Jump to content

Is Hubble law Euclidean?


Recommended Posts

In light travel distances this would mean that if we consider three points aligned in space, the distance between the two extremes is not the sum of the two sub-segments..

 

Let me explain. The idea is that the Hubble coefficient in the light travel distance (LTD) framework is expressed as a function of LTD: the recession speed is Ho*c*T where T is the distance in LTD and c the celerity of light.

The Hubble coefficient is in the Euclidean framework is expressed as a function of Eclidean distances: the recession speed is He*y where y is the Euclidean distance.

The rationale is to find a time-varying Hubble coefficient in the Euclidean framework, such that Ho*c*T = He(t)*y.

 

Because space is expanding, the expansion is added to the overall distance the photon has to travel to reach the observer, therefore we obtain two differential equations describing the distance between the observer and the photon:

(1) In the LTD Framework at any time t > 0:

dy/dt = -c + Ho*c*(Tb - t) with boundary conditions: y(Tb-T) = yo, and y(Tb) = 0, where Tb is the time of the Big Bang and y the Euclidean distance.

(2) In the Euclidean Framework:

dy/dt = -c + He(t)*y

Now let us consider the nth order Hubble coefficient in the Euclidean framework of the form He = n/t with t the time from the Big Bang.

Therefore the differential equation becomes:

dy/dt = -c + n/t*y (this is a first order non homogeneous differential equation).

 

By solving the math for both differential equations we find that the solution is the same when n=2 (read the paper http://fr.calameo.com/books/00014533338c183febd92 for the details about the math). This is the proof that an apparently steady Hubble in the LTD framework is equivalent to a time-varying Hubble coefficient in the Euclidean framework of order two (i.e. Universe expanding at a steady acceleration).

 

Now using kinetics, I will show why a second order time-varying Hubble coefficient means that the Universe is expanding at a steady accelerated pace in the Euclidean framework. Assuming that the time-varying

Euclidean Hubble behaves according to the same Hubble law than in the LTD framework, leads to He(t) =v(t)/x(t), where v(t) is the Universe expansion velocity and x(t) the scale factor. Let us consider the following form for the time-varying Hubble coeficient in the Euclidean framework: He(t) =n/t, where n is the order of the time-varying Euclidean Hubble, and t the time from the big bang.

 

(1) Considering the simple scenario of a Universe expanding at a constant velocity, we obtain: v(t) = v and x(t) = v * t. Hence, the time-varying Hubble coefficient from equation is He(t) =1/t. A first order time-varying Hubble coefficient is recognised.

 

(2) Considering a Universe expanding at a steady accelerated rate, and with an initial expansion velocity null, we obtain: v(t) = a*t and x(t) = a*t^2/2, where a is the expansion acceleration. Hence, the time-varying Hubble coefficient from is He(t) = 2/t. A second order time-varying Hubble coefficient is recognised.

 

This is to show that a second order time-varying Hubble coefficient in the Euclidean framework, means that the Universe is expanding at a steady accelerated pace.

 

This would have great implication on the calculation of the age of the Universe!

Link to comment
Share on other sites

The Hubble Law, maybe more accurately described as the Hubble formula, seems to be totally Euclidean as you suggest. The B factor in this formula is the only thing complicated about it. It transforms redshifts into distances and is based upon Lorentz Transforms of moving time frames away from the observer. Although simple in form, this factor was quite difficult to derive.

 

The Dark Energy hypothesis is based upon distance calculations using the Hubble formula. If it is off in its formulation by as little as 11%, then Dark Energy would not be needed to explain type 1a supernova observations and maybe not anything else.

 

The Hubble formula must necessarily be flat since Riemann geometry and GR are not part of its formulation. The Hubble formula relates to a straight-line path of EM radiation even though we know that galaxies are not in straight line paths as we observe them, due to the effects of intervening gravity, so that at least some galaxies would be accordingly at closer distances than what their redshifts indicate.

Link to comment
Share on other sites

The geometry of space-time need not be Euclidean to get Hubble's law. All one needs is that the 3-space be locally homogeneous and isotropic. This leads to the FRW cosmologies.

Link to comment
Share on other sites

The geometry of space-time need not be Euclidean to get Hubble's law. All one needs is that the 3-space be locally homogeneous and isotropic. This leads to the FRW cosmologies.

 

Tis true, granted! The Hubble formula does not involve the geometry of space in its formulation but calculates distances independent of such geometry.

Edited by pantheory
Link to comment
Share on other sites

Tis true, granted! The Hubble formula does not involve the geometry of space in its formulation but calculates distances independent of such geometry.

 

Well it does require the 3-space be locally homogeneous and isotropic.

Link to comment
Share on other sites

ajb,

Well it does require the 3-space be locally homogeneous and isotropic.

I agree but imagining that 3 space is not homogeneous and isotropic within the observable universe, I think would be an exercise in futility, but that's just me :)

Edited by pantheory
Link to comment
Share on other sites

I agree but imagining that 3 space is not homogeneous and isotropic within the observable universe, I think would be an exercise in futility, but that's just me :)

 

Well, we all need to think mathematically here. The cosmological principle translates mathematically into the statement that the 3-space is homogeneous and isotropic. (We also assume that a universe is connected).

 

If we were to lose the cosmological principle, then one would be forced to consider such things. In particular one could have "cosmologies" that are not FRW ;)

Link to comment
Share on other sites

Well, we all need to think mathematically here. The cosmological principle translates mathematically into the statement that the 3-space is homogeneous and isotropic. (We also assume that a universe is connected).

 

If we were to lose the cosmological principle, then one would be forced to consider such things. In particular one could have "cosmologies" that are not FRW ;)

Again I agree, but have never considered such cosmologies seriously since, in my opinion, they would be contrary to Occam's Razor, but again that may be only me :)

.

Edited by pantheory
Link to comment
Share on other sites

Again I agree, but have never considered such cosmologies seriously since, in my opinion, they would be contrary to Occam's Razor, but again that may be only me.

 

Well not really against Occam's razor, but certainly not consistent with what we know about the large scale structure of the Universe. It could be possible that on very large scales the cosmological principle is not so good, it is of course not good on the galactic sale. But that is just wild speculation, based on no facts that I am aware of. :lol:

Link to comment
Share on other sites

Well not really against Occam's razor, but certainly not consistent with what we know about the large scale structure of the Universe. It could be possible that on very large scales the cosmological principle is not so good, it is of course not good on the galactic sale. But that is just wild speculation, based on no facts that I am aware of. :lol:

 

As you might realize, if it cannot be explained verbally in a relatively simple and logical manner, I probably will not be too fond of it concerning theory -- but then again that's just me :rolleyes: My belief (and theory) is that on the largest scale beyond the observable universe, that the matter densities as well as the field, will eventually start to fade off gradually into nothingness but this seemingly will be forever unknowable :(

Edited by pantheory
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.