The universe expands

Nowadays standard cosmology assumes the empty space itself between the galaxies is expanding and drives them away from each other. The newly formed empty space distinguishes itself in nothing from the old, already present space and will expand at precisely the same rate.

We assume space is uniformly expanding and the rate of expansion is constant in time. The universe is steady in expansion, it always looks as expanding as ever, no matter WHEN you look to the sky.

After a certain time delta-t the distance between any two points of space has doubled. After another time delta-t the distance between any two points of space has doubled again. The doubling every time delta-t is an exponential expansion. When followed backwards in time the galaxies half their speed every interval of time delta-t. They will finally merge but they will never actually come to a complete standstill with respect to each other.

In our idealized expansion our galaxies have no own movement, no proper motion with respect to the expanding frame.

This was the view until I started to find my own conclusions.

Standstill

You are in your laboratory on earth. When you look at the sky you see the galaxies receding from you in all directions and receding faster when further away. If there was a galaxy in every possible place then every possible velocity would be possessed by precisely one galaxy. All galaxies seem to recede from you. They recede slower when more nearby and when such nearby they actually reached you, their speed is zero. You are standing still with respect to the universe.

Distance is measured by redshift. When the distribution of galaxies over space is uniform and curvature is zero, then the number of galaxies of a certain redshift is proportional to r (r = distance to you). You can count this. The frame of standstill then is the frame that at each of its points has such a velocity that the counted distribution is maximal isotropic (in all directions the same). One could call this the Galaxy Frame of Expansion, in brief the Galaxy frame.

The overall curvature might be zero, denoted by = 1. But observations indicate the distribution is not completely uniform and observed large scale streams of galaxies might spoil the simple picture further.

The 2.7 kelvin background radiation (CMB or Cosmic Microwave Background) provides another frame of standstill. We are immersed in the background radiation as in a gas and one can stand still relative to the gas. That is, the frame of standstill is the frame with, at each of its points, a velocity such that the CMB radiation in all directions resembles the best a Planck curve; a velocity such that the CMB is as isotropic as possible with respect to its Planck curve spectrum. Call this the CMB frame. Mind eventually present large scale streamsof the photongas might spoil the simple picture.

The CMB frame is by no means Lorentz invariant. It is possible to move relative to the background radiation with such a speed that in front of you its temperature is enlarged to 10 K. Oké, behind you the temperature drops to virtually zero, but the average overall temperature will be significantly higher than 2.7 K. The background radiation isn't isotropic anymore.

There is only one relativistic invariant radiation field: intensity frequency^3, see THE FIELD OF ALL POSSIBLE FREQUENCIES.

There should be a 2 kelvin neutrino background radiation and a 1 kelvin gravitational wave background radiation, providing still more frames of locally standstill. (Call them the Neutrino frame and the Gravitational wave frame.)

At the very first moments after the Big Bang the elementary particles that will become the frames of galaxies, CMB, neutrino and gravitational waves, are thought to drag each other along in one big expanding motion. When one after the other frame separates, they might evolve a little different from then on. One can separate the frames in observation and if one does, is there already something found? What are the relative velocities of the frames at our location? For observations, what is taken the best as our location, the Sun, or our Galaxy as a whole?

Velocity

Let's enter a rocket frame passing nearby with a constant and high (but not relativistic) speed, e.g. 100,000 km/sec. Your friends at home in the earthly laboratory see that in the sky there is precisely one galaxy having the same velocity as you now. (We neglect for the moment the voids and assume for the moment that everywhere you look there is a galaxy available.) It is in front of the rocket on the straight line of the rocket’s track and it is further away when the rocket frame velocity is higher. In your rocket you observe this galaxy as the only one standing still relative to you.

Will you in your rocket observe the distributions of the galaxy velocities - spherical up until now - be distorted? Egg-shaped or crushed to disk-shape or what?

No matter in what galaxy you are, you know you will see precisely the same picture of the sky: all galaxies recede from you as if you are in the center of the universe *). Observed from the rocket, the distant galaxy standing still to you now is the new center from where all galaxies seem to recede. You know you would see if you would go there AND you are already standing still with respect to that galaxy. As observed in the rocket the expansion remains the same but it is off-center now.

*) This is philosophically remarkable: for centuries people have thought the earth is in the center of the universe and now when people actually see that they are, nobody believes it anymore!

Let’s call the new expansion center your homebase. It is the place where you in your current state of motion would be standing still. Standstill can be defined as the case when you and your homebase coincide.

The Globular Cluster

Globular clusters generally look perfectly point-symmetric. If the cluster had any angular momentum, it would showed itself in some disc-shape properties in the cluster, I suppose. This can be tested by computer simulation. We take the globular cluster to have zero angular momentum. This fits in with the picture that the cluster did arise from a single cloud of gas that had no angular momentum.

When disordered motion, as in globular star clusters, is transformed into an ordered expansion by translocating stars, keeping their impulse vector lengths and directions the same, the result looks like a fireworks expansion. In a fireworks expansion as it is meant here, the velocity of one piece of firework is proportional to the distance to the expansion center. It looks very much like the expanding universe described above. Each possible velocity has its unique proper place. This space then is a version of the set of all possible velocities.

Let's start to convert the globular cluster to a fireworks expansion. First we make sure to stand still relative to the mass center of the globular cluster. Next we choose an expansion center and we choose a sphere around it of points of 1 m/s. Although an unnecessary step, we choose a convenient moment in the history of the globular cluster, where all stars have convenient velocities. Now we translocate each star of the globular cluster to its proper place in the fireworks expansion, keeping velocity properties (magnitude and direction) the same while translocating. We swiftly determine the mass center of the new fireworks star cluster and call it our home base. The home base has a certain velocity in the chosen fireworks expansion coordinate system. We speed up our frame of reference until our system stands still with respect to the home base. The fireworks expansion now expands from the homebase, the expansion center coinciding with the mass center.

