Sign of matter Lagrangian term in curved space In field theory the (matter) Lagrangian $\mathcal{L}_m$ is uncertain upto an overall constant multiplying factor (i.e. $\mathcal{L}_m$ and $a\mathcal{L}_m$ yield the same field equation(s) on extremizing the action). Going into curved space the total Lagrangian is $\mathcal{L}=\mathcal{L}_{EH}+\mathcal{L}_m$ (here $\mathcal{L}_{EH} \sim R$ is the Einstein-Hilbert part) with appropriate replacement of $\partial \text{ (or D) by }\nabla$.
How does one fix $\mathcal{L}=\mathcal{L}_{EH}+\mathcal{L}_m$ OR $\mathcal{L}=\mathcal{L}_{EH}- \mathcal{L}_m$ OR $\mathcal{L}=\mathcal{L}_{EH}+a\mathcal{L}_m$ ?
 A: Normally, when combining Lagrangians, we often leave the constant multiplying factor to be determined by experiment.
For example, if $\mathcal{L}_{k}$ is the kinetic term (for a system of charges and the electromagnetic field), and we choose to describe the electromagnetic coupling by $\mathcal{L}_{int} = A_\mu J^\mu$, then we combine them as
$$\mathcal{L}_{EM} = \mathcal{L}_{k} + e\mathcal{L}_{int}$$
where $e$ is an undetermined constant. This constant may be absorbed into $J^\mu$, effectively redefining the electric charge. The redefined electric charge is ultimately what makes a difference, and it may only be obtained experimentally (assuming you have a real system of charges in mind).
For the specific case of the Einstein-Hilbert action,
$$\mathcal{L}_{EH} = \frac{1}{2\kappa} R$$
the factor $a$ (using the reciprocal of what is specified in the question) in the combination $$\mathcal{L} = \mathcal{L}_{m} + a\mathcal{L}_{EH}$$ may simply be absorbed into $\kappa$ to so that $\kappa / a$ is what effectively matters, and it is this that is obtained experimentally, leading to the measured value for the gravitational constant $$ G = \frac{\kappa}{a} \frac {c^4}{8\pi}$$
In other words, the $\kappa$ of your inital $\mathcal{L}_{EH}$ does not matter. The overall multiplying constant in $\mathcal{L}$ alone determines the physics.
