Undergrad Hubble's Law and Star Velocity

[Gege01]

TL;DR

The universe is still expanding faster! The maximum expansion speed can reach the speed of light C! Thereafter, the expansion rate slowed down.

Hubble's Law and Star Velocity
By using Hubble's law and the principle of velocity superposition of relativity, we can deduce when the distance of star from the observation point is (photometric distance)r, its velocity V can be expressed as:
V = Csin (Hr/C)
Therefore, the following conclusions are drawn:

The universe is still expanding faster! The maximum expansion speed can reach the speed of light C! Thereafter, the expansion rate slowed down.
The maximum radius of the universe is 3.14 times that of today.

 

[phyzguy]

Most of what you've said here is incorrect. I highly recommend the attached "Expanding Confusion". Read it carefully and if you have questions, come back and ask.

 

[Dale]

V = Csin (Hr/C)

This equation appears to be dimensionally inconsistent. Where did you get it?

 

[Gege01]

Thank you very much!

 

[Gege01]

This equation appears to be dimensionally inconsistent. Where did you get it?

There is a functional relationship between star velocity and distance:

V=V(r)

 

[Gege01]

There is a functional relationship between star velocity and distance:

V=V(r)
View attachment 245993

It can satisfy Hubble's law :

V(dr’) =Hdr’
The true velocity V(r+dr) of B, measures by viewer of o, is the superposition of the velocity V(r) of the star A and the velocity V(dr') of the star B relative to A.

 

[Gege01]

The relationship between dr’ and dr is:

dr’=γdr
Also, according to the differential definition:

V（r+dr）-V(r)≡dV(r)
We get

After the above formula, wei have

 

[Gege01]

Most of what you've said here is incorrect. I highly recommend the attached "Expanding Confusion". Read it carefully and if you have questions, come back and ask.

Thank you very much!

 

[Dale]

View attachment 245994
It can satisfy Hubble's law :

V(dr’) =Hdr’
The true velocity V(r+dr) of B, measures by viewer of o, is the superposition of the velocity V(r) of the star A and the velocity V(dr') of the star B relative to A.
View attachment 245995

Oh, I see. I thought your Hr was the Hubble radius so Hr/c would have units. But you meant H is the Hubble constant and r the radius so Hr has units of velocity and Hr/c is dimensionless.

Next time please be clear about the meaning of your variables.

 

[Gege01]

Thank you for your reminder.

 

[Dale]

We get

View attachment 245997
After the above formula, wei have

View attachment 245998

Note that the last formula only follows from the previous one for $r\le \frac{\pi}{2} \frac{c}{H}$. So you cannot use it to claim:

The maximum expansion speed can reach the speed of light C! Thereafter, the expansion rate slowed down.

Your claim is based on applying a formula in a region where it is not valid. I am not certain that the rest of the derivation is correct, but at least that part has clear mathematical limitations.

 

[Gege01]

Yes, some of the conclusions are not very rigorous.

 

[Gege01]

However, it is reasonable in the scope of r≤πc/(2H).

 

[Dale]

However, it is reasonable in the scope of r≤πc/(2H).

I am not sure. With none of the variables being clearly defined it is hard to follow. Unless this has been derived in the professional scientific literature (not only Chinese) then it is highly suspect.

 

[Gege01]

Based on the following facts That Hubble's law is valid at short distances and at small speeds:

V(dr') = Hdr'

 

[Gege01]

Although we can't guarantee that Hubble's law holds at large distances and high speeds, we can still assume that special relativity is satisfied.

The true velocity v (r + dr) of b, measured by the observer of o, is the superposition of the velocity v (r) of a star and the velocity v (dr') of B star relative to a.

 

[Ibix]

The relationship between dr’ and dr is:

dr’=γdr

This appears to me to be attempting to relate the distance, $dr$, between two nearby objects as measured by local observers to the distance, $dr'$, measured by observers who see the objects receding at $Hr$. It seems to be simply the special relativistic formula for length contraction, and therefore assumes a single inertial reference frame covering a spatial patch of radius at least $r$. This seems to me unlikely to be valid at cosmological distances.

