In relativity, the symmetric energy-momentum tensor is given by
$$
T^{ij}
$$
where $T^{00}$ is the energy density and $\frac{1}{c}T^{10}$ is the momentum density. Thus 
$$ \left(\frac{1}{c}T^{00}dV, \frac{1}{c}T^{10}dV\right)^{T}$$ 
is four-momentum.
Under a Lorentz transformation, this should transform like a four-vectors where
$$ \frac{1}{c}T^{00}dV= \left[\frac{1}{c}T'^{00}dV'+\frac{v}{c^2}T'^{10}dV'\right] \left( 1-\frac{v^2}{c^2}\right)^{-1/2}\\dV=dV'\sqrt{1-\frac{v^2}{c^2}}$$
After simplifications, we have
$$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10} \right] \left( 1-\frac{v^2}{c^2}\right)^{-1}$$
But if we apply the transformation to the tensor directly we get
$$ T^{00}= \left[T'^{00}+\frac{v}{c}T'^{10}+\frac{v^2}{c^2}T\
^{11} \right]\left( 1-\frac{v^2}{c^2}\right)^{-1}$$
What makes the difference? I think the first is wrong but have no idea why. Thanks.