Differential operator has changed into another one of laplacian operator. How this change occured?

48 Views Asked by At

$$ \begin{align} \text{Given equation}~:\nabla^2\mathbf E&={\partial\mathbf E\over\partial\mathrm{t}}+{\partial^2\mathbf E\over\partial\mathrm{t^2}}~~\text{where}~~E_z=0\\ \text{Equation which I can't understand now}~:{\partial^2 E_x\over\partial\mathrm{z^2}}&={\partial E_x\over\partial\mathrm{t}}+{\partial^2 E_x\over\partial\mathrm{t^2}} \end{align} $$

The book says that second equation can be obtained taking$~x,y~$axes such that$~E_x,H_y~$only remain with direction of proceed is taken to$~z~$axis.

Currently I can't comprehend this statement.

So in the first place with my memory of def of laplacian operator, the eqn1 is completely same as following equation.

$$\left({\partial^2 E_x\over\partial\mathrm{x^2}},{\partial^2 E_y\over\partial\mathrm{y^2}},{\partial^2E_z\over\partial\mathrm{z^2}}\right)=\left({\partial E_x\over\partial\mathrm{t}},{\partial E_y\over\partial\mathrm{t}},{\partial E_z\over\partial\mathrm{t}}\right)+\left({\partial^2 E_x\over\partial\mathrm{t^2}},{\partial^2 E_y\over\partial\mathrm{t^2}},{\partial^2E_z\over\partial\mathrm{t^2}}\right)\tag{1}$$

So the eqn2 is relavent with the first element of the vector above, that is

$${\partial E_x\over\partial\mathrm{x^2}}={\partial E_x\over\partial\mathrm{t}}+{\partial^2 E_x\over\partial\mathrm{t^2}}\tag{2}$$

Why$~z~$can be replaced for$~x~$?

ADD

I thought that coefficients of original equation is needless here so I removed it above and I wrote down the true given equations in the book.

$$\nabla^2 \mathbf E=\epsilon\mu{\partial^2 \mathbf E\over\partial\mathrm{t^2}}+\sigma\mu{\partial \mathbf E\over\partial\mathrm{t}}$$

$$ {\partial^2 E_x\over\partial\mathrm{z^2}}=\sigma\mu{\partial E_x\over\partial\mathrm{t}}+\epsilon\mu{\partial^2 E_x\over\partial\mathrm{t^2}} $$

And the solution for the above equation is as follows.

$$ E_x=E_0 \exp\left(- \sqrt{\mu\sigma\omega/2}z \right) \cos\left(\omega t - \sqrt{\mu\sigma\omega/2} z \right) $$

1

There are 1 best solutions below

0
On

Too long for a comment.

As far as I can tell you are asking how the equations \begin{align} {\partial^2 E_x\over\partial\mathrm{z^2}}&={\partial E_x\over\partial\mathrm{t}}+{\partial^2 E_x\over\partial\mathrm{t^2}}\,,\\ {\partial^2 E_y\over\partial\mathrm{z^2}}&={\partial E_y\over\partial\mathrm{t}}+{\partial^2 E_y\over\partial\mathrm{t^2}}\,,\\ {\partial^2 E_z\over\partial\mathrm{z^2}}&={\partial E_z\over\partial\mathrm{t}}+{\partial^2 E_z\over\partial\mathrm{t^2}} \end{align} follow from \begin{align} \nabla^2\mathbf E&={\partial\mathbf E\over\partial\mathrm{t}}+{\partial^2\mathbf E\over\partial\mathrm{t^2}} \end{align} when --as your book seems to say-- $E_x,E_y,E_z$ only depend on $z$.

  • This is totally trivial as $\mathbf{E}$ is the vector $(E_x,E_y,E_z)$.

  • Unfortunately you have not answered my question what book that exactly is which is why I am now voting to close your question.