# Can the Polyakov action be derived from the point particle action?

I was wondering if it feasible to think of the Polyakov action as a generalisation of the point-particle action $$S_0 = \frac{1}{2}\int d\tau(\eta^{-1}\dot{X}^\mu\dot{X}_\mu - \eta m^2)$$ by considering the Lagrangian $$L_0(\tau) = \frac{1}{2}(\eta^{-1}(\tau)\dot{X}^\mu(\tau)\dot{X}_\mu(\tau) - \eta(\tau) m^2)$$ as the Lagrangian density $$\mathcal{L}_1(\tau, \sigma) = \frac{1}{2}(\eta^{-1}(\tau, \sigma)\dot{X}^\mu(\tau, \sigma)\dot{X}_\mu(\tau, \sigma) - \eta(\tau, \sigma) m^2) + T(\tau,\sigma)$$ of the Polyakov action, and I'm guessing a potential $T$ would be needed as the 'point' is no longer free.

I have tried working the other direction, differentiaing the Lagrangian of the Polyakov action and performing some integration by parts to obtain $$\mathcal{L}_1 = \frac{\partial S_p}{\partial \sigma} = \frac{1}{4\pi\alpha'} \sqrt{h}(h^{\gamma\delta}\frac{\partial h_{\gamma\delta}}{\partial\sigma}\frac{\partial h^{\alpha\beta}}{\partial \sigma}\partial_\alpha X^\mu \partial_\beta X_\mu+h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu+h^{\alpha\beta}\partial_\alpha \frac{\partial X^\mu}{\partial \sigma}\partial_\beta X_\mu)$$ where $X^\mu X_\mu \equiv g_{\mu\nu}X^\mu X^\nu$.

However I am not entirely sure how to proceed from here - I guess my problem is that I expect the resulting Lagrangian to be independent of $\sigma$ but I'm unsure how to deal with this.

If anybody could point me to a paper which does something along these lines, or could elucidate as to why this is not possible then I would be very grateful.