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I have a two-component field:

$$\phi(\vec{x}) = \left( \begin{array}{c} \phi_1(\vec{x}) \\ \phi_2(\vec{x}) \end{array} \right)$$

with $\phi^T = (\phi_1, \phi_2)$. And I am trying to evaluate:

$$(\vec{\nabla}\phi)^T(\vec{\nabla}\phi)$$

I'm pretty sure the result of that expression is supposed to be a scalar, but I can't figure out why. $\phi$ is a 2x1 matrix, which means $\vec{\nabla}\phi$ must be the result of a matrix multiplication between a Nx2 and 2x1 matrix. You would think N would be 3, since $\vec{\nabla} = \frac{\partial}{\partial x}\vec{x} + \frac{\partial}{\partial z}\vec{z} + \frac{\partial}{\partial z}\vec{z}$, but then why are there 2 columns? Not sure what to do here so any assistance would be quite helpful.

(I'll mention that the full thing I'm trying to evaluate is $\mathcal{L} = \frac{1}{2}\left(\dot{\phi}^T\dot{\phi}-m(\vec{\nabla}\phi)^T(\vec{\nabla}\phi)\right)$. $\mathcal{L}$ is a scalar, right?)

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    $\begingroup$ It's not straight forward notation, but my guess is that $\nabla\phi$ sound be interpretet as the vector with six components, $\tfrac{\partial}{\partial x^1}\phi_1, \tfrac{\partial}{\partial x^2}\phi_1, \dots, \tfrac{\partial}{\partial x^3}\phi_2$ and then $(\vec{\nabla}\phi)^T(\vec{\nabla}\phi):=\sum_{i\in\{1,2,3\}}\sum_{j\in\{1,2\}}(\tfrac{\partial}{\partial x^i}\phi_j)^2$ $\endgroup$ – Nikolaj-K Nov 27 '13 at 10:39
  • $\begingroup$ More naturally, you would consider $\vec{\nabla}\phi$ to be a 3-component vector of 2-component vectors. It's not good to have the same index stand for two things, because it gets messy. Mathematically, of course, it's equivalent. $\endgroup$ – lionelbrits Nov 27 '13 at 12:58
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When you have a matrix $\Phi = \begin{pmatrix} \phi_1\\ \phi_2\end{pmatrix}$, with one column and two rows, and its transpose matrix $\Phi^T = \begin{pmatrix} \phi_1 & \phi_2\end{pmatrix}$, with one row and two columns, the product of the two matrix $\Phi^T \Phi$ is a matrix $P$ with one column and one row:

$P =\Phi^T \Phi = \begin{pmatrix} \phi_1 & \phi_2\end{pmatrix}\begin{pmatrix} \phi_1\\ \phi_2\end{pmatrix} = (\phi_1^2+\phi_2^2)$

Because this matrix $P$ has one row and one column, it may considered as a scalar.

For your particular problem, you have :

$(\vec{\nabla}\phi)^T.(\vec{\nabla}\phi) = \sum\limits_{i=1}^N ({\partial_i}\phi)^T({\partial_i}\phi) = \sum\limits_{i=1}^N ((\partial_i \phi_1)^2 + (\partial_i \phi_2)^2)$

The first equality comes from the definition of the inner product and the gradient, and the second equality comes from the definition of the transpose operation $T$, and the manipulation of these matrices, as seen at the beginning of the answer.

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