Use Maxwell's Equations to derive one law from another

In summary, 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.
  • #1
Cursed
39
0

Homework Statement



Derive equation (1) from equation (2):

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

Homework Equations



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

The Attempt at a Solution



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

Now I'm stuck at the continuity equation. How do I prove that [tex]\int-\nabla \cdot J = \nabla \cdot D[/tex]? (What I want to do is take the integral of both sides.)
 
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  • #2
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.
 
  • #3
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?


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

What do I do then? I know that [tex]\Phi = \frac{d}{dt} \int B \cdot ds[/tex], but using that doesn't seem right.
 
  • #4
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...
 
  • #5
Even if I use the divergence theorem and Stokes' Theorem, I get stuck:

[tex]\nabla \times H = J + \frac{\partial D}{\partial t}[/tex]

[tex]\int\int_s (\nabla \times H) \cdot ds = \int\int_s J \cdot ds + \int\int_s \frac{\partial D}{\partial t} \cdot ds[/tex]

[tex] \oint_c H \cdot dl = \int\int_s J \cdot ds + \int\int\int_v (\nabla \cdot D) \cdot ds[/tex]
 
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  • #6
Cursed said:
Even if I use the divergence theorem and Stokes' Theorem, I get stuck:

[tex]\nabla \times H = J + \frac{\partial D}{\partial t}[/tex]

[tex]\int\int_s (\nabla \times H) \cdot ds = \int\int_s J \cdot ds + \int\int_s \frac{\partial D}{\partial t} \cdot ds[/tex]

[tex] \oint_c H \cdot dl = \int\int_s J \cdot ds + \int\int\int_v (\nabla \cdot D) \cdot ds[/tex]

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.
 
  • #7
[tex] \oint_c H \cdot dl = \int\int_s J \cdot ds +\frac{\partial (\int_v (\nabla \cdot D) \cdot dV)}{\partial t}[/tex]

You forgot the partial derivative with respect to time.

ehild
 
  • #8
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.
 
  • #9
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:

[tex]\Phi_{D,S} = \iint_S D \cdot ds[/tex]

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

1. What are Maxwell's Equations?

Maxwell's Equations are a set of four equations that describe the fundamental laws of electricity and magnetism, known as electromagnetism. They were developed by James Clerk Maxwell in the 19th century and are used to understand and predict the behavior of electric and magnetic fields.

2. Why is it important to use Maxwell's Equations?

Maxwell's Equations are important because they are the foundation of our understanding of electromagnetism. They allow us to mathematically describe and predict the behavior of electric and magnetic fields, which are essential for many technologies such as electricity, electronics, and telecommunications.

3. How do you derive one law from another using Maxwell's Equations?

To derive one law from another using Maxwell's Equations, you can use mathematical operations such as differentiation and integration to manipulate the equations. This allows you to rewrite one equation in terms of the other, thus showing the relationship between the two laws.

4. What is an example of deriving one law from another using Maxwell's Equations?

An example of deriving one law from another using Maxwell's Equations is using Faraday's Law to derive Ampere's Law. By taking the curl of Faraday's Law, you can obtain Ampere's Law, which describes the relationship between electric currents and magnetic fields.

5. What are some real-world applications of using Maxwell's Equations to derive one law from another?

Some real-world applications of using Maxwell's Equations to derive one law from another include designing and optimizing electromagnetic devices, such as antennas and motors. These equations also play a crucial role in the development of technologies such as wireless communication, radar, and MRI machines.

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