Page:BatemanElectrodynamical.djvu/16

 Putting

we see that &lambda;² is positive, and therefore

is negative. This shows that $$\frac{\partial(x',y',z')}{\partial(x,y,z)}$$ and $$\frac{\partial t'}{\partial t}\frac{\partial(x',y',z',t')}{\partial(x,y,z,t)}$$ must have the same sign. Hence, if $$\frac{\partial t'}{\partial t}$$ is positive, $$\frac{\partial(x',y',z')}{\partial(x,y,z)}$$ must have the same sign as the Jacobian. Accordingly, a transformation which changes a right-handed system of axes into a right-handed system must have a positive Jacobian; a transformation which changes a right-handed system of axes into a left-handed system must have a negative Jacobian.

The sign of &theta; may now be determined from equation (A). Since

it is necessarily positive. Consequently &theta; must have the same sign as $$\frac{\partial(x',y',z')}{\partial(x,y,z)},$$ and therefore the same sign as the Jacobian.

We can now obtain the formulae of transformation in the two possible cases.

(i) When the Jacobian is positive,$$\theta=+1$$, and the formulae of transformation are

$$\begin{array}{rl} E_{x}= & E'_{x}\frac{\partial(y',z')}{\partial(y,z)}+E'_{y}\frac{\partial(z',x')}{\partial(y,z)}+E'_{z}\frac{\partial(x',y')}{\partial(y,z)}\\ \\ & \qquad-H'_{x}\frac{\partial(x',t')}{\partial(y,z)}-H'_{y}\frac{\partial(y',t')}{\partial(y,z)}-H'_{z}\frac{\partial(z',t')}{\partial(y,z)}\\ \\-H_{x}= & E'_{x}\frac{\partial(y',z')}{\partial(x,t)}+E'_{y}\frac{\partial(z',x')}{\partial(x,t)}+E'_{z}\frac{\partial(x',y')}{\partial(x,t)}\\ \\ & \qquad-H'_{x}\frac{\partial(x',t')}{\partial(x,t)}-H'_{y}\frac{\partial(y',t')}{\partial(x,t)}-H'_{z}\frac{\partial(z',t')}{\partial(x,t)}\\ \\\rho w_{x}= & \rho'w'_{x}\frac{\partial(y',z',t')}{\partial(y,z,t)}+\rho'w'_{y}\frac{\partial(z',x',t')}{\partial(y,z,t)}+\rho'w'_{z}\frac{\partial(x',y',t')}{\partial(y,z,t)}-\rho'\frac{\partial(x',y',z')}{\partial(y,z,t)},\\ \\-\rho= & \rho'w'_{x}\frac{\partial(y',z',t')}{\partial(x,y,z)}+\rho'w'_{y}\frac{\partial(z',x',t')}{\partial(x,y,z)}+\rho'w'_{z}\frac{\partial(x',y',t')}{\partial(x,y,z)}-\rho'\frac{\partial(x',y',z')}{\partial(x,y,z)}.\end{array}$$