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Stringy corrections of Einstein's vacuum field equations

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From string theory, the vacuum field equations obtain correction of the order $O[\alpha'R]^n$ such that they can be written as

$$ R_{\alpha\beta} -\frac{1}{2}g_{\alpha\beta}R + O[\alpha'R] = 0 $$

where $\alpha' = \frac{1}{2\pi T_s}$, when including just the first order term for example.

What is the physical interpretation of these corrections, what do they look like more explicitely, and how can they be obtained?

asked May 27, 2013 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
What exactly are the correction terms? Surely they cannot be scalors.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user MBN
@MBN have just heard that there are such corrections, but I dont know how they look explicitely :-/. So I have put this into the question too.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dilaton
Are these the corrections you get when you compute the effective action (to some chosen number of loops)? Having done that, you write down the "corrected" equations of motion.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user twistor59
@twistor59 you mean that one starts with some bare action at a high (around Planck) energy scale, the RG flows it down to a scale where classical GR can be applied, and the additional terms the effective action has obtained lead to these corrections of the corresponding EOM?

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dilaton
@Dilaton Sort of I think; I meant the stuff that's described in section 7.3.1 and 7.3.2 here

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user twistor59
Thanks @twistor59, it seems to be exactly the business desribed there :-).

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dilaton
See also: How do Einstein's field equations come out of string theory?

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dimensio1n0

1 Answer

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As Witten explains in his NOTICES OF THE AMS article (please see also his more recent lecture), the fully quantum string theory is characterized by two coupling constants (or in the language of deformation quantization: two deformation parameters). The string coupling $g_s$ and the string tension $\alpha^{'}$. In perturbation theory, one gets dependence of the string amplitudes on powers of $g_s$ (or equivalently in $\hbar$) through the genus expansion.

The dependence of the amplitudes $\alpha^{'}$ is obtained once one takes into account that in the presence of background fields, the string Lagrangian is not free, it is described by a sigma model. If we compute the quantum correction to this sigma model we get terms with more and more derivatives multiplied by more powers of the of the string tension (as in chiral perturbation theory). When the quantum corrections to the trace of the energy momentum tensors are calculated then here also terms depending on powers of $\alpha^{'}$ will appear and the condition of vanishing of the beta function will results Einstein's equations with correction terms proportional to $\alpha^{'}$.

Please see equations 3.7.14. in Polchinsky's first volume, where the beta functions are given to the first power of $\alpha^{'}$.

Witten explains that for a while, the work on string theory concentrated on finding candidates of $\alpha^{'}$ deformed theories (as conformal field theories), then $\hbar$-quantize them as in ordinary quantum theory.

But, as Witten explains, after the discovery of the full set of string dualities and the role of membranes, it was realized that in order to fully quantize string theory, the two quantizations or two deformations($\hbar$, $\alpha^{'}$) must be perfomed together.

This route has profound consequences, for example, it leads to the conclusion that the string full quantum theory should be in the realm of noncommutative geometry, because in the presence of a $B$-field and brane boundary conditions, the position-position commutation relations will obtain $\alpha^{'}$ deformation and become noncommutative. As a consequence, the ordinary uncertainty relation will get $\alpha^{'}$ deformation and turns into a generalized (minimal scale) uncertainty, in which the position uncertainty has a nonvanishing minimum:

$\Delta x \geqslant \frac{\hbar}{\Delta p}+ \alpha^{'}\frac{\Delta p}{\hbar}$

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user David Bar Moshe
answered May 28, 2013 by David Bar Moshe (3,505 points) [ no revision ]

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