 

[Dale]

we can still assume that special relativity is satisfied

No we can’t. Special relativity is only satisfied for flat spacetime. The FLRW spacetime is not flat.

The true velocity v (r + dr) of b, measured by the observer of o, is the superposition of the velocity v (r) of a star and the velocity v (dr') of B star relative to a.

What is dr’. What are a, b, and B? Which observer is o?

 

[Gege01]

 

[Gege01]

No we can’t. Special relativity is only satisfied for flat spacetime. The FLRW spacetime is not flat.What is dr’. What are a, b, and B? Which observer is o?

Although FLRW spacetime is not flat，We can still think that special relativity can be satisfied in a very small space-time range.

 

[Gege01]

O is the position of any observer. A is a star whose distance from the observer is r.

 

[Gege01]

B Distance Observer Distance is r+dr

The distance of B measured by point A observer is dr'

 

[Dale]

Although FLRW spacetime is not flat，We can still think that special relativity can be satisfied in a very small space-time range.

Sure, but “the maximum radius of the universe” hardly qualifies as “very small”.

 

[Mordred]

What your seeing as expansion >c is a consequence of separation distance between the observer (us) and the objects beyond Hubble horizon. In point of accuracy the expansion rate is approximately 70 km/s/Mpc far slower than c. If you situate an observer at every Mpc this rate will be the same in our universe today.
Due to Hubble law $v_{resessive}=H_OD$ the greater the distance the greater the resessive velocity the separation distance gives the appearance of exceeding c. However this is merely a consequence of the separation distance and application of Hubble's law nor a true velocity.

If you prefer to apply Hubble's law in terms of radius simply replace D with r. V=Hr. SR isn't needed to apply this as pointed put it's a simple consequence of separation distance and expansion rate.

 

[Gege01]

Sure, but “the maximum radius of the universe” hardly qualifies as “very small”.

When we study specific problems, for example, the distance between A and B is R and r+dr, then, in the range of R to r+dr, it is a small area. Special relativity is satisfied

 

[Mordred]

The radius of the observable universe being roughly 46 billion light years isn't small. Think about how far into the past you are looking at when you see the surface of last scattering at z=1100. Does that sound like something SR handles ?

Edit: I should specify in this scenario as certainly the FRWL metric applies GR and SR. Example being it's light cone and worldline applications.

 

[Gege01]

The radius of the observable universe being roughly 46 billion light years isn't small. Think about how far into the past you are looking at when you see the surface of last scattering at z=1100. Does that sound like something SR handles ?

Edit: I should specify in this scenario as certainly the FRWL metric applies GR and SR. Example being it's light cone and worldline applications.

If the universe is finite but unbounded, it means that the universe is smaller than the observable universe. In this case, galaxies that look far away may be illusions of neighboring galaxies, which are illusions created by light circling the universe. This hypothesis is very difficult to detect because galaxies are different at different ages, or even completely different.

 

[Mordred]

No it doesn't the observable portion is certainly finite as the cosmological event horizon sets an observational limit. The full universe could very well be finite or even infinite.
We simply do not know as we cannot observe or measure the entire universe. We however do know it is far larger than our observable portion.

The mirror effect you described was once studied but observational evidence of objects at extreme distances have different metalicity percentages than objects nearby. Then there is also different temperatures involved. As you approach the CMB the temperature rises.

Observational data doesn't rely solely on distance estimates but also includes thermodynamic changes due to expansion.

 

[Gege01]

No it doesn't the observable portion is certainly finite as the cosmological event horizon sets an observational limit. The full universe could very well be finite or even infinite.
We simply do not know as we cannot observe or measure the entire universe. We however do know it is far larger than our observable portion.

If we use the formula in this post, some data may be changed.

 

[Gege01]

This conclusion is based on the following assumptions:
1. Hubble's law is valid at small distances and velocities.
2. Special relativity can be satisfied in a very small space-time range.
3. There is a functional relationship between star velocity and distance:
V=V (r)