# Maxwell's equations

1. Apr 9, 2005

### robert25pl

I want to make sure that I understand this good.
Given E and B are possible in a region of free space (J=0) only if $$\triangledown \times E=0$$ and $$\triangledown \cdot B = 0$$

2. Apr 9, 2005

### dextercioby

That $\vec{B}$ needs to be stationary (time independent)...Else $\vec{E}$ would not be a purely potential-derived field.

Daniel.

3. Apr 9, 2005

### robert25pl

I have this two equations:

$$E=3\sin(3z-6t) \vec{k}$$
$$B=- \frac{1}{15} \sin(3z-6t) \vec{j}$$

So what should I do first?

4. Apr 9, 2005

### dextercioby

Verify whether such a field configuration satisfies the eqn-s

$$\nabla\cdot\vec{B}=0$$

$$\nabla\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}$$

Isn't this what u're supposed to do?

Daniel.

5. Apr 9, 2005

### Staff: Mentor

That depends on the question you're supposed to be answering.

6. Apr 9, 2005

### robphy

"Free space" means $$\rho={\color{red}0}$$ and $$\vec\jmath={\color{red}\vec 0}$$.
So, in two of Maxwell's Equations, this means that
$$\nabla \cdot \vec E=\rho ={\color{red}\ 0}$$ and
$$\nabla \times \vec B=\vec \jmath + \frac{\partial \vec E}{\partial t}={\color{red}\ \vec 0} + \frac{\partial \vec E}{\partial t}$$.
Of course, [from the other two equations] we must always have
$$\nabla \cdot \vec B= 0$$ and
$$\nabla \times \vec E= -\frac{\partial \vec B}{\partial t}$$

Last edited: Apr 9, 2005
7. Apr 9, 2005

### Nylex

What?! When we were taught Maxwell's equations in free space, we were told that: $$\nabla \cdot \vec E=\frac{\rho}{\epsilon_{0}}$$, as free space meant in air not in any sort of medium.

8. Apr 9, 2005

### robphy

So, maybe term I should have used is "source-free".

9. Apr 9, 2005

### robert25pl

$$\nabla\cdot\vec{B}=0$$ I verified that

$$\nabla\times\vec{E}=\left|\begin{array}{ccc}\vec{i }& \vec{j} &\vec{k}\\\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ 0 & 0 & 3\sin (3z-6t) \end{array} \right|$$

$$\triangledown \times E$$ gave me 0 and $$-\frac{\partial \vec{B}}{\partial t} =- \frac{2}{5} \cos(3z-6t) \vec{j}$$

I think this wrong?

Last edited: Apr 9, 2005
10. Apr 9, 2005

### jdavel

robert,

You say:

divB = 0, I agree

curlE = 0, I agree

-dB/dt = (-2/5)sin(3z - 6t)j, I think you've got a mistake here

11. Apr 9, 2005

### dextercioby

Should be a "cos" there.

Daniel.

12. Apr 9, 2005

### robert25pl

$$\frac{\partial \vec{B}}{\partial t} =-\frac{2}{5} \cos(3z-6t) \vec{j}$$
So $$\nabla \times \vec E= -\frac{\partial \vec B}{\partial t}$$
are nor equal and they are not possible in region of space?

13. Apr 9, 2005

### dextercioby

That's that.It is not possible.They should identically solve every equation from Maxwell's system...

Daniel.