- #26

mysearch

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**Response to #25**

Marcus, informed comments are no intrusion and the points you raise do home in on any aspect on which I am slightly confused about. I realise that there is still some debate over whether k=0 or is just very close to zero, but I wanted to just focus on the case of k=0.

First, could I just clarify your terminology; are you defining an ‘Watch out! the k=0 case refers to SPATIAL flatness it does not mean that you have 4D flatness or that you can fit the picture to a Minkowski frame----or even more radically to an Euclidean frame.

*Euclidean frame’*as 3-D plus absolute time, whilst a `

*Minkowski frame*` is essentially 4-D spacetime of special relativity?

Again, to clarify, the FRW metric is often presented in spherical coordinates as:It does NOT mean that the FRW "reduces to the flat spacetime metric of SR"

[tex]- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2} + r^2 \mathrm{d}\theta^2 + r^2 \sin^2 \theta \, \mathrm{d}\phi^2 \right)[/tex]

If I consider a radial path, I can reduce the complexity to:

[tex]- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1-k r^2}\right)[/tex]

If I set k=0, the equation appears to reduce to:

[tex]- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2[/tex]

It was my understanding that today, a(t)=1, which seemed to imply that:

[tex]- c^2 \mathrm{d}\tau^2 = - c^2 \mathrm{d}t^2 + \mathrm{d}r^2[/tex]

Which appears equivalent to SR spacetime metric at a given point in time.

So my basic question is what is the difference between the formal definition ofpeople are often confused about this and get the mistaken notion that they can fit all or part of a Friedmann model universe onto an SR frame, in cases when this is inappropriate. the confusion partly arises from there being two different meanings of the word flat, namely spatial flat and 4D flat

*'spatial flat'*and

*'4D flat'*?

Many thanks