Does Non-Uniform Gravity Transform E into B Fields?

[Twigg]

Does a static uniform E field in space appear to be a mix of E and B fields to an observer on Earth (ignoring the Earth's motion)?

My GR is not so great, but would the rigorous way to calculate the new fields be to parallel transport the Maxwell 2-form $F_{\mu\nu}$ from free space, where it would only have non-zero values in the first row and top column, to the Earth frame by path integrating the covariant derivative obtained from the Schawrzschild metric? Am I making this more difficult than necessary? And is this a reasonable exercise for me to attempt? Thanks in advance!

 

[PeterDonis]

Does a static uniform E field in space appear to be a mix of E and B fields to an observer on Earth (ignoring the Earth's motion)?

First you need to be clearer about what a "static uniform E field" is. To me that means something like the field inside a capacitor; but how do you propose to put the Earth inside a capacitor? If you mean the static field due to a point charge, that field is not "uniform" in any sense of that word I'm aware of.

Assuming you mean a static field due to a point charge, the answer to your question is no; a static gravitational field does not make a static E field non-static. See further comments below.

would the rigorous way to calculate the new fields be to parallel transport the Maxwell 2-form $F_{\mu\nu}$ from free space, where it would only have non-zero values in the first row and top column, to the Earth frame by path integrating the covariant derivative obtained from the Schawrzschild metric?

No. The fully rigorous procedure would be to solve the combined Einstein-Maxwell equations with an appropriate assumption for the sources. But we already know that there is a static solution to these combined equations: the Reissner-Nordstrom metric (which is treated in all GR textbooks I'm aware of, and also can be easily found online). This is the metric for a charged black hole, not a charged particle in the field of the Earth, but it illustrates that you can have a static gravitational field and a static E field coexisting, without either one changing the other to non-static.

 

[Twigg]

Thanks for the reply! My apologies for being unclear. Two observers look at a test charge between the plates of a charged parallel plate capacitor floating in space far from the Earth. Observer A is near the capacitor and the capacitor's rest frame, while Observer B is located on Earth. See attached. My question is what kind of field does Observer B see if Observer A sees a uniform electric field between the plates? Let me know if there's anything that needs clarification.

No. The fully rigorous procedure would be to solve the combined Einstein-Maxwell equations with an appropriate assumption for the sources. But we already know that there is a static solution to these combined equations: the Reissner-Nordstrom metric (which is treated in all GR textbooks I'm aware of, and also can be easily found online). This is the metric for a charged black hole, not a charged particle in the field of the Earth, but it illustrates that you can have a static gravitational field and a static E field coexisting, without either one changing the other to non-static.

I'm confused why we need to use a new metric if the dominant source of gravity is the earth. Am I wrong to think that any curvature induced by the energy stored in the capacitor is negligible (given a non-ridiculously-large capacitance and voltage)?

 

[PeterDonis]

what kind of field does Observer B see if Observer A sees a uniform electric field between the plates?

A uniform electric field between the plates. The only thing that will change is that the energy stored in the field will seem larger to Observer B, because of gravitational blueshift.

I'm confused why we need to use a new metric if the dominant source of gravity is the earth.

I wasn't clear about the scenario you were describing. For the case of an ordinary sized capacitor, you are correct that the gravitational effect of the capacitor is negligible and we can just use the Schwarzschild metric.

 

[Twigg]

A uniform electric field between the plates. The only thing that will change is that the energy stored in the field will seem larger to Observer B, because of gravitational blueshift.

That is surprising but cool. Thanks! Sorry again for the poor explanation in the original post.

About the method I would use to derive this result, it seems as though the Einstein-Maxwell equations aren't necessary in this case on the same grounds that we don't need the Reissner-Nordstrom metric (the electromagnetic field due to the capacitor has negligible effect on the metric). Am I correct in that the only field equations needed are the (covariant) Maxwell equations? How would I calculate the field seen by Observer B? In my earlier post my thoughts were to take the covariant derivative (using the Schwarzschild metric to compute the Christoffel symbols) of the Maxwell 2-form and integrate $\nabla_{\sigma} F_{\mu\nu}$ along a parametrized world line from Observer A to Observer B. In light of the confusion with the problem statement, may I ask again if this is a valid approach?

