Relativistic form of the displacement current using the Biot-Savart Law

[tade]

The Biot-Savart law which describes a magnetic field created by a displacement current: $$\frac{dB}{dV}=\frac{\mu_0\epsilon_0}{4\pi}\frac{\frac{∂E}{∂t}×r}{r^2}$$

What's the relativistically co-variant form of this equation?

Is the introduction of speed of light propagation delays enough, or are there other factors involved?

 

[pervect]

The Biot-Savart law which describes a magnetic field created by a displacement current: $$\frac{dB}{dV}=\frac{\mu_0\epsilon_0}{4\pi}\frac{\frac{∂E}{∂t}×r}{r^2}$$

What's the relativistically co-variant form of this equation?

Is the introduction of speed of light propagation delays enough, or are there other factors involved?

The covariant formulation of electromagnetism in general is

$$\partial_a F^{ab} = u_0 J^b$$
$$\partial_a G^{ab} = 0$$
$$G^{ab} = \frac{1}{2} \epsilon^{abcd} \, F_{cd}$$

here $F^{ab}$ is the Farday tensor, and $G^{ab}$ is the hodges dual of the Faraday tensor.

See https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism
and https://en.wikipedia.org/wiki/Electromagnetic_tensor

The later has more information on what the Farday tensor (and it's dual) are, in terms of the possibly more familiar E and B fields.

If you're familiar with the covariant derivative, feel free to replace $\partial_a$ with $\nabla_a$.

I don't believe there is any way to separate out the Biot-Savart laws in particular in a covariant manner - if one adopts a particular frame of reference, the Biot Savart law is one piece of a bigger picture, the bigger picture is the covariant formulation of electromagnetism.. The electric fields $E^a$ and the magnetic fields $B^a$ are not tensors themselves, they are parts of a larger tensor. The covariant formulation of electromagnetism combines E and B into a larger entity, the Farday tensor F.

 

[tade]

The covariant formulation of electromagnetism in general is

$$\partial_a F^{ab} = u_0 J^b$$
$$\partial_a G^{ab} = 0$$
$$G^{ab} = \frac{1}{2} \epsilon^{abcd} \, F_{cd}$$

here $F^{ab}$ is the Farday tensor, and $G^{ab}$ is the hodges dual of the Faraday tensor.

See https://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism
and https://en.wikipedia.org/wiki/Electromagnetic_tensor

The later has more information on what the Farday tensor (and it's dual) are, in terms of the possibly more familiar E and B fields.

If you're familiar with the covariant derivative, feel free to replace $\partial_a$ with $\nabla_a$.

I don't believe there is any way to separate out the Biot-Savart laws in particular in a covariant manner - if one adopts a particular frame of reference, the Biot Savart law is one piece of a bigger picture, the bigger picture is the covariant formulation of electromagnetism.. The electric fields $E^a$ and the magnetic fields $B^a$ are not tensors themselves, they are parts of a larger tensor. The covariant formulation of electromagnetism combines E and B into a larger entity, the Farday tensor F.

thanks, so, if I want to calculate the relativistically accurate B-field in my rest frame, using the displacement current field, do I still use the Biot-Savart law, but with the displacement current of r/c seconds ago?

 

[vanhees71]

First of all one should state that the OP is wrong. You cannot apply Biot-Savart's formula to this situation, because Biot-Savart only applies to magnetostatics, i.e., time-independent electromagnetic fields and charge-current distributions.

For time-dependent problems to be fully consistent you have to use the retarded potentials (or equivalently Jefimenko's solutions for the em. field which are also using the retarded propagator of the wave equation). Only then everything is also consistent with relativity as it must be!

In addition one should stress that the formulation "a displacement current is the cause of a magnetic field" (looking at Ampere-Maxwell's Law) or "a time-dependent magnetic field causes a electric field". What we sloppily call "electric field" and "magnetic field" are in reality the components of the one and only electromagnetic field with respect to some (within SRT usually inertial) reference frame.

