|NEWTON EINSTEIN KIEKENS GRAVITATION|
|THE EXPANSION OF THE UNIVERSE|
The expansion of the universe
The universe expands
Here we assume the empty space between the galaxies itself 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.
In this page we assume further 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.
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 square (r = distance to you). You can count this. The frame of standstill then is the frame where the counted distribution is maximal isotropic (in all directions the same).
The overall curvature might be zero, denoted by = 1, but the distribution is not uniform and large scale streams of galaxies might spoil the simple picture further.
The 2.7 kelvin background radiation 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. But large scale “streams”of the “gas” spoil the simple picture.
There should be a 2 kelvin neutrino background radiation and a 1 kelvin graviton background radiation, providing still more frames of standstill.
Now you enter a rocket frame passing nearby with a constant and high (but not relativistic) speed, e.g. 1000 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. It is 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 velocity distributions of the galaxies - 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. (Which is philosophically remarkable: for centuries people have thought the earth IS in the centre of the universe and now when people actually SEE that they are, nobody believes it anymore!) Observed from the rocket, the distant galaxy standing still to you now is the new centre from where all galaxies seem to recede. You know you would see if you would go there. As observed in the rocket the expansion remains the same but it is off-centre now.
Let’s call the new expansion centre 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.
Mind this frame of standstill can be defined too when the acceleration is very small, or even when the acceleration is zero.
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 NEKG.
Well, this is not my strongest point.
NDER ONSTRUCTION; eventually.
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: all velocity states where you coincide with you homebase.
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. 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.
Accelerated expansion and static curvature
Assume space is flat (no curvature) and expands accelerated.
Imagine eight galaxies happen to be on the corner of a rectangular box with relative small height and depth but large length. When the universe expands and the galaxies are driven apart, the box preserves its shape while expanding. Three particles 1, 2 and 3 (like protons or atoms) propagate collectively along parallel lines through empty space on the edges of the box.
Since the box preserves its shape its edges will remain parallel. The paths of the particles lie along those edges and thus will remain parallel too. The velocities of the particles remain parallel. That is, with respect to the box. The corner-galaxies of the box are free-floating and fall along with the expansion of the universe; hence the box is a co-expanding frame of reference.
Fig.2 time t = 0
But the atoms of the earth resist the expansion. Our meter sticks don’t expand along with the universe. When the space between the particles 1, 2 and 3 expands, they curve away from each other. After a time delta-t the distance between particles 1, 2 and 3 has increased: as observed in the earthly frame of reference particle 1 as well as particle 2 must have accelerated relative to particle 3 and have a velocity in the y direction now. As observed from the earth, measured with earthly meter sticks, the particles follow curved paths.
The three particles experience themselves as moving straight forward with constant speed. As far as concerned themself a particle draw out straight tracks; it are the others that are accelerating, deviating away from him.
Fig.3 time t = delta-t
Negatively curved space is in principle static - it is not automatically expanding or anything the like. In negative curved space, straight lines that are parallel at some place deviate from each other when followed further on the track. This look like the deviating tracks of particles in a flat, accelerated expanding space.
And light? We had particles so far, but what about light speed? Locally light always moves with light speed where it obeys all demands of special relativity. But localities separated as far as galaxies recede from each other by the expansion of the universe. ”A” (fig. 1) is a star precisely in the middle of an edge connecting two corner-galaxies. A photon leaves one of the two corner-galaxies and heads for the other one. When it has reached A it knows it is half way down. But the other half takes a little longer because that half had expanded a little in the mean time, elongating the travelling time. The total covered distance - which is the distance between the two galaxies - at the photon’s departure is smaller that at its arrival. Speed as covered distance divided by the occupied time becomes a little bit meaningless when distances are expanding all the time. Locally light always travels at light speed and the localities might recede from each other. That’s all there is to light speed in an accelerated expanding universe.
Wave 4 is an endless array of photons propagating on the same line. They all have the same wavelength and phase, the next photon steps precisely in the wavepattern of its predecessor. It is a ray of light, propagating perpendicular to the tracks of particles 1, 2 and 3. When particle 1 and 2 are one wavelength apart they keep one wavelength apart. The distance between 1 and 2 increases and the wavelength of 4 increases by the same amount. Wave 5 at time t= delta-t is a new wave equal to wave 4 at time t = 0.
Here we end our description of a uniform accelerating universe. In the previous pages of the storyline THE EXPANSION OF THE UNIVERSE is suggested the acceleration of the expansion of the universe is caused by backward time evolving, dark antimatter universes, as much as ours. There dark mechanics rules. If so, then the acceleration of the expansion of the universe is certainly not uniform at details as large as galaxy clusters. And then it is the question how much of the reasoning of this page will hold up.
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Local gravitation (locally uniform gravitation) and acceleration are indistinguishable: “laboratory experiments in a rocket standing still on the earth surface are indistinguishable from the same experiments in the same rocket in empty space (no gravitation) with its engine at one G propulsion”.
Gravitation described as positive spacetime curvature has a preferred frame of standstill (C-60 molecules uncontracted) that will not be found in its free-space accelerated rocket counterpart. Thus in GR in a uniform gravitational field straight lines remain parallel, even when the field is strong.