Question on an example in Griffiths' book, Not a homework

[yungman]

The is example 10.1 in page 417. The example is the find the current density from given condition:

$$V=0,\;\;\hbox{ and } \;\;\vec A=\frac{\mu_0 k}{4c}(ct-|x|)^2 \hat z \;\hbox { for x = +ve and }\;\; \vec A=0 \;\;\hbox { for x = -ve.}$$

$$c=\frac 1 {\sqrt{\mu_0 \epsilon_0}}$$

From this, you get:

$$\vec E= -\frac {\partial \vec A}{\partial t} \;=\; -\frac {\mu_0 k}{2} (ct-|x|)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \nabla X \vec A = ^+_- \hat y \frac{\mu_0 k}{2c}(ct-|x|)$$

$$\nabla X \vec B = -\hat z \frac {\mu_0 k}{2c} = \mu_0 \vec J + \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t} = \mu_0 \vec J -\hat z \mu \epsilon_0 \frac{\partial^2 \vec A}{\partial t^2} = \mu_0 \vec J - \hat z \frac{\mu_0 k c}{2}$$

$$\Rightarrow \vec J = -\hat z \frac {\mu_0 k}{2c} + \hat z \frac{\mu_0 k c}{2}$$


The book claimed $\vec J = 0$ without going through the steps.

I cannot get that, please help.

Thanks

Alan

 

[dextercioby]

The $\vec{B}$ doesn't look correct. Can you check your differentiations ?

 

[yungman]

The $\vec{B}$ doesn't look correct. Can you check your differentiations ?

Thanks for the reply.

$$\vec A=\frac{\mu_0 k}{4c}(ct-|x|)^2 \hat z$$

$$\hbox{ If x = +ve }\; |x|=x \;\Rightarrow \;\frac { d |x|}{dx}=1.\;\;\;\hbox { If x = -ve }\; |x|=-x \;\Rightarrow \; \frac { d |x|}{dx}=-1$$

I don't know how to put it in determinant form use Latex so I just give the result:


$$\vec B = \nabla X \vec A = -\hat y \frac {\mu_0 k}{4c} [2(ct-|x|)] [-(^+_-1)]$$

$$\hbox { Where }\;[-(^+_-1)]\; \hbox{ stand for -1 if x is possitive and +1 for x is negative. }$$

$$\vec B = \nabla X \vec A = ^+_- \hat y \frac{\mu_0 k}{2c}(ct-|x|)$$

Where +ve where x is possitive and -ve when x is negative.

The book gave this for B also. I think this is correct. I skip typing in a lot of step to avoid boring people with the detail.


I just change the $\;c = \frac {1}{\sqrt { \mu \epsilon }} \; \hbox { to }\;\frac {1}{\sqrt { \mu_0 \epsilon_0 }}$. This tell you that the dielectric is vacuum or air and is lossless. $\vec J \;\hbox { is free current density }\;\Rightarrow\; \vec J = \sigma \vec E =0$ when $\sigma = 0$. Maxwell's equation cannot verify that!

 

[yungman]

Can anyone help, it really don't make sense!?

 

[qbert]

looks to me like you forgot the 1/c² factor of the dE/dt term
in the curl(B) equation.

 

[yungman]

looks to me like you forgot the 1/c² factor of the dE/dt term
in the curl(B) equation.

I cannot see anything wrong, here is a more detail derivation:

$$\vec E= -\frac {\partial \vec A}{\partial t} \;=\; -\frac {\mu_0 k}{2} (ct-|x|)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \nabla X \vec A = ^+_- \hat y \frac{\mu_0 k}{2c}(ct-|x|)$$

$$\frac{\partial \vec E}{\partial t} = -\hat z \frac{\partial^2 \vec A}{\partial t^2}= -\hat z \frac {\mu_0 k}{2} \frac{\partial (ct-|x|)}{\partial t} = -\hat z \frac{\mu_0 k c}{2}$$

$$\nabla X \vec B = -\hat z \frac {\mu_0 k}{2c} = \mu_0 \vec J + \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t} = \mu_0 \vec J -\hat z \mu_0 \epsilon_0 \frac{\partial^2 \vec A}{\partial t^2} = \mu_0 \vec J - \hat z \frac{\mu_0^2 \epsilon_0 k c}{2}$$

I did forget to put back $\mu_0 \epsilon_0$ in the original post but that still will not make $\mu_0 \vec J$ equal zero. Most of the equations are in the book also. My confident with these equation is very high. It is more about something I totally miss and is not here.

Given the medium is characterized by $\mu_0 \epsilon_0$, you cannot have current density anywhere except at the discontinue at x=0 where there is surface current density. The problem is Putting the equations into the Maxwell's equation does not yield J=0 in the Maxwell's curl equation as shown. The only thing that can be wrong is that I should not try to put this into the Maxwell's curl equation like this. And if so, I really want to hear the reason why.

Thanks

Alan

 

[dextercioby]

Come on, Alan. This is an extract of what you wrote:

$$-\hat z \frac {\mu_0 k}{2c} = [...] = \mu_0 \vec J - \hat z \frac{\mu_0^2 \epsilon_0 k c}{2}$$

So

$$\mu_0 \vec{J} = -\hat z \left(\frac{\mu_0 k}{2c} - \frac{\mu_0 \mu_0 \epsilon_0 k c}{2}\right)$$

But $\mu_0 \epsilon_0 c = \frac{1}{c}$

So, finally

$$\mu_0 \vec{J} = -\hat z \left(\frac{\mu_0 k}{2c} - \frac{\mu_0 k }{2c}\right) = 0$$

 

[yungman]

Come on, Alan. This is an extract of what you wrote:

$$-\hat z \frac {\mu_0 k}{2c} = [...] = \mu_0 \vec J - \hat z \frac{\mu_0^2 \epsilon_0 k c}{2}$$

So

$$\mu_0 \vec{J} = -\hat z \left(\frac{\mu_0 k}{2c} - \frac{\mu_0 \mu_0 \epsilon_0 k c}{2}\right)$$

But $\mu_0 \epsilon_0 c = \frac{1}{c}$

So, finally

$$\mu_0 \vec{J} = -\hat z \left(\frac{\mu_0 k}{2c} - \frac{\mu_0 k }{2c}\right) = 0$$

I need to find a hole and jump in. I am blind, I was stuck for over a day.

Thanks

Alan