One thing people often criticise about String Theory as a model for particle physics 1 is that it has a lot of parameters (infinitely many?), i.e. there is not just one vacuum, there is a whole family of them.

Let's consider the following few-point: we consider everything we know about particles and their properties, the fundamental forces and their dynamics as data. And we consider the family of String Theories as a candidate for a model of this data. Let's also assume that there is a competing model2 that fits the data well, but not as well as some of the variations of String Theory.

In Bayesian inference based Data Analysis there is a phenomenon one discovers when objectively comparing models that resembles what is commonly known as Occam's Razor in Philosophy. Let me emphasise though that there is nothing philosophical about it, in a nutshell the posterior probabilities just turn out to be lower for models with similar capability of fitting the data when they have more parameters (see e.g. this book for an account of this standard technique, if anyone has an alternative online resource at hand a comment would be appreciated).

Given the infinite amount of parameters of String Theory would a rational Bayesian agent3 have to reject String Theory in favour of any other reasonably fitting model? What would a positive conclusion to the former question imply for the justification of treating String Theory as a particle physics theory in general?

Disclaimer: I virtually know nothing about String Theory (some QFT though) and I don't mean any offence or to devalue the hard and ingenious work of String Theorists!

1 There are of course heaps of other uses of String Theory, ranging from applications in condensed matter physics to pure mathematics.

2 I don't really know if there are and as far as I know currently we are also unable to quantify "how well" any of these models fit the data, but for the sake of the argument let's stick with these assumptions.

3 Someone purely deciding based Bayesian inference methods who has all the necessary prior information and data.

  • $\begingroup$ Related: physics.stackexchange.com/q/15/2451 , physics.stackexchange.com/q/5057/2451 and links therein. $\endgroup$ – Qmechanic Nov 19 '16 at 9:14
  • $\begingroup$ Have you ever heard of "the only game in town"? That is, string theory is the only known self-consistent quantum gravity capable of producing realistic low-energy physics. Alternatives are either inconsistent or incapable of producing known physics mess, or phenomenological models (like Standard model + GR) that are known to break down at some point. Also strings don't have infinite number of parameters - landscape consists of finite number of vacua, and each of them gives some effective low energy QFT with fixed parameters. $\endgroup$ – OON Nov 19 '16 at 13:05
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    $\begingroup$ I think this question is unnecessarily verbose/long. $\endgroup$ – innisfree Nov 19 '16 at 13:25
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    $\begingroup$ What do you mean "given the infinite amount of parameters of String Theory"? It has drastically less parameters than standard quantum field theory (all string theories give QFTs, but not all QFTs come from string theories), so every objection you make based on this claim applies equally well to QFT. $\endgroup$ – ACuriousMind Nov 19 '16 at 16:08

In Bayesian inference, one judges a model by calculating its relative plausibility in light of data. This would involve calculating so-called evidences, e.g., $$ \mathcal Z \equiv p(\text{data}\,|\,\text{ST}) $$ I don't think we understand string theory (ST) well enough to calculate this evidence. A Bayesian might not consider ST as a well-defined model, for that reason.

However, a commenter notes the 'only game in town argument'. In Bayesian statistics, if you have a single hypothesis $H$, and no alternatives, $$ p(H) = p (H\,|\,\text{data})=1 $$ It's hard to call this scientific, though, since we didn't update our belief with data.

  • $\begingroup$ I don't think think we understand the Standard Model well enough to compute these evidences. Effective field theories are not models in the Bayesian sense. $\endgroup$ – user1504 Nov 19 '16 at 13:30
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    $\begingroup$ For both sentences, why do you think that? $\endgroup$ – innisfree Nov 19 '16 at 13:34
  • $\begingroup$ Unitarity is generically violated in effective field theories. If you're being strict, such theories are not computing probabilities. $\endgroup$ – user1504 Nov 19 '16 at 13:38
  • $\begingroup$ Unitarity is generically violated at $\sqrt s \gtrsim \Lambda $ cutoff? That doesn’t stop accurate calculations below cutoff. $\endgroup$ – innisfree Nov 19 '16 at 13:45
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    $\begingroup$ You asked about a Bayesian analysis of ST? I'm telling you why it's impossible at this stage. $\endgroup$ – innisfree Nov 19 '16 at 22:48

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