Stress energy tensor and the covariant derivative of the 4-momentum Another basic question. I have usually seen the stress energy tensor $T^{ij}$ described as the flow of the 4-momentum field $p^i$ along direction $x^j$ in spacetime with $p^0$ as energy and $x^0$ as time as is standard in relativity. At the level I have been reading this is usually not defined further, but it sounds to me somewhat like
$$T^i\,_j = \nabla_j p^i$$
or equivalently
$$T^{ij} = g^{jk} \nabla_k p^i$$
but I have never seen it written that way.
Is it correct? If not what is the relationship between $T^{ij}$ and $\nabla_j p^i$? If it is correct what properties of the 4-momentum field are needed to show the symmetry and continuity of the stress-energy tensor if it is defined this way?
EDIT
After some thought $T^{ij} = g^{jk} \nabla_k p^i$ does not seem right since 
$$ T^{0i} = p^i = T^{i0} $$
I am still curious if there is some other relationship.
 A: You're definitely on the right track but the relationship that you're looking for will depend upon which way you want to model your matter. Dust and radiation are the two models that work the best and are almost equivalent in the answers they produce. Of course the general definition of $ T_{\mu\nu} $ is “the flux of four-momentum $p_{\mu}$ across a surface of constant $ x_{\nu}$”. Since a dust cloud is a collection of moving particles with a somewhat fixed velocity in the inertial reference frame, we can figure out the ( particle ) flux with the velocity and the number of particles by defining the four velocity, $U_{\nu}$, and the number density of the particles $ n $. By this, we can determine the ( four ) flux as $ N_{\nu} = nU_{\nu} $.
Don't forget since our $0^{th}$ index is always full of fun, the $ N_{0} $ component is the number density of the particles, and the non-zero indices correspond to flux in the direction of whichever index you're dealing with.
We're almost there I promise! From that, define your energy density in terms of what we have which is just the particle number density, $n$, and the particle mass to give our $ T_{00} $ term energy density term: $\rho = nm $. Typically, it would be $ nmc^{2} $, but we take units such that $ c = 1 $. The zero index term of the four momentum using these units comes to $ \frac {E} {c^{2}} $ which you should notice to be $ m $. 
We now have all the constituents to create the entire stress-energy tensor, we have a component for it, and the relation between our four momentum and flux that brought us that first component. The rest can then be generalized as follows: $$ T_{\mu\nu} = p_{\mu} N_{\nu}=nmU_{\mu}U_{\nu}=\rho U_{\mu}U_{\nu} $$
Where we have either the tensor product between our momentum and flux ( for particles, we can make one massive particle as well and it be just as simple ) vectors, or the tenor product of our velocities with a scale factor $\rho = nm$.
Taking the mixed stress-energy tensor and $ T^{i}_{j} = \nabla_{j}p^{i} $, the connection $ \nabla_{j}$ acting on a vector $ p^{i}$, reduces to $ \frac {\partial p^{i}}{\partial x_{j}}$. As far as I know, that doesn't have any real physical significance.
All in all though, the equation above is the one that you're most likely going to want to go with. Introducing the $\nabla_{j}$ pokes at taking covariant derivatives of tensors, which requires an extra term in the answer from the regular partial differentiation called ( you guessed it ) the affine connection, but when you make $ \nabla_{j}T^{ik} = 0 $ or equivalently $ T^{ik}_{\:\:\:;j} = 0 $ we have conservation of energy and momentum!
All of this was referenced directly from these lecture notes(particularly lecture 1), if I come off as confusing or unclear. The author does a far better job than I.
