How does it make sense for the universe to have started from a big bang?

Question

It has been said that the Big Bang started from a singularity. Think about a balloon radially growing over time. Fix a time $t_0, t_1 > 0$, and let $M_0, M_1$ be two balloons at time $t_0, t_1$ respectively. I can find a two-parameter diffeomorphism $\phi(t_0, t_1): M_0 \rightarrow M_1$.

However, I cannot find a diffeomorphism if I let $t_0 = 0$ and $t_1 > 0$, i.e. $\phi(0, t_1): \{*\} \rightarrow M_1$. In what sense should I interpret a homotopy between initial state (big bang) and final state (the current universe)? Is it even true that the Big Bang started from a singularity?

Answer 1

The singularity at the start of the universe in the Big Bang model is not supposed to be understood as part of the smooth manifold of spacetime, precisely for this reason.

The time function on spacetime does not actually assign a "point" to $t = 0$. It's undefined (otherwise spacetime wouldn't be a Lorentzian manifold), and the same is true if you take an FLRW universe and try to keep the initial spatial slice 3d - since the scale factor goes to zero, the manifold is not Lorentzian there.

If you want to model the initial singularity of the Big Bang as part of spacetime, you need to consider more general models of spacetime than a Lorentzian manifold.

Answer 2

A “singularity” is shorthand for “a place where our current understanding of the laws of physics breaks down”. If we use our current model of the laws of physics (specifically general relativity) to extrapolate the state of the early universe backwards in time, we find that density and temperature are unbounded - loosely speaking, we reach a state of infinite density and temperature. This is not physically possible. Therefore something else must happen - but at the moment we are not sure exactly what that is. It seems likely that we will need to have a theory of quantum gravity to explain the very first moments of the universe.

Answer 3

You're asking a mathematical question. Mathematics is a human invention. Sometimes it maps well onto the phenomena we see. Spacetime geometry is a case where, in the portion of the universe accessible to observation, the math and the phenomena are in decent agreement.

Part of this is interpreting the observations as indicating that the universe was once in a much hotter and denser state than it is at the present time. That's really what we mean by the "big bang". But we really don't know enough physics to reliably extrapolate back to the "beginning", whatever that was.

Mathematical models of physical phenomena are prone to singularities, but when the theoretical singularities should be accessible to observation, they are never seen. This is a limitation of using math to model the universe.

Answer 4

There are several problems with this argument.

First, cosmological models having a singularity just means that they blow up as $t\to 0^+$ and aren't defined at $t=0$. That's interpreted as a limitation of the models, not evidence that real-world quantities were infinite at $t=0$. Several other answers have already covered this.

Second, even inasmuch as you can complete the topological manifold and treat the model as having a $t=0$, the completion usually isn't a single point, because no single point can be in the causal past of the whole space at early $t$ (this is the horizon problem). Even if $k>0$, meaning the space at every $t>0$ has a finite volume and it goes to zero at $t\to 0$, the completion is still topologically a 3-sphere, not a point.

Third, even if you could complete the manifold with a single point, that isn't obviously a problem. If you put polar coordinates on Euclidean space, there is no homotopy from the space at $r=0$ to the space at any $r\ne 0$. To be fair, I don't think anything closely analogous to that Euclidean example can happen with Lorentzian signature, for the handwaving reason that at the central point all directions would have to be timelike, which is inconsistent with the assumed signature. But to prove that you have to make a coordinate-independent argument, not an argument that depends on a $t$ coordinate and associated foliation.

I think there are (not widely accepted) models in which the universe starts with a Euclidean signature, so that $t=0$ can be a mere coordinate singularity like the $r$ of the previous paragraph. I don't know how the transition from one signature to another is dealt with.

Answer 5

Physicists feel a great aversion to infinity-valued physical quantities. They call them singularities and usually say that the physics fails there. A mathematician would never say the mathematics fails, because the value of $1/x$ function at $x=0$ is infinite. The inverse of this function is clearly defined and equal to zero there.

Coming to your example: imagine a two dimensional universe with constant positive Gaussian curvature $K$. It is two-dimensional manifold called the two-sphere ($S^2$). There are only two intrinsic properties characterizing it, the Gaussian curvature $K$ and the total surface area $A$. They are not independent, but related to each by Gauss-Bonnet theorem:

\begin{equation} \int_{A}~K~dA~=~A\cdot K~=~4\pi \end{equation}

Due to this equation, in case of the surface area A going to zero, the curvature K diverges to infinity. However, their product remains finite being $4\pi$. Moreover, $S^n$-spheres are not contractible to a point (they have no interior). A sphere with vanishing surface area remains topologically a sphere despite of its infinite curvature.

Your picture with balloon parametrized by the time I would replace with a fiber bundle $R^{+}\times S^2$ with fiber $S^2$ and base its surface area $A$. In conclusion, the places with infinite values of some physical quantities are interesting ones and not just abhorrence, or a proof that the underlying theory has its limit there. See for example Is the Planck force a truly "Planck unit"?.