0
$\begingroup$

I am studying Maxwell's equations and their use to derive a wave equation to derive the behaviour of electromagnetic waves in vacuum. In the case of plane waves, EM fields can be described by:

$\vec E_z = \vec E_0e^{i(kz - wt)}$

$\vec B_z = \vec B_0e^{i(kz - wt)}$

$ \nabla^2E = \mu_o\epsilon_0\frac{\partial^2 E}{\partial t^2} $

$ \nabla^2B = \mu_o\epsilon_0\frac{\partial^2 B}{\partial t^2} $

where $z$ is an arbitrary direction of propagation.

The book I am using as my guide is Introduction to Electrodynamics by Griffiths. In the book, the author writes the following: Relation between E and B

However, I do not understand what is the reasoning behind equation 9.45. How does the Maxwell equation involving the curl of the electric field generate that piece of work?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

You insert into the Maxwell equation $$\vec \nabla \times \vec E=-\frac {\partial \vec B}{\partial t} \tag 1$$ the wave solutions $$\vec E=\vec E_0 exp (i\vec k \vec r - i\omega t)$$ and $$\vec B=\vec B_0 exp (i\vec k \vec r - i\omega t)$$ In the cross product of equ. (1), you can move the exponential function to the nabla. Thus you have a gradient of the wave exponential giving $$\vec \nabla exp(i\vec k \vec r)=i\vec k expi(\vec k \vec r)$$ and you obtain $$\vec k \times \vec E=\omega\vec B$$ This should answer your question.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks for the answer. However, I do not understand what you mean by "move the exponential function to the $\nabla$". Could you please clarify your explanation a bit more or point me towards a reference with a detailed account of your reasoning? $\endgroup$
    – user63248
    Commented Mar 5, 2018 at 0:00
  • $\begingroup$ @daljit97 - Simply put, when you have a cross product of two vectors (here $\vec \nabla \times \vec E_0 exp (i\vec k \vec r)$ ) and one vector has a numerical factor (here $ exp (i\vec k \vec r)$), you can move this factor to the other vector (here in front of the $\vec \nabla$) without change of the cross product. Thus you can do this, take the gradient of the exponential and then move the remaining exponential back. $\endgroup$
    – freecharly
    Commented Mar 5, 2018 at 0:12
  • 1
    $\begingroup$ @daljit97 $\vec a \times f \vec b=\vec a f\times \vec b$ where f is a number. $\endgroup$
    – freecharly
    Commented Mar 5, 2018 at 0:18
  • $\begingroup$ @daljit97 You can find the cross product for the nabla operator here: en.wikipedia.org/wiki/Del#Product_rules $\endgroup$
    – freecharly
    Commented Mar 5, 2018 at 0:31
  • 1
    $\begingroup$ @daljit97 You move in the cross prodcut the $exp(\vec k \vec r -\omega t)$ back to the $\vec E_0$ giving $\vec E_0exp(\vec k \vec r -\omega t) $ again. Then you have left $$i \vec k \times \vec E_0exp(\vec k \vec r -\omega t)=i\omega \vec B_0 exp(\vec k \vec r -\omega t)$$ and you can divide this by $i$. In the end you can also divide by the expontial wave factor $exp(\vec k \vec r -\omega t)$. $\endgroup$
    – freecharly
    Commented Mar 5, 2018 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.