Formal difference between electrostatics and magnetostatics

[Hypercube]

In his book on EM, Griffiths states:

Formally, electro/magnetostatics is the régime $$\frac{\partial \rho}{\partial t}=0, \hspace{0.25in} \frac{\partial \vec{\boldsymbol{J}}}{\partial t}=\boldsymbol{0}$$

He explains how in electrostatics charges do not move, or (more specifically), charge density does not change. It must remain fixed. I understand that. Second equation is regarding magnetostatics, which also makes sense. But then (few paragraphs down) he says:

More generally, since $\frac{\partial \rho}{\partial t}=0$ in magnetostatics...

Wait. I thought the first of the two equation applies to electrostatics, and the second one to magnetostatics?

So my question essentially comes down to which one(s) of the above equations applies to electrostatics and which one(s) to magnetostatics. Thank you.

 

[Dale]

But then (few paragraphs down) he says:

More generally, since ∂ρ∂t=0∂ρ∂t=0\frac{\partial \rho}{\partial t}=0 in magnetostatics...

In the intervening few paragraphs does he use the continuity equation to derive this for the magnetostatic case also?

 

[Hypercube]

In the intervening few paragraphs does he use the continuity equation to derive this for the magnetostatic case also?

Hi Dale,

Thank you for taking the time to respond.

Intervening few paragraphs are a qualitative discussion with two main points:
1. Electrostatics and magnetostatics are idealisations: artificial worlds that "only exist in textbooks";
2. Moving point charge (on its own) cannot constitute a steady-state current.

Then he makes that second statement that I mentioned in OP:

When a steady current flows in a wire, its magnitude I must be the same all along the line; otherwise, charge would be piling up somewhere, and it wouldn't be a steady current. More generally, since $\frac{\partial \rho}{\partial t}=0$ in magnetostatics, the continuity equation becomes $$\nabla \cdot \boldsymbol{J}=0$$

And that is the end of the chapter.

 

[Dale]

otherwise, charge would be piling up somewhere,

This is the key statement. It is in principle possible to have $\partial J/\partial t = 0$ with $\partial \rho /\partial t \ne 0$. But with the continuity equation you would get that $\rho$ would steadily increase without bound over time. This “charge piling up” is usually rejected on physical grounds, so then we get $\partial \rho/\partial t=0$ for magnetostatics also.

 

[Hypercube]

This is the key statement. It is in principle possible to have $\partial J/\partial t = 0$ with $\partial \rho /\partial t \ne 0$. But with the continuity equation you would get that $\rho$ would steadily increase without bound over time. This “charge piling up” is usually rejected on physical grounds, so then we get $\partial \rho/\partial t=0$ For magnetistatics also.

I understand, I think. To summarise, would you agree with the following:

In electrostatics: $$\frac{\partial \rho}{\partial t}=0, \hspace{0.25in} \boldsymbol{J}=\boldsymbol{0}, \hspace{0.25in} \frac{\partial \boldsymbol{J}}{\partial t}=\boldsymbol{0}$$

In magnetostatics: $$\frac{\partial \rho}{\partial t}=0, \hspace{0.25in} \boldsymbol{J}=\boldsymbol{F}(x,y,z), \hspace{0.25in} \frac{\partial \boldsymbol{J}}{\partial t}=\boldsymbol{0}$$

 

[Dale]

Yes. I would agree

 

[Hypercube]

Excellent, thank you for clarifying this for me.