Struggles with the Continuum - Part 6 - Comments

[john baez]

john baez submitted a new PF Insights post

Struggles with the Continuum - Part 6

Continue reading the Original PF Insights Post.

 

[jedishrfu]

Thanks!

 

[bhobba]

Another wonderful paper - many thanks to John.

I would like to mention a series of lectures I enjoyed very much and re-watch them every now and then with equal enjoyment:


It examines perturbation theory, extracting finite answers from divergent series, and some of the things touched on in the paper. The most startling for me was the origin of quantisation which naturally comes out of this approach. Fascinating. Oh - he touches on its relation to renormalisation - not in depth though.

Thanks
Bill

 

[strangerep]

Hi John,

[...] Physicists have repeated the calculation summing over diagrams with ≤2 loops, and again found a Landau pole.

I'm not familiar with the 2-loop calculation. Does the Landau pole appear at the same energy as for 1-loop, or does it move?

 

[anorlunda]

Almost everything in this Continuum series of Insights articles is way over my head. Nevertheless, I manage to learn something, and to gain a glimpse of the outstanding problems in this field when I read them. I can only attribute that to excellent writing.

Kudos and thank you John Baez.

 

[vanhees71]

One cannot stress this enough. These Insights are written in a way to explain the issues discussed "as simple as possible but not simper". It's really great!

 

[maline]

The argument goes as follows. If these power series converged for small positive αα, they would have a nonzero radius of convergence, so they would also converge for small negative αα. Thus, QED would make sense for small negative values of αα, which correspond to imaginary values of the electron’s charge. If the electron had an imaginary charge, electrons would attract each other electrostatically, since the usual repulsive force between them is proportional to e2e2. Thus, if the power series converged, we would have a theory like QED for electrons that attract rather than repel each other. However, there is a good reason to believe that QED cannot make sense for electrons that attract. The reason is that it describes a world where the vacuum is unstable. That is, there would be states with arbitrarily large negative energy containing many electrons and positrons. Thus, we expect that the vacuum could spontaneously turn into electrons and positrons together with photons (to conserve energy).

If we are considering imaginary values of charge, then logically we must also consider imaginary values for the EM fields. A classical plane wave with imaginary amplitude would have a (real-valued) Poynting vector opposite the direction of propagation- the wave radiates negative energy. I am guessing that the QED equivalent would be a negative-energy photon. Now if all of the electrons & positrons have only imaginary charge, I don't see how real-valued fields could be generated. It would seem that in such a pair production, the only photons that could be produced would have negative energy, so energy conservation should rule out this scenario!

 

[maline]

@john baez , is my question above meaningful?

 

[TGlad]

(excuse in advance if this question is too newbie)
I was reading Penrose's Road to Reality, chapters 7, where he discusses how some complex functions have Taylor expansions that are only correct within a small domain. The expansion converges within that domain but diverges outside of it. This is usually when the complex function has singularities, or is multi-sheeted, like y = log z. To get a description of the whole domain you need analytic continuation (patch together Taylor expansions around different points). The same is true for more general Reimannian surfaces (chapter 8).

Anyway, my question is whether this property has anything to do with the QED problem. It sounds quite similar to me. Maybe the latter iterations are somehow actually sampling from outside the domain of convergence.

 

[maline]

Anyway, my question is whether this property has anything to do with the QED problem. It sounds quite similar to me. Maybe the latter iterations are somehow actually sampling from outside the domain of convergence.

In this power series, the "complex valued independent variable" is the fine-structure constant alpha. By the definition of a power series, or any function for that matter, all the terms must be evaluated for the same value of the variable. The important question is whether the actual value, about 1/137, is within the domain of convergence. Dyson argued that it must not be, because domains of convergence are symmetrical around the origin. If the series converges for 1/137, it must also converge for alpha=-1/137, That would imply that in a hypothetical universe with imaginary-charged electrons, we would have finite values for all the measurements that are finite in real life. But in such a universe the vacuum should instantly generate infinite numbers of electrons, positrons, & photons, so it seems that the series should diverge. The same argument implies that the domain of convergence should simply be zero.

