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I'm trying to understand basic electrodynamics at a fundamental level. It's very confusing. I have heard that there are only four forces: gravity, EM, and two nuclear forces. So it appears that the electrostatic force is therefore part of the entire EM field.

Does that mean that an $E$ field cannot exist without a corresponding B and H field? In other words, if something is creating or inducing an E field, does it also create those two magnetic fields as well?

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    $\begingroup$ I think given the two different answers you received, it may be helpful to clarify what you mean by "an $E$ field cannot exist without a correspond $B$ and $B$ field". Do you mean within a fixed frame of reference? Allowing changes in frames of reference, etc. $\endgroup$ Sep 3, 2023 at 22:59
  • $\begingroup$ @SillyGoose Definitely fixed frame of reference, but i did not consider relative movement of that frame of reference (relative to the E field). $\endgroup$
    – DrZ214
    Sep 3, 2023 at 23:41

3 Answers 3

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In a world where the laws of Newton hold (i.e. the "non-relativistic" world), an electric field $E ≠ 0$ and magnetic field $B = 0$ would have the same appearance in all moving frames of reference. In that sense, for Newtonian physics, one could have a world where $B = 0$.

In a world governed by Special Relativity, the same remains true - but only for frames of reference moving in a direction collinear with $E$. In frames of reference moving in a direction perpendicular to $E$ at a speed $v$, the fields would be $E' = E/\sqrt{1 - (v/c)^2}$ and $B' = (vE/c^2)/\sqrt{1 - (v/c)^2}$, with $B'$ being in a direction perpendicular to both the moving observer and $E$, where $c$ is the speed of light. In particular, if $E$ were oriented north, the observer were moving east, then the observer would see $B'$ pointing up.

So, a moving observer can see an electric field as, partly, being a magnetic field. Therefore, it is not possible to have just electric fields without also having magnetic fields in a world governed by Special Relativity.

In Special Relativity, the quantities $BE \cos{Θ}$ (where $Θ$ is the angle between the $B$ and $E$ field) and $B^2 - E^2/c^2$ are independent of the observer's frame of reference. This would also be true in Newtonian physics, except that the latter invariant would be $B^2$, instead of $B^2 - E^2/c^2$.

Edit: it's also worth saying something about the $D$ and $H$ fields, now that I see some discussion about them.

If, the displacement field $D$ is parallel to $E$ for a stationary observer, with the ratio $D/E = ε$ (the permittivity) and if $B$ and $H$ are both 0, then for observers moving collinearly with $E$, the fields $D$, $E$, $B$ and $H$ remain unchanged.

For observers moving perpendicular to $E$ at a speed $v$, the ratio $D'/E' = ε$ continues to hold intact, while $H' = D' v$, with $H'$ in the same direction as already noted for $B'$ in the case of Special Relativity. The relations $H' = D' v$ and $D' = ε E'$ - in this case - hold for such observers irrespective of the distinction between Newton and Special Relativity.

In particular, that means a moving observer will see a non-zero $H'$ field, if moving perpendicularly to $E$.

The quantities $DH \cos{Θ}$ and $D^2 - H^2/c^2$ are the invariants, where, this time, $Θ$ is the angle between the $D$ and $H$ vectors. For non-relativistic theory, $D^2$ is the invariant, rather than $D^2 - H^2/c^2$.

So, even in non-relativistic physics, it is not possible to consistently maintain $H = 0$. A purely electric contribution to $H$ arises for moving observers. Maxwell failed to note this in his treatise in the 1870's, though he considered the possibility in his earlier 1860's papers. The correction was made by Thomson, I think, in the 1890's, and was directly observed c. 1900 by a husband and wife team.

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  • $\begingroup$ Okay but what about the H field? Is it somehow equivalent to the B field in both cases? $\endgroup$
    – DrZ214
    Sep 3, 2023 at 23:39
  • $\begingroup$ Yeah, I noticed the oversight. So I added further comments about that. Thanks. $\endgroup$
    – NinjaDarth
    Sep 3, 2023 at 23:49
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Yes and no.

