Physics Asked by Yuzuriha Inori on April 16, 2021
The Maxwell’s Equations are one of the most famous sets of equations physics have ever known. But just as different sets of equations are applicable to different frames of reference, where are Maxwell’s Equations applicable? To be more specific:
Are Maxwell’s Equations valid in only inertial frames of reference?
If not, then how can we reformulate the equations so that it is valid in any accelerating frame with an arbitrary acceleration?
As they are conventionally written Maxwell's equations are valid only in inertial frames of reference in flat spacetime. This is because the derivatives in the equation are not covariant derivatives and therefore don't apply when the coordinate system is curved.
It is possible to write Maxwell's equations in arbitrary coordinate systems though it gets somewhat complicated. The trick is to note that Einstein's equivalence principle tells us that acceleration is locally indistinguishable from gravity, and therefore the treatment of Maxwell's equations in accelerating frames is the same as formulating them in curved spacetime.
In principle all we need to do is replace all physical quantities by tensors, and replace normal derivatives by covariant derivatives. However the process of doing this makes the equations look very different. The details are described in the Wikipedia article Maxwell's equations in curved spacetime. Specifically note that the introduction to this article states:
The electromagnetic field also admits a coordinate-independent geometric description, and Maxwell's equations expressed in terms of these geometric objects are the same in any spacetime, curved or not. Also, the same modifications are made to the equations of flat Minkowski space when using local coordinates that are not Cartesian. For example, the equations in this article can be used to write Maxwell's equations in spherical coordinates.
So this approach is just as useful for curved (e.g. non-inertial) coordinates in flat spacetime as it is for curved spacetimes.
Answered by John Rennie on April 16, 2021
The best way to answer that is to use, as a point of reference, the Maxwell-Minkowski(-Einstein-Laub) (or MMEL for short), posed independently in 1908 by Einstein & Laub, and by Minkowski; and to express them using differential forms, since this is actually how Maxwell himself expressed them in his papers in the 1850's and 1860's and (to a large degree) in the treatise itself.
The MMEL form of Maxwell's equations are:
Field-Potential Equations (and resulting "Bianchi Identities"):
? = -∇φ − ∂?/∂t, ? = ∇×? ⇒ ∇·? = 0, ∇×? + ∂?/∂t = ?
Field Equations (and resulting "Continuity Equation"):
∇·? = ρ, ∇×? − ∂?/∂t = ? ⇒ ∇·? + ∂ρ/∂t = 0
Force and Power Density Laws:
? = ρ? + ?×?, P = ?·?
Constitutive Relations:
? + α?×? = ε(? + β?×?), ? − α?×? = μ(? − β?×?)
In Relativity αβ > 0, with β/α = c²; while in the non-relativistic version α = 0. In both cases β can be normalized to 1. The case β = 0, α ≠ 0 gives you the version of the equations that would be suitable for a "Carrollian" universe (one where c → 0). The Einstein-Laub and Minkowski 1908 treatments dealt only with the case αβ > 0, and both muddied their formulations by using different notations and conventions that needlessly and wrongly mixed in c's and 4π's with the various quantities and terms shown above.
Finally, there was also the:
Resistivity Law:
? − β?ρ = σ(? + β?×? − αβ??·?)/√(1 − αβG²).
which Einstein-Laub and Minkowski both also treated 1908. This assumes that the frame of "resistivity", where it reduces to ? = σ? is the same as the frame of isotropy, where the constitutive laws reduce to (? = ε?, ? = μ?). Both Minkowski and Einstein-Laub tacitly made that assumption - as did Maxwell. The non-relativistic form of this equation is:
? − ?ρ = σ(? + ?×?).
The constitutive relations are suitable for isotropic media. As written, they are frame dependent and select out a frame in which the relations become isotropic.
? = ε?, ? = μ?
The ? = ? case was referred to, in the 19th century, as the "stationary" form of Maxwell's equations. Maxwell used the vector ? and d?/dt more or less interchangeably. Even in the section in his treatise where he laid out his equations, he wrote down an equation once with ? and then elsewhere in the same section with d?/dt. He was not clear on what this "velocity" actually was, and his treatment was confused and inconsistent.
