmology, or am I just putting too
much weight on semantics and I should move on?
Thanks - shvk
Josef Matz
"Spoonfed" <jonatha...@spoonfedrelativity.com> schrieb im Newsbeitrag
news:1124510348.9...@o13g2000cwo.googlegroups.com...
And if the real universe is infinite? Then you need no extra dimensions at
all, also no curvature
in time as GR assumes it to explain the universe expansion.
Ben Rudiak-Gould
> Ben Rudiak-Gould wrote:
>> there's no net force, as you did above, but one can also argue that
>> there's a net force with any magnitude and direction you please. [...
>> invalid argument]
>
> Your argument is invalid because you assume that an infinite set can be
> partitioned sensibly however you choose and that they are "equivalent".
> Infinite sets have subtleties....
That was my point.
> In Newtonian gravitation the gravitational force at any point for such a
> universe is zero. This is clear and unambiguous both from the
> translational symmetry and from solving the Poisson equation for the
> potential.
Are you saying that phi = constant is a solution to del^2 phi = k, where
k > 0? I'm pretty sure it isn't. On the other hand, phi(r) = k/6 |r-r0|^2 is
a solution (for arbitrary r0), and if you work out the corresponding field,
you'll find that it's the nonzero field I described in my last post.
>> In GR the problem doesn't arise because of locality: the geometry is
>> only affected by mass within the visible universe, which is finite.
>
>Not true. In GR one must model the entire universe, not just the portion
>visible to one specific observer located at one specific location in
>space and time.
?! We have no idea what's going on beyond the Hubble horizon, and no reason
to believe that we aren't in an inflationary bubble of finite size. We don't
have to know in order to model the local geometry.
>> Because GR is local, and local (finite) arrangements of matter
>> gravitate together, it must be the case that an infinite homogeneous
>> arrangement of matter will also gravitate together in GR.
>
> No. See my other posts in this thread. The FRW solutions do not all do
> this; only class C does so.
I think we're just talking past each other here. When I said "gravitate
together" I didn't mean "get closer together with time." My notion of
gravitating together is time-symmetric. You could say that either the matter
is getting closer together as t increases or it's getting closer together as
t decreases (or you're at a critical point).
-- Ben
Ben Rudiak-Gould
> So you're saying, if you have a test mass, and you partition the
> universe into an infinite number of concentric spherical shells, the
> test mass should be attracted to the center of those shells?
Yes, I'm saying there's an argument to that effect, and also arguments
giving contradictory answers. None of them is correct: rather, the problem
simply doesn't have a well defined answer.
> As a check, you could re-do the problem with cartesian instead of
> spherical coordinates, subdividing the universe into an infinite number
> of cubes instead of an infinite number of concentric spheres. In this
> case, I'm sure you would find that no matter where you placed the
> origin, the calculated force would equal zero.
Actually I think you would get something akin to integral{-inf to inf} x dx,
which is *not* equal to zero; rather, it is undefined.
Take a look at the "Riemann series theorem", which is the discrete version
of this problem:
http://mathworld.wolfram.com/RiemannSeriesTheorem.html
> Let me see if I understand:
> Finite subsections of the infinite universe must pull together because
> they are finite. Since the finite sections are not "visible" to each
> other, they do not halt each other from shrinking. Moreover, the
> infinite universe, as a whole must shrink because it is made of an
> infinite number of these shrinking finite parts?
Exactly. Except that it's a bit iffy to talk about an infinite universe
shrinking, but I can't fault you for that since I did the same thing. :-)
-- Ben
Spoonfed
I accept your point that k and the cosmological constant are not the
same, though I'm not sure what the cosmological constant is.
Spoonfed
Ben Rudiak-Gould wrote:
> Spoonfed wrote:
> > So you're saying, if you have a test mass, and you partition the
> > universe into an infinite number of concentric spherical shells, the
> > test mass should be attracted to the center of those shells?
>
> Yes, I'm saying there's an argument to that effect, and also arguments
> giving contradictory answers. None of them is correct: rather, the problem
> simply doesn't have a well defined answer.
>
> > As a check, you could re-do the problem with cartesian instead of
> > spherical coordinates, subdividing the universe into an infinite number
> > of cubes instead of an infinite number of concentric spheres. In this
> > case, I'm sure you would find that no matter where you placed the
> > origin, the calculated force would equal zero.
>
> Actually I think you would get something akin to integral{-inf to inf} x dx,
> which is *not* equal to zero; rather, it is undefined.
>
> Take a look at the "Riemann series theorem", which is the discrete version
> of this problem:
>
> http://mathworld.wolfram.com/RiemannSeriesTheorem.html
>
I stand corrected. So you would get an undefined answer in the
cartesian case, just as you would in the spherical coordinates. It
would still depend on what order you did the integration in. If you do
the integral in order of symmetrical pairs around the test particle,
you would get zero, but if you started integrating symmetrically from
some other point, the answer would converge to another value.
That is actually somewhat of a relief that we can get the same (albeit
ambiguous) answer whether you use cartesian or spherical coordinates.
I still think, though, that even if the observer is not AT the test
particle, he should still do the integral FROM the test particle to
calculate the force ON the test particle. It would more sense to pair
off the particles in a symmetrical pattern around it when doing the
integral. It would seem rather artificial to be pairing off particles
which pull it in the same direction.
--Jonathan
Tom Roberts
> Tom Roberts wrote:
>>In GR one must model the entire universe, not just the portion
>>visible to one specific observer located at one specific location in
>>space and time.
>
> meant by the "entire universe" ?
In a GR model it clearly means the entire manifold. That is, a
geodesically complete manifold.
