91Ó°ÊÓ

The Galilean coordinate transformations along the x-direction are$x\text{'}=x-vt$ and $t\text{'}=t$. Using the chain rule, we obtain

$$\begin{gathered} \frac{\mathrm{∂}}{\mathrm{∂}x}=\frac{\mathrm{∂}{x}^{\mathrm{′}}}{\mathrm{∂}x}\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}+\frac{\mathrm{∂}{t}^{\mathrm{′}}}{\mathrm{∂}x}\frac{\mathrm{∂}}{\mathrm{∂}{t}^{\mathrm{′}}}=\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}, \\ \frac{\mathrm{∂}}{\mathrm{∂}t}=\frac{\mathrm{∂}{x}^{\mathrm{′}}}{\mathrm{∂}t}\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}+\frac{\mathrm{∂}{t}^{\mathrm{′}}}{\mathrm{∂}t}\frac{\mathrm{∂}}{\mathrm{∂}{t}^{\mathrm{′}}}=−v\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}+\frac{\mathrm{∂}}{\mathrm{∂}{t}^{\mathrm{′}}} \end{gathered}$$

Applying the chain rule again gives the second derivatives,

$$\begin{gathered} \frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^2}=\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}2}}, \\ \frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^2}=v^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}2}}+\frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^{\mathrm{′}2}}−2v\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}}\mathrm{∂}{t}^{\mathrm{′}}} \end{gathered}$$

Substituting the transformed derivatives into the wave equation in S, we obtain the new form of the wave equation in S,

$\begin{array}{r}\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{x}^{\mathrm{′}2}}−\frac{1}{c^2}\left[v^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}2}}+\frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^{\mathrm{′}2}}−2v\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}}\mathrm{∂}{t}^{\mathrm{′}}}\right]E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)=0\\ \left(1−\frac{v^2}{c^2}\right)\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{x}^{\mathrm{′}2}}+\frac{2v}{c^2}\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{x}^{\mathrm{′}}\mathrm{∂}{t}^{\mathrm{′}}}−\frac{1}{c^2}\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{t}^{\mathrm{′}2}}=0\end{array}$

where we expressed the electric field in the new coordinates $(x\text{'},t\text{'})$, since the old coordinates$(x\text{'},t\text{'})$ are functions of these coordinates.

We will follow the same procedure, but we use the Lorentz coordinate transformations instead. By use of the chain rule, we obtain

$\begin{array}{r}\frac{\mathrm{∂}}{\mathrm{∂}x}=\frac{\mathrm{∂}{x}^{\mathrm{′}}}{\mathrm{∂}x}\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}+\frac{\mathrm{∂}{t}^{\mathrm{′}}}{\mathrm{∂}x}\frac{\mathrm{∂}}{\mathrm{∂}{t}^{\mathrm{′}}}=\mathrm{γ}\left(\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}−\frac{v}{c^2}\frac{\mathrm{∂}}{\mathrm{∂}{t}^{\mathrm{′}}}\right)\\ \frac{\mathrm{∂}}{\mathrm{∂}t}=\frac{\mathrm{∂}{x}^{\mathrm{′}}}{\mathrm{∂}t}\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}+\frac{\mathrm{∂}{t}^{\mathrm{′}}}{\mathrm{∂}t}\frac{\mathrm{∂}}{\mathrm{∂}{t}^{\mathrm{′}}}=\mathrm{γ}\left(−v\frac{\mathrm{∂}}{\mathrm{∂}{x}^{\mathrm{′}}}+\frac{\mathrm{∂}}{\mathrm{∂}{t}^{\mathrm{′}}}\right)\end{array}$

where $\mathrm{γ}=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$. Applying the chain rule again gives the second derivatives,

$\begin{array}{r}\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^2}={\mathrm{γ}}^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}2}}−\frac{2{\mathrm{γ}}^{2}v}{c^2}\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}}\mathrm{∂}{t}^{\mathrm{′}}}+\frac{{\mathrm{γ}}^2{v}^2}{c^4}\frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^{\mathrm{′}2}},\\ \frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^2}={\mathrm{γ}}^2{v}^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}2}}−2{\mathrm{γ}}^{2}v\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}}\mathrm{∂}{t}^{\mathrm{′}}}+{\mathrm{γ}}^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^{\mathrm{′}2}}\end{array}$

Substituting the transformed derivatives into the wave equation in S, we obtain the new form of the wave equation in S’,

$\left[{\mathrm{γ}}^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}2}}−\frac{2{\mathrm{γ}}^{2}v}{c^2}\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}}\mathrm{∂}{t}^{\mathrm{′}}}+\frac{{\mathrm{γ}}^2{v}^2}{c^4}\frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^{\mathrm{′}2}}\right]E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)−\frac{1}{c^2}\left[{\mathrm{γ}}^2{v}^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}2}}−2{\mathrm{γ}}^{2}v\frac{{\mathrm{∂}}^2}{\mathrm{∂}{x}^{\mathrm{′}}\mathrm{∂}{t}^{\mathrm{′}}}+{\mathrm{γ}}^2\frac{{\mathrm{∂}}^2}{\mathrm{∂}{t}^{\mathrm{′}2}}\right]E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)=0$

Cancelling the cross-derivative terms and grouping the other terms, we obtain

${\mathrm{γ}}^2\left(1−\frac{v^2}{c^2}\right)\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{x}^{\mathrm{′}2}}−\frac{{\mathrm{γ}}^2}{c^2}\left(1−\frac{v^2}{c^2}\right)\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{t}^{\mathrm{′}2}}=0$

Using the fact that $\mathrm{γ}=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$, or ${\mathrm{γ}}^2=\left(1-\frac{v^2}{c^2}\right)=1$, the equation reduces to

$\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{x}^{\mathrm{′}2}}−\frac{1}{c^2}\frac{{\mathrm{∂}}^{2}E\left(x^{\mathrm{′}},t^{\mathrm{′}}\right)}{\mathrm{∂}{t}^{\mathrm{′}2}}=0$

which has exactly the same form as that in S, but expressed in the new transformed coordinates$(x\text{'},t\text{'})$ of S’. Comparing the two forms of the wave equation for electromagnetic waves in a vacuum in S and S’ with the general wave equation, we see they describe electric-field disturbance (electromagnetic waves) which propagates along the x-axis (or the x’-axis) with the wave speedc in both frames. In other words, the light (which is an electromagnetic phenomenon) propagates through vacuum with the speedc in both frames.