# Are electrical fields affected by gravity?

1. Oct 12, 2008

### rcgldr

Are electrical fields affected by the space time curvature of gravity under GR?

2. Oct 12, 2008

### tim_lou

I was just reading this and this answer should be definitely yes! The electromagnetic differential forms involve the metric specifically. In fact, maxwell's equation becomes
$$\nabla_\beta F^{\alpha\beta}=J^\alpha$$
compare to the flat space equation:
$$\partial_\beta F^{\alpha\beta}=J^\alpha$$

The equation explicitly involves the covariant derivative, which involves terms that contain derivatives of the metric.

3. Oct 12, 2008

### Bible Thumper

It'd be freaky-weird if gravity didn't affect electric fields!

4. Oct 12, 2008

### atyy

Space time curvature is also affected by electrical fields.

5. Oct 12, 2008

### Bible Thumper

I thought the structure of space was determined by the presence of matter.

6. Oct 12, 2008

### tim_lou

matter field and the metric field couple together. Curvature is determined from matter, and matter flows through geodesics. The equations are coupled in a mess just like maxwell's equation + Lorentz force law.

7. Oct 12, 2008

### atyy

Newton: mass generates gravity.
Maxwell: electromagentic fields have energy.
Einstein I: energy has mass.

Hence Einstein II: energy, including electromagentic energy, generates gravity.

There are many sensible definitions of "energy" and "mass" in relativity, and one has to state exactly which is being used in any serious discussion - so take the above only as a heuristic!

8. Oct 12, 2008

### Phrak

Hi, tim_lou. Do you have the vacuum wave equation in F in covariant form at your disposal?

Last edited: Oct 12, 2008
9. Oct 13, 2008

### Naty1

The above posts seem rational yet in what I believe is a closely related thread here,
How does light slow in the presence of gravity?,
I seemed to have gotten a rather different set of explanations. Can anyone reconcile the very different apparent responses? In the referenced thread I have so far interpreted the view to be: lightspeed isn't affected by gravity, only by the reference frames used...light appears to slow only when curvature of space is significant....non local reference frames and cosmic distances....

I would have thought here in this thread, in local reference frames, the curvature of gravity would normally be so minor so gravity would have virtually no effect on EM fields...

(As you can tell, I am still stumped....)

Here atyy posts :
I hadn't thought about that but it,too, seems rational....How does this come about and what's the effect? Does this mean we can "turn gravity off" via an electric field...likely not, of course, but its a natural question given the post statement...

10. Oct 13, 2008

### Phrak

Yes, the effect due to gravity is small. A beam of light, or an electromagnetic wave, is deflected 1.75 arc seconds in a grazing path with the sun. Effects in Earth's gravity are substantially less.

If the ultimate question is about arcing radio signals over the Earth's horizon, then the gravitational effects are negligible.

Last edited: Oct 13, 2008
11. Oct 13, 2008

### George Jones

Staff Emeritus
Electromagnetism couples to gravity and spacetime curvature through Einstein's equation, $G = 8 \pi T$, where $G$,the Einstein tensor, is a geometric quantity and $T$, to which electromagneticfields contribute, is the energy-momentum tensor.

12. Oct 13, 2008

### tim_lou

Can't really cough it up just on top of my head...

wouldn't it just be solving for F using
$$\nabla_\beta F^{\alpha\beta}=0$$
$$G^{\mu\nu}=8\pi GT^{\mu\nu}=8\pi G \left(-\frac{1}{4}F^{\mu}\,\!_{\alpha}F^{\alpha\nu} + g^{\mu\nu} F^{\alpha\beta}F_{\alpha\beta}\right)$$

13. Oct 15, 2008

### Naty1

Essentially the gravitational field accelerates an electromagnetic field orthogonally to the direction of electromagnetic (EM)propagation...hence the EM curves due to warped space but the speed remains c in appropriate reference frames. Shortly after relativity was proposed the experimentally observed bending of light rays around the sun by Eddington confirmed Einstein's views and dashed other theories which did not predict such bending.

14. Oct 15, 2008

### Phrak

Dunno. It could be conservation of charge, for all I know. It would be nice to see all the various tensors all layed out in covariant form. When it comes down to it, all of Maxwell's equations on curved spacetime, are about the behavior of a single vector field, and it's derivatives on a pseudo Riemann manifolf having Lorentz metric. The vector being the 4 vector potential.

In differential forms, the vacuum wave equation isn't simple to derive, though simply stated, d*F=0, where * is the Hodge star operator, effectively the antisymmetric tensor.

Unfortunately, the differential form masks the connection coefficients, which is what is needed here, to show that covariant form is different from the noncovariant form.

It should be sufficient to show that the covariant and noncovariant forms of the Poynting vector (BxE) are not equal in curved spacetime. But, then again, the problem being, one would need the covariant form of the Poynting vector to start with!

Last edited: Oct 15, 2008
15. Oct 16, 2008

### Naty1

From THE RIDDLE OF GRAVITATION by Peter Bergmann:

16. Oct 17, 2008

### atyy

17. Oct 17, 2008

### Naty1

I think the simple answer might be: all (force) fields are affected by gravity.... strong,weak,EM......(as are energy,mass,pressure....)
is this correct???..I'm not sure about the nuclear forces...