Solution by Separation of Variables

The differential equation (11) above is commonly solved by the technique of separation of variables. The basic idea behind this technique is to assume, initially, that the variable of interest can be separated into factors which depend separately on the coordinates. In this case, we assume that $B$ can be expressed as

$$B = R(r)Z(z)T(t)$$ (12)

(again recall that $\phi$ does not enter because of symmetry). I substitute this equation into the differential equation, and after dividing once by $B$ I arrive at

$${Z^{\prime\prime}\over Z} + \left({R^{\prime\prime}\over R}... ... - {1\over r^2}\right) = {1\over c^2}{T^{\prime\prime}\over T}$$ (13)

where the derivatives are with respect to their individual variables. Now the interesting thing about this technique is that each of the terms is a function of a different variable,

$$\left(z{\rm -only}\right) + \left(r{\rm -only}\right) = \left(t{\rm -only}\right).$$ (14)

Therefore, each of the terms must be constant. If one of the terms was not constant, then there would be no way that the other terms could compensate, since they depend on different variables. Thus we can rewrite the equation as

$$(-k^2) + (-\lambda^2) = (-\omega^2)/c^2$$ (15)

where $k$, $\lambda$ and $\omega$ are constants, and

$${Z^{\prime\prime}\over Z} = -k^2$$ (16)

$${R^{\prime\prime}\over R} + {R^{\prime}\over r R} - {1\over r^2} = -\lambda^2$$ (17)

$${T^{\prime\prime}\over T} = -\omega^2.$$ (18)

It is worth asking why I inserted negative values rather than positive values. The answer is that I could have, but these would lead to solutions which diverge at infinite radius or time, and thus cannot be physical.