 

[PeterDonis]

it seems as though the Einstein-Maxwell equations aren't necessary in this case

Yes, because the electromagnetic field is so localized that you can ignore spacetime curvature when describing it; you just have an ordinary capacitor in a small patch of spacetime that can be approximated as flat. So you just have the ordinary description of a capacitor without having to worry about how it interacts with gravity.

In my earlier post my thoughts were to take the covariant derivative (using the Schwarzschild metric to compute the Christoffel symbols) of the Maxwell 2-form and integrate $\nabla_{\sigma} F_{\mu\nu}$ along a parametrized world line from Observer A to Observer B. In light of the confusion with the problem statement, may I ask again if this is a valid approach?

No. The EM field is zero except in a small patch of spacetime where the capacitor and whatever is charging it (a circuit with a battery, for example) is located, and, as above, that small patch of spacetime can be treated as flat.

 

[Twigg]

No. The EM field is zero except in a small patch of spacetime where the capacitor and whatever is charging it (a circuit with a battery, for example) is located, and, as above, that small patch of spacetime can be treated as flat.

Just to make sure I understand, is what you're saying that both observers see the same electric field (as opposed to uniform electric fields with different magnitudes or directions)? I'm still not sure I see why this is. I agree that a small neighborhood around the capacitor and Observer A can be treated as flat space, and I agree that a small neighborhood around Observer B can be treated as flat space, but the spacetime between is curved. Aren't these two distinct local flat spaces connected by the curved spacetime described by the Schwarzschild solution? What makes them identical?

Thanks again!

 

[Nugatory]

What makes them identical?

They're both locally flat.

They are far enough apart that we will notice curvature effects (such as gravitational redshift/blueshift) if we move between them, but that has no bearing on the local physics within each patch - both patches are flat, both patches display the same locally flat physics.

One way of seeing this is to remember that the E and B fields at a point are defined by their action on a charged test particle: $\vec{F}=q(\vec{E}+\nabla\times\vec{B})$. How does a test particle between the capacitor plates behave according to an observer near the capacitor? Does it appear any different to the earth-bound observer?

 

[Twigg]

They're both locally flat.

They are far enough apart that we will notice curvature effects (such as gravitational redshift/blueshift) if we move between them, but that has no bearing on the local physics within each patch - both patches are flat, both patches display the same locally flat physics.

One way of seeing this is to remember that the E and B fields at a point are defined by their action on a charged test particle: $\vec{F}=q(\vec{E}+\nabla\times\vec{B})$. How does a test particle between the capacitor plates behave according to an observer near the capacitor? Does it appear any different to the earth-bound observer?

I agree with you that the same laws of physics (specifically Maxwell's equations) apply in each region (though we only need it in the neighborhood of the capacitor), but I would like to know if the curvature effects from moving between observers behave in a way comparable to a Lorentz transformation in turning an electric field into a combination of electric and magnetic fields. Also I would like to know how I would derive the transformation which would send the electric field seen by one observer to the field seen by the other.

 

[PeterDonis]

is what you're saying that both observers see the same electric field (as opposed to uniform electric fields with different magnitudes or directions)?

No. I said the energy stored in the capacitor will look larger to Observer B (who is much deeper in the gravity well than the capacitor and Observer A). That implies that the electric field will also look larger in magnitude, since the energy and the field magnitude are related. But there is no easy way for Observer B to directly measure the electric field inside the capacitor, which is why I focused on the energy storage, which is easier for Observer B to measure (for example by looking at the energy of light emitted by the capacitor as it discharges).

I would like to know if the curvature effects from moving between observers behave in a way comparable to a Lorentz transformation in turning an electric field into a combination of electric and magnetic fields.

I've already answered that: no.

I would like to know how I would derive the transformation which would send the electric field seen by one observer to the field seen by the other.

The easiest way is to transform the energy, which is just the gravitational redshift/blueshift formula, and then calculate what electric field magnitude would produce the new energy. (Basically, the energy is the field magnitude squared, with some constant factors that depend on your choice of units.)

Notice that this transformation has nothing at all to do with Maxwell's Equations in curved spacetime.