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Consider a flat Robertson-Walker metric.

When we say that there is a singularity at $t=0$, clearly it is a coordinate dependent statement. So it is a "candidate" singularity.

In principle there is "another coordinate system" in which the corresponding metric has no singularity as we approach that point in the manifold.

However, we know that Big Bang is "a true" singularity, but how should we test that?

Is it intuitively self-evident, or should we check rigorously all scalars based on the Ricci tensor? If so "which order of scalar" goes to infinity at that point called Big Bang?

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    $\begingroup$ "In principle there is "another coordinate system" in which the corresponding metric has no singularity as we approach that point in the manifold. " - Is there? Just because the usual statement about the $t = 0$ singularity is coordinate dependent, this does not mean that such a coordinate system exists. In particular, if the singularity is a true singularity, the "point $t = 0$" doesn't belong to the manifold at all - think of a cone (that becomes singular at its tip) - it is only a manifold if you exclude the tip. $\endgroup$ – ACuriousMind Nov 25 '14 at 18:46
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    $\begingroup$ I believe showing that any one scalar blows up would be enough to prove a singularity. To prove that a point wasn't a singularity, you would either need to check all the scalars to show that they're well behaved, which is a potentially infinite task, or find a different coordinate system where the singularity vanishes. $\endgroup$ – G. Paily Nov 25 '14 at 18:50
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The singularity comes from the scale factor $a(t)$:

$$ds^2 = -dt^2 + [a(t)]^2 ( dr^2 + r^2 d \Omega^2)$$

By solving the Friedmann equations for the scale factor we know that:

$$a(t) = a_0 t^{\lambda}$$

where $\lambda$ is some positive number that depends on the matter-radiation ratio of the universe. At $t=0$ the scale factor becomes $a(0)=0$. So at $t=0$ the spacial part of the metric becomes zero. You can check that scalars will blow up by showing that the volume element $\sqrt{-g} ~d^4x \,$ gives nonsense. At $t=0,$ $g=det(g_{\mu \nu}) = 0,$ meaning the volume element is zero. This is not anything that can be fixed by a coordinate transformation.

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