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 the 4-momentum. Under a Lorentz transformation, this should transform like 4-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 Lorentz 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 accounts for the difference? I think the first is wrong but have no idea why.
4 Answers
One really doesn't talk about components of the Stress-Energy tensor being momentum densities. Instead, you think about projections on vectors. So, the energy density observed by a timelike observer with 4-velocity $u^{a}$ is given by
$$j^{a}=T^{a}{}_{b}u^{b}$$
Now, when you boost, you have two choices. You can either boost to a new reference frame, and still look at the density observed with our original observer, which then means you have to boost $u^{a}$ as well:
$$\begin{align} j^{\prime}{}^{a} &= \Lambda^{a}{}_{c}\Lambda_{b}{}^{d}T^{c}{}_{d}\left(\Lambda^{b}{}_{e}u^{e}\right)\\ &= \Lambda^{a}{}_{c}\Lambda_{e}{}^{d}T^{c}{}_{d}u^{e} \end{align}$$
OR, we could measure the stress energy tensor as observed by someone who has a four velocity $v^{a}$ which has the same components as $u^{a}$, but in the boosted frame (i.e., $v^{a} = \Lambda^{-1}{}^{a}{}_{c}u^{c}$). Then, we have:
$$\begin{align} j^{\prime}{}^{a} &= \Lambda^{a}{}_{c}\Lambda_{b}{}^{d}T^{c}{}_{d}\left(v^{b}\right)\\ &= \Lambda^{a}{}_{c}\Lambda_{b}{}^{d}T^{c}{}_{d}\left(\Lambda^{-1}{}^{b}{}_{e}u^{e}\right)\\ &=\Lambda^{a}{}_{c}\delta_{e}{}^{d}T^{c}{}_{d}u^{e}\\ &=\Lambda^{a}{}_{c}T^{c}{}_{d}u^{d}\\ \end{align}$$
As you can see, the two factors differ by a factor of the lorentz transformation, and most critically, you need the two factors of the lorentz transformation to mix in the $T_{11}$ component into the momentum density.
Let $\textbf{u},\textbf{w}$ be the four velocities of observers at rest in the original and primed frames, respectively. Let $\Delta V$ be a small volume in $\textbf{u}$'s frame, and let $\Delta p^{\mu}$ be the total amount of four momentum contained in $\Delta V$. $\Delta p^{\mu}$ is also expressible as $T^{\mu0}\Delta V=T^{\mu\alpha}(u_{\alpha}\Delta V)$ - the flux of $T\cdot\textbf{d}x^{\mu}$ through an oriented (in the direction of $\textbf{u}$) hypersurface $\Delta V$.
When you perform a Lorentz transformation on the volume $\Delta V$ to obtain $\Delta V'$ the components of the volume's orientation also transform. The total four momentum in $\Delta V'$, oriented with $\textbf{u}$, is: $$\Delta p'^{\mu}=T'^{\mu\alpha}(u'_{\alpha}\Delta V')=T^{\sigma \rho}\Lambda^{\mu'}_{\sigma}\Lambda^{\mu'}_{\rho}\Lambda^{\beta}_{\mu'}u_{\beta}\Delta V'=\left(\frac{\Delta p^{\sigma}}{\Delta V}\right)\Lambda^{\mu'}_{\sigma}\Delta V'$$ So that the four momentum density flowing through the hypersurface $\Delta V$ oriented by $\textbf{u}$ is indeed a four vector. This is not the same object as $T '^{\mu0}\Delta V'=T'^{\mu \alpha}w'_{\alpha}\Delta V'$ - the total four momentum flowing through a hypersurface (of volume $\Delta V'$) oriented by $\textbf{w}$. $T^{00}$ is $\textbf{u}=\textbf{e}_{0}$ component of the momentum flux through $\textbf{u}$ hypersurface, this corresponds to your last equation. The quantity on the right hand side of the preceding eqaution is the $\textbf{u}=\textbf{e}_{0}$ component of momentum flux through the $\textbf{w}$ hypersurface.
For an observer with 4-velocity $u^\mu$, the density of 4-momentum is $u_\nu T^{\mu\nu}$. If $u^\nu = (1,0,0,0)$, then indeed the 4-momentum is $p^\mu = T^{\mu 0}$ and this is a 4-vector. Now let primed indices correspond to components in some other frame. Then $$ p^{0'} = L^{0'}_\mu p^\mu = L^{0'}_{\mu} T^{\mu 0}.$$ But it is not the case that $$p^{0'} \overset{!}{=} T^{0'0'}.$$ You can see this from the covariant expression $$p^{\mu'} = u_{\nu'}T^{\mu' \nu'}.$$ Since after a boost along the $1$-axis, $u_\nu' = (\gamma, -\gamma v, 0,0) \neq (1,0,0,0)$, what is true is that $$p^{0'} = \gamma T^{0'0'} - \gamma vT^{1'0'}.$$
This should cancel the additional term that confused you.
I build up on Brian Trundy's answer, giving a more explicit derivation. Indeed, when you do a Lorentz boost your orientation changes and you cannot think of $dV$ transforming as a simple length. You need to think of it as the component of a four-vector $dV^\mu=u^\mu dV$, where $u^\mu$ is the four-velocity. In a static reference frame, the four-velocity is $u^\mu=(c,0)$ (let me assume one spatial dimension for simplicity), so the energy is just
$$dE=\frac{1}{c}dp^0=\frac{1}{c}T^{0\nu}dV_\nu=\frac{1}{c}T^{00}dV$$
Now, indeed, if you transform this object as a vector you get
$$\frac{1}{c}T^{00}dV=\frac{1}{c}T^{0\nu}dV_\nu=\left[\frac{1}{c}T'^{0\nu}dV'_\nu-\frac{v}{c^2}T'^{1\nu}dV'_\nu\right]\left(1-\frac{v^2}{c^2}\right)^{-1/2}$$
And the transformation of the volume element four-vector is $$dV'^\nu=\left(1-\frac{v^2}{c^2}\right)^{-1/2}(dV,-v\,dV)$$
So in the end we have $$\frac{1}{c}T^{00}dV=\frac{1}{c}\left[T'^{00}-2\frac{v}{c}T'^{01}+\frac{v^2}{c^2}T'^{11}\right]\left(1-\frac{v^2}{c^2}\right)^{-1}dV$$
Which coincides with the transformation of the stress-energy tensor as a $2$-tensor, the factor of $2$ appearing due to the symmetry $T^{\mu\nu}=T^{\nu\mu}$.
