# Use Maxwell's Equations to derive one law from another

#### Cursed

1. Homework Statement

Derive equation (1) from equation (2):

(1) $$\nabla \cdot D = \rho_f$$
(2) $$\nabla \times H = J + \frac{\partial D}{\partial t}$$

2. Homework Equations

[PLAIN]http://img198.imageshack.us/img198/4645/maxwell.png [Broken]

3. The Attempt at a Solution

$$\nabla \times H = J + \frac{\partial D}{\partial t}$$
$$\nabla \cdot (\nabla \times H) = \nabla \cdot (J + \frac{\partial D}{\partial t})$$
$$0 = \nabla \cdot J \frac{\partial \rho_f}{\partial t}$$ (This is assuming no source or sink.)
$$-\nabla \cdot J = \frac{\partial \rho_f}{\partial t}$$

Now I'm stuck at the continuity equation. How do I prove that $$\int-\nabla \cdot J = \nabla \cdot D$$? (What I want to do is take the integral of both sides.)

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#### I like Serena

Homework Helper
Hi Cursed! Welcome to PF!

Your 2 equations are 2 different laws of Maxwell.
One can not be derived from the other (otherwise Maxwell wouldn't have defined them as separate laws).

Is it possible that the problem asks to derive the integral form of the law?
That is something you can do.

#### Cursed

Well, the problem asks for us to derive the equations as they are, but if integrals are necessary, I can try that. I just didn't know if I had to use integrals or not.

How would I go about starting that?

$$\nabla \times H = J + \frac{\partial D}{\partial t}$$
$$\int H \cdot dl = I_f,s + \frac{\partial \Phi}{\partial t}$$

What do I do then? I know that $$\Phi = \frac{d}{dt} \int B \cdot ds$$, but using that doesn't seem right.

#### I like Serena

Homework Helper
For equation (1) you would need Gauss's theorem for integrals.
For equation (2) you would need Stoke's theorem for integrals.

Are you familiar with those?
You can find them e.g. on wikipedia...

#### Cursed

Even if I use the divergence theorem and Stokes' Theorem, I get stuck:

$$\nabla \times H = J + \frac{\partial D}{\partial t}$$

$$\int\int_s (\nabla \times H) \cdot ds = \int\int_s J \cdot ds + \int\int_s \frac{\partial D}{\partial t} \cdot ds$$

$$\oint_c H \cdot dl = \int\int_s J \cdot ds + \int\int\int_v (\nabla \cdot D) \cdot ds$$

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#### I like Serena

Homework Helper
Even if I use the divergence theorem and Stokes' Theorem, I get stuck:

$$\nabla \times H = J + \frac{\partial D}{\partial t}$$

$$\int\int_s (\nabla \times H) \cdot ds = \int\int_s J \cdot ds + \int\int_s \frac{\partial D}{\partial t} \cdot ds$$

$$\oint_c H \cdot dl = \int\int_s J \cdot ds + \int\int\int_v (\nabla \cdot D) \cdot ds$$
Ah, right, you know the basics of those theorems. :)
However, you're not applying them quite right yet.

For starters, they are never both applicable at the same time.

The thing is, Gauss and divergence work on a volume and the closed surface around that volume. It is not applicable here.
Stokes and curl work on a surface and the close loop curve around that surface.

Now if you look at the surface integral of the current density J through the surface, you need to realize that this is actually the total current going through the surface.
Equivalently you need to interpret the surface integral over ∂D/∂t.
Do you see?

You just did equation (2). Actually equation (1) is a little easier.
Again you need to consider what the integral over the volume actually is.

Cheers!

Edit: btw, I'm still not sure that I understand your problem statement properly.
If it is what I think you're saying, they would be asking for the reverse, that is, to show that the equations can be derived from their equivalent integral form.

#### ehild

Homework Helper
$$\oint_c H \cdot dl = \int\int_s J \cdot ds +\frac{\partial (\int_v (\nabla \cdot D) \cdot dV)}{\partial t}$$

You forgot the partial derivative with respect to time.

ehild

#### Cursed

It says to derive the Electric Gauss Law (1) from Generalized Ampere's Law (2).

What is the appropriate way to apply Stokes' theorem? I'm confused.

Thanks for both of your help, by the way. I greatly appreciate it.

#### I like Serena

Homework Helper
It says to derive the Electric Gauss Law (1) from Generalized Ampere's Law (2).
I'm sorry, but it can't be done as I explained before.
Your problem statement must be off.

Now if you want to derive your equations from:

we can start talking again.

What is the appropriate way to apply Stokes' theorem? I'm confused.

Thanks for both of your help, by the way. I greatly appreciate it.
Your formula for Φ is off.
It should be:

$$\Phi_{D,S} = \iint_S D \cdot ds$$

That should also give you a hint how to complete the way to apply Stokes' theorem....

Btw, I appreciate your thanks. All too often we don't get any. :)

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