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M theory

  1. Apr 17, 2009 #1
    How does M theory force gravity as a necessity when the standard model couldn't fit in gravity???
  2. jcsd
  3. Apr 18, 2009 #2


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    Why should the standard model "fit in gravity"?

    M-theory should have some supergravity theory as low energy limit. Are you asking if the standard model fits into some supergravity theory?
  4. Apr 18, 2009 #3
    String theory didn't "force" gravity; although never intended to represent gravity, when a spin 2 massless particle appeared it was later discovered: wow, a graviton!!!! And string theory is not part of the standard model...
  5. Apr 18, 2009 #4
    I think it works like this:

    The standard model describes the forces using gauge groups. When you construct your forces this way, you cannot include gravity without the math breaking. This is because the gravitational force (the graviton, rather) is a "spin 2 boson", and a spin 2 particle cannot be "renormalized", which is a mathematical procedure you have to be able to do for the standard model to be usable.

    M theory describes the forces using vibrational modes of strings. When you construct your forces this way, not only can you have a spin 2 particle, you always have a spin 2 particle. Every string theory you construct has a spin 2 boson as one of its vibrational modes, and this is the graviton.
  6. Apr 18, 2009 #5
    Was it proven that "spin-2 interactions" can not be renormalized ? Was it proven that string theory is finite to all orders ?

    It's not ironic, it's a serious question. As far as I know, none have been proven, but I only follow as an amateur. After periods of hopes, there has been for a few decades a general belief that no practical definition (allowing computations) of "spin-2 gravitation" would be perturbatively renormalizable, but even this belief has been challenged again lately.
  7. Apr 19, 2009 #6


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    It has been proven that the Einstein-Cartan action is not renormalizable in perturbarion theory. Each order in perturabtion theory produces new types in infinities and counter terms.

    There are recent results that certain supergravity models might be finite to all orders w/o infinities (they cancel in each order due to the large symmetry).

    It has not been proven that string theory is finite to all orders. In is known hat certain infinities are absent, but as far as I know it is not clear if no new infinitires could appear at higher genus.

    It has not been proven that summing the prturbaruon series produces a finite answer. I guess that the seroies is (as usual) an asymprtotoc series which diverges eventually. One reason is that you simply neglect non-perturbatibe aspects. This is comparable to QCD: the series itself is divergent (as you can see from the scaling of running coupling constant);
  8. Apr 19, 2009 #7
    I will say, it has always seemed a bit weird to me that we match the problem "if we add a graviton to the standard model, no one anymore knows how to calculate anything" with the solution "we will switch to String Theory, where no one knows how to calculate anything". Did we solve the problem here or just hide it behind a bigger one...?
  9. Apr 19, 2009 #8


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    My feeling is that string theory is currebtly the most ambitious program to unify all interactions; nevertheless it has some serious problems:
    - no central equation of principle for the whole program
    - perturbative finiteness only wishful thinking
    - unified theory (M-theory) not rigorously constructed
    - no background independence (= requires flat space-time!)
    - not falsifiable by experiment

    But there is currently no alternative! Loop Quantum Gravity (which is in my opinion very successfull) addresses only gravity.
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