Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I would like to expand on what I mean by the title of this question to focus the answers.

Normally whenever a theory (e.g. General Relativity) replaces another (e.g. Newtonian Gravity) there is a correspondence requirement in some limit. However there is also normally some experimental area where the new larger theory makes predictions which are different from the older theory which made predictions of the same phenomena. This is ultimately because the newer theory has a deeper view of physics with its own structures which come into play in certain situations that the old theory didn't cover well. Additionally the newer theory will make predictions based on its novel aspects which the older theory did not consider. I know that String Theory is quite rich in this regard, but am not interested in that here. Nor am I concerned as to whether experiment has caught up, as I know that ST (and Quantum Gravity in general) is not easy to test.

So for the GR to Newtonian example an answer to this question would be: bending of light rays; Mercury perihelion movement - GR had a different results to Newton. What would not count as an answer would be new structures which GR introduces like Black Holes or even gravitational curvature per se.

So does ST have anything like Mercury perihelion movement waiting to be experimentally verified, and thus "improving" on GR within GR's own back yard?

share|improve this question
    
See my answer here: physics.stackexchange.com/q/70060 –  Dimensio1n0 Jul 5 '13 at 5:19
    
And also joshphysics's answer here: physics.stackexchange.com/q/54317 –  Dimensio1n0 Jul 5 '13 at 5:34

2 Answers 2

up vote 10 down vote accepted

String theory implies new physics in - and only in - the quantum regime. In particular, at distances that are very short - comparable to the string scale or Planck scale - there are new effects. The black holes decay (while they preserve the information), effective actions have higher-derivative terms, e.g. $R^2$, there are strings, branes, fluxes, extra dimensions, and so on.

The new physics at the Planck scale - which is very far - is almost certainly not testable by the naive 19th century observations such as the Mercury perihelion's precession. This is not a problem of string theory in any sense: it is a tautological consequence of the questions that string theory addresses - namely the behavior of the Universe at the most fundamental scale - and any other theory that addresses the same questions inevitably shares the inaccessibility of the phenomena by direct tests.

When you only look at the classical limit or classical physics, string theory exactly agrees with general relativity. In some sense, this is true even in the quantum regime: string theory is the only consistent quantum completion of general relativity.

share|improve this answer
1  
I think the higher derivative terms are a good answer to the question - if you had sensitive enough probes, even without using any new "stringy" physics, you'd be able to detect them. Also - they are a result of classical string theory ($\alpha'$ corrections). –  user566 Feb 24 '11 at 19:36
1  
Motl: So this is saying that String Theory doesnt replace General Relativity - in its predictive and testable aspects - even in principle, at the larger-than-Planck scale? –  Roy Simpson Feb 24 '11 at 19:45
    
@Moshe: I dont understand what the higher derivative terms are doing in this answer, but in other theories higher derivative terms correspond to non-local effects. If "local" means Planck scale, then "non-local" could mean "greater than Planck scale" - perhaps much greater. Thus these effects would be measurable - was that your point? –  Roy Simpson Feb 24 '11 at 20:26
1  
@Roy: Higher derivative terms are not "non-local", they correspond to additional terms beyond the Einstein-Hilbert action, which modify in a small and specific way any solution of GR. One example is $R^2$ terms, which by dimensional analysis are suppressed by a power of high mass scale (string scale in this case). Those occur with calculable coefficients in string theory. Generally, GR occurs automatically at distances larger than Planck length in string theory, with corrections that are small in that regime or become more dramatic and qualitative at short distances. –  user566 Feb 24 '11 at 20:32
2  
@Raindrop yes, is the short answer. –  anna v Feb 28 '13 at 7:28

The higher derivative terms are there just like in semiclassical gravity simply because when you introduce quantum mechanics, any term that is not forbidden by some symmetry principle or other physical criteria must be present, albeit Planck suppressed. So it's not s that urprising that string theory predicts them. The interesting thing is that there are additional objects beyond the usual spectrum of GR. Things like the dilaton, supersymmetric objects (gravitino et al) and so on.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.