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If I understand correctly, the electrostatic approximation assumes that all charges are stationary (i.e the charge density is constant in time, and current density is zero). The magnetostatic approximation assumes the current density is constant in time. Are these the correct definitions?

I have also seen electrostatics being characterized by an electric field which is constant in time. Is this supposed to be some obvious consequence of the above definition?

Looking at Maxwell's equations, the divergence of $E$ is fixed, but the curl of $E$ might vary in time if $B$ varies in time. Even in empty space (an electrostatic situation by defualt) we have the wave solution to Maxwell's equations, in which both $E$ and $B$ vary in time.

So I'm guessing most authors are just sloppy when it comes to nomenclature, and when they say electrostatics, they really mean quasi-electrostatics, i.e electrostatics in the quasistatic approximation where the deplacement current $j_D$ is neglected from Ampere's law.

In quasi-electrostatics, we definitely can't have the wave solution in empty space anymore. But is it still possible that the $B$ field varies in time? The magnetic field equations reduce to $curl(B)=0$ and $div(B)=0$. How do we know it's not possible to find some time-varying $B$ field which always has zero curl and zero divergence?

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This are Maxwell Equations: $$ \begin{align} \nabla\cdot\mathbf E = \frac{\rho}{\epsilon_0}, \quad\quad &\nabla\cdot\mathbf B = 0\\ \nabla\times\mathbf E = -\frac{\partial\mathbf B}{\partial t}, \quad\quad &\nabla\times\mathbf B = \mu_0\left(\mathbf J + \epsilon_0\frac{\partial\mathbf E}{\partial t}\right)\\ \end{align} $$

  • In the electrostatic case: $\nabla\times\mathbf E = 0$.

  • In the magnetostatic case: $\nabla\times\mathbf B = \mu_0\mathbf J$.

So, In the static cases, we do not have the other field changing. That is, in the electrostatic case, magnetic fields cannot change. In the magnetostatic case, eletric fields cannot change.

An electric or magnetic field is static when it is uncoupled. Not to be confused with spatially-constant field, or non-time-varying field. On normal conditions (Maxwell Equations), the sources which generates the electric field are $\mathbf B$ and $\rho$. That is, the electric field depends on what the magnetic field is doing, that is, they are coupled.

The whole point of the static approximations is to neglect field coupling. In the electrostatic case, a magnetic field can never generate an electric field. Meaning, the only sources of electric fields are $\rho$. The same is valid for the magnetostatic case: Only $\mathbf J$ generates $\mathbf B$, that is, $\mathbf E$ cannot produce $\mathbf B$.

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  • $\begingroup$ So why does wikipedia define the static cases as the charge density or current density being static? $\endgroup$ – Joshua Benabou Jul 13 '17 at 16:30
  • $\begingroup$ @JoshuaBenabou It does not. Where have you saw this? $\endgroup$ – Physicist137 Jul 13 '17 at 16:32
  • $\begingroup$ I'm still confused. You're telling me the electrostatic field, the B field is static, while according to wikipedia, its the E field that must be static... $\endgroup$ – Joshua Benabou Jul 13 '17 at 17:23
  • $\begingroup$ @JoshuaBenabou, in the electrostatic case, its the E-Field which is static (obviously, as the name suggests). The E-Field is static when its uncoupled with the B-Field. That happens when the B-Field is not time-varying. Perhaps you are confusing static fields with time-varying fields. These are different things. $\endgroup$ – Physicist137 Jul 13 '17 at 18:14
  • $\begingroup$ What is the difference between a time-varying field and a static field? $\endgroup$ – Joshua Benabou Jul 13 '17 at 20:55
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It appears that you've put the cart in front of the horse: the fact that the field is static means that the charges are stopped or moving slowly and that charge and current densities are mostly constant.

The time varying Maxwell Equations (which couple the $\mathbf E$ and $\mathbf B$ fields) are : $$ \begin{align} \nabla \times \mathbf E &= -\frac{\partial\mathbf B}{\partial t} \\ \nabla\times\mathbf B &= \mu_0\left(\mathbf J + \epsilon_0\frac{\partial\mathbf E}{\partial t}\right)\\ \end{align} $$

In physics, we set time derivatives to zero for "static" cases. This may indicate that those derivatives are actually zero, or it may indicate that they are simply too small to consider.

$$\frac{\partial \mathbf B }{\partial t} = \frac{\partial \mathbf E }{\partial t} = 0$$

meaning electrostatics studies $ \{ \nabla \times \mathbf E = 0, \nabla \cdot \mathbf E = \frac{\rho}{\epsilon_0} \}$ and magneto-statics studies $\{\nabla \cdot \mathbf B = 0, \nabla \times \mathbf B = \mu_0 \mathbf J\} $

There are no waves in electro or magneto-statics. If a field has zero curl and zero divergence, it should be obvious that it is constant (see Kelvin-Stokes Theorem).

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  • $\begingroup$ Though this answer is all correct, one could mislead in thinking that static E-Fields means $\partial\mathbf E / \partial t = 0$, which is not true. In fact, static E-Fields implies $\partial\mathbf B / \partial t = 0$. Indeed, is fairly easy to find an electrostatic example in which E-Fields are changing in time. =). Same for static B-Fields which imply $\partial\mathbf E / \partial t = 0$. =). $\endgroup$ – Physicist137 Jul 13 '17 at 18:48
  • $\begingroup$ @Physicist137 I'd love to see an example of what you're thinking of - I'm not coming up with it off the top of my head. $\endgroup$ – user121330 Jul 13 '17 at 19:05
  • $\begingroup$ Oh. Well, as you pointed out, looking at maxwell equations, we find that in the electrostatic case that $\nabla\times\mathbf E = 0$, which Faraday Law implies $\partial \mathbf B / \partial t = 0$, meaning, by $\nabla\times\mathbf B$ equation, I can have $\mathbf E(\mathbf r, t) = \mathbf at + \mathbf b$, where $\mathbf a$, $\mathbf b$ depends on position. With that in mind, simplest example I can think off, imagine a sphere of radius $R$ being charged linearly: $Q(t) = q_0 + \alpha t$. This shows an electrostatic example with changing electric field. =]. $\endgroup$ – Physicist137 Jul 13 '17 at 19:21
  • $\begingroup$ @Physicist137 Which would mean that we have $\nabla \times \mathbf B = \mu_0\mathbf J + \frac{\mathbf a (\alpha)}{c^2}$ which helps find the (obviously much smaller) $\mathbf B$ field. I usually use electrostatics to find $\mathbf E$ fields. Cool problem. $\endgroup$ – user121330 Jul 13 '17 at 19:34
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I would say the definition is that the fields should be constant, not the charges. After all, an electromagnetic wave in a vacuum has constant charges and currents, and it's surely not electrostatic.

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