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In chapter 7 of David Tongs' string theory lectures, the low-energy effective action of string theory is presented, and given by eq.(7.16):

$$S=\frac{1}{2\kappa^2_0}\int d^{26}X\sqrt{-G}e^{-2\Phi}\left[R-\frac{1}{12}H_{\mu\nu\lambda}H^{\mu\nu\lambda}+4\partial_{\mu}\Phi\partial^{\mu}\Phi\right].\tag{7.16}$$

If we make a dimensional analysis in natural units of mass $M$ for the first term, we will see that because that $$ds^2=G_{\mu\nu}dX^\mu dX^\nu $$ and $$\left[ds^2\right]=M^{-2}$$ thus $$\left[G_{\mu\nu}\right]=M^0 \Rightarrow \left[\sqrt{-G}\right]=M^0 .$$ The gravitational constant $\kappa^2_0$ is given in terms of string length $l_s$ thus: $$\kappa^2_0 \sim l_s^{24}\Rightarrow \left[\kappa^2_0\right]=M^{-24}$$

In many sources, the mass dimension of a Ricci scalar is given by: $$\left[R\right]=M^2$$
and $$\left[d^{26}X\right]=M^{-26}.$$ So by the basic demand that the action is dimensionless: $$\left[S\right]=M^0$$

we infer that: $$\left[e^{-2\Phi}\right]=M^0.$$

Now, looking at the third term in the effective action, recall that $$\left[\partial_{\mu}\right ]=\left[\partial^{\mu}\right]=M^1.$$ Thus we infer that: $$\left[\Phi\right]=M^0$$ but we know that $\Phi$ is a massless spinless scalar field, and the mass dimensions of a scalar field in general $D$ spacetime dimensions, is given by: $$\left[\Phi\right]=M^{\frac{D-2}{2}}.$$ In our case $D=26$ and thus we expect: $$\left[\Phi\right]=M^{12}.$$ what is the source of this mismatch? and if there is a real mismatch, why don't we scale it with $\alpha'=l_s^2$ to fit the scalar field dimensions so eventually, the action will take the form: $$S = \frac{1}{2\kappa^2_0}\int d^{26}X\sqrt{-G}e^{-2(\alpha')^{6}\Phi}\left[R-\frac{1}{12}H_{\mu\nu\lambda}H^{\mu\nu\lambda}+4(\alpha')^{12}\partial_{\mu}\Phi\partial^{\mu}\Phi\right] $$ instead?

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  • $\begingroup$ Well, OP has correctly shown that the dilaton $\Phi$ in the string frame action (7.16) is dimensionless, and it does not have the standard engineering dimension of a scalar field. $\endgroup$
    – Qmechanic
    Dec 24, 2023 at 19:06
  • $\begingroup$ @Qmechanic I showed it on the target space picture not in the worldsheet picture $\endgroup$ Dec 24, 2023 at 19:31

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