A much more consistent picture of "cause and effect" is to say that electric charge-current distributions (described by the four-vector field $(j^{\mu})=(c \rho,\vec{j})$) are the causes for the electromagnetic field described by the four-potential via the retarded solution (in Lorenz gauge). Then everything turns out to be local an really causal in the sense of Minkowski space as it must be for a relativistic field theory.

 

[tade]

First of all one should state that the OP is wrong. You cannot apply Biot-Savart's formula to this situation, because Biot-Savart only applies to magnetostatics, i.e., time-independent electromagnetic fields and charge-current distributions.

For time-dependent problems to be fully consistent you have to use the retarded potentials (or equivalently Jefimenko's solutions for the em. field which are also using the retarded propagator of the wave equation). Only then everything is also consistent with relativity as it must be!

So let's say I know the displacement current field and I want to calculate the B-field. Is it ok to use the Biot-Savart law but with delays of t = r/c ?

 

[vanhees71]

Already the question doesn't make sense. If you only know the displacement current, i.e., essentiall $\partial_t \vec{E}$ you have not enough information to deduce the electromagnetic field. You need all Maxwell equations, i.e., in Heaviside-Lorentz units
$$\partial_{\mu} F^{\mu \nu}=j^{\nu}, \quad \partial_{\mu} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$
For a complete determination of the fields you need the four-current $j^{\mu}$ and the initial values $F_{\mu \nu}(t=0,\vec{x})$ of the fields.

 

[tade]

Already the question doesn't make sense. If you only know the displacement current, i.e., essentiall $\partial_t \vec{E}$ you have not enough information to deduce the electromagnetic field.

Understood. Still, is the method correct if we want to calculate the component of the B-field that's due to the $\partial_t \vec{E}$ field?

 

[pervect]

The displacement current would contribute to the B-field in a non-covariant formulation, one would need to add the displacement current to the non-displacement current to get the entire B-field in a non-covariant formulation.

The covariant formulation is something different - I tried to outline it earlier. I'm not quite sure why you keep asking about the covariant formulation, then asking questions that would apply to a non-covariant formulation.

 

[sweet springs]

So let's say I know the displacement current field and I want to calculate the B-field. Is it ok to use the Biot-Savart law but with delays of t = r/c ?

I thinks so. Please try it on your case and tell us whether it works.

 

[tade]

The displacement current would contribute to the B-field in a non-covariant formulation, one would need to add the displacement current to the non-displacement current to get the entire B-field in a non-covariant formulation.

The covariant formulation is something different - I tried to outline it earlier. I'm not quite sure why you keep asking about the covariant formulation, then asking questions that would apply to a non-covariant formulation.

hmm, yeah, I might've been mixing the co-variant and non-co-variant stuff.
Nevermind, let's just talk about accuracy. Otherwise I'll confuse myself

Understood. Still, is the method correct if we want to calculate the component of the B-field that's due to the $\partial_t \vec{E}$ field?

I wonder if this method gives relativistically accurate figures for that component of the B-field. Calculating for the perspective of my rest frame.

 

[tade]

I thinks so. Please try it on your case and tell us whether it works.

lol, the work for all the co-variant formulations was established decades ago

 

[sweet springs]

lol, the work for all the co-variant formulations was established decades ago

Formula in your post #1 includes distant source magnitude distance of which is r.
Familiar co-variant formulations are form of derivatives that includes local quantity only.
I am not sure both co-variant and non local formula are so familiar that you can find in general texts. Lienard-Wiechert treatment might be what you are seeking.
If you find any uneasiness in my post #9 that would contradict co-variance or TOR, please let me know.

 

[vanhees71]

Understood. Still, is the method correct if we want to calculate the component of the B-field that's due to the $\partial_t \vec{E}$ field?

This simply doesn't make any sense. The unique solution is given by the retarded fields!

 

[greswd]

This simply doesn't make any sense. The unique solution is given by the retarded fields!