A stationary charge produces an $E$ field but no $B$ or $H$ field. So yes.

If you run past the charge, it is no longer stationary. A moving charge produces a $B$ field in addition to an $E$ field. So no.

You are used to thinking of $E$ and $B$ as two separate vectors. This works, but together they form a more complex tensor object. A tensor is a generalization of a vector.

To get the flavor of why this is done, think of a $2$D velocity. If you throw an object straight up, you see a vertical component of velocity, but no horizontal component. If you run by the object as it is thrown, now you do see a horizontal component.

You can think of these as two separate quantities and do physics. But if you combine them into a vector, you can do vector addition and coordinate transformations. Using vectors makes patterns clear that are hard to see when you keep components separate.

You can do coordinate transformations with the Electromagnetic field tensor to see how electric and magnetic fields change when you choose a new frame of reference.

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  • $\begingroup$ Thank you but A moving charge produces an H field in addition to an E field. So no. And what about the B field here? Is it also produced? I might ask a separate q about this to see if B or H can exist independently of one another. $\endgroup$
    – DrZ214
    Sep 3, 2023 at 23:43
  • $\begingroup$ In vacuum, $B = \mu_0 H$. In a magnetic material, atoms are tiny magnets. An external magnetic field can align them so that the material produces its own magnetic field in addition to the external field. $B$ and $H$ account for the field from the material differently. See the Wikipedia Magnetic field for more. Also see MAGNETS: How Do They Work? for more on magnetic materials. $\endgroup$
    – mmesser314
    Sep 3, 2023 at 23:52
  • $\begingroup$ This is the best answer and easiest to understand so i accepted it. My understanding is that if a B field exists, then an H field must also exist, and vice versa. There are no exceptions to that. And if an E field exists, then the other two may or may not exist depending on frame of reference. $\endgroup$
    – DrZ214
    Sep 4, 2023 at 14:34
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$\vec{B}$ and $\vec{H}$ are not two different fields. $\vec{H}$ is (at least in Griffiths notation) the appropriately defined magnetic field in magnetized materials.

If you have an electrostatic situation (such as the ideal, textbook example of a point charge at rest), then certainly you can have an electric field $\vec{E}$ with no non-zero magnetic field. One can see this by recalling the Maxwell equations and observing that there are no sources of magnetic field (no moving charges and no dynamic electric fields).

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  • $\begingroup$ "no non-zero" (... magnetic field): instead of a double negation, it might be easier to understand if you replace it with a simple "no" (... magnetic field) $\endgroup$
    – hydra4jh
    Sep 3, 2023 at 22:49
  • $\begingroup$ There is a magnetic field with value zero everywhere @hydra4jh. I find stating "no magnetic field" confusing for this reason. $\endgroup$ Sep 3, 2023 at 22:50
  • $\begingroup$ What is the difference between "no magnetic field" and "a magnetic field with zero magnitude everywhere"? $\endgroup$
    – hydra4jh
    Sep 3, 2023 at 22:53
  • $\begingroup$ Depends how you define "no magnetic field" @hydra4jh. I define it as there being no field whatsoever as in the concept of $\vec{B}$ doesn't exist and does not permeate all space and time. $\endgroup$ Sep 3, 2023 at 22:57
  • $\begingroup$ Does sound "zero magnetic field" better to you? That´s semantically equivalent to "no non-zero magnetic field". If you want to state that there is a need for the concept of a magnetic field in the example you provided, then you should clarify that in another sentence. For me it is not obvious why you need the concept for a system of a resting observer and a resting charge. That specific system can be completely described by Coulombs Law, i.e. you need no magnetic field <=> there is no magnetic field. If any kind of movement happens, then you need one. $\endgroup$
    – hydra4jh
    Sep 3, 2023 at 23:23

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