Maxwell's equations and Coulomb's law

In summary, Feynman stated that Maxwell's first two equations in electrostatics are equivalent to Coulomb's law and the superposition principle. However, for a specific charge distribution, Coulomb's law and the superposition principle can be used to determine a unique field. On the other hand, for Maxwell's equations, satisfying the first two equations can result in infinitely many solutions by adding the gradient of a harmonic field. This is possible because there is a uniqueness theorem for the first two Maxwell's equations. However, this theorem depends on appropriate boundary conditions, which may not be satisfied by adding the gradient of a harmonic field. The proof of this theorem can be found in most electromagnetic textbooks. The boundary conditions for a known charge distribution can
  • #1
kof9595995
679
2
In Feyman's lectures on physics, he said Maxwell's first 2 equations in electrostatics, namely curl E =0 and div E=rho/epsilon, is equivalent to Coulomb's law and superposition principle,
But for a particular charge distribution, we can always use Coulomb's law and superposition principle to determine one unique field, and when it comes to Maxwell's equation-if we just want to satisfy the 2 equations-we can add any gradient of a harmonic field to the field we get using Coulomb's law and superposition principle,thus we can get infinitely many solutions.
So how can it be that they are equivalent??
 
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  • #2
Adding a gradient won't satisfy the divE equation.
There is a uniqueness theorem for the divE and curlE dquations.
 
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  • #3
but I'm adding the gradient of a harmonic field,say, h, then laplacian h=0,which means
div(grad h)=0, then div(E+grad h)=div E+0=div E, which still satisfies the first equation.Am I correct?
 
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  • #4
"There is a uniqueness theorem for the divE and curlE equations. "
could you give some more details about the theorem? Thanks.
 
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  • #6
Any localized charge distribution has a field that tends to zero at infinity, while any harmonic function is not even bounded at infinity. One of the boundary conditions is violated, so the gradient of a harmonic function may not be added.
 
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  • #7
kof9595995 said:
"There is a uniqueness theorem for the divE and curlE equations. "
could you give some more details about the theorem? Thanks.
The uniqueness theorem depends on appropriate boundary conditions atyy suggests.
The proof is in most EM textbooks.
 
  • #8
But what's the boundary condition for a known charge distribution?
I cannot see how Maxwell's equations require that "localized charge distribution has a field that tends to zero at infinity"
 
  • #9
Knowing the charge (or current) distribution isn't enough to determine the electric field in the general case. Because electromagnetic waves can travel far from their sources, you could reasonably specify an incoming light wave at the boundary. In electrostatics, we assume there are no light waves by definition of statics, and the boundary conditions are set by eg. whether you have a conductor etc present.
 
  • #10
kof9595995 said:
But what's the boundary condition for a known charge distribution?
I cannot see how Maxwell's equations require that "localized charge distribution has a field that tends to zero at infinity"

Since the [itex]\nabla\times\vec{E}=0[/itex], we can ignore incoming and outgoing electromagnetic waves.

Choose some point as the center of a Gaussian Sphere of radius r so that all of the charge is contained within the sphere. Then the first of Maxwell's equations says, roughly, that:

[itex]\nabla \cdot\vec{E}=\frac{Q_{enc}}{\epsilon _0}\sim 4\pi r^2 |\vec{E}|[/itex]

Since the sphere is contains all of the charge, [itex]Q_{enc}[/itex] is constant, so that:

[itex]|\vec{E}|\sim \frac{Q_{enc}}{4\pi \epsilon _0r^2}\propto \frac{1}{r^2}[/itex]

So that the electric field tends to 0 at infinity. Another way to see this is through the multipole expansion, but I won't go into that right now.
 
  • #11
but if we don't know the field is radical ,we cannot write E*4pi*r^2=Q/epsilon, right?
 
  • #12
for example, in empty space there's a sphere (radius R) with a constant charge density rho, in electrostatics,then the electric field should be uniquely determined, but what's the boundary condition in this case?
 
  • #13
That the field at infinity is zero.
 
  • #14
but how do you know that without Column's law?I mean, just by Maxwell's equations.
 
  • #15
kof9595995 said:
but how do you know that without Column's law?I mean, just by Maxwell's equations.
You are correct that Max alone does not uniquely specify the E field, but you haven't understood the posts. Max + Boundary Conditions do uniquely specify the E field.
 
  • #16
Yeah, I just cannot figure out what the boundary condition is in my example.
 
  • #17
"That the field at infinity is zero. "
"but how do you know that without Column's law?I mean, just by Maxwell's equations. "
I just cannot figure out what the boundary condition is in my example, I don' understand where the condition "That the field at infinity is zero. " comes from.
 
  • #18
In the case of a point charge, Gauss's law plus isotropy of space gives you Coulomb's law, from which you can see that the field goes to zero at infinity.

In the case of a charge distribution with no symmetry, it's hard to use Gauss's law, so one has to use "common sense" or "physical sense" to make the boundary condition at infinity the same as that of a superposition of point charges.

You can see that Maxwell's equations alone don't contain all the physics in classical electrostatics when a conductor is present. There we need "common/physical sense" to tell us that the conductor is equipotential. In some cases with a conductor, the boundary conditions plus uniqueness enables us to solve the problem with the "method of images" in which we solve the problem by replacing it with a different problem.

So the boundary conditions are determined by additional physics you choose according to the situation at hand, just like initial conditions.
 
  • #19
atyy said:
In the case of a point charge, Gauss's law plus isotropy of space gives you Coulomb's law, from which you can see that the field goes to zero at infinity.

But if we must take isotropy of space into account, does it mean divE= rho/epsilon and curlE=0 indeed say less than Column's law?
 
  • #20
kof9595995 said:
But if we must take isotropy of space into account, does it mean divE= rho/epsilon and curlE=0 indeed say less than Column's law?

Hmm, yes and no. Do you count the principle of superposition as part of Coulomb's law?
 
  • #21
Yes, but Feynman said divE= rho/epsilon and curlE=0 just "say no more and no less than" Column's law+superposition principle
 
  • #22
RF oversimplified by leaving out the BC. He liked to make sweeping statements without too much care about the details.
 
  • #23
kof9595995 said:
Yes, but Feynman said divE= rho/epsilon and curlE=0 just "say no more and no less than" Column's law+superposition principle

Hmm, perhaps he should have left out "no more and no less". There's a web page with errata for the lectures, you can try sending your comments there.
 

1. What are Maxwell's equations and Coulomb's law?

Maxwell's equations and Coulomb's law are fundamental laws in the field of electromagnetism. They describe the behavior of electric and magnetic fields and how they interact with each other and with charged particles.

2. Who discovered Maxwell's equations and Coulomb's law?

Maxwell's equations were developed by James Clerk Maxwell in the 19th century. Coulomb's law was discovered by Charles-Augustin de Coulomb in the 18th century.

3. What is the relationship between Maxwell's equations and Coulomb's law?

Coulomb's law is a simplified version of one of Maxwell's equations, which describes the force between two stationary charged particles. Maxwell's equations expand upon this to include dynamic interactions between electric and magnetic fields.

4. How are Maxwell's equations and Coulomb's law used in everyday life?

Maxwell's equations and Coulomb's law have many practical applications in everyday life, such as in the design and operation of electronic devices like computers and cell phones, as well as in the transmission and distribution of electricity.

5. Are Maxwell's equations and Coulomb's law still relevant today?

Yes, Maxwell's equations and Coulomb's law are still widely used today and are considered to be some of the most important laws in physics. They have been extensively tested and continue to accurately describe the behavior of electric and magnetic fields.

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