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?
 A: When Carroll says "the equation", he means the equation with four terms that you wrote down. The summation convention is explicit in the placement of indices. Now on to the physical intuition.
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:
$$ \frac{d\rho_A}{dt} = \text{flux of energy into point A}$$
