I have read this question on the Dilaton, but I am a little confused with the distinction between the Dilaton and the Radion.

I definitely have the feeling that these two scalar fields are different particles.

I am fimiliar with the Diliton being related to the coupling constant in String Perturbation Theory.

$$g_s = e^{\langle \phi\rangle}$$

Moreover, the Dilaton is related to the size of a compactificated dimension. This was covered in our Bosonic String Theory lectures (I cannot link from outside the network).

On the other hand, the Radion is usually the name given to the $g_{55}$ (or $g_{zz}$ in the notes referenced) component of the metric tensor in a Kaluza-Klein theory, and too is related to the size of the compactified dimension

$$ \hat{g}_{zz} = \exp(2 \beta \phi) $$

giving a four-dimensional effective field theory

$$ \mathcal{L} = \sqrt{-\hat{g}}\hat{\mathcal{R}} = \sqrt{-g}\left(\mathcal{R} – \frac{1}{2}(\partial \phi)^2 \frac{-1}{4} \exp \left(-2 (D-1) \alpha \phi \right) \mathcal{F}^2 \right) $$

Furthermore, in Randal-Sundrum model I have seen the scalar field called a Radion, even though here we explicitly avoid compactification.

Finally, in the Cyclic Model of the Universe I have heard the moduli scalar field which 'measures' the distance of seperation between the two branes the Radion.

I have been studying the original paper on the Alternative to Compactification and a review on the Cyclic Model of the universe, as well the lecture notes on Kaluza-Klein Theory by C. Pope to try to learn about these things.

User1504 mentions that they are the same in M-Theory and Type IIA string theory, but I am afraid that I have not studied Superstring theory or beyone yet.

So to reiterate, my question is, can anyone give me a discription of the difference between the Dilaton and the Radion?

## Best Answer

Dilaton is the generic name of the Goldstone Boson (GB) associated with spontaneous breaking of scale invariance. Any model that break scale invariance spontaneously will give rise to a dilaton. N=4 SYM for example has a moduli space and any modulus away form the origin will break conformal invariance, and a massless dilaton would appear. The Radion is just the 5D holographic realization of a 4D CFT that break conformal symmetry spontaneously.

As for any GB, the dilaton $\sigma(x)$ provides a non- linear realization of scale invariance $x\rightarrow x^\prime=x/\lambda$, $$ \sigma(x)\rightarrow \sigma^\prime(x)= \sigma(\lambda x)+\log\lambda $$ where the effective lagrangian reads

$$ L=\frac{f^2}{2}(\partial\chi)^2- k f^4\chi^4+\frac{(\partial\chi)^4}{\chi^4}+\ldots $$ with $f$ is the analog of the pion decay constant and $\chi$ is just shorthand for $\chi=e^{\sigma}$. Since $\chi$ transforms as a field with dimension $\Delta=1$, $$ \chi(x)\rightarrow \lambda\, \chi(\lambda x) $$ the action $\int d^4 x L$ is left invariant under the scale transformation. In practice, any term with dimension 4 in the lagrangian must be written down. The only difference wrt to an ordinary GB is that the symmetry is a spacetime symmetry so that acts also on $x$. Analogous expressions hold for the full conformal group tang simply forbids certain operators that are allowed by scale invariance only. Notice that a pure quartic potential is allowed in principle (although it would require a curved space for consistently be non-vanishing\ldots)

Let's move now to the radion that arise in extra dimensional theory. To show that it is just a special realization of a dilaton from extra dimensions, we derive its effective action and matches to the one for the dilaton. Let's take the Randall Sundrum model that is a slice of 5D AdS space

$$ ds^2=\frac{L^2}{z^2}\left(dx^2_4-dz^2\right) $$ between an UV brane and an IR brane at $x=z_{UV}$ and $z_{IR}$ respectively. $L$ is the AdS curvature. Of course, one of the isometries of AdS is $$ x\rightarrow x/\lambda\,,\qquad z\rightarrow z/\lambda $$ which is broken softly e.g. by the IR brane that gets moved $z_{IR}\rightarrow z_{IR}/\lambda$. There are various ways to get the effective action for the radion. One possibility would be looking at the metric perturbations and disentangle the spin-2 (graviton) from the spin-0 (radion) since both have zero modes. But, actually, a faster and more intuitive way comes from remembering that the radion is supposed to give the size of the extraD. In fact, it is clear that $z^{-1}_{IR}$ that breaks the scale isometry has to be identified with the vev of $\chi$. We can thus integrating out the bulk of the extra dimension but letting the IR brane free to move and bend, so that you end with a 4D effective action for $z_{IR}$ $$ L=-\frac{1}{M_*^3}\int_{z_{IR}}^{z_{UV}} dz \sqrt{|g|} R+\mbox{boundary terms} $$ where the boundary terms contain the localized actions on the branes and the Gibbons-hawking term. The resulting action is exactly the one for a dilaton up to identifying $\chi\sim z^{-1}_{IR}$. For example, the bulk and the boundary terms give rise to an exact quartic potential (after using the bulk eom's that link the 5D cc $\Lambda _5$ to the 5D curvature $L$, and the 5D Planck mass $M_*$) $$ V=k z_{IR}^{-4}\qquad k\propto\Lambda_{5}L-V_{IR} $$ that is indeed the only non trivial potential that a scale invariant theory can admit. If the 4D space is actually flat then it results $k=0$ by consistently tuning the bulk cosmological constant $\Lambda_5$ and the localized IR potential to be the same. In this way the radion has a flat direction since the potential vanishes. Analogously, the derivative terms are also dictated by scale invariance and reproduce the action for the dilaton. Again, this is not surprising since the isometry we have been looking above is the 5D dual of a scale transformation. In this gauge/gravity duality, the IR brane represents a vev of an operator of very high dimension. Moving the IR brane around, that is rescaling the vev, is what a dilaton/radion excitation does.