Expansion; it couldn't be this simple, could it?

[Stonius]

I was thinking of this explanation for the expansion of the universe. I'm sure there is either a big reason why it's not possible, or someone else has thought of it first, but here goes.

Let's say I throw a ball at 60km/hr. That falls back to earth some tens of meters away.

Now say I had a magic box that could 'turn down' the earth's gravity. I throw the ball as before. Then while it's in flight, I turn the earth's gravity down. That same ball with the same momentum sails high into the sky. If I turn it down again it *seems to accelerate until it leaves the gravitational pull of the earth altogether. *Seems to, only it's just the gravity of the earth that is being reduced.

Then consider the fact that at some point as the universe expanded, the size of the *observable universe diverged from the size of the *total universe.

From any one point in space, it would appear that the observable universe was losing mass - you can't be gravitationally linked to something that is not in your observable universe. Essentially, the universe you are in becomes less massive, like turning down the gravity. The more matter that is lost, the more the expansion seems to be accelerating, when in effect, the momentum is the same. Only the gravitational force acting on the universe as a whole is decreasing (like with my magic 'turn down the gravity' box).

In fact, the cause of the big bang could ultimately be explained by a divergence between the observable and total universes in the primordial 'atom' in a similar way to which the weak nuclear force is involved in nuclear decay. All you need is a force that doesn't quite have enough range to keep a glob of matter together, so some of it escapes, in this case, releasing energy and leading to rapid and accelerating expansion.

It's interesting to ponder, anyway.

Markus

[iborg]

Hi Markus

I would think that the cosmic microwave background at least implies that neither mass nor energy have left the observable universe.

I have come across the idea that gravity may weaken over extreme distances. I don't know what became of the idea.

It was a suggestion in regard to the Voyager spacecraft as they are not quite where they were originally expected to be.

Philip

[Stonius]

Why is that? Couldn't there still be CMB under this model, depending on how the divergence between observable and total universe developed?

[iborg]

Hi

My understanding is that the CMB is directly and immediately resulting from the Big Bang itself. Not something generated later. Observed later of course, and much altered over time.

Effectively, there is no before that point in time, or outside the location the CBM originates from. I expect that neither of these statements are actually accurate, but, probably close enough for now.

I did read an article about how we are currently well positioned in time, to determine the origin of the universe. This relates to your 'Why is that?' question.

If you are interested, I will try to find it.

Philip

[Stonius]

Maybe I don't see it - but why would my :)crackpot:) idea require the CMB to be generated later? I see it as happening the same as the current model.

We already have a discrepancy between the observable universe and the total universe (however big that is). I'm simply suggesting this discrepancy and the reducing mass of the observable universe as a driver for the accelerated expansion we observe.

[iborg]

My reasoning is that you said "Then consider the fact that at some point as the universe expanded, the size of the *observable universe diverged from the size of the *total universe."

I would change this statement to "will expand".

I think that the CMB indicates that the total universe has not yet expanded beyond the observable universe. The CMB is, so to speak, the edge of the total universe.

I think this is one the points that the article mentioned makes.

[Stonius]

I thought that the CMB came from the recombination/transparency event ie, after the opaque era? Which is to say, some time after the actual big bang itself?

Markus

[leon]

Now say I had a magic box that could 'turn down' the earth's gravity. I throw the ball as before. Then while it's in flight, I turn the earth's gravity down. That same ball with the same momentum sails high into the sky. If I turn it down again it *seems to accelerate until it leaves the gravitational pull of the earth altogether. *Seems to, only it's just the gravity of the earth that is being reduced.

There is you answer

Leon :thumbsup:

[Stonius]

It's an analogy to illustrate that a body in motion in a changing gravitational field would appear to be accelerating or decelerating. Surely you'll allow me that by way of illustrating the main point? I'm not saying such a magic box is required or actually exists.

[iborg]

Hi
My boo boo, I thought it was from earlier than recombination.

I'll try thinking again.

Philip

[sjastro]

Hello Markus,

I'm afraid your idea doesn't work as it cannot explain the linear relationship between distance and recession velocity.
The linearity can only be explained if the space between stationary objects expands rather than the objects themselves moving through space.

If an object recedes from the observer in space at an increasing acceleration at non relativistic velocities you get a rather messy mathematical equation of the format.

x=vt+(dv/dt)t^2/2!+(d^2v/dt^2)t^3/3!+(d^3v/dt^3)t^4/4!
+.......

x is the distance, v the recession velocity, t is time, (d^nv/dt^n) are the increasing higher order derivatives of velocity with respect to time.