As you may verify, directly, by substitution, when the condition α = βεμ holds true, (and provided that αβG² ≠ 1) then the constitutive relations are equivalent to the isotropic form and the vector ? become superfluous.
The explicit expression for (?,?) in terms of (?,?) is:
? = ε? + K ?×(? - α?×?),
? = ?/μ + K ?×(? + β?×?),
where K = (βε - α/μ)/(1 - αβG²) and only applies if αβG² ≠ 1. Otherwise, there is a residual ?-dependence - even in Relativity (αβ > 0) and even in the limit (εμ → α/β) of a vacuum! I don't know if anyone else has ever noted this feature of the MMEL equations before, but it's there and might lead to some interesting experimental results, if a way can be found to test it.
The condition α = βεμ can never hold in non-relativistic theory since εμ > 0 for both the vacuum and ordinarily for isotropic media. Consequently, Maxwell and others in the 19th century were faced with a conundrum that they never resolved.
Part of the reason that the situation here was so muddied was because Maxwell made a mistake that was only much later corrected by Thomson: he failed to note the −?×? term and he mistakenly thought ? and ? transformed the same way under boosts (that is: not at all). Thomson added the extra term in the 1890's.
In addition, Maxwell put the ?×? term in with the field-potential relations, and actually wrote ?' = -∇φ − ∂?/∂t + ?×?, instead; which muddied the issue on the true nature of those equations. As a consequence, his version of the constitutive laws and resistivity law looked more like this:
? = ε?', ? = μ?, ? = σ?'
again, failing to include not just the −?×? but also the −?ρ term because of his flawed and confused analysis of the behavior of the various quantities under a Galilean boost.
In the form that Maxwell actually wrote the fields and their components, he used differential forms, writing them down as what we today would denote:
?·d?, ?·d?, ?·d?, ?·d?, ?·d?, ?·d?, φ
(the last one being the electric potential, written as a 0-form; and also tacitly, he had ρdV but I don't think he ever explicitly wrote it down as such.)
... where d? = (dx, dy, dz), d? = (dy∧dz, dz∧dx, dx∧dy), dV = dx∧dy∧dz.
Although he did not explicitly write the differential forms in Grassmannian form as actual wedge products, he did on at least one instance make use of the anti-symmetry law dx∧dy = -dy∧dx, in the treatise. Also, he generally expressed them as differential forms only when in the context as kernels to integrals, though sometimes wrote the differential forms as such in their own right.
(He also wrote the differential forms in vector form, as I just did here, on a few occasions, but with different notation; e.g. ?·d? rather than ?·d?.)
When they are written this way, the resulting objects - the differential forms themselves - are invariant under all transformations of the spatial coordinates and can be used for arbitrary coordinate frames in place of the (x,y,z) Cartesian grid.
In addition to the errors pointed out above, another error in Maxwell's treatment was his failure to study or note the behavior of these differential forms under boosts. He went to pains to note how components transformed, but never bothered to study the transform properties of the differential forms. Had he done so, he would have noticed that ?·d? actually mixes in with ?·d?∧dt; since under a Galilean boost d? → d? − ?dt, d? transforms as d? → d? + ?×d?∧dt. Consequently, he failed to notice that not only should the 1-forms ?·d?, ?·d? actually be written as 2-forms with the time coordinate ?·d?∧dt and ?·d?∧dt, but that also the electric potential should be written as a 1-form with dt as well
And finally, his biggest failure - brought large on by the 2 gaffes that he made that I pointed out above - is that he didn't note that the differential forms paired off as
A = ?·d? − φdt,
F = ?·d? + ?·d?∧dt,
G = ?·d? − ?·d?∧dt,
Q = ρdV − ?·d?∧dt.
The opportunity to make this discovery clearly presented itself, also, in the section where he did dimensional analysis on the various quantities. In that section, he failed to do any dimensional analyses on the differential forms themselves, instead focusing only on the components. Had he done so, he would have noticed that both A and F have the dimensions of "magnetic charge" (his [m] dimension), while G and Q have the dimensions of "electric charge" (his [e] dimension) while the dimensions for [φ] = [?·d?] = [m]/[dt], [?·d?] = [?·d?] = [e]/[dt] all clearly show the absence of the "dt".
