Wednesday, May 23, 2012 ... Deutsch/Español/Related posts from blogosphere

Sheldon Cooper's revenge to Stephen Hawking: Hawking made a boo boo

Yesterday, we discussed an interesting new paper by Hartle, Hawking, and Hertog. It claimed that because of some mysterious maths of the Wheeler-DeWitt equation, a theory with a negative cosmological constant may accelerate the cosmic expansion just like as if it were a positive cosmological constant.

Fun update: A reader, yaya jon, has pointed out that in the new version of the Hartle-Hawking-Hertog paper, your humble correspondent and HB are thanked for the innocent sign error whose impact so far looks isolated (but I still believe that there must be some other errors unless the paper shows some really important loophole in the Ehrenfest "theorem" way of thinking about the evolution in quantum gravity).
I said it couldn't be right: there had to be a sign error. But I didn't know where the error was. A reader named "test" or "HB" has quickly filled the gap. I just verified that HB's remark is right. So I will alert Jim Hartle – the only author whom I have talked to for a long enough time – and send him a link to this blog entry. I am sure he will be happy!

My message to Hartle et al. is the following:

Prof Hartle, Hawking, and Herzog, it's an honor and privilege to meet you, Sirs. (We know.) I wanna thank you for taking time to see my blog. (Our pleasure.)

I enjoyed reading your paper very much. You clearly have the brilliant minds. (We know.) Your thesis that the accelerated Universe is an anti de Sitter space exposing its cosmological constant in the backwards way is fascinating. (Thank you. Came to us one morning in the shower.)

That's nice. Too bad it's wrong. (What do you mean "wrong"?) You made an arithmetic mistake on page 7, equation (2.4). It was quite a boner. (No, no, that that that can't be right! We don't make arithmetic mistakes.)

Are you saying I do? (No, no, no, of course not. I was thinking. Oh gosh Golly. We made a boo boo. And we submitted it to the arXiv so it's visible to The Reference Frame readers. Collapse.)

Great. Three more fainters.

A video that inspired my message. TBBT, sitcom, CBS.

Now, let's be a bit more specific. The equation (2.2) on page 7 of their paper defines the action \(I\) as follows:\[

I[N(\lambda), a(\lambda)] = \eta\int \dd \lambda \,N \left[
\frac 12 G(a) \zav{\frac{a^\prime}{N}}^2 + {\mathscr V}(a)

\] In equation (2.3), they also tell us that\[

G(a)\equiv -a,\qquad {\mathscr V}(a) \equiv -\frac 12 (a+ \frac{1}{{\ell}^2} a^3)

\] Finally, the equation (2.4) claims to write down the constraint we may obtain from varying the action \(I\) in equation (2.2) with respect to \(N\). Let's just do it. First, we ignore (i.e. divide by) the prefactor \(\eta\) which is universal. Second, we vary with respect to \(N\). There are two things we must vary: the first term is proportional to \(G(a)\) while the second term is proportional to \({\mathscr V}(a)\).

Both of these terms are multiplied by \(N\) so the variation with respect to \(N\) just erases this \(N\). That would give us the constraint\[

0 = \frac 12 G(a) \zav{\frac{a^\prime}{N}}^2 + {\mathscr V}(a) + \dots

\] However, we shouldn't forget that the term proportional to \(G(a)\) also has an extra \(1/N^2\) whose derivative is \(-2/N^3\). So it also reduces the power of \(N\) by \(1\) but with an extra prefactor of \(-2\). Adding the prefactors, we get \((1-2)=(-1)\) as the total prefactor for the term proportional to \(G(a)\), something we should have known immediately because the total power of \(N\) in this term is \(N^{-1}\).

So the right variation is\[

0 = -\frac 12 G(a) \zav{\frac{a^\prime}{N}}^2 + {\mathscr V}(a) + \dots

\] But this subtlety switching the signs wasn't the actual point where the mistake was generated. The real mistake fully boils down to the term proportional to \({\mathscr V}(a)\). Let's use the equation (2.3) to rewrite the last displayed equation above. After we divide the equation by the omnipresent \((a/2)\), we obtain\[

\zav{\frac{a'}{N}}^2 - 1 - \frac{a^2}{{\ell}^2} = 0

\] Now, compare this result with their equation (2.4). Their term \({a^2}/{{\ell}^2}\) has the opposite, plus sign! This sign is wrong because the relative sign between the second term \(-1\) and the last term has to be \(+1\) because the signs of both terms in \({\mathscr V}(a)\) are the same. But they got the opposite sign!

Needless to say, this last term is exactly the cosmological constant term because, as they write below equation (2.3), \[

\frac{1}{{\ell}^2} = -\frac{\Lambda}{3}.

\] So by a sign error, they reverted the sign of the cosmological constant and unless (2.4) is just an isolated equation with a typo that isn't used later (which could be the case because e.g. (2.5) and (2.6) seem to be fixed again), all the other remarkable claims in the paper probably boil down to this single sign error.

They may also boil down to another sign error or another error. It seems to me now that they're doing something similar to my March 2012 blog entry, Reality of complexified fields, and with the kind help by some readers (who pointed out that one must insist on the positivity of the kinetic terms etc. to avoid ghosts, a rule I may have violated), I think that I convinced myself that the signs can't be switched this easily.

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reader Unknown said...

Hi Lubos,

In their latest modification to the paper, the authors thanked you for pointing out the typo in their work that you were discussing.
So what is the situation now in your opinion? I mean in terms of the reliability of the results.
Honestly, I'm no expert at all in this field, but my understanding is that one of the challenges facing the AdS/CFT correspondence is that in an AdS space the cosmological constant is negative. And people showed that a dS/CFT correspondence is not achievable.
So if these results are true, then what does that mean for the AdS/CFT?

Best regards,

reader Luboš Motl said...

Dear Yaya, thanks for your news! New pdf.

It's fun that who is thanked is "Lubos Motl and blogger HB". In reality, the blogger is myself and HB is a commenter who could very well be their colleague in Oxford, Cambridge, Santa Barbara, or anything else.

It's clear that there are several equations that are unaffected by the typo but I haven't verified the whole paper. If it's true, it's very important because it shows that some of the steps in the naive argument that this flip of the C.C. is impossible has a loophole.

However, my belief that this is possible is limited and I would probably switch to a disbelief if it were claimed that this switching effect affects AdS/CFT or can be seen in AdS/CFT. That's a theory about states whose asymptotics respect the AdS-like shape so there's no room for dS-like accelerated expansion in it, I think.


reader James said...

Even after correcting some typos I don't see how they got the minus sign in (3.12)? If I take \theta_c as in (3.11) then the scale factor a(u) is purely real when y=y_r. Squaring it gives me something strictly positive.

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