Interpretation of a term in the Maxwell stress tensor With no magnetism, the $xx$ component of the Maxwell stress tensor $T$ is
$$T_{xx} = \frac{1}{2}(E_x^2 - E_y^2 - E_z^2)$$
I can see why there should be a $+E_x^2$ term, but intuitively I don't see why $E_y$ or $E_z$ should be relevant at all. Is there a physical way to see this?
 A: When there is no $\vec B$ field, the force density $\vec f =\rho\vec E +\vec J \times \vec B$ equals $\epsilon_0\left( \vec \nabla \cdot \vec E \right)\vec E.$
And if there is also no $\vec B$ field in a time interval then $\vec \nabla \times \vec E=\vec 0.$ This last part is important, because the time rate of change of $\vec B$ is essential. But when $\vec \nabla \times \vec E$ is zero, we get that $\epsilon_0\vec E \times \left(\vec \nabla \times \vec E \right)=\vec 0$ and then $\epsilon_0\left( \vec \nabla \cdot \vec E \right)\vec E$=$\epsilon_0\left( \vec \nabla \cdot \vec E \right)\vec E-\epsilon_0\vec E \times \left(\vec \nabla \times \vec E \right).$
Which equals $\epsilon_0\vec \nabla \cdot \left( \vec E \otimes \vec E -(E^2/2)\mathbb{1}\right).$ 
In short, the stress energy tensor is something whose divergence needs to give the change in momentum density. The xx component is just one of the three terms a divergence needs to take the derivative of to give the force density.  Focusing on one component makes no sense.
When you have $\vec \nabla \cdot \vec J =-\partial \rho / \partial t$ you know that all three components of $\vec J$ are needed to get $\rho$ similarly $T_{xx}, T_{yx} $ and $T_{zx}$ are all needed to get  $f_x.$ There is no reason to focus on the $T_{xx}$ in isolation because if $\mathscr P_x$ is the density of x momentum then $\partial_x T_{xx}+\partial_y T_{yx}+\partial_z T_{zx}=\partial_t\mathscr P_x.$
Just as $\vec J$ measures the flow of charge, so does $T_{xx}\hat x+T_{yx}\hat y+T_{zx}\hat z$ measure the flow of $\mathscr P_x.$
You asked for a way to see it physically. So note that $T_{xx}$ measures the flow of $\mathscr P_x$ in the x direction. To claim it doesn't depend on $E_y$ or $E_z$ is a rather wild (and incorrect) claim. All we really know about the stress of the electromagnetic field is that the divergence of it is the time rate of change of the momentum density. When there is no magnetic field in a time interval, the only momentum is mechanical momentum so we need the divergence to be the force density. But the force density includes the the charge density which depends on the divergence of $\vec E$ so depends on all three components of $\vec E.$ So we fully expect all three components of $\vec E$ to matter.
So we want the divergence to be the force density. The divergence will give parts that equal $\epsilon_0\left( \vec \nabla \cdot \vec E \right)\vec E$ because it always does. And it will give parts that happen to be zero, simply because the curl of the electric field has to be zero because you insisted on considering a case with no magnetic field.
So consider $T_{xx}=(E_{xx}^2-E_{yy}^2-E_{zz}^2)/2$ as well as $T_{yx}=E_yE_x$ and $T_{yx}=E_zE_x$ then the divergence gives a term like $\rho E_x$ but also gives terms like $-E_y\partial_xE_y$ and $-E_z\partial_xE_z$ (from the $T_{xx}$ term) and terms like $E_y\partial_yE_x$ and $E_z\partial_zE_x$ (from $T_{yx}$ and $T_{zx}$). So we needed the terms in $T_{yx}$ and $ T_{zx}$ to get the full charge density (since only those terms get partials in y and z), but then we need extra terms in $T_{xx}$ to counter the terms that appeared. In our example not that the curl of $\vec E$ being zero makes them cancel exactly.
So since each term $T_{xx},$ $T_{yx},$ and $T_{zx}$ has a physical meaning (the flow of $\mathscr P_x$ in the x, y, and z directions respectively), the physical meaning is exactly that. We have to have some flow of $\mathscr P_x$ in the x direction to compensate for the flow of $\mathscr P_x$ in the y and z directions so that the net flow is exactly the transfer to the mechanical momentum.
