(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x -
vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in
a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2}
{\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric
field in the wave. Show that by using the Galilean transformation the wave
equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} )
{\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x',
t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t')
\over \partial{t'^2}} = 0$$ This has a different form than the wave equation
in \(S\). Hence the Galilean transformation \(violates\) the first relativity
postulate that all physical laws have the same form in all inertial reference
frames. (\(Hint\): Express the derivatives \(\partial/\partial{x}\) and
\(\partial/\partial{t}\) in terms of \(\partial/\partial{x'}\) and
\(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of
part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and
show that in frame \(S'\) the wave equation has the same form as in frame \(S\):
$${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x',
t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light
in vacuum is c in both frames \(S\) and \(S'\).