The differential forms, themselves, are invariant under ALL coordinate transforms involving (x,y,z,t) and the equations
Field-Potential Equation (and resulting "Bianchi Identity"):
dA = F, dF = 0
Field Equation (and resulting "Continuity Equation"):
dG = Q, dQ = 0
are all invariant under ALL coordinate transforms and therefore apply to all frames of reference. They make no reference to the metrical or causal structure of the underlying chrono-geometry and are thus equally applicable to both the relativistic and non-relativistic forms. They would apply equally well, even if our 4 dimensions were all space-like and the universe were a 4+0 dimensional timeless space.
They're diffeomorphism-invariant.
The same also holds true for the force and power laws, though their expression as differential forms is a bit more subtle:
W ≡ ?·d? − P dt, W(Δ) ≡ (Δ ˩ W) dt∧dV = (?·Δ? − P Δt) dt∧dV
where the density W(Δ) is expressed as a function of a vector field written as
Δ ≡ Δ?·∇ + Δt ∂/∂t = Δx ∂/∂x + Δy ∂/∂y + Δz ∂/∂z + Δt ∂/∂t
In terms of this, the force and power laws may be written as
W(Δ) = Δ˩F ∧ Q
which is invariant under ALL diffeomorphisms. The contraction in component form works out to the following
Δ˩F = Δ˩(?·d? + ?·d?∧dt) = ?·(Δ?×d?) + ?·(Δ?dt − d?Δt)
or collecting terms:
Δ˩F = -(Δ?×? + Δt?)·d? + (Δ?·?)dt
and already in that you can see the shadow of the expressions used in the force and power laws.
Maxwell's treatment of the force law was completely mangled and did not age well or at all (and Maxwell himself never aged at all either, since he died early). He got the power law correct, and would have found the correct form of the force law had he simply applied the Galilean transforms to them, but missed the opportunity because of the errors and gaffes that he made, that I mentioned above.
In contrast, the constitutive laws and resistivity laws break diffeomorphism-invariance and even boost-invariance. Boost-invariance can only be salvaged for the constitutive laws in a relativistic vacuum and requires that α = βεμ, which in turn requires αβ > 0.
In the case of the constitutive laws, it can be written in the language of differential forms, but requires an additional operation which calls out both the metric (here that means: α and β), the constitutive coefficients (ε and μ) and - unless α = βεμ, the velocity vector ?. A ?-dependent, metrical, "duality" operator is required. It is independent of ? only for the in-vacuo relations in the Relativistic setting, but still has dependence on the metric.
The explicit expression for the constitutive law and required operator is:
G = ?[?] F,
?[?] ≡ ?[?]⁻¹ ∘ ? ∘ ?[?],
?(d?, d?∧dt) = (-1/μ d?∧dt, ε d?),
?[?](d?, d?∧dt) = (d? - β?×d?∧dt, d?∧dt + α?×d?).
Again: note that this requires αβG² ≠ 1, else ?[?] is singular and ?[?]⁻¹ is ill-defined. For the relativistic vacuum when αβG² ≠ 1, the ?-dependence is superfluous and the operator ?[?] reduces to ? and ?, in turn, reduces to a multiple of the duality operator for Minkowski geometry.
So, the answer to your question is yes and no; and the "yes" part includes "it depends on how you segregate the equations and what form you write them in."
Yes: Field-Potential Equations, Field Equations, if using differential forms
in which case: they make no reference to the causal or metric structure of the underlying chrono-geometry at all and apply equally across the board to all paradigms, both relativistic as well as non-relativistic.
Cosmetically Yes: Constitutive Laws
require an additional metrically-dependent (and even frame-dependent) operator.
Maybe Cosmetically Yes: Resistivity Laws
requires additional non-invariant objects or operators to be explicitly brought out.
As they are traditionally written (in the form I wrote them at the top) they make an explicit but mostly inessential reference to an inertial Cartesian frame. As Maxwell wrote them, they make an explicit (but mostly inessential) reference to an inertial frame suitable for general spatial coordinate grids in place of the (x,y,z) Cartesian grid.
Answered by NinjaDarth on April 16, 2021
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