 

[bhobba]

Maybe the latter iterations are somehow actually sampling from outside the domain of convergence.

The answer is finite answers can be extracted from divergent series - see the videos I linked to.

Just as a start consider the following series 1 + s + s^2 + s^3 ++++++.

Complex analysis shows its divergent if s > 1. But consider the following:

S = 1 + s + s^2 + s^3 ++++++ = 1 + (s + s^2 + s^3 ++++) = 1 + s(1 + s + s^2 +++++) = 1 + sS or S-sS = 1 or S = 1/1-s. This gives an actual value if s is not 1. For example if s =-1 you have 1 - 1 + 1 ... =1/2.

Its called generic summation:
http://math.stackexchange.com/questions/633887/divergent-series-intuition

For more detail of this interesting issue see the videos.

It may seem like a mathematical quirk, but as John says it leads to results consistent with experiment so there is obviously somethin going on.

Thanks
Bill

 

[maline]

@bhobba, once you're on the thread, would you mind responding to my question re. Dyson's argument against the possibility of actual convergence (that it would imply convergence for negative values of alpha i.e. imaginary electron charge, & in such a universe infinite numbers of pairs could be produced)?

If we are considering imaginary values of charge, then logically we must also consider imaginary values for the EM fields. A classical plane wave with imaginary amplitude would have a (real-valued) Poynting vector opposite the direction of propagation- the wave radiates negative energy. I am guessing that the QED equivalent would be a negative-energy photon. Now if all of the electrons & positrons have only imaginary charge, I don't see how real-valued fields could be generated. It would seem that in such a pair production, the only photons that could be produced would have negative energy, so energy conservation should rule out this scenario!

 

[bhobba]

@bhobba, once you're on the thread, would you mind responding to my question re. Dyson's argument against the possibility of actual convergence (that it would imply convergence for negative values of alpha i.e. imaginary electron charge, & in such a universe infinite numbers of pairs could be produced)?

Sorry - but I don't know enough about it to respond.

Thanks
Bill

 

[nikkkom]

"If these power series converged for small positive alpha, they would have a nonzero radius of convergence, so they would also converge for small negative alpha. Thus, QED would make sense for small negative values of alpha [which is problematic since it describes unstable vacuum]."

I don't understand the above logic. Why is it a problem that QED with negative alpha results in a physically nonsensical picture? Why would anyone care? This is not a model of our Universe.

 

[maline]

Why is it a problem that QED with negative alpha results in a physically nonsensical picture?

The problem is that a convergent power series would imply that QED should work, i.e. give finite results, in that nonsensical universe.

 

[nikkkom]

The problem is that a convergent power series would imply that QED should work, i.e. give finite results, in that nonsensical universe.

And that is a problem... why?

The vacuum in that fictitious Universe would decay, creating more and more electrons. The electrons would feel finite forces and have finite charges at any given time, there just would be more and more of them as time progresses, which is clearly not what we see in our Universe. But why should I care? I didn't expect QED with negative alpha to describe our Universe, thus the mismatch is *expected*.

 

[stevendaryl]

And that is a problem... why?

I think the reasoning is that

1. If the power series for QED were convergent, then the behavior of particles and fields would be an analytic function of $\alpha$.
2. If the behavior was an analytic function of $\alpha$, then the behavior for $\alpha$ slightly less than zero would be only slightly different from the behavior for $\alpha$ slightly greater than zero.
3. But the behavior for $\alpha < 0$ is nothing like the behavior for $\alpha > 0$.

In other words, if the vacuum is stable for $\alpha > 0$ but unstable for $\alpha < 0$, then that implies that something weird happens at $\alpha = 0$.

 

[stevendaryl]

I think the reasoning is that

1. If the power series for QED were convergent, then the behavior of particles and fields would be an analytic function of $\alpha$.
2. If the behavior was an analytic function of $\alpha$, then the behavior for $\alpha$ slightly less than zero would be only slightly different from the behavior for $\alpha$ slightly greater than zero.
3. But the behavior for $\alpha < 0$ is nothing like the behavior for $\alpha > 0$.