I did mention some form of retardation

So let's say I know the displacement current field and I want to calculate the B-field. Is it ok to use the Biot-Savart law but with delays of t = r/c ?

 

[vanhees71]

For the potentials in Lorenz gauge the retarded solution is stated very simply as (in Heaviside-Lorentz natural units with $c=1$)
$$A^{\mu}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{j^{\mu}(t-|\vec{x}-\vec{x}'|,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
This simplifies to the Coulomb- and Biot-Savart Law if and only if the problem is stationary, i.e., if $j^{\mu}=(\rho,\vec{j})$ is time-independent.

For (one form of) the fields directly, see

https://en.wikipedia.org/wiki/Jefimenko's_equations

 

[greswd]

For the potentials in Lorenz gauge the retarded solution is stated very simply as (in Heaviside-Lorentz natural units with $c=1$)
$$A^{\mu}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{j^{\mu}(t-|\vec{x}-\vec{x}'|,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
This simplifies to the Coulomb- and Biot-Savart Law if and only if the problem is stationary, i.e., if $j^{\mu}=(\rho,\vec{j})$ is time-independent.

For (one form of) the fields directly, see

https://en.wikipedia.org/wiki/Jefimenko's_equations

Thanks; the B-field is just take the curl of $A^{\mu}$ right?

 

[tade]

For (one form of) the fields directly, see

https://en.wikipedia.org/wiki/Jefimenko's_equations

Jefimenko's equations , unlike Maxwell's, obscure the symmetrical relationship between the E and B fields.

Do you know of a formula for the Biot-Savart Law which includes the displacement current? Cos I'm not sure how to do ∂J/∂t for a moving point charge, thanks

 

[vanhees71]

Sigh... There is NO Biot-Savart law which includes the displacement current. The displacement current is no causal source of fields (or field components). There is no "symmetrical relationship between E and B fields" either. There's one field with components $\vec{E}$ and $\vec{B}$ with respect to a reference frame and the retarded propagator solving the field equations with given sources $J^{\mu}$ given in #15.

A moving point charge is described by its current density
$$J^{\mu}(t,\vec{x})=\int_{-\infty}^{\infty} \mathrm{d} \lambda q \frac{\mathrm{d} y^{\mu}(\lambda)}{\mathrm{d} \lambda} \delta^{(4)}[x-y(\lambda)],$$
where $y(\lambda)$ is the world line of the particle. Plugging this into the only correct formula in #15, using the retarded propagator, leads to the Lienard-Wiechert potentials and their derivatives to the corresponding em. field. You an of course as well use the Jefimenko equations to get these fields directly, but that's mathematically even more cumbersome than first calculating the potential.

 

[tade]

Sigh... There is NO Biot-Savart law which includes the displacement current.

I was just trying to work around ∂J/∂t for point charges.

There is no "symmetrical relationship between E and B fields" either.

With the exception of the minus sign, $∇×B=\frac{1}{c^2}\frac{∂E}{∂t}$ and $∇×E=-\frac{∂B}{∂t}$ have similar forms.

 

[sweet springs]

With the exception of the minus sign, ∇×B=1c2∂E∂t∇×B=1c2∂E∂t∇×B=\frac{1}{c^2}\frac{∂E}{∂t} and ∇×E=−∂B∂t∇×E=−∂B∂t∇×E=-\frac{∂B}{∂t} have similar forms.

The former can have source term of current j. The latter does not have source term. Not symmetric.
The former with current and
div \mathbf{E}=\frac{\rho}{\epsilon_0} form a pair.

The latter and
div \mathbf{B}=0 form a pair.

 

[tade]

The former can have source term of current j. The latter does not have source term. Not symmetric.

You can count the magnetic current $j_m$

https://en.wikipedia.org/wiki/Magnetic_monopole#In_Gaussian_cgs_units

It's the similar forms that led to the speculation of magnetic monopoles in the first place.

Anyway, Jefimenko's equations obscure this similarity.

