# Thread: mathematical question from physics

1. ## mathematical question from physics

$\vec{E}=f_1(z-ct)\hat{x}+f_2(z-ct)\hat{y}+0\hat{z}$
$f_1(u)$ and $f_2(u)$ are functions of "u"
u=z-ct
i have the formula
$\nabla \times \vec{E}=-\frac{db}{dt }$
$\nabla\times\vec{E}=|\begin{array}{ccc}\hat{x} & \hat{y} & \hat{z}\\\frac{{d}}{dx} & {\frac{{d}}{dy}} & \frac{{d}}{dz}|=\\f_{1}(z-ct) & f_{2}(z-ct) & 0\end{array}\hat{-x}\frac{{df_{2}}}{dz}-\hat{y}\frac{{df_{1}}}{dz}+\hat{z0}=-\frac{{d\overrightarrow{B}}}{dt}$

i want to find B
so i need to integrate the left side by t
in order to get B
but f_1 f_2 are function of u
how to make a result
?

and then i need to show that $B\bullet E=0$
i cant get 0

2. Hmm. I get a slightly different cross product from you:

$\nabla\times\mathbf{E}=\left|\begin{matrix}\hat{x} &\hat{y} &\hat{z}\\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z}\\ f_{1}(z-ct) &f_{2}(z-ct) &0\end{matrix}\right|$

$=\hat{x}\left(-\frac{\partial}{\partial z}\,f_{2}(z-ct)\right)-\hat{y}\left(-\frac{\partial}{\partial z}\,f_{1}(z-ct)\right)+\hat{z}(0)$

$=-\hat{x}\,\frac{\partial}{\partial z}\,f_{2}(z-ct)+\hat{y}\,\frac{\partial}{\partial z}\,f_{1}(z-ct)=-\hat{x}f_{2}'(z-ct)+\hat{y}f_{1}'(z-ct).$

Now then, Faraday's Law gives us that

$-f_{2}'(z-ct)=-\frac{\partial B_{x}}{\partial t},$

$f_{1}'(z-ct)=-\frac{\partial B_{y}}{\partial t},$ and

$0=-\frac{\partial B_{z}}{\partial t}.$

Just integrate all three of these equations w.r.t. time, and you should be good to go. What do you get?