Explain the mathtematical proof of transverse nature of EM waves.

Considered a plane electromagnetic wave travelling in the X-direction and an elementary rectangular parallelopiped OABCDEFG placed with its edge parallel to the three axes, as in figure.The wave front of the electromagnetic wave is in the YZ plane and ABCD is a portion of it at time t. The values of electric field and magnetic fields will be zero to the right of the face ABCD , but to the left of this face the value of the two fields depends on x and t but not on y and z ,since we are considering a plane wave.

To establish transverse nature of electromagnetic waves ,we will establish that $\overrightarrow{E }and \overrightarrow{B}$ are perpendicular to the direction of the wave propagation.
  If the rectangular parallelopiped does not enclose any charge ,the total electric flux across it must be zero,according to gauss's law i.e.
  $$\begin{gathered} \oint \overrightarrow{E}.d\overrightarrow{s}=0 \\ or \left({\int }_{ABCD}\overrightarrow{E}.d\overrightarrow{s}+{\int }_{OGFE}\overrightarrow{E}.d\overrightarrow{s}\right)+ \left({\int }_{BGFC}\overrightarrow{E}.d\overrightarrow{s}+{\int }_{OEDA}\overrightarrow{E}.d\overrightarrow{s}\right)+ \left({\int }_{DCFE}\overrightarrow{E}.d\overrightarrow{s}+{\int }_{OABG}\overrightarrow{E}.d\overrightarrow{s}\right)=0 \\ Since \overrightarrow{E} does not depend on y and z,therefore ,the contribution to the total flux coming from the faces \\ normal to y and z axes will cancel out in pairs i.e. \\ \left({\int }_{BGFC}\overrightarrow{E}.d\overrightarrow{s}+{\int }_{OEDA}\overrightarrow{E}.d\overrightarrow{s}\right)=0 \\ and \left({\int }_{DCFE}\overrightarrow{E}.d\overrightarrow{s}+{\int }_{OABG}\overrightarrow{E}.d\overrightarrow{s}\right)=0 \\ From the above equations, we have \\ \left({\int }_{ABCD}\overrightarrow{E}.d\overrightarrow{s}+{\int }_{OGFE}\overrightarrow{E}.d\overrightarrow{s}\right)=0 \\ If E_x and E_x\text{'} are the values of x-components of electric field on face ABCD and OGFE respectively. \\ S be the area of each of the faces ,then \\ {\int }_{ABCD}\overrightarrow{E}.d\overrightarrow{s}=E_{x}S \\ and {\int }_{OGFE}\overrightarrow{E}.d\overrightarrow{s}=-E_x\text{'}S \\ Negative sign arises due to opposite direction \\ {E}_{x}S-E_x\text{'}S=0 or \left(E_x-E_x\text{'}\right)S=0 \\ {E}_{x}S=E_x\text{'}S \end{gathered}$$
Since S is not equal to zero E_x and E_x ' predicts that the field is static .But a static field cannot propagate a wave of finite wavelength,therefore,E_x- E_x ' =0 i.e. no component of the electric field is parallel to the direction of the wave propagation . It means, the electric field is perpendicular to the direction of propagation of electromagnetic wave.
This shows transverse nature of EM wave.