在前面我们得到x方向的坐标变换：
\( x=x’+vt’\)
\(t=t’+x’v/c^2\)
根据上面两个式子，可以得到反变换：
\(x’=\gamma^2 (x-vt)\)
\(t’ =\gamma^2 (t-xv/c^2)\)
假设函数f(x,t)符合上述变换，有：
\(\frac{\partial}{\partial t} = \frac{\partial x’}{\partial t} \frac{\partial}{\partial x’} + \frac{\partial t’}{\partial t} \frac{\partial}{\partial t’}\)
\( = -\gamma^2 v \frac{\partial}{\partial x’} + \gamma^2 \frac{\partial}{\partial t’} = \gamma^2 \left( -v \frac{\partial}{\partial x’} + \frac{\partial}{\partial t’} \right) \)
\(\frac{\partial}{\partial x} = \frac{\partial x’}{\partial x} \frac{\partial}{\partial x’} + \frac{\partial t’}{\partial x} \frac{\partial}{\partial t’}\)
\( = \gamma^2 \frac{\partial}{\partial x’} - \gamma^2 \frac{v}{c^2} \frac{\partial}{\partial t’} = \gamma^2 \left( \frac{\partial}{\partial x’} - \frac{v}{c^2} \frac{\partial}{\partial t’} \right) \)
有：
\(\frac{\partial^2 f}{\partial t^2} = \left( \gamma^2 \right)^2 \left( -v \frac{\partial}{\partial x’} + \frac{\partial}{\partial t’} \right)^2 f = \gamma^4 \left( v^2 \frac{\partial^2}{\partial x’^2} - 2v \frac{\partial^2}{\partial x’ \partial t’} + \frac{\partial^2}{\partial t’^2} \right) f\)
\(\frac{\partial^2 f}{\partial x^2} = \left[ \gamma^2 \left( \frac{\partial}{\partial x’} - \frac{v}{c^2} \frac{\partial}{\partial t’} \right) \right]^2 f = \gamma^4 \left( \frac{\partial}{\partial x’} - \frac{v}{c^2} \frac{\partial}{\partial t’} \right)^2 f\)
于是可得：
\(\frac{\partial^2 f}{\partial t^2} - c^2 \frac{\partial^2 f}{\partial x^2} = \gamma^2 \left( \frac{\partial^2 f}{\partial t’^2} - c^2 \frac{\partial^2 f}{\partial x’^2} \right)\)
如果\(\frac{\partial^2 f}{\partial t^2} - c^2 \frac{\partial^2 f}{\partial x^2} =0 \)，可得：
\(\left( \frac{\partial^2 f}{\partial t’^2} - c^2 \frac{\partial^2 f}{\partial x’^2} \right) =0\)
所以对无源线性波动方程，没有\(\gamma\)也能直接变换
如果是洛伦兹变换，则可以得到：
\(\frac{\partial^2 f}{\partial t^2} - c^2 \frac{\partial^2 f}{\partial x^2} = \left( \frac{\partial^2 f}{\partial t’^2} - c^2 \frac{\partial^2 f}{\partial x’^2} \right)\)
所以洛伦兹变换是上面变换的变种，使得变换变得更容易计算
上面是匀速的情况。
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考虑弦的振动导致的波动，也是类似延迟累积导致波动：