# Simple argument for a linear cosmology

1. Oct 16, 2012

### johne1618

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 travelling 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

$a(t) \propto t$

Last edited: Oct 16, 2012
2. Oct 16, 2012

### Staff: Mentor

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.

3. Oct 16, 2012

### johne1618

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 $t$.

4. Oct 16, 2012

### Garth

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

5. Oct 16, 2012

### George Jones

Staff Emeritus

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

6. Oct 16, 2012

### Staff: Mentor

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.

7. Oct 17, 2012

### johne1618

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.

Last edited: Oct 17, 2012
8. Oct 17, 2012

### Staff: Mentor

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.

9. Oct 18, 2012

### twofish-quant

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. Oct 18, 2012

### Garth

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. Oct 18, 2012

### johne1618

Hi Everyone,

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

As George Jones says

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

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