Is the Big Bang singularity a property of spacetime — or a matter of how you keep the books?
The Big Bang singularity — go back in time and both density and curvature diverge, and physics loses its meaning. It is one of the deepest problems in cosmology. Yet the conformal transformation we have been running since Episode 3 turned the \(c\cdot t=\text{const}\) universe into exact Minkowski spacetime. Flat spacetime has no singularity. One transformation, and the greatest problem in cosmology disappears — or so it looks. Is that true? Today's answer is "half true". And the half that did not vanish is probably the more interesting one.
Compute, in the original picture, the quantity that measures curvature (the scalar curvature). In a flat universe with \(a\propto t\) we have \(\ddot a=0\) and \(\dot a/a=1/t\), so
Plugging in today's values gives the very small number \(R=3.5\times10^{-52}\ \mathrm{m^{-2}}\), but as \(t\to0\) it is infinite. That is the Big Bang singularity. Meanwhile the metric we built in Episode 3 has
$$d\tilde s^2=-c_0^2d\eta^2+dx^2\qquad\Longrightarrow\qquad \tilde R=0\ \ (\text{everywhere, always})$$Flat as a board. And, as Episode 5 showed, conformal time can be taken all the way to \(\eta\to-\infty\) only when \(w=-1/3\). That is —
\(c\cdot t=\text{const}\) (\(w=-1/3\))
$$\eta=t_0\ln\frac{t}{t_0}:\qquad t\in(0,\infty)\ \longrightarrow\ \eta\in(-\infty,+\infty)\quad\text{(all of it)}$$Compare: radiation domination (\(w=1/3,\ a\propto t^{1/2}\))
$$\eta\propto t^{1/2}:\qquad t\in(0,\infty)\ \longrightarrow\ \eta\in[0,+\infty)\quad\text{(only half)}$$For radiation an edge remains at \(\eta=0\) and you cannot go beyond it — the singularity stays as an "edge of spacetime". Only \(c\cdot t=\text{const}\) pushes the edge out to infinity. This is the face of Episode 5's "the summit takes infinite time because the potential is quadratic".
Minkowski spacetime is geodesically complete: no particle and no light ray falls off the edge of spacetime in finite time. A comoving observer runs along the straight line \(x=\text{const}\) from the infinite past to the infinite future. There is no beginning.
The idea that "the singularity is an artefact of a choice of coordinates" is not a whim. Wetterich's 2013 paper, introduced in Episode 2, closes its abstract like this.
Then in 2014 Bars, Steinhardt and Turok showed that lifting a theory to a Weyl-invariant form allows one to construct geodesically complete cosmologies and to pass analytically through a big crunch to a big bang. Changing the gauge to erase the "edge" — the same trick we performed here.
Up to this point everything reads pleasantly. The problem is what comes next.
Put the two equations side by side once more.
$$R=\frac{6}{c^2t^2}\ \ (\text{diverges})\qquad\text{and}\qquad \tilde R=0\ \ (\text{flat everywhere})$$These describe the same universe at the same instant. A quantity that looks like a property of spacetime itself has gone from infinite to zero merely because we swapped rulers.
There is no contradiction. Curvature is not conformally invariant. Recall the transformation formula from Episode 4 —
$$R=\phi^{-2}\Big[\tilde R-6\tilde\Box\ln\phi-6(\tilde\partial\ln\phi)^2\Big]$$The right-hand side carries a row of terms besides \(\tilde R\). So \(R\ne0\) is perfectly possible with \(\tilde R=0\). Curvature is a quantity with a dimension (length to the power \(-2\)), so of course it is. Quantities with units depend on the ruler — the thing the previous series said from the first page is at work here too.
✗ "After the transformation \(\tilde R=0\), so there is no singularity."
✓ "Whether there is a singularity must be judged with a dimensionless quantity."
Let us restate what a singularity was. The problem is not that curvature becomes infinite in itself. The problem is that the scale over which spacetime curves becomes smaller than the quantum size of matter — at which point neither classical spacetime nor a classical particle means anything any more.
So the ratio to build is this.
