Cosmology, The Big Bang and Entropy

I went to Sean Carroll's talk yesterday at Sydney University. As many of you are aware, Sean is one of the bloggers on Cosmic Variance and is touring Australia for his talks. He also has an upcoming book titled From Here to Eternity: The Quest for the Ultimate Theory of Time.

Sean talked about all the big ideas that face Cosmologists these days and in particular Entropy and its relation to the evolution of our Universe starting from a state of low entropy / high order progressing towards a state of high entropy / low order. I didn't like his interpretation of time (particularly the "arrow of time") as mentioned in my previous post and as for the state of the Universe, this is ambiguous because we cannot observe the Universe from the outside hence there is no entropy for the Universe either.

Cosmologists have a fairly good idea what the state of the Universe was 1 second after the Big Bang however at "t = 0" (time cannot be defined here as there is no matter at this stage hence no clocks), no one knows as General Relativity breaks down however before the Big Bang? Sean speculated on an idea that our Universe may have begun from a single quantum fluctuation from the inherent energy of the vacuum, ie we live in a baby Universe an offshoot from another Universe. There was quite a good turnup and the lecture theatre was pretty full, all in all enjoyed the talk.

Still experimenting with $\LaTeX$ in Blogger using this script. Testing...

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\[\int \csc^2x\, dx = -\cot x+ C.\]

\[\ P_{r-j}=\begin{cases} 0& \text{if $r-j$ is odd},\\ r!\,(-1)^{(r-j)/2}& \text{if $r-j$ is even}. \end{cases}\]

\[\qquad \lim_{\alpha\to \infty} {\sin\alpha \over \alpha} = 0\]

\[\root n \of {\prod_{i=1}^n X_i} \leq {1 \over n} \sum_{i=1}^n\]

\[\ \cfrac{1}{\sqrt{2}+\cfrac{1}{\sqrt{2}+\cfrac{1}{\sqrt{2}+\dotsb }}}\]

\[\ \nabla \cdot \mathbf{D} = \rho_f \]
\[\ \nabla \cdot \mathbf{B} = 0 \]
\[\ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t} \]
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\[\ \boxed{m &= \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}}\]