# Could there be equivalence between anisotropic space and the presence of a field? [duplicate]

What if we are so used to the curvature of space caused by mass and the range of its effects that we totally ignore the possibility of the existence of "opposite" curvature1, i.e. objects that bend space opposite to mass and cause repulsion by creating local space anisotropy?

What if we are so used to space curvature caused by mass that we invent a force that has no charge to explain possible effects of opposite curvature (repelling force - diamagnetism)?

What if the oscillating electric field has the ability to cause tiny local (anisotropy) fluctuations in space and its effects on objects are interpreted as "magnetic field"?

Could magnets be just an example of objects with anomalous gravitational fields due to their ability to distort the isotropy of space?

I guess my question boils down to:

How do we know that the force at the poles of a magnet is not gravitational (large but local curvature caused by anisotropy) with opposite signs, rather than what we call magnetic?

Can space curvature arise from something else different than mass?

What would change on the r.h.s of the Einstein's equation (components of stress-energy tensor) if we assume a connection with torsion?

1. Same size of the volume element $$dV$$, but with stretch in one element, say $$dx$$ and proportional compression in the other two $$dy$$, $$dz$$.

• @dmckee Apologies for wasting your time (sincere!), but isn't your comment a prove of the first paragraph? I don't deny that there is an area of study, but possible failure of recognition of existence in real life, as existence under false description. – Ziezi Nov 13 '17 at 3:52
• Can space-time curvature arise from something else different than mass [or energy or pressure]?" Yes, it is possible and yes it can be the "opposite" curvature. However, it has nothing to do with magnets or magnetic charges or electromagnetism at all. So the "curvature" part of your question is not necessarily nonsense. However, the part of magnets causing the opposite curvature does seem to be without a good ground. – safesphere Nov 13 '17 at 6:31
• Related: physics.stackexchange.com/q/4784/2451 , physics.stackexchange.com/q/261246/2451 and links therein. – Qmechanic Nov 13 '17 at 9:42
• @Ziezi You changed your question without changing the content. This is a strong indicator of you not understanding what you are saying. – BB681 Nov 15 '17 at 18:12
• @Ziezi The volume you are talking about is not dxdydz as you are talking about coupling to spacetime, where is your time in the volume element you provide? An oscillating electric field is most definitely not caused by fluctuations of space-time. A variation in charge volume is not a variation in space-time. You should start from the beginning instead of jumping to this stuff. We've all done it, but get the basics down first. – BB681 Nov 16 '17 at 17:39

Here is one such construction: Take the action $$S=\int_M d^dx \left\{\frac{1}{2q_m^2}B_\mu B^\mu +\frac{1}{2}\mathcal{D}_\mu \phi\mathcal{D}^\mu\phi \right\}$$
where the magnetic field is $B_\rho = -\frac{1}{2}\epsilon_{\rho\sigma\lambda}G^{\sigma\lambda}$ with G being the field strength tensor. The field equation with appropriate bounds for this action implies the existence of a monopole. If you do the math, you will find that the magnetic charge is $\frac{4\pi}{q_m}$.