Let us compute it in both pictures.
① The expanding picture
The scale of spacetime is the Hubble radius \(c_0/H=c_0t\), and the Compton wavelength \(\hbar/(mc_0)\) is constant. Hence
$$N=\frac{c_0t}{\hbar/(mc_0)}=\frac{mc_0^2\,t}{\hbar}$$② The transformed picture (Minkowski)
Geometry is flat, so it supplies no length standard. Instead the gradient of the dilaton gives the only scale: since \(\phi=e^{\eta/t_0}\), \(\tilde L=c_0|\phi/(d\phi/d\eta)|=c_0t_0\) (a constant). Meanwhile the Compton wavelength shrinks as the mass grows: \(\tilde\lambda_C=\hbar/(\tilde mc_0)=\hbar t_0/(t\,mc_0)\). Hence
$$N=\frac{c_0t_0}{\hbar t_0/(t\,mc_0)}=\frac{mc_0^2\,t}{\hbar}$$Exactly the same expression. One says "spacetime shrinks", the other says "atoms swell" — but take the ratio and they cannot be told apart.
And as \(t\to0\), \(N\to0\). The ratio breaks in both pictures. However flat the transformed spacetime is, that fact cannot be moved.
The time at which \(N=1\) is \(t=\hbar/(mc_0^2)\), that particle's Compton time. Let us compute it for three masses.
| Mass | \(mc^2\) | Time at which \(N=1\) | How the transformed picture says it |
|---|---|---|---|
| Electron | 511 keV | \(1.29\times10^{-21}\) s | The electron's Compton wavelength reaches 13.8 billion light years |
| Proton | 938 MeV | \(7.0\times10^{-25}\) s | Ditto (for the proton) |
| Planck mass | \(1.22\times10^{19}\) GeV | \(5.39\times10^{-44}\) s | = the Planck time, exactly |
Look at the last row. The time at which \(N=1\) for the Planck mass is \(\hbar/(M_{\rm Pl}c^2)=5.39\times10^{-44}\) seconds — the Planck time itself. So this dimensionless test correctly points at "the moment quantum gravity is needed". And that moment is the same whichever picture you compute in.
The previous series' bonus episode 2 said "absolute values are bookkeeping; only differences are physics". One notch stronger, it becomes this — whether there is a singularity is not written in spacetime; it is written in a dimensionless ratio.
What section 04 showed applies to the naive model \(c\cdot t=\text{const}\). Wetterich's model has different matter content (a "cosmon" field with a specific potential), and he argues there that the curvature scalar stays almost constant throughout every epoch. The calculation here does not refute his conclusion. What is demonstrated here is the method — that "it became flat under the transformation" is not enough, and you must go down to dimensionless quantities to check.
Also, \(N\to0\) is not a proof that a singularity exists. All it proves is that the classical description cannot be used in that regime. What actually happens there remains open, as a question for quantum gravity — the same door left ajar in the finale of the previous series.
Conformally transform \(c\cdot t=\text{const}\) and spacetime becomes Minkowski with \(\tilde R=0\), geodesically complete. Moreover, only for \(w=-1/3\) does conformal time run to \(-\infty\) and cover Minkowski space entirely (radiation covers half). The "edge of spacetime" that is the Big Bang genuinely disappears. Wetterich's "the singularity is related to a singular choice of field coordinates" and Bars et al.'s geodesically complete cosmologies in Weyl gauge both stand on this fact.
But curvature carries units and is not conformally invariant, so \(\tilde R=0\) proves nothing. Build the dimensionless ratio \(N=mc^2t/\hbar\) (scale of spacetime ÷ Compton wavelength) and you get literally the same expression in both pictures, and it breaks in both as \(t\to0\). For the Planck mass \(N=1\) at \(5.39\times10^{-44}\) s — the Planck time exactly. Geometry vanished; the ratio remained. The conformal transformation was not an answer but a tool for putting the question into the right shape.
Print / PDF: ⌘+P (Ctrl+P on Windows). On screen the slider changes the gauge and shows that the crossing time does not move. "Show answer" opens the solutions.