In other words, if the vacuum is stable for $\alpha > 0$ but unstable for $\alpha < 0$, then that implies that something weird happens at $\alpha = 0$.

But I guess I'm not completely convinced, myself. In general, if you have a physics problem with a parameter $\lambda$, some quantities might be analytic in $\lambda$, and others might not. So the fact that QED completely falls apart with $\alpha < 0$ means that not everything is analytic in $\alpha$, but does it mean that the quantities of interest fail to be analytic in $\alpha$?

 

[maline]

So the fact that QED completely falls apart with α<0α<0\alpha < 0 means that not everything is analytic in αα\alpha, but does it mean that the quantities of interest fail to be analytic in αα\alpha?

What do you mean by "the quantities of interest"? I thought the question was whether a particular power series converges. According to this argument, it does not. What other quantities are relevant?

Once you are here, would you perhaps respond to my question in post 7 above? In a universe with only imaginary-charged electrons, how could positive-energy photons, corresponding to real EM field values, be generated to balance the negative energy of the pairs?

 

[nikkkom]

If the behavior was an analytic function of $\alpha$, then the behavior for $\alpha$ slightly less than zero would be only slightly different from the behavior for $\alpha$ slightly greater than zero.

This is not true in many cases.
Trivial examples: behavior of 1/x is quite different for x>0 and x<0. Behavior of exp(1/x) even more so.

 

[maline]

behavior of 1/x is quite different for x>0 and x<0. Behavior of exp(1/x) even more so.

These functions are not analytic at zero, and therefore cannot be expressed as power series in x.

 

[stevendaryl]

This is not true in many cases.
Trivial examples: behavior of 1/x is quite different for x>0 and x<0. Behavior of exp(1/x) even more so.

$1/x$ and $exp(1/x)$ are not analytic at $x=0$, so those are not counterexamples. The definition of a function $f(x)$ being analytic at $x=0$ is that there is a convergent Taylor series of the form $f(x) = a + b x + c x^2 + ...$. [edit] (With only positive powers of $x$)

 

[nikkkom]

$1/x$ and $exp(1/x)$ are not analytic at $x=0$, so those are not counterexamples. The definition of a function $f(x)$ being analytic at $x=0$ is that there is a convergent Taylor series of the form $f(x) = a + b x + c x^2 + ...$. [edit] (With only positive powers of $x$)

My point was not about functions being analytic.

My point is that qualitative differences are more likely to appear when you go from 0.0001 to -0.0001 than when you go from 0.5001 to 0.4999. Often, sign matters a lot.

 

[stevendaryl]

My point was not about functions being analytic.

My point is that qualitative differences are more likely to appear when you go from 0.0001 to -0.0001 than when you go from 0.5001 to 0.4999. Often, sign matters a lot.

Right, but the sign change making a huge difference is a sign (no pun intended) that things are not analytic in whatever parameter you're considering. Not analytic implies that whatever it is can't be expanded in a convergent power series.

 

[Strum]

I do not know much about this stuff but according to the wiki article on Landau poles (https://en.wikipedia.org/wiki/Landau_pole), some russian guy recently showed that the beta function does not blow up in QED. Isn't this in contrast with the calculations discussed in the text?

Also somewhat related, in the appendix of Robert Helling's http://homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf he shows that in ordinary $\phi^{4}$ theory with a weird sign we have solutions in 1d to the equations of motion given by $\phi \sim tanh(x)$. He then makes the comment

We see that the action scales like 1/g. Thus, in a path integral (which gives the classical theory in the $\hbar \rightarrow 0$ saddle point approximation), these solutions contribute with a weight $\exp(-\frac{1}{g}$ which is exponentially small for small g as we assume it in the perturbative analysis. In fact, due to the essential singularity of this expression at $g = 0$, all $g$-derivatives vanish at $g= 0$ and thus in a Taylor expansion (as we assume it for the perturbative Ansatz), this function is indistinguishable from 0. Thus, these “solitonic” solutions are missed by a perturbative expansion. It is believed that these missed solutions are the reason that the perturbative expansion cannot be convergent to the exact solution

Is there any way to make these comments a bit more concrete? A simple calculation for example.