# I Hubble's Law and Star Velocity

#### Gege01

Summary
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.

Related Special and General Relativity News on Phys.org

#### 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.

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#### Dale

Mentor
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

Mentor
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.

#### Dale

Mentor
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

Mentor
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

Mentor
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

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

Mentor
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.

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#### 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

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