3
$\begingroup$

Consider the wave equation for linearly $x$ polarized waves travelling in the $\pm z$ directions:

$$\frac{\partial^2\vec E_x}{\partial t^2}=c^2\frac{\partial^2\vec E_x}{\partial z^2}\tag{1}$$ The general solution to equation $(1)$ is $$\vec E_x=\vec E_{+}(q)+\vec E_{-}(s)$$ where $\vec E_{+}$ and $\vec E_{-}$ are arbitrary functions. $$q=z-ct$$ & $$s=z+ct$$

Calculate the general form of the magnetic field in terms of $\vec E_{+}$ and $\vec E_{-}$


I am stuck at the very beginning.

I have the solution to this question, but the problem is I cannot understand the author's solution.

So instead I will ask questions about the author's solution which is as follows:

We can obviously write $$\vec B_y=\vec B_{+}(q)+\vec B_{-}(s)\tag{a}$$ and $$\frac{\partial \vec B_y}{\partial t}=-\frac{ \partial \vec E_x}{\partial z}\tag{b}$$ then $$\frac{\partial q}{\partial t}\Bigg |_z\cdot\frac{d\vec B_{+}}{dq}+\frac{\partial s}{\partial t}\Bigg |_z\cdot\frac{d\vec B_{-}}{ds}=-\left(\frac{\partial q}{\partial z}\Bigg |_t\cdot\frac{d\vec E_{+}}{dq}+\frac{\partial s}{\partial z}\Bigg |_t\cdot\frac{d \vec E_{-}}{ds}\right)\tag{c}$$ $$\implies -c\frac{d\vec B_{+}}{dq}+c\frac{d\vec B_{-}}{ds}=-\frac{d\vec E_{+}}{dq}-\frac{d\vec E_{-}}{ds}\tag{d}$$ therefore $$\vec B_y=\frac{1}{c}\left[\vec E_{+}(q)-\vec E_{-}(s)\right]\tag{e}$$


I understand completely how $(\mathrm{d})$ follows from $(\mathrm{c})$.

I don't understand why you "can obviously write $\vec B_y=\vec B_{+}(q)+\vec B_{-}(s)$"; What is the origin of this equation: $(\mathrm{a})$? It is far from obvious to me that you can write $\vec B_y=\vec B_{+}(q)+\vec B_{-}(s)$.

Also, what is the origin of equation $(\mathrm{b})$? What does it mean? Is it a reformulation of one of Maxwell's equations?

Lastly, how does $(\mathrm{c})$ follow from $(\mathrm{b})$? I note that the author is using the chain rule here, but I'm not sure about the logic.


If anyone could help me by giving hints or explanations to any of the questions I have raised then I would be most grateful.


EDIT:

Thanks to @Farcher I now understand part $(\mathrm{a})$ and was able to write an answer of my own elaborating on parts $(\mathrm{b})$ and $(\mathrm{c})$.


EDIT 2:

The only part of the authors solution which I still don't understand is how $(\mathrm{e})$ follows from $(\mathrm{d})$.

Rearranging $(\mathrm{d})$ we have that $$c\left(\frac{d\vec B_{-}}{ds}-\frac{d\vec B_{+}}{dq}\right)=-\left(\frac{d\vec E_{-}}{ds}+\frac{d\vec E_{+}}{dq}\right)\tag{d}$$ we know that $$\vec E_x=\vec E_{+}(q)+\vec E_{-}(s)$$ and $$\vec B_y=\vec B_{+}(q)+\vec B_{-}(s)$$

Rearranging $(\mathrm{d})$ further I find that

$$c\frac{d\vec B_{-}}{ds}+\frac{d\vec E_{-}}{ds}=c\frac{d\vec B_{+}}{dq}-\frac{d\vec E_{+}}{dq}\tag{d}$$

My first thought was to integrate both sides but since the LHS depends on $s$ and the RHS on $q$ it cannot be correct to write

$$\int \left(c\frac{d\vec B_{-}}{ds}+\frac{d\vec E_{-}}{ds}\right)ds=\int \left(c\frac{d\vec B_{+}}{dq}-\frac{d\vec E_{+}}{dq}\right)dq\tag{?}$$

So I am stuck at this point.

