Simple argument for a linear cosmology

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The discussion centers on the derivation of a linear cosmology model using the Robertson-Walker metric, suggesting that the scale factor a(t) is proportional to time t. Participants debate the validity of integrating the metric equations and whether the assumptions lead to a meaningful conclusion about the universe's expansion. The Milne model is mentioned as a potential framework for a linearly expanding universe, but it is noted that current observations contradict this model. The conversation highlights the complexities of cosmological models and the need for further exploration of alternative theories, such as the Milne-Dirac model. Ultimately, the consensus acknowledges the elegance of the linear cosmology idea while recognizing its shortcomings in light of observational evidence.
johne1618
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Starting with the Robertson-Walker metric

\large ds^2 = -dt^2 + a^2(t) [ \frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2\theta d\phi^2]

Consider a light ray emitted at the Big Bang traveling radially outwards from our position.

Therefore we have:

ds = 0
d\theta = d\phi = 0

Substituting into the above metric we have

dt = a(t) \frac{dr}{\sqrt{1-kr^2}}

Integrating both sides and assuming a(0)=0 we have

t = a(t) \int{\frac{dr}{\sqrt{1-kr^2}}}

Thus we must have a linear cosmology

a(t) \propto t
 
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Why do you expect the integral to be constant? If you integrate from 0 to t, I would expect that the integral depends on the integration limits.
 
mfb said:
Why do you expect the integral to be constant? If you integrate from 0 to t, I would expect that the integral depends on the integration limits.

On the left hand side I integrate up to the present age of the Universe t.

On the right hand side I integrate up to some co-moving co-ordinate limit which does not depend on t.
 
johne1618 said:
On the left hand side I integrate up to the present age of the Universe t.

On the right hand side I integrate up to some co-moving co-ordinate limit which does not depend on t.

Hi johne,

You have to integrate both from the same event (e.g. the Big Bang) to the same event, t, r, θ and \phi in the Robertson-Walker metric are all coordinates of the same event.

Whereas your argument does not work (I think you could have guessed that as somebody else would have come up with it a long time ago!) why are you so intrigued by the linearly expanding or freely coasting model?

Note to get such a model dynamically one either has to have an empty universe - the Milne model (obviously not the real universe but it could be the asymptotic limit of an open or DE dominated universe), or a Milne-Dirac model in which the gravitational attraction of matter is counteracted by the repulsion of anti-matter and the universe is divided into matter and anti-matter regions. (As discussed here recently), or by the universe having an equation of state of p = - \frac{1}{3}\rho.

Garth
 
johne1618 said:
dt = a(t) \frac{dr}{\sqrt{1-kr^2}}

does not lead to
johne1618 said:
t = a(t) \int{\frac{dr}{\sqrt{1-kr^2}}}

it leads to

\int \frac{dt}{a\left(t\right)} = \int \frac{dr}{\sqrt{1-kr^2}}
 
johne1618 said:
On the left hand side I integrate up to the present age of the Universe t.

On the right hand side I integrate up to some co-moving co-ordinate limit which does not depend on t.
y=x
dy = dx
##\int_0^a dy=\int_0^b dx##
a=b for all arbitrary, real a,b?

In addition, see the previous post.
 
Garth said:
Whereas your argument does not work (I think you could have guessed that as somebody else would have come up with it a long time ago!) why are you so intrigued by the linearly expanding or freely coasting model?

I just think that the linear cosmology model is really neat and elegant.

Maybe I am wrong but here is a slight rephrasing of my argument:

An element of proper distance ds is given by

ds = a(t) \frac{dr}{\sqrt{1-kr^2}}

By considering a light ray we have

dt = a(t) \frac{dr}{\sqrt{1-kr^2}}

Combining these expressions we have

ds = dt

s = t

Thus the maximum proper distance s, the current size of the Universe, is equal to the age of the Universe.

Cosmic time is the size of the Universe.
 
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To get the first equation, you set dt=0. To get the second equation, you set ds=0. The combination of both gives a meaningless result.
 
johne1618 said:
I just think that the linear cosmology model is really neat and elegant.

It is neat and elegant. Unfortunately our best observations show that it's wrong. Search for Milne model for the details.

There is some work on the Milne-Dirac universe model, that I personally don't think is "totally nuts." It might be a good idea if you did some reading on that.
 
  • #10
twofish-quant said:
It is neat and elegant. Unfortunately our best observations show that it's wrong. Search for Milne model for the details.

There is some work on the Milne-Dirac universe model, that I personally don't think is "totally nuts." It might be a good idea if you did some reading on that.

My only comment is that the 'standard' GR model does not work either, that is it does not work without the inventions of Inflation, non-bayonic Dark Matter and Dark Energy; none of which have been discovered in laboratory physics. (But then with these hypothetical entities it does fit with observations very well) Discover the Inflaton, the non-baryonic DM particle(s) and positively identify DE and then we shall know what we are talking about.

Perhaps the Linearly Expanding or Coasting Cosmology model (or indeed another model such as MOND) would fit if an equal amount of work and inventiveness were applied to them!

Just a thought...
Garth
 
  • #11
Hi Everyone,

After thinking about it again I realize I am talking nonsense!

As George Jones says

dt = a(t) \frac{dr}{\sqrt{1-kr^2}}

leads to

\int \frac{dt}{a\left(t\right)} = \int \frac{dr}{\sqrt{1-kr^2}}

John
 

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