Independence of Gauss's law and Ampere's law

[loom91]

Hi,

I'm just learning about Maxwell's equations in high-school and was playing around with them. Supposedly they are 4 independent and self-sufficient equations that when connected with the Lorentz force law will predict classical electrodynamics in its entirety. But then, it appears to me that Gauss's law can be derived from Ampere's law! How can this be possible?

Ampere's law with Maxwell's correction:-

$$\int_{\partial S} \vec{B} \cdot \vec{dl} = \mu i_S + \mu \epsilon \frac{d\Phi_{E,S}}{dt}$$

Now let us take the limit as S becomes a closed surface. In this limit, the line integral of B vanishes (a closed surface has no boundary), while on the right side we have the net current coming out of the closed surface plus the electric flux over the closed surface. Note the the net current coming out = - rate of change of charge enclosed by the surface. This gives us

$$\epsilon d\Phi_{E,S} = dQ_S$$

Integrating, we get

$$\Phi_{E,S} = \frac {Q_S}{\epsilon}$$

Which is Gauss's law for electricity.

How can this happen? Maxwell's equations are supposed to be independent! The only way out I can personally see is to take Gauss's law as simply the boundary condition that the net flux through a closed surface enclosing zero net charge is zero, a condition that is used to do the integral in the above derivation. But this boundary condition is nowhere near as strong as Gauss's law itself! What is the matter?

Thanks for your help. I will also appreciate it if you kept this discussion within the boundary conditions of my knowledge (for example, I don't understand the relativistic formulation of classical electrodynamics)

Molu

 

[lalbatros]

You used charge conservation to get the result.
So your conclusion arises not because of Ampère law alone, but because of Ampère law combined with charge conservation.

You could as well derive charge conservation from Ampère and Gauss.
This is precisely the purpose of Maxwell's correction.

 

[Meir Achuz]

Maxwell's great contribution was to show that it was necessary to add a term
$$\sim\partial_t{\vec D}$$ to the curl H equation.
He did this using the div D equation.
These two Maxwell equations are connected by this.
You have just worked Maxwell's introduction of the
$$\partial_t{\vec D}$$ term backwards.
Note you have written the integral versions of Maxwell's differential equations.

 

[loom91]

You used charge conservation to get the result.
So your conclusion arises not because of Ampère law alone, but because of Ampère law combined with charge conservation.

You could as well derive charge conservation from Ampère and Gauss.
This is precisely the purpose of Maxwell's correction.

Ah, I see now. A very subtle point, I would have never realized that on my own. I was actually replacing Gauss's law with the continuity equation. Conservation of charge is so deeply ingrained in my mentality that I never considered that it was actually a postulate itself. Thanks a lot for pointing this out, I feel I've a deeper understanding of electrodynamics after this small exercise.

But one minor point still remains. The continuity equation alone was not enough, I did have to add the boundary condition. Does this mean that the continuity equation is actually weaker than Gauss's law?

Molu

 

[loom91]

Maxwell's great contribution was to show that it was necessary to add a term
$$\sim\partial_t{\vec D}$$ to the curl H equation.
He did this using the div D equation.
These two Maxwell equations are connected by this.
You have just worked Maxwell's introduction of the
$$\partial_t{\vec D}$$ term backwards.
Note you have written the integral versions of Maxwell's differential equations.

Sorry, that went over my head. What does a partial sign used alone mean? That smells of relativistic notation

Molu

 

[cristo]

What does a partial sign used alone mean?

$$\partial_t \vec{D}\equiv\frac{\partial \vec{D}}{\partial t}$$

 

[loom91]

$$\partial_t \vec{D}\equiv\frac{\partial \vec{D}}{\partial t}$$

Oh, I see.

Molu

 

[lalbatros]

"The continuity equation alone was not enough, I did have to add the boundary condition."

Go to the differential form of the equations and you will see no boundary conditions needed.