Page:BatemanConformal.djvu/6

Accordingly, from a solution V = F(x, y, z, w) of the equation

$\left({\frac {\partial V}{\partial x}}\right)^{2}+\left({\frac {\partial V}{\partial y}}\right)^{2}+\left({\frac {\partial V}{\partial z}}\right)^{2}+\left({\frac {\partial V}{\partial w}}\right)^{2}=0,$

we may derive a second solution

$\upsilon =F\left({\frac {x}{r^{2}}},\ {\frac {y}{r^{2}}},\ {\frac {z}{r^{2}}},\ {\frac {w}{r^{2}}}\right),$

and from the solution U = f(x, y, z, w) of

${\frac {\partial ^{2}U}{\partial x^{2}}}+{\frac {\partial ^{2}U}{\partial y^{2}}}+{\frac {\partial ^{2}U}{\partial z^{2}}}+{\frac {\partial ^{2}U}{\partial w^{2}}}=0,$

we may derive another solution

$u={\frac {1}{r^{2}}}f\left({\frac {x}{r^{2}}},\ {\frac {y}{r^{2}}},\ {\frac {z}{r^{2}}},\ {\frac {w}{r^{2}}}\right).$

Putting w = ict, where c is the velocity of light and t is the time, the equations take the well known form^[1]

$\left({\frac {\partial V}{\partial x}}\right)^{2}+\left({\frac {\partial V}{\partial y}}\right)^{2}+\left({\frac {\partial V}{\partial z}}\right)^{2}={\frac {1}{c^{2}}}\left({\frac {\partial V}{\partial t}}\right)^{2},$ ${\frac {\partial ^{2}V}{\partial x^{2}}}+{\frac {\partial ^{2}V}{\partial y^{2}}}+{\frac {\partial ^{2}V}{\partial z^{2}}}={\frac {1}{c^{2}}}{\frac {\partial ^{2}V}{\partial t^{2}}},$

and the transformation may be written

$X={\frac {x}{r^{2}-c^{2}t^{2}}},\ Y={\frac {y}{r^{2}-c^{2}t^{2}}},\ Z={\frac {z}{r^{2}-c^{2}t^{2}}},\ T={\frac {t}{r^{2}-c^{2}t^{2}}},$

where now

$r^{2}=x^{2}+y^{2}+z^{2}$

The study of this transformation will be taken up later.

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1. It may be mentioned here that Lorentz's fundamental equations of the electron theory, viz.,

   ${\frac {\partial \gamma }{\partial y}}-{\frac {\partial \beta }{\partial z}}=4\pi u,\ {\frac {\partial Q}{\partial z}}-{\frac {\partial R}{\partial y}}={\frac {\partial \alpha }{\partial t}},$ $u={\frac {\partial f}{\partial t}}+\rho \xi ,\ P=4\pi c^{2}f,$ ${\frac {\partial \rho }{\partial t}}+{\frac {\partial }{\partial x}}(\rho \xi )+{\frac {\partial }{\partial y}}(\rho \eta )+{\frac {\partial }{\partial z}}(\rho \zeta )=0$ ${\frac {\partial f}{\partial x}}+{\frac {\partial g}{\partial y}}+{\frac {\partial h}{\partial z}}=\rho ,\ {\frac {\partial \alpha }{\partial x}}+{\frac {\partial \beta }{\partial y}}+{\frac {\partial \gamma }{\partial z}}=0,$

   may be reduced to a symmetrical form by writing s = ict and putting

   $\alpha +i{\frac {P}{c}}=p,\ \beta +i{\frac {Q}{c}}=q,\ \gamma +i{\frac {R}{c}}=r.$

   The four mutually orthogonal vectors (A, B, C, D) whose components are respectively

   (0, r, -q, -p), (-r, 0, p, -q), (q, -p, 0, -r), (p, q, r, 0)

   satisfy the equations

   $div\ A=4\pi \rho \xi ,\ div\ B=4\pi \rho \eta ,\ div\ C=4\pi \rho \zeta ,\ div\ D=4\pi \rho ic,$

   where

   $div\ M\equiv {\frac {\partial M_{1}}{\partial x}}+{\frac {\partial M_{2}}{\partial y}}+{\frac {\partial M_{3}}{\partial z}}+{\frac {\partial M_{4}}{\partial s}}$ if $M\equiv \left(M_{1},\ M_{2},\ M_{3},\ M_{4}\right)$

   Again, if we put

   $4\pi \rho \xi =X,\ 4\pi \rho \eta =Y,\ 4\pi \rho \eta =Z,\ 4\pi \rho ic=S,$

   and introduce four new vectors $A_{1},\ B_{1},\ C_{1},\ D_{1}$ whose components are respectively

   (S, -Z, Y, -X) (Z, S, -X, -Y) (-Y, X, S, -Z) (X, Y, Z, S),

   we find

   $div\ A_{1}=\nabla ^{2}p,\ div\ B_{1}=\nabla ^{2}q,\ div\ C_{1}=\nabla ^{2}r,\ div\ D_{1}=0,$

   where

   $\nabla ^{2}\phi \equiv {\frac {\partial ^{2}\phi }{\partial x^{2}}}+{\frac {\partial ^{2}\phi }{\partial y^{2}}}+{\frac {\partial ^{2}\phi }{\partial z^{2}}}+{\frac {\partial ^{2}\phi }{\partial s^{2}}}.$

   Finally, if X, Y, Z, S can be derived from a potential function n so that

   $X=-{\frac {\partial n}{\partial x}},\ Y=-{\frac {\partial n}{\partial y}},\ Z=-{\frac {\partial n}{\partial z}},\ S=-{\frac {\partial n}{\partial s}},$

   we can form four mutually orthogonal vectors θ, Φ, ψ, χ whose components are respectively

   (n, r, -q, -p). (-r, n, p, -q), (q, -p, n, -r), (p, q, r, n),

   and the equations then take the simple form

   ${\begin{array}{ccccccccc}div\ \theta &=&div\ \phi &=&div\ \psi &=&div\ \chi &=&0,\\\\\nabla ^{2}p&=&\nabla ^{2}q&=&\nabla ^{2}r&=&\nabla ^{2}n&=&0,\\\\div\ A_{1}&=&div\ B_{1}&=&div\ C_{1}&=&div\ D_{1}&=&0.\end{array}}$