exploding at the sound of logic

[michel123456]

-------------------

I didn't expect to be such a difficult question to answer.

 

Let me put it in another way:

 

From the wiki article it says that

 

The accelerating universe is the observation that the universe appears to be expanding at an increasing rate. In formal terms, this means that the cosmic scale factor has a positive second derivative,^[1] so that the velocity at which a distant galaxy is receding from the observer is continuously increasing with time.^[2]

 

What I say is that a scale factor by itself introduces a distance continuously increasing with time. There is no need for an "increasing scale factor".

Am I wrong?

[Strange]

There is no need for an "increasing scale factor".

 

That is what they mean by "expanding at an increasing rate".

As you can tell from the fact they go on to say "In formal terms, this means that the cosmic scale factor has a positive second derivative".

[michel123456]

 

That is what they mean by "expanding at an increasing rate".

As you can tell from the fact they go on to say "In formal terms, this means that the cosmic scale factor has a positive second derivative".

Right.

What i say is that "expanding at an increasing rate" is a direct effect of scaling. The second derivative is inherent (or maybe I am wrong here, that was my question)

Now, if the data show that something more is needed to coincide with the observation, then it is not about an acceleration of expansion anymore, it is about an increase of acceleration of expansion, a third derivative.

[Strange]

What i say is that "expanding at an increasing rate" is a direct effect of scaling.

 

No, that is not what it means. A constant scaling factor means that (apparent) recessional velocity is proportional to distance (Hubble's Law) as some basic arithmetic will demonstrate.

 

If you get things receding faster than that (for a given distance) then the scaling factor has increased and expansion has accelerated.

[michel123456]

 

No, that is not what it means. A constant scaling factor means that (apparent) recessional velocity is proportional to distance (Hubble's Law) as some basic arithmetic will demonstrate.

 

If you get things receding faster than that (for a given distance) then the scaling factor has increased and expansion has accelerated.

But the Hubble's law is a kind of acceleration. When velocity increases with distance (and thus with time) it corresponds to an acceleration.

 

And accordingly, if you need more than that, you are talking about an increasing of acceleration.

[Strange]

But the Hubble's law is a kind of acceleration. When velocity increases with distance (and thus with time) it corresponds to an acceleration.

 

Acceleration requires a force. If you are in free fall, for example you are not accelerating even though your speed relative to the ground is increasing.

 

Edit: as we are talking about GR, the word "acceleration" means proper acceleration.

https://en.wikipedia.org/wiki/Proper_acceleration

 

 

And accordingly, if you need more than that, you are talking about an increasing of acceleration.

 

If you want to change the way words are used and view it that way, you can I suppose. But, not surprisingly, you will get confused when you encounter the standard descriptions.

Edited December 28, 2015 by Strange

[michel123456]

 

Acceleration requires a force. If you are in free fall, for example you are not accelerating even though your speed relative to the ground is increasing.

 

 

If you want to change the way words are used and view it that way, you can I suppose. But, not surprisingly, you will get confused when you encounter the standard descriptions.

You are correct. I am confused.

[Strange]

You are correct. I am confused.

 

The increasing speed of separation with distance is just basic arithmetic/geometry; there is no force involved. Therefore there is no acceleration.

 

Edit: as we are talking about GR, the word "acceleration" means proper acceleration.

https://en.wikipedia.org/wiki/Proper_acceleration

Edited December 28, 2015 by Strange

[michel123456]

 

The increasing speed of separation with distance is just basic arithmetic/geometry; there is no force involved. Therefore there is no acceleration.

We have bumped onto definitions.

To me a change of velocity is always an acceleration, be positive or negative.

The fact that a force is involved or not has no importance for the definition of acceleration.

.

Edited December 28, 2015 by michel123456

[Strange]

We have bumped onto definitions.

To me a change of velocity is always an acceleration, be positive or negative.

The fact that a force is involved or not bares no matter.

 

Then you are not talking about physics.

[michel123456]

 

Then you are not talking about physics.

So, how do you call a change of velocity when no force is involved?

[Strange]

So, how do you call a change of velocity when no force is involved?

