Proving identity related to Maxwell Stress Tensor on boundary (electrostatic) Maxwell Stress Tensor is defined as $T_{ij}=\epsilon(E_i E_j-\frac{1}{2}|E|^2\delta_{ij})$, where $i=1,2$ and $|E|^2=E_1^2+E_2^2$ (consider two-dimensional for simplicity).
The boundary conditions on an interface are $[\epsilon\ \mathbf{n}\cdot\mathbf{E}]^I_{II}=0$ (continuity of the normal component of displacement field) and $[\ \mathbf{t}\cdot\mathbf{E}]^I_{II}=0$ (continuity of tangential component of electric field), where the notation $[...]^I_{II}$ denotes the difference between medium I and II across the boundary.
A paper I'm reading states that $[\mathbf{t}\cdot\mathbf{T}\cdot\mathbf{n}]^I_{II}=0$. How to show this is true? I tried writing out all terms and do the matrix multiplication, but I don't know how to deal with the product of E fields inside the tensor.
 A: It's probably easiest to see this if we ditched the indices and wrote this in matrix notation. The matrix that represents $T_j^i$ (note the upper index) is given by:
$$\mathbf{T}  = \epsilon \mathbf{E}\mathbf{E^T} - \frac{1}{2} (\mathbf{E^T}\mathbf{E})\mathbf{I}$$
Now, as a tensor, 
$$\mathbf{t}\cdot\mathbf{T}\cdot\mathbf{n} = T_{ij}t^i n^j = T_j^i t_i n^j$$
using Einstein summation convention. The equality between the terms on the far ends means that we can evaluate that product in the paper like so:
$$\Bigr[\mathbf{t^T} \left(  \epsilon \mathbf{E}\mathbf{E^T} - \frac{1}{2} (\mathbf{E^T}\mathbf{E})\mathbf{I}\right)\mathbf{n}\Bigr]_{II}^I = \Bigr[(\mathbf{t^T}\mathbf{E})(\epsilon \mathbf{E^T}\mathbf{n}) - \frac{1}{2}(\mathbf{E^T}\mathbf{E})\mathbf{t^T}\mathbf{n}\Bigr]_{II}^I$$ 
$$\equiv \Bigr[ (\mathbf{t}\cdot\mathbf{E})(\epsilon\mathbf{n}\cdot\mathbf{E}) - \frac{1}{2}(\mathbf{E}\cdot\mathbf{E})(\mathbf{t}\cdot\mathbf{n})\Bigr]_{II}^I = 0$$
where we switched from the matrix representations to the actual vectors in the last line.