> I have seen cosmologists claim -the
> universe- is infinite, and still basically desrribed by the FW metric.
Only if class A or B holds. But measurements are consistent with class B.
> Others claim it is finite, and also basically decribed by the FW
> metric.
That's class C.
> However, mathematicians will tell you that there is no such
> thing as "the universe", as a universal set poses logical
> inconsistencies.
Math is not physics. "universal set" is not the universe.
> Is there a common physics definition of "universe"?
Sure. Everything there is.
Tom Roberts tjro...@lucent.com
Tom Roberts
> Tom Roberts wrote:
>>k is not the cosmological constant, it is a discrete topological
>>parameter for the FRW manifolds.
>
> same, though I'm not sure what the cosmological constant is.
The Einstein field equation is:
G + L*g = T
Where G is the Einstein curvature tensor, L is the cosmological
constant, g is the metric tensor, and T is the energy-momentum tensor
(units with 8*pi*k=1 where k is Newton's gravitational constant).
The FRW manifolds are spatially-homogeneous solutions to the EFE for a
constant mass density and zero cosmological constant.
Tom Roberts tjro...@lucent.com
Spoonfed
You say FLRW manifolds are solutions when L=0. So, specifically, the
FLRW manifolds are the possible solutions to the differential equation
(G = T)?
Most people are saying that the cosmological constant is one. Ben
Rudiak-Gould mentioned that zero is way outside the error bars. Does
this mean that the FLRW metric has been entirely invalidated? Is
there a working solution to (G + g = T)?
Secondly, I disagree that the FLRW metric implies a constant mass
density, even in its most simple form. Assuming k=0 and a(t)=1, you
can check my work from equations 46 and 52 in the document:
http://www.phys.washington.edu/users/dbkaplan/555/lecture_03.pdf
Putting these two values into equation 52, and switching to cartesian
coordinates yields:
ds^2=dt^2-(dx^2+dy^2+dz^2)
This is nothing more than the definition of spacetime interval between
two events in free space. It says nothing about whether there is a
constant mass density throughout the universe.
On the other hand, I begin to see one effect of gravity on the
cosmological scale: I begin to see the possibility that particles and
photons coming from far away may have to go further because they must
pass through the gravitational field of huge numbers of tiny particles
along their way.
I actually believe that the regions between galactic superstructures
contain an almost crystal lattice of Baryonic particles with density
equal to the average density of the galactic superstructures. In these
regions, perhaps there could be an increase in the path length for
light from a distant galaxy without any change in direction. However,
I don't see any need for adding non-baryonic dark matter to the mix to
explain what we see.
I think most of the data is well explained by Special Relativity, and
only a few minor details need to take into account gravity. And more
importantly, no details require dark energy or dark matter to resolve.
I've posted more about my idea here
http://groups.google.com/group/sci.physics.relativity/msg/540a99d4a69391fe
I think Ben may have been right, that my assumption is that the
cosmological constant is zero or close to zero, (but varying with
direction) if it involves the extra path length due to traveling
through "undisturbed" dark regions of the universe. I've heard that
there only appears to be 1% or 10% of the necessary baryonic matter to
give Omega=1 anyway. But, this is assuming we're talking about
something similar to the same thing, which seems a big if.
Spoonfed
Spoonfed wrote:
> On the other hand, I begin to see one effect of gravity on the
> cosmological scale: I begin to see the possibility that particles and
> photons coming from far away may have to go further because they must
> pass through the gravitational field of huge numbers of tiny particles
> along their way.
>
> I actually believe that the regions between galactic superstructures
> contain an almost crystal lattice of Baryonic particles with density
> equal to the average density of the galactic superstructures. In these
> regions, perhaps there could be an increase in the path length for
> light from a distant galaxy without any change in direction. However,
> I don't see any need for adding non-baryonic dark matter to the mix to
> explain what we see.
One other note: The redshift shouldn't be affected by this increase in
distance. The light would fall into each gravitational well, gaining
energy, but it must lose an equal amount of energy climbing out.
This seems, to me, the main question of the original thread. The
standard model seems to say that the photon would lose more energy
climbing out of a gravitational well than it would gain falling in,
thereby being redshifted. How would this happen? Where does the
energy lost by the photon go?
penguinista
> The Einstein field equation is:
>
> G + L*g = T
>
> Where G is the Einstein curvature tensor, L is the cosmological
> constant, g is the metric tensor, and T is the energy-momentum tensor
> (units with 8*pi*k=1 where k is Newton's gravitational constant).
>
> The FRW manifolds are spatially-homogeneous solutions to the EFE for a
> constant mass density and zero cosmological constant.
>
>
> Tom Roberts tjro...@lucent.com
I think I need a math upgrade to follow that. my differential equation
background doesn't quite cover it.
Tom Roberts
> Tom Roberts wrote:
>>The FRW manifolds are spatially-homogeneous solutions to the EFE for a
>>constant mass density and zero cosmological constant.
>
> FLRW manifolds are the possible solutions to the differential equation
> (G = T)?
Yes.
> Most people are saying that the cosmological constant is one. Ben
> Rudiak-Gould mentioned that zero is way outside the error bars. Does
> this mean that the FLRW metric has been entirely invalidated? Is
> there a working solution to (G + g = T)?
I dunno. As I said before I am not up-to-date on this.
> Secondly, I disagree that the FLRW metric implies a constant mass
> density, even in its most simple form.
I misspoke, and meant "spatially constant". That is, a spatially
homogeneous manifold. Yes indeed, mass density chages over time for all
FRW manifolds.
Tom Roberts tjro...@lucent.com
mlut...@wanadoo.fr
universe, the range of gravity is infinite.
Marcel Luttgens.