 

Apparently, this is called "coordinate acceleration": https://en.wikipedia.org/wiki/Proper_acceleration

[michel123456]

You should maybe edit your comment in post #33

And in post #31

[Strange]

Done.

[Mordred]

[latex]v_{recessive}=H_{o}d[/latex]. Recessive velocity as Strange pointed out is an apparent not a true velocity no inertia is involved. Think of it this way....Take a galaxy then surround that galaxy with a homogeneous and isotropic medium. That medium being the interstellar medium. Now we know f=ma from the three laws of inertia.

 

So ask yourself this if the medium surrounding all galaxies is uniform meaning there is no pressure difference on any angle or direction. Then f=ma cannot be applied. Instead the increased distance can only be accounted for by and increase in space itself. As f=ma isn't involved then it would be inaccurate to treat the galaxies appparent velocity based on inertia.

 

Also the term explosion is also inappropriate except as a measure of rate of change. Explosions start at a point and radiate outward. This isn't the case with expansion. There is now origin point or direction involved. Expansion is homogeneous and isotropic. No preferred direction or location. (I mentioned this due to the thread title lol)

Edited December 30, 2015 by Mordred

[michel123456]

[latex]v_{recessive}=H_{o}d[/latex]. Recessive velocity as Strange pointed out is an apparent not a true velocity no inertia is involved. Think of it this way....Take a galaxy then surround that galaxy with a homogeneous and isotropic medium. That medium being the interstellar medium. Now we know f=ma from the three laws of inertia.

 

So ask yourself this if the medium surrounding all galaxies is uniform meaning there is no pressure difference on any angle or direction. Then f=ma cannot be applied. Instead the increased distance can only be accounted for by and increase in space itself. As f=ma isn't involved then it would be inaccurate to treat the galaxies appparent velocity based on inertia.

 

Also the term explosion is also inappropriate except as a measure of rate of change. Explosions start at a point and radiate outward. This isn't the case with expansion. There is now origin point or direction involved. Expansion is homogeneous and isotropic. No preferred direction or location. (I mentioned this due to the thread title lol)

What I said is that the formula [latex]v_{recessive}=H_{o}d[/latex] describes a change in velocity AKA an acceleration in my simplistic vocabulary. It is already a second derivative. If the universe expands more than that, it means one has to introduce the acceleration of an acceleration, a 3rd derivative.

[Strange]

But, as this is model is based on general relativity, you have to use the appropriate definition of acceleration, not the "common sense" one.

Edited January 1, 2016 by Strange

[michel123456]

But, as this is model is based on general relativity, you have to use the appropriate definition of acceleration, not the "common sense" one.

In this view you must also change the definition of velocity and finally the definition of distance.

[Strange]

In this view you must also change the definition of velocity and finally the definition of distance.

 

Well, relativity uses generalizations of these (4-velocity and space-time interval) if that is what you mean.

[Mordred]

What I said is that the formula [latex]v_{recessive}=H_{o}d[/latex] describes a change in velocity AKA an acceleration in my simplistic vocabulary. It is already a second derivative. If the universe expands more than that, it means one has to introduce the acceleration of an acceleration, a 3rd derivative.

I think the problem your having is that recessive velocity used in Hubbles law is inaccurate. Hubble didn't know why galaxies were receding. He only knew that the greater the distance the more apparent it became.

 

We're stuck using velocity as a result. However none of the galaxies gain momentum in any fashion due to expansion. The separation distance does increase. However it does so without imparting momentum.

 

You can however accurately consider it an apparent acceleration. As long as you realize it's not a true acceleration or velocity.

[michel123456]

I think the problem your having is that recessive velocity used in Hubbles law is inaccurate. Hubble didn't know why galaxies were receding. He only knew that the greater the distance the more apparent it became.

 

We're stuck using velocity as a result. However none of the galaxies gain momentum in any fashion due to expansion. The separation distance does increase. However it does so without imparting momentum.

 

You can however accurately consider it an apparent acceleration. As long as you realize it's not a true acceleration or velocity.

Ok that is a clear explanation.

So we have an apparent acceleration and an apparent velocity that do not involve momentum, caused by a scaling factor that is not provoked by any force.

And on the top of that, we have an increase of the apparent acceleration that is supposed to be the result of a force and thus must involve momentum. Is that it?