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.
 A: No it's not: infact if you have a non-diagonal metric tensor it's not just about $g_{00}$. For example in Kerr space time you obtain $E/m=-g_{00}\dot{t} - g_{03}\dot{\phi}$.
A: Yes that is true and the verification is a routine calculation. In SR the energy of a particle with 4-momentum $p^{\mu}$ is simply $E = p^0$ relative to some inertial frame $\mathcal{O}$. 
Let's say an observer at rest in $\mathcal{O}$ has a 4-velocity $u^{\mu}$. Then in the coordinates of $\mathcal{O}$ we simply have $u^{\mu} = (1,0,0,0)$ so we can write $E = -p^{\mu}u_{\mu}$, the negative sign coming from the Lorentzian signature. Now this is a Lorentz invariant expression so it must hold in all frames. 
In particular, let's now go to curved space-time and consider a particle with 4-momentum $p^{\mu}$. Let's also consider an observer $\mathcal{O}$ (I'm using observer and frame interchangeably) with a local Lorentz frame $e_{\hat{\alpha}}$. We know that the metric in this frame is just $\eta_{\hat{\alpha}\hat{\beta}}$ so, as long as we are working with calculations that involves taking no derivatives, we are effectively just doing SR. In other words the relation $E = -p^{\mu}u_{\mu}$ carries over unscathed. 
As a quick aside, calculations involving taking only one derivative that are still equivalent to SR calculations require a freely falling coordinate system around an event.
Your question regarding Killing fields requires a little bit of care. Note that $u^{\mu}$ by itself can never in general be a Killing field except in flat space-time. This is because $u^{\mu}$ has to be normalized to proper time but the time-like Killing field $\xi^{\mu}$ will not be normalized as such. The conserved energy is actually $\mathcal{E} = -p^{\mu}\xi_{\mu}$, not $E = -p^{\mu}u_{\mu}$ even if $p^{\mu}$ is tangent to a geodesic and $u^{\mu} = \gamma \xi^{\mu}$. 
In fact, for the latter we have $\frac{dE}{d\tau} = -p^{\mu}p^{\nu}\nabla_{(\mu}u_{\nu)}$ and in general $\nabla_{(\mu}u_{\nu)} \neq 0$. If $u^{\mu}$ is tangent to a Killing field then in general only $h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}\nabla_{(\alpha}u_{\beta)} = 0$ i.e. the congruence is Born-rigid; here $h_{\mu\nu}$ is the 3-metric relative to the congruence.
The difference between $\mathcal{E}$ and $E$ is that the former is the energy of the particle as measured by an observer at spatial infinity whereas the latter is the energy as measured by a local observer. Even if we have a congruence of local observer that are stationary (follow the orbits of a time-like Killing field) $E$ will not be conserved along the particle worldline. 
Roughly speaking, the reason for this is that, as mentioned earlier, $E$ is basically just the resultant of an SR (i.e. local) measurement. It doesn't know anything about the gravitational field in and of itself. Gravitational energy cannot be localized. $\mathcal{E}$ on the other hand is the resultant of a global measurement since it is measured at spatial infinity. Global measurements do know about the gravitational field.             
Finally, in practice $E$ does not contribute to the space-time curvature as defined because it is the energy of a test particle. In principle it would of course contribute but then we get into the technical issues of the space-time geometry due to a point particle. 
A: By definition, energy is that quantity that is conserved as a result of time invariance. So yes, in GR the energy of a particle is given by your formula (multiplied by the mass). It does act as a source of curvature; after all, the things we usually think of as producing gravity, like the Earth and the Sun, are themselves made out of particles. 
Edit: Actually, there is a subtlety: the energy as defined by your formula is a scalar, but there is no reason to expect it to be. I would define the energy of a particle as the time component of $-m u_\mu$. In strange coordinate systems a time coordinate might not be easy to define: I'll leave that issue to someone more knowledgeable about the subject. 
A: One has to be very careful here. In General Relativity, there is no general definition of energy or particles.  
The reason is as follows: 
The definition cited is based on the naive particle interpretation of quantum field theory. It is well understood that in the context of cosmological models and more general spacetimes, one simply does not have a time-translation symmetry because of a lack of timelike Killing vector field. Therefore, the very definition of particles is undefined for general curved spacetimes, and only defined in the context of general relativity for asymptotically flat spacetimes. The reason is that for the particle interpretation to work, one needs to be able to decompose the quantum field into positive and negative frequency parts, which in itself depends heavily on the presence of a such a time translation symmetry in either an asymptotically flat spacetime or a Minkowski spacetime.  The problem of course is that, our universe, or any spatially homogeneous and non-static universe, that is, one that does not contain a global timelike Killing vector field is necessarily \emph{not} asymptotically flat. 
Namely, consider a spacetime $(\hat{M}, \hat{g}_{ab})$. Let this spacetime have the following three properties:


*

*There exists a function $\omega \geq 0 \in C^3$, such that $g_{ab} = \omega^2 \hat{g}_{ab}$,

*on the boundary $\omega = 0$, and $\omega_{,a}  \neq 0$,

*Every null geodesic intersects the boundary in two points.


These spacetimes are called asymptotically simple. However, if we now associate the metric tensor, $g_{ab}$  with the Einstein field equations, the existence of these three conditions implies that the spacetime is asymptotically flat. It is only under these three conditions, for which one can in a meaningful way talk about ``particles'' or energy or energy of such particles! 
Just a very important caveat that is often missed in such discussions nowadays. :-)
A: 
Assume that a particle is moving with four-velocity $u^\mu$ through a spacetime with metric...

Actually, there's a problem with that. A particle moves through space. Not through spacetime. Spacetime models space at all times. So it's static. You can plot worldlines in it to depict the motion through space over time, but the particle isn't actually moving up that worldline.   

But the energy of a particle need not remain constant in the general case - is that true?

No. The energy remains constant. Conservation of energy applies. You must know this, because you surely know that if you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². It doesn't increase by 512keV/c², or by a zillion tonnes. There is no magical mechanism by which the descending photon gains energy. Gravity is not a force in the Newtonian sense. The photon looks blueshifted because you and your clocks go slower when you're lower. It looks like it's gained energy, but it hasn't. Instead, you've lost it, check out the mass deficit. The situation in special relativity isn't totally different. If you accelerate towards the source, the photons appear to be blueshifted. But they haven't changed one jot. Instead you have.   

If $E=-mt_\mu u^\mu$ is indeed the particle mass/energy then does it act as a source of spacetime curvature

Yes. This is how you know the particle's energy doesn't increase. The net gravitational field of two bodies doesn't increase as they fall towards one another. 
