# “low-energy effective action” but in what sense?

In string theory, consistency with Weyl invariance imposes dynamics on the background fields through the vanishing of the beta functions. Those dynamics can also be derived from the so-called low energy effective action: $$S = \frac{1}{2\kappa_0^2}\int d^{26} X\; \sqrt{-G}\; \mathrm{e}^{-2\Phi}\,(R-\frac{1}{12}H_{\mu\nu\lambda}H^{\mu\nu\lambda}+4 \partial_{\mu}\Phi\partial^{\mu}\Phi)$$ (at least in bosonic string theory)

Maybe I shouldn't worry over lexical denomination, but I find this naming of "low-energy" a bit obscure. In what sense is it used? Would it be because the background fields are supposed to emerge from the fundamental strings? Or because we forget the massive excited states of the string (with masses around the Planck scale and irrelevant for low energy phenomenology)?

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It's a standard terminology – and set of insights – not only in string theory but in quantum field theories or anything that can be approximated by (other) quantum field theories at... low energies.

Such a low-energy action becomes very accurate for the calculation of interaction of particles (quanta of the fields) of low energies, in this case $E\ll m_{\rm string}$. Equivalently, the frequencies of the quanta must be much smaller than the characteristic frequency of string theory. The previous sentence may also be applied in the classical theory: the low-energy effective action becomes accurate for calculations of interactions of waves whose frequency is much lower than the stringy frequency or, equivalently, whose wavelength is much longer than the string scale, $\lambda\gg l_{\rm string}$.

Low-energy effective actions may completely neglect particles whose mass is (equal to or) higher than the characteristic energy scale, in this case $m_{\rm string}$, because such heavy particles can't be produced by the scattering of low-energy particles at all – so they may be consistently removed from the spectrum in this approximation.

The scattering of the light and massless particles that are kept may be approximately calculated from the low-energy effective action and this approximation only creates errors that are proportional to positive powers of $(E/m_{\rm string})$ so these errors may be ignored for $E\ll m_{\rm string}$. You may imagine that there are corrections in the action proportional to $\alpha'$ or its higher powers that would make the effective action more accurate at higher energies but become negligible for low-energy processes.

There are lots of insights – conceptual ones as well as calculations – surrounding similar approximations and they're a part of the "renormalization group" pioneered mainly by Ken Wilson in the 1970s. In particular, by "low-energy effective actions", we usually mean the Wilsonian effective actions. But they're pretty much interchangeable concepts to the 1PI (one-particle-irreducible) effective actions, up to a different treatment of massless particles.

It is impossible to teach everything about the renormalization group and effective theories in a single Stack Exchange answer. This is a topic for numerous chapters of quantum field theory textbooks – and for whole graduate courses. So I just conclude with a sentence relevant for your stringy example: string theory may be approximated by quantum field theories for all processes in which only particles much lighter than the string mass are participating and in which they have energies much smaller than the string scale, too. If that's the case, predictions of string theory for the amplitudes are equal to the predictions of a quantum field theory, the low-energy effective field theory, up to corrections proportional to powers $(E/E_{\rm string})$.

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 Thanks for the time you put in this detailed answer Lubos. I am familiar with the concept of renormalization group and effective field theory. My interrogation was about arguments specific to the derivation of this effective action, or differences between in an "effective field theory"-description in string theory in comparison to quantum field theory. But you answered that. – Just_a_wannabe Sep 28 '12 at 14:07