Use Maxwell's Equations to derive one law from another

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Homework Help Overview

The discussion revolves around deriving Electric Gauss's Law from Generalized Ampere's Law using Maxwell's equations. Participants are exploring the relationships between these laws and the necessary mathematical tools for the derivation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the possibility of deriving one equation from another, questioning the validity of such a derivation. Some express uncertainty about whether integrals are necessary for the derivation process. Others mention the use of Gauss's and Stokes' theorems and the challenges faced when applying them correctly.

Discussion Status

There is an ongoing exploration of the problem, with participants providing insights into the application of mathematical theorems. Some guidance has been offered regarding the interpretation of integrals and the context of the equations, but no consensus has been reached on the feasibility of the derivation itself.

Contextual Notes

Participants note that the problem statement may be unclear or misinterpreted, leading to confusion about the derivation process. The discussion includes references to assumptions about the applicability of certain theorems and the nature of the equations involved.

Cursed
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Homework Statement



Derive equation (1) from equation (2):

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

Homework Equations



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

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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Hi Cursed! Welcome to PF! :smile:

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.
 
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.
 
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...
 
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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Cursed said:
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! :smile: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.
 
\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
 
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.
 
Cursed said:
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:

[URL]http://upload.wikimedia.org/math/3/0/c/30c5316a93681d924ba53461f05ab521.png[/URL]​

we can start talking again. :smile:



Cursed said:
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... :wink:


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