The impulse (mass of cluster times speed of mass center) and angular momentum of the original globular cluster is zero. The impulse and angular momentum of the new fireworks star cluster is zero too. Only the energy still can differ. Now enlarge (or reduce) the whole picture as in a three dimensional photocopier until the energy (kinetic plus potential energies) of the stars in the original globular cluster and in the new fireworks star cluster are the same. This multiplication is just another translocation of stars, their velocity vectors remaining constant while translocating.

Is there another way? Suppose you translocate stars while maintaining its impulse, angular momentum and energy, thus constantly obeying conservation laws, does this work? Regard distance r between mass m of the star you translocate and the mass center of the globular cluster. To obey angular momentum conservation law mvr = constant, the velocity magnitude must change while translocating. After translocation, r has increased by a factor b, r(before) times b = r(after). Since r in mvr has increased by factor b, you must divide v by b in order to keep mvr at constant. That is, the component of v perpendicular to r(start) has to be divided by b; the velocity component along r(start) remains unchanged.

Since the velocity has changed now, the impulse mv changes too, as does the kinetic energy mv/2 and the potential energy GMm/r, where M mainly is the mass of the globular cluster below the translocated star. The shell around the mass center above the star tends to cancel itself out, see A sphere of free falling clocks, page 1 of NEG (storyline NEWTON EINSTEIN GRAVITATION).

Well, this is not my strongest point.

NDER   ONSTRUCTION,

eventually.

Sphereworld

Larry Niven has written a book named Ringworld. There they’d build a huge ring that rotates around a star, the star is in the center of the ring. The rotation is such the centrifugal force provides a gravitation on the inner surface of the ring like on the earth. Something the like can be done with the expansion of the universe. Suppose you chain a large circle of galaxies in such a way they cannot fly away from each other no more. The expansion of the universe tries to draw them further apart. The chain effectively prevents this. The result is a force on the chain that is a kind of gravitation. Instead of galaxies on a circular chain we build a solid sphere just somewhere in empty space. Take its magnitude such that, when standing on its inner surface, the experienced gravitational force is 1 g. The sphere doesn’t rotate, no centrifugal forces arise.

When you start with a first sphere of a certain diameter, you construct on its surface a second sphere. Then you dig up the parts of the first sphere and use them to build the third sphere on the surface of the second sphere. And so on until the desired magnitude is reached.

It will take some time.

You stand on the inner surface of the sphere. What forces act on you? For convenience, assume there are no stars in the volume in the sphere, or other matter. How large is the gravitational force F excerted on you by the mass M of the sphere? Inside a perfect sphere holds F = 0, outside the sphere F is as if M is concentrated in the sphere’s centre. (Gauss’ Theorem, holds too in GR).

Well, since you reside on the inner surface: F = 0. But in case you want to get out, let’s calculate.

The radius of the sphere is r, the gravitational constant = G.
F exerted by M on 1 kg of you-just-outside-the-sphere
= G * 1 * M / r (Newton’s gravitational law)
= ( G / r ) * 4 r * U,
where U is the mass of a unit column of the sphere, a column of 1 square meter outer surface times the depth, or thickness, of the sphere
= 4 G U
= 8.4 * 10^-10 * U

The gravitational force of the sphere on you is proportional to its thickness (and mass density) and is independent of its size.

Make U as small as possible for practical use, e.g. construct the sphere of a kind of spider web with still stronger and lighter treads and larger holes, but not that large that one of our spaceships would immediately fall through it.

On the inside surface of the sphere one feels a one-g gravitation, which in fact is the accelerated expansion of the universe. A spaceship is standing there on a large hatch and the hatch is opened to the outside. The spaceship, free floating now, leaves the sphere by the one-g of the expansion of the universe at the sphere’s surface. Suppose one let fall the whole sphere along with the spaceship. One has to accelerate the sphere then. The parts of the world on the inner surface of the sphere in the neighbourhood of the ship will be weightless, exerting no force anymore on the sphere’s inner surface. On the other hand, at the other side of the sphere one expects a doubled gravitational force to be felt there. These combined forces make the sphere to counteract acceleration. The acceleration of the sphere will be undone and the sphere is free floating again.

(In fact correcting disturbances travel with light speed or slower along the sphere’s surface, effectively causing the surface to vibrate.)

Summary for an accelerated expanding universe

1) Is there a preferred system of standstill in the universe?

Yes, there is: the set of all velocity states where you coincide with you homebase. The preferred system of standstill expands, but locally a standing-still frame can always be found.

2) Is there a force acting to bring you to a standstill with respect to this preferred system?

No, there isn’t. When you have a velocity you will maintain to have that velocity. It is taken along with the expansion of the universe and increases with it. Everywhere you are the constant value of your speed will be added to the zero standstill velocity (see 1) in that location.

Especially a small velocity increment dv will be maintained while taken along with the expansion and so a lorentz invariant field of all possible velocities can be constructed in the expanding space of the universe too.

3) Is there a force opposing acceleration?

Yes, there is, for each coherent object occupying some space that expands, see Static curvature of space tends a gas to cool down in the column at the right. However the force is very tiny for the small spaces occupied by the elementary particles, it in principle keeps them at constant speed when no other forces are acting on them.

4) Is this force explaining the first law of Newton, A mass where no forces are acting upon, maintains its velocity?

No, I don’t think so. It could if it wasn’t for the weakness of the force. But the principle is worth remembering, it might work in other cases.