Could someone please explain to me how I can obtain the result $$\bbox[5px,border:2px solid red]{\vec B_y=\frac{1}{c}\left[\vec E_{+}(q)-\vec E_{-}(s)\right]}\tag{e}$$ from $$\bbox[yellow]{-c\frac{d\vec B_{+}}{dq}+c\frac{d\vec B_{-}}{ds}=-\frac{d\vec E_{+}}{dq}-\frac{d\vec E_{-}}{ds}}\tag{d}\,\,?$$

$\endgroup$
  • $\begingroup$ Let's not have posts look like revision histories. $\endgroup$ – AccidentalFourierTransform Oct 29 '18 at 15:54
  • $\begingroup$ @AccidentalFourierTransform I was always told not to edit a question in such a way that the existing answers are rendered obsolete, henceforth, this is the purpose of updating the question with an EDIT so that the given answers are still in context. Are you seriously telling me that all these years I have been doing the wrong thing and should erase all edits? If that is so, then I will comply. I don't really care whether the edits are there or not I just want this question answered. $\endgroup$ – BLAZE Oct 29 '18 at 16:11
  • $\begingroup$ If you feel like your "EDIT" is drastic enough to change the question, it should not be an edit, it should be a new question. If the "EDIT" is not too drastic, it should be incorporated smoothly into the post. Either way, updating the question with "EDIT" at the end is very much frowned upon, especially when the update changes the focus of the question and comprises more than 50% of its length/content. $\endgroup$ – AccidentalFourierTransform Oct 29 '18 at 16:14
  • $\begingroup$ @AccidentalFourierTransform "it should be a new question" - but all the background and context is already in this question. Making a new question would mean that I would have to copy all of the question again and then just change the part at the end. Is this really the way to handle these situations? $\endgroup$ – BLAZE Oct 29 '18 at 16:18
  • 1
    $\begingroup$ Yes, it is perfectly OK to have multiple posts with the same background and context, provided they ask different questions. So let me reiterate, to make sure we are on the same page: if the update changes the question, it is OK and encouraged to create a new thread. If the update does not change the question, you should not relegate it to the end of the post, but incorporate it in the post directly. $\endgroup$ – AccidentalFourierTransform Oct 29 '18 at 16:21
2
$\begingroup$

(a) An electromagnetic wave has a magnetic field $B_+/B_-\,\hat y$ oscillating exactly in phase with and at right angles to an electric field $E_+/E_- \, \hat x$ both of which are at right angles to the direction of propagation $\hat z$.

(b) is Faraday's law in differential form $\dfrac{\partial \vec B}{\partial t}= -\vec \nabla \times \vec E$

(c) is the application of the chain rule.

$\endgroup$
5
$\begingroup$

For my own reference (and others if they're interested) I'm going to expand on what @Farcher wrote in his answer for parts $(\mathrm{b})$ and $(\mathrm{c})$:

To show part $(\mathrm{b})$ from Faraday's Law: $$\begin{align}\frac{\partial \vec B_y}{\partial t}&= -\vec \nabla \times \vec E_x\\&=- \begin{vmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ E_x & 0 & 0 \\ \end{vmatrix}\\&=-\left(\frac{\partial E_x}{\partial z }\hat j-\frac{\partial E_x}{\partial y}\hat k\right)\\&=-\frac{\partial \vec E_x}{\partial z}\end{align}$$

as the $\hat k$ component disappears since $E_x$ does not depend on $y$ so it's derivative is zero.

The $\vec B_y$ only has a $\hat j$ component as it is oscillating in the $y$ direction. After taking the curl of the electric field only the $\hat j$ component survived; so this makes perfect sense as the vector components must match for equality to hold.

Hence, we conclude that $$\frac{\partial \vec B_y}{\partial t}=-\frac{\partial \vec E_x}{\partial z}$$ which is indeed equation $(\mathrm{b})$


For part $(\mathrm{c})$ we have $$\vec B_{y}=\vec B_{y}(\vec B_{+},\vec B_{-})$$ $$\vec B_{+}=\vec B_{+}(q) \qquad\text{and}\qquad \vec B_{-}=\vec B_{-}(s)$$ $$q=q(z,t) \qquad\text{and}\qquad s=s(z,t)$$

So the appropriate tree diagram which connects the dependent variables at the top to the independent variables at the bottom is:

Tree diagram

From the tree diagram we see that

$$\frac{\partial\vec B_y}{\partial t}=\frac{\partial\vec B_y}{\partial\vec B_{+}}\cdot\frac{d\vec B_{+}}{d q}\cdot \frac{\partial q}{\partial t}+\frac{\partial\vec B_y}{\partial\vec B_{-}}\cdot\frac{d\vec B_{-}}{d s}\cdot \frac{\partial s}{\partial t}$$

Now since $$\frac{\partial\vec B_y}{\partial\vec B_{+}}=\frac{\partial\vec B_y}{\partial\vec B_{-}}=1$$ and $$\frac{\partial q}{\partial t}=-c\,,\qquad \frac{\partial s}{\partial t}=c$$ therefore I can write $$\fbox{$\color{blue}{\frac{\partial\vec B_y}{\partial t}=-c\frac{d\vec B_{+}}{d q}+c\frac{d\vec B_{-}}{d s}}$}$$

which is the LHS of $(\mathrm{c})$


The RHS of $(\mathrm{c})$ is completely analogous to the method used to get the LHS. But for reference I'm going to write out the steps explicitly.

Similar to before we have $$\vec E_{x}=\vec E_{x}(\vec E_{+},\vec E_{-})$$ $$\vec E_{+}=\vec E_{+}(q) \qquad\text{and}\qquad \vec E_{-}=\vec E_{-}(s)$$ $$q=q(z,t) \qquad\text{and}\qquad s=s(z,t)$$

So the appropriate tree diagram for this case is:

2nd Tree diagram

and from it we see that

$$-\frac{\partial\vec E_x}{\partial z}=-\left(\frac{\partial\vec E_x}{\partial\vec E_{+}}\cdot\frac{d\vec E_{+}}{d q}\cdot \frac{\partial q}{\partial z}+\frac{\partial\vec E_x}{\partial\vec E_{-}}\cdot\frac{d\vec E_{-}}{d s}\cdot \frac{\partial s}{\partial z}\right)$$

Now since $$\frac{\partial\vec E_x}{\partial\vec E_{+}}=\frac{\partial\vec E_x}{\partial\vec E_{-}}=1$$ and $$\frac{\partial q}{\partial z}=\frac{\partial s}{\partial z}=1$$ therefore I can write $$\fbox{$\color{red}{-\frac{\partial\vec E_x}{\partial z}=-\frac{d\vec E_{+}}{d q}-\frac{d\vec E_{-}}{d s}}$}$$

which is the RHS of $(\mathrm{c})$.

$\endgroup$
3
+100
$\begingroup$

Let $c=1$ to simplify the notation.

We know that $$ \boldsymbol B'_-(s)+\boldsymbol E'_-(s)=\boldsymbol B'_+(q)-\boldsymbol E'_+(q) $$ where the prime denotes differentiation.

Note that $s$ and $q$ are independent variables. The l.h.s. depends on $s$ only, and the r.h.s. on $q$ only. Thus, both sides must in fact be constants: $$ \begin{aligned} \boldsymbol B'_-(s)+\boldsymbol E'_-(s)&=\boldsymbol\alpha\\ \boldsymbol B'_+(q)-\boldsymbol E'_+(q) & =\boldsymbol \alpha \end{aligned} $$ for some constant vector $\boldsymbol\alpha$. These equations are trivial to integrate: $$ \begin{aligned} \boldsymbol B_-(s)+\boldsymbol E_-(s)&=\boldsymbol \alpha s+\boldsymbol\beta\\ \boldsymbol B_+(q)-\boldsymbol E_+(q) & =\boldsymbol \alpha q+\boldsymbol\gamma \end{aligned} $$ for some integration constants $\boldsymbol\beta,\boldsymbol\gamma$.

From this, we obtain $$ \boldsymbol B_y=(\boldsymbol E_+(q)-\boldsymbol E_-(s))+z\boldsymbol \alpha'+\boldsymbol\beta' $$ with $\boldsymbol\alpha'=2\boldsymbol\alpha$ and $\boldsymbol\beta'=\boldsymbol\beta+\boldsymbol\gamma$.

This is the most general solution consistent with your equations. The boundary conditions, which you didn't specify, presumably set $\boldsymbol\alpha'=\boldsymbol\beta'=\boldsymbol 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.