# Page:BatemanConformal.djvu/6

1908.]
75
The conformal transformations of a space of four dimensions.

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

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

we may derive a second solution

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

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

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

we may derive another solution

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

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

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

and the transformation may be written

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

where now

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

The study of this transformation will be taken up later.

1. It may be mentioned here that Lorentz's fundamental equations of the electron theory, viz.,
 ${\displaystyle {\frac {\partial \gamma }{\partial y}}-{\frac {\partial \beta }{\partial z}}=4\pi u,\ {\frac {\partial Q}{\partial z}}-{\frac {\partial R}{\partial y}}={\frac {\partial \alpha }{\partial t}},}$ ${\displaystyle u={\frac {\partial f}{\partial t}}+\rho \xi ,\ P=4\pi c^{2}f,}$ ${\displaystyle {\frac {\partial \rho }{\partial t}}+{\frac {\partial }{\partial x}}(\rho \xi )+{\frac {\partial }{\partial y}}(\rho \eta )+{\frac {\partial }{\partial z}}(\rho \zeta )=0}$ ${\displaystyle {\frac {\partial f}{\partial x}}+{\frac {\partial g}{\partial y}}+{\frac {\partial h}{\partial z}}=\rho ,\ {\frac {\partial \alpha }{\partial x}}+{\frac {\partial \beta }{\partial y}}+{\frac {\partial \gamma }{\partial z}}=0,}$

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

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

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

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

where

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

Again, if we put

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

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

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

we find

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

where

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

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

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

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

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