The equation is clearly not linear for distance and recession velocity.

Regards

Steven

[Stonius]

Hi Steven, sorry for the late reply - I've been away. Also, I've been trying to get what you're saying - I'm not particularly mathematical. So let me see if I have this right?

So you're saying my idea would not be linear as is observed in nature, but would instead be, what, logarithmic?



Which is interesting because if spacetime expands only in the absence of matter, then you would assume that the gravitational force has limited range, or falls off in a way we don't understand, which would have implications for dark matter, one would have thought.



Why non-relativistic? Isn't the whole point of the Observable universe's event horizon that it recedes from us at relativistic velocities? That's why we can never see past it?



So why can't we just use a = Δv / Δt if we're talking non-relativistic velocities, or a 4-vector for relativistic velocities? I'm not sure why the higher order derivatives are necessary, or why you begin by multiplying Velocity and time instead of dividing delta time by delta velocity. Sorry for my ignorance. Which acceleration equation are you using?

Best,

Markus

[sjastro]

Hello Markus,



The mathematics is to show the non linearity of recession velocity vs distance for an object travelling through space if acceleration is present without having to know the specific values for acceleration.
This will become clearer below.



In the early Universe there was no matter in the form of atoms, molecules, stars, and galaxies but gravity still existed.
Where as gravity requires the presence of mass in the Newtonian model, mass and energy are the requirements for gravity in General Relativity.
In the early Universe most of the energy was carried by photons and neutrinos travelling at relativistic velocities in a radiation dominated Universe which was followed by a matter dominated Universe (including dark matter) which gave way to a dark energy dominated Universe from about six billion years ago.
The difference of each stage of the Universe relates to the how expansion varies as a function of time.
https://en.wikipedia.org/wiki/Scale_factor_(cosmology)


As strange as it appears objects that have a recession velocity due to space expansion that exceed c, the speed of light, are no longer “relativistic” because the equations for special relativity don’t apply.
In this case if an object is travelling at a recession velocity v which exceeds c and a photon is emitted by the object back towards the observer at a speed c, the observer will only see the object if the vector equation is v-c<c.
In the case of your example for objects moving in space it is mathematically simpler to deal with Newtonian time and slower speeds without having to worry about time dilation.
Mathematically this is a good approximation for velocities up to about 90% the speed of light.



The equation a = Δv / Δt is only valid if the acceleration is constant.
If the acceleration is not constant as your idea suggests then Δv / Δt will vary at any given location.

The equation x=vt+(dv/dt)t^2/2!+(d^2v/dt^2)t^3/3!+(d^3v/dt^3)t^4/4+...... is known as the Taylor expansion of a function.
Many functions can be broken down into this series and expressed to any given degree of accuracy depending on the number of terms in the expansion.

Consider the equation
dv/dt=d2x/dt2=0 which is the equation for zero acceleration.
Integrating this equation gives dx/dt=v where v is the velocity.
Integrating further gives x=vt, where x is the position at time t for an object moving at constant velocity v.
Note this is the first term in the Taylor expansion.

Since there is no v^2 or higher term (or any term other than the power of one) there is a linear relationship between x and v at any time t.
Despite the linear relationship, it bears no resemblance to the recession velocity distance relationship in Hubble’s law as it has a zero gradient due to the constant velocity.

Incorporating acceleration into the picture doesn’t help either.
Consider the equation dv/dt=d2x/dt2=a where a is uniform or constant acceleration.
Integrating gives dx/dt=at+v
Integrating again gives x=(a)t^2/2+vt
=vt+(a)t^2/2
=vt+(dv/dt)t^2/2!
Note we now have the first two terms in the Taylor expansion.
If the acceleration is not constant the Taylor expansion incorporates more terms, the higher the accuracy the more terms are required.

Eliminating t from the equation dv/dt=a for constant acceleration shows the non linearity between v and x.

dv/dt=(dx/dt)(dv/dx) by the chain rule
=vdv/dx
=d/dv(1/2v^2)dv/dx
=d(1/2v^2)/dx

Hence d(1/2v^2)/dx=a
Integrating gives 1/2v^2=ax+u^2/2 where u is the initial velocity.
Hence v^2= 2ax+u^2
Clearly there is no linear relationship between v and x because of the v^2 term.
Any constant acceleration will destroy the linear relationship between v and x.

A variable acceleration which includies higher terms in the Taylor expansion will also result in non linearity.
Hence we don’t need to specifically know the acceleration to indicate the non linear relationship between v and x.

For objects moving in space, it is impossible to reproduce Hubble’s law

Hope this clears it up.

Steven