Range of validity of Maxwell's equations with charges

  • Thread starter fluidistic
  • Start date
  • #1
fluidistic
Gold Member
3,662
104

Main Question or Discussion Point

Maxwell's equations with charges can be written as the following (in the cgs system):

[tex]\frac{\partial \vec E}{\partial t} =c \vec \nabla \wedge \vec B-4\pi \vec J[/tex].
[tex]\frac{\partial \vec B}{\partial t} =-c \vec \nabla \wedge \vec E[/tex].
[tex]\vec \nabla \cdot \vec E =4 \pi \rho[/tex].
[tex]\vec \nabla \cdot \vec B =0[/tex].

If I'm right, these equations are valid only where charges are present. It means, almost nowhere! For instance consider a positively charged table. [tex]\rho(t,\vec x)[/tex] is of course not continuous since we're dealing with charges. In other words it is zero everywhere when there's no charge. If we assume the electron has a volume, then the equations would be valid within the volume of the electron. So in an neutral atom, these equations are valid less than 0.001% of the space (only where the proton and the electron are).
I'm just curious if I'm right. In an affirmative case, why do we bother with them and why don't we only use Maxwell's equation in vacuum?
In the latter case, I could imagine having a lot of boundary equations (precisely the places enclosing the charges) and I could only use Maxwell's equation in vacuum, taking into account all the boundary equations not to lose any information about the charges.
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,965
326
Maxwell's equations are valid even when no charges are present. In that case, you can set [tex]\rho=0[/tex] and you can see that the equations are still there, and the E and B fields also interact.

We use this formulation because it's more general than the formulation without charges.

Because the equations are differential equations, even if some terms equal zero, the E and B fields need not equal zero. Just like if dx/dt=0 it doesn't mean x=0 for all time.
 
  • #3
fluidistic
Gold Member
3,662
104
Maxwell's equations are valid even when no charges are present. In that case, you can set [tex]\rho=0[/tex] and you can see that the equations are still there, and the E and B fields also interact.

We use this formulation because it's more general than the formulation without charges.

Because the equations are differential equations, even if some terms equal zero, the E and B fields need not equal zero. Just like if dx/dt=0 it doesn't mean x=0 for all time.
Ah ok, thanks.
I'm getting it. I feel I'm learning a lot, it's incredibly nice!
 
  • #4
Pengwuino
Gold Member
4,989
15
You need to consider that charges create fields throughout your space, not simply where the charges are. Infact, in classical EM, the fields diverge at the actual charges. For example your coulomb field blows up as your field point approaches your charges.

When you formulate a problem, you define a space you're working in. Maxwell's equations will be valid in the entire space. You don't have a new equation for every infinitesimal volume in the space, it's simply one set of equations for the whole space given some charge distribution inside.

Think of a gravitational potential. It's defined for your entire volume of space, not just where the mass is.
 

Related Threads on Range of validity of Maxwell's equations with charges

Replies
15
Views
6K
Replies
2
Views
2K
Replies
11
Views
2K
Replies
10
Views
3K
Replies
95
Views
11K
Replies
7
Views
2K
Replies
23
Views
14K
Replies
2
Views
742
Replies
4
Views
541
  • Last Post
Replies
4
Views
3K
Top