 

[sweet springs]

Re:post #21 I find in wiki you refer saying that "With magnetic monopoles". Magnetic monopoles have not been found yet.

 

[tade]

re:post #21 I find in wiki you refer saying that "With magnetic monopoles". Monopole has not been found yet.

Yes, of course. The similarity exists regardless. A changing magnetic field can induce a circulating electric field and vice versa.

 

[vanhees71]

Well, this is an arbitrary split of Maxwell's equations into components. In manifest covariant form you just have the equations (in Heaviside-Lorentz units)
$$\partial_{\mu} F^{\mu \nu} = \frac{1}{c} j^{\nu}, \quad \partial_{\mu} ^{\dagger} F^{\mu \nu} = \frac{1}{2} \partial_{\mu} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$

Dirac made a point out of this when thinking about magnetic monopoles, i.e., the question what happens if you make the Maxwell equations symmetric under duality transformations, i.e., what if there's a fundamental magnetic charge in addition to the fundamental electric charge. Then there'd indeed be a kind of "duality symmetry", because there's a source term in the 2nd equation too, i.e.,
$$\partial_{\mu} ^{\dagger}F^{\mu \nu} =\frac{1}{c} j_{\text{mag}}^{\nu}.$$
So far no magnetic monopoles have been observed, but there's no big obstacle to formulate an extended electrodynamics with magnetic charge, and in the quantum context leads to the interesting possibility to explain why electric charge is quantized (while there's no reason for this observational fact within the Standard Model).

Nevertheless, nowhere in electrodynamics with or without magnetic monopoles does it make sense to think of the displacement current (or its dual in the symmetric theory) as a source term of field component.

It's a very interesting exercise, however, to follow this point of view and reformulate the solutions of Maxwell's equations in this sense. You'll see that this is formally possible but leads to quite complicated non-local interpretations of electromagnetic interactions which are at some tension with causality. It doesn't make sense to spoil the consistency of local relativistic field theory just to insist on the weird idea that the "displacement current" (based on an old-fashioned aether-theoretical mechanical model Maxwell originally used to derive his equations; Maxwell abandoned these models himself partially since he realized that it's much more complicated to think in terms of these models rather than just taking his equations themselves and thus the fields as the fundamental dynamical quantities; there was however still some confusion left concerning the question whether the fundamental quantities are the fields or the potentials since the gauge-theoretical basis of electromagnetism hasn't been fully understood yet, but that's another story).

 

[sweet springs]

Re:post #23 For your information I find in wiki https://en.wikipedia.org/wiki/Electromagnetic_tensor an explanation of what I said about the pairs.

 

[vanhees71]

Yes, of course. The similarity exists regardless. A changing magnetic field can induce a circulating electric field and vice versa.

As I said, just try to follow this idea mathematically. You'll see it's a mess compared to the very clear standard interpretation as the only causal sources of the em. field being the charge-current distributions (this also holds if you extent Maxwell's theory to speculative magnetic monopole charge-current distributions). For a thorough discussion, see

https://arxiv.org/abs/1103.3529
https://doi.org/10.1119/1.3533223

 

[tade]

As I said, just try to follow this idea mathematically. You'll see it's a mess compared to the very clear standard interpretation as the only causal sources of the em. field being the charge-current distributions

Actually, I have always held the Jefimenkoan view. The "symmetrical/similar" Maxwell equations can only lead to that. My earlier answer to sweet springs was just to get my point across to him, believe me (although this might remind one of Trump)

You'll see that this is formally possible but leads to quite complicated non-local interpretations of electromagnetic interactions which are at some tension with causality. It doesn't make sense to spoil the consistency of local relativistic field theory just to insist on the weird idea that the "displacement current" (based on an old-fashioned aether-theoretical mechanical model

Jefimenko's equations are partly derived from the "symmetrical/similar" Maxwell equations, and they can certainly be rearranged to explicitly show that "symmetrical/similar" form.

I don't think that would spoil anything. And sadly I lack the mathematical expertise to rearrange them myself.