Page:BatemanElectrodynamical.djvu/28

 We shall now obtain some further properties of a general space time transformation.

When an expression has the same form in the dashed letters as in the original ones it will be convenient to say that it is an invariant.

Let us suppose that the bilinear form

is an invariant. If we multiply it by itself four times according to Grassmann's rule, we obtain the invariant

$$\left|\begin{array}{cccc} A & H & G & U\\ H & B & F & V\\ G & F & C & W\\ U & V & W & D\end{array}\right|\begin{array}{c} dx\ dy\ dz\ dt\ \delta x\ \delta y\ \delta z\ \delta t.\end{array}$$

Putting $$dx=\delta x,\dots,$$ and denoting the determinant by &Delta;, we have the invariant

Now, let

be an integral invariant of the third order.

Multiplying by (1) and rejecting the factor $$\sqrt{\Delta}dx\ dy\ dz\ dt$$, we obtain an invariant

where

{{MathForm2|(5)|$$\left.\begin{array}{l} \sqrt{\Delta}v_{x}=Au_{x}+Hu_{y}+Gu_{z}+Uu_{t}\\ \sqrt{\Delta}v_{y}=Hu_{x}+Bu_{y}+Fu_{z}+Vu_{t}\\ \sqrt{\Delta}v_{z}=Gu_{x}+Fu_{y}+Cu_{z}+Wu_{t}\\ \sqrt{\Delta}v_{t}=Uu_{x}+Vu_{y}+Wu_{z}+Du_{t}\end{array}\right\} $$}}

Multiplying (3) and (4), we obtain the invariant

hence the quantity in the square brackets divided by &Delta; is an invariant

Conversely, if we are given an integral form of the first order (4), we may obtain a reciprocal integral form of the third order (8), by multiplying