Does the Dilaton field change particle mass?

[jcap]

The low-energy effective action of the bosonic string is given by:
$$S=\frac{1}{2k_0^2}\int d^{26}X\sqrt{-G}e^{-2\Phi}\Big(\mathcal{R}-\frac{1}{12}H_{\mu\nu\lambda}H^{\mu\nu\lambda}+4\partial_\mu\Phi\partial^\mu\Phi\Big)$$
where $H_{\mu\nu\lambda}=\partial_\mu B_{\nu\lambda}+\partial_\nu B_{\lambda\mu}+\partial_\lambda B_{\mu\nu}$.

There are three fields: the space-time metric $G_{\mu\nu}$, the anti-symmetric tensor field $B_{\mu\nu}$ and the scalar dilaton field $\Phi$.

Does the scalar dilaton field $\Phi$ change the effective string coupling and therefore the mass of massive particles (but leaves massless particles unaffected)?

 

[AndyW]

Yes, the scalar dilaton field $\Phi$ does indeed change the effective string coupling and therefore the mass of massive particles. This is because the dilaton field is responsible for determining the overall strength of the string coupling, which in turn affects the mass of particles.

In the low-energy effective action of the bosonic string, the dilaton field appears in the exponential term $e^{-2\Phi}$. This term acts as a coupling constant for the rest of the action, and thus any changes in the dilaton field will directly affect the effective coupling.

In particular, if the dilaton field increases, the effective coupling will decrease, resulting in a decrease in the mass of massive particles. On the other hand, if the dilaton field decreases, the effective coupling will increase, leading to an increase in the mass of massive particles.

This effect is not seen in massless particles because they do not have a mass term in the action, and thus are not affected by changes in the coupling.

Overall, the dilaton field plays a crucial role in determining the strength of the string coupling and can significantly impact the masses of particles in the low-energy effective action of the bosonic string.