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Has anyone experimentally verified that the measured values of the electric and magnetic fields do not change due to gravity, that there is no post-Newtonian effect on the strength of either $E$ or $B$?

I understand why in general relativity there should be no effect. The electromagnetic field strength tensor is a rank-2 antisymmetric tensor. Write it using covariant derivatives:

$$\begin{align*}F_{\mu \nu}&=\partial_\mu A_\nu - \Gamma^\sigma_{\mu \nu} A_\sigma - \partial_\nu A_\mu + \Gamma^\sigma_{\nu \mu} A_\sigma\\&=\partial_\mu A_\nu - \partial_\nu A_\mu\end{align*}$$ For a torsion-free connection, the connection is symmetric for the two lower indices. All terms from the metric drop out no matter if space-time is flat or curved.

I am interested in the experimental conformation because I work on an approach to physics that does not use tangent spaces, metric tensors, and connections. Instead the 4D manifold is given the power to multiply and divide other elements in the manifold instead of being passive anchor points for tangent spaces. My efforts indicate that the $E$ field still does not change its value at different heights in a gravitational field. For my proposal, the same cannot be said for the magnetic field which is predicted to change on the order of $B \rightarrow B'=B(1 + 4 G M/c^2 dh)$. A magnetic field quantifies electric charges in motion relative to an observer. Motion is always effected by gravity. It may be reasonable to hope to see the electric field unaltered, but the magnetic field changed by gravity. Such a test would be a deep technical challenge if done on Earth, needing a measurement of $B$ to one part in $10^{14}$ if done over a difference of a hundred meters on Earth.

  • Why do you think there might be such an effect if GR does not predict any? I don't think anyone is going to undertake a "deep technical challenge" just to confirm that no such effect exists. – sammy gerbil Jul 27 '16 at 01:58
  • It is very difficult to challenge, let alone prove, that Nature uses a certain type of mathematical machinery such as a Christoffel symbol of the second kind. Yet this looks like a chance to do so. Test of GR are always like this due to the crazy weakness of gravity. – sweetser Jul 27 '16 at 03:18
  • I wonder what will change in the two lines of your calculations if the Levi-Chivita connection and its components (Christoffel symbols) are replaced with ones that allow torsion instead of the presence of magnetic field? For more context take a look at this wild claim: https://physics.stackexchange.com/q/368540/ – Ziezi Nov 17 '17 at 14:19

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The question is ambiguous: it depends on what counts as "Post-Newtonian". Taken literally, anything involving electromagnetism is post-Newtonian, since Newton's speculations about electromagnetism (and, also, about a short-range attractive force binding atoms together, strong enough to overcome electrical repulsion) were confined to the Scholium section of the Principia.

Taken less literally, there is an ambiguity centering on the term "light speed". There are two senses of the term: the speed of electromagnetic radiation, which I'll label as $V = \sqrt{1/(με)}$, versus the finite non-zero speed of relativistic theories, which I'll label as $c$.

An electromagnetic field in a vacuum in the relativistic world has $V = c$. In a non-vacuum - both in relativity and in non-relativistic theory, $V$ can be anything else, except that relativity is normally assumed to also have the constraint $V ≤ c$. In a non-relativistic vacuum, no such finite non-zero value for $V$ exists.

This was one of the big mysteries Maxwell posted in the close of his treatise: what is $V$ being taken with respect to in such settings as the apparent (near-)vacuum of outer space? And what medium is giving $V$ its value? The equations he posed, consequently, were actually in two forms, not one: the "stationary equations" (which ultimately led to those we use today), in the reference frame for $V$ and the "moving equations" (which are now encapsulated by the Maxwell-Minkowski constitutive law, separately arrived at by Einstein & Laub in and by Minkowski, both in 1908).

To arrive at a "Newtonian limit", naively one would take $c → ∞$. But what about $V$? That's the ambiguity!

So, you have two senses of "Newtonian", if you allow electromagnetism to be counted as "Newtonian": one with $V = c → ∞$ and the other with $V < c → ∞$, where $V$ stays finite. The latter sense precludes any such thing as a vacuum; so actually you need to do something with a background medium.

In the following, in place of $c$, I will use $α = 1/c²$, so that the relevant limit becomes $α → 0$.

The Reissner–Nordström solution, for a non-rotating electrically charge spherical source, has a line element of the following form: $$\frac{dr^2}{1 + 2αU} + r^2\left(dθ^2 + (\sin θ dφ)^2\right) - c^2 (1 + 2αU) dt^2,$$ where the gravitational potential per unit mass, $U$, is given by $$U = -\frac{GM}{r} + \frac{μ_0}{4π} \frac{GQ^2}{r^2}.$$ The source has mass $M$, charge $Q$ and is located at $r = 0$, where the coordinates $(r,θ,φ)$ are spherical coordinates, and $t$ is the time.

There's your ambiguity. The $μ_0/(4π)$ part is normally presented as $$\frac{1}{4πε_0c^2} = \frac{α}{4πε_0}.$$ But, if the $c$ is actually $V$, then it is actually $$\frac{1}{4πε_0V^2} = \frac{μ_0}{4π}.$$

In the first sense of the solution, the non-relativistic limit $α → 0$ is $$U → -\frac{GM}{r}.$$ In the second sense of the solution, the non-relativistic limit is: $$U → -\frac{GM}{r} + \frac{μ_0}{4π} \frac{GQ^2}{r^2},$$ as indicated above, whose extra term speaks of an anti-gravity tidal force proportional to the square of the charge.

Is the extra term "Newtonian" or "post-Newtonian"? More interestingly: can it actually be measured? [A sound suddenly appears of experimentalists all over the world scrambling.] Meanwhile, something that might be worth looking at, though I haven't done so in any detail yet:

"Experimental Indications of Electro-Gravity"
https://arxiv.org/pdf/physics/0509068.pdf

NinjaDarth
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