Particle energy in general relativity?

Assume that a particle is moving with four-velocity $u^\mu$ through a spacetime with metric $g_{\mu\nu}$ ($-+++$ signature).

Let us also assume that there exists a time-like vector field $t^\mu=(1,0,0,0)$.

Is it generally true that the energy $E$ of a particle is given by the following expression:

$$\begin{eqnarray*} E &=& -mt_\mu u^\mu\\ &=& -mg_{\mu\nu} t^\nu u^\mu\\ &=& -mg_{00}u^0\\ &=& -mg_{00}\frac{dt}{d\tau}? \end{eqnarray*}$$

In the special case that $t^\mu$ is a Killing vector and $u^\mu$ is tangent to a geodesic then $E$ is a constant of motion of the particle. But the energy of a particle need not remain constant in the general case - is that true?

If $E=-mt_\mu u^\mu$ is indeed the particle mass/energy then does it act as a source of spacetime curvature given that it depends on the $00$-component of the spacetime metric ifself? I suppose that is not a problem.