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 When projecting into the space of three-dimensional ($$x, y, z$$), then the first six components form a three-dimensional tensor ($$T^{3}$$), which are transforming as the squares and products of ($$x, y, z$$); the following three components of $$T^{4}$$ are forming a $$V^{3}$$, the tenth a scalar $$S^{3}$$.

The four components of the operator "lor" transform as components of $$V_{I}^{4}$$, we can deduce – from a four-dimensional scalar $$\varphi$$, given as a function of $$x, y, z, u$$ – a $$T^{4}$$ which is twice differentiable with respect to $$x, y, z, u$$:

$\frac{\partial^{2}\varphi}{\partial x^{2}},\ \frac{\partial^{2}\varphi}{\partial y^{2}},\ \frac{\partial^{2}\varphi}{\partial z^{2}},\ \frac{\partial^{2}\varphi}{\partial y\ \partial z},\ \frac{\partial^{2}\varphi}{\partial z\ \partial x},\ \frac{\partial^{2}\varphi}{\partial x\ \partial y};\quad\frac{\partial^{2}\varphi}{\partial x\ \partial u},\ \frac{\partial^{2}\varphi}{\partial y\ \partial u},\ \frac{\partial^{2}\varphi}{\partial z\ \partial u};\quad\frac{\partial^{2}\varphi}{\partial u^{2}}.$|undefined

A $$S^{4}$$ is given, being a quadratic homogeneous function of $$x, y, z, u$$:

the 10 coefficients:

$c_{11},\ c_{22},\ c_{33},\ c_{23},\ c_{31},\ c_{12};\ c_{14},\ c_{24},\ c_{34};\ c_{44},$

form a four-dimensional tensor.

In the electrodynamics of moving bodies, the equations of the momentum and energy apply:

In order to give to these four equations a more symmetrical from, we put: