# How terms disappear in conservation of stress-energy tensor?

If gravity is not present the stress-energy tensor has vanishing divergence so that using cartesian coordinates for spacetime we have $$\partial_\mu T^{\mu \nu} = 0.$$

For above equation Carroll writes in his general relativity introduction book that

The equation with $\nu = 0$ corresponds to conservation of energy, while $\partial_\mu T^{\mu k} = 0$ expresses conservation of kth component of momentum.

Now I understand that for index $\mu = 0$ time derivative of energy and momentum density appear on the left side of the equation, but if I understand correctly partial derivative is defined as giving tensor with one more down index so that the left side is contraction of indices so that for fixed $\nu$ there should be four terms on left side. So for example for $\nu = 0$ the equation would give $$\partial_0T^{00} + \partial_1T^{10} + \partial_2T^{20} + \partial_3T^{30} = 0.$$ How does Carroll's statement handle these extra three terms? Are they zero for some obvious reason I do not see or have I misunderstood the whole tensor notation?

The first thing to notice is that this equation allows flux of energy. So the statement of energy conservation is not:$$\frac{dE}{dt} = 0$$ But instead: $$\frac{dE_{bulk}}{dt}=\text{flux of energy into the bulk}$$ Equivalently, we can look at a local version of the law which only deals with one point A in the bulk, and we write this version as: $$\frac{d\rho_{A}}{dt}=\text{flux of energy into point A}$$ Now let's go back to your equation with four terms: $$\partial_0 T^{00}+\partial_1 T^{10}+\partial_2 T^{20}+\partial_3 T^{30}=0$$ If we take the bulk to be a fluid and A is a point in the fluid. Then $T^{00}$ is defined as the energy density $\rho$ of the fluid (in natural units) at point A. $T^{i0}$ (where $i=1,2,3$) is defined to be the energy flux through the $x^i$ surface. What does that mean? Well, take $T^{10}$ as an example. That means in each unit of time, there is $T^{10}$ amount of energy that flows in the $x^1$ direction through a patch of unit area. So when we take the derivative of it at point A, we can visualize it this way:
Suppose you have two plates of unit area that are perpendicular to the $x^1$ direction. The $x^1$ coordinate of plate 1 is $x$, and the $x^1$ coordinate of plate 2 is $x+\Delta x$, while A is at $x+\Delta x/2$. So we can say: $$\partial_1T^{10} \approx \frac{T^{10}_{Plate2}-T^{10}_{Plate1}}{\Delta x}$$
How do we interpret this? Well, imagine you are sitting at point A and enclosed in a little box with plates 1 and 2 as walls. Then $T^{10}_{Plate_2}$ flows out of the box, and $T^{10}_{Plate_1}$ flows into the box (due to the minus sign). So if we take the limit as the box shrinks, the derivative represents the flux of energy out of point A in the $x^1$ direction, as expected. Similar arguments works for other components. Now your equation reads: $$\frac{d\rho_A}{dt}+\text{flux of energy out of point A in x^1 direction} + \text{flux of energy out of point A in x^2 direction} + \text{flux of energy out of point A in x^3 direction}=0$$
If we use the fact that:$$\text{flux of energy out of point A} = - \text{flux of energy into point A}$$ We finally arrive at the result that we set out to show: $$\frac{d\rho_A}{dt} = \text{flux of energy into point A}$$