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If we have Einstein's field equation $$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R=kT_{\mu\nu}$$ could we generalize it to $$R_{\mu\nu} - \frac{1}{2}m_{\mu\nu}R=kS_{\mu\nu}$$ where $S_{\mu\nu}$ is the source of the curvature and $m_{\mu\nu}=\eta_{\mu\nu}+f_{\mu\nu}$ where $f_{\mu\nu}$ is the perturbation caused by the force. Could we write one of these equations for each force, solve the equation for $f_{\mu\nu}$, sum up all of the $f_{\mu\nu}$ into one perturbation and add it to $\eta_{\mu\nu}$ to get one metric $g_{\mu\nu}$.

For example, since the electromagnetic field is extremely similar to the gravitational field, we could solve for the acceleration and find Poisson's equation. Using Einstein's "derivation" of general relativity one could find that

$$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R=\frac{2q}{m\epsilon_0 c^2}Q_{\mu\nu}$$

Where $$g_{\mu\nu}=\eta_{\mu\nu}+A_{\mu\nu}$$ $A_{\mu\nu}$ is the electromagnetic pertubation and $$Q_{00}=\rho_Q$$ $$Q_{ij}=\frac{1}{c}v^i v^j\rho_Q$$ $$Q_{i0}=\frac{\vec{J}}{c}$$ and $\rho_Q$ is the charge density

These are all the components since the tensor is symmetric.

With this you can derive Maxwell's equations. But is this a valid approach, to describing how electromagnetism curves space-time?

There is a related question https://physics.stackexchange.com/qu/148028/ but this is about describing the forces using a Yang-Mills approach i.e. a curvature in the connection. In this question I am excluding gauge curvature and only talking about space time curvature.

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    $\begingroup$ Does this answer your question? Can all fundamental forces be fictitious forces? $\endgroup$ Apr 13, 2020 at 17:15
  • $\begingroup$ The electromagnetic, weak, and strong forces cannot be described in this way. When you write equations that cannot be found anywhere in the literature, that's a personal theory. $\endgroup$
    – G. Smith
    Apr 13, 2020 at 17:36
  • $\begingroup$ @G.Smith Why is it no, that's why I asked the question, to find out why. Where can I found out why this doesn't work? $\endgroup$ Apr 13, 2020 at 18:19
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    $\begingroup$ Based on what evidence? Right now the SM doesn't completely work. $\endgroup$
    – user196418
    Apr 13, 2020 at 19:06
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    $\begingroup$ I'd say the answer is more, no one has done it successfully that I know of to date. But at the same time, the SM isn't completely correct and quantum gravity doesn't quite work yet. So basically its something we can't give a complete answer to because we don't know. But like I said, gravity doesn't play well with the other forces, so maybe its unique affect from curvature, is the reason it hasn't been formulated quantum-mechanically, in which case, the answer is no. Because the space-time curvature is unique to gravity and not the other forces. $\endgroup$ Apr 13, 2020 at 23:27

2 Answers 2

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There are some reasons why this will not work.

The obvious point is that we can write gravity as curvature because gravity is universal, i.e. gravitationaly interactions act in the same way on everything (roughly speaking). In the end, that's a reason why it makes sense to consider gravity as a property of spacetime itself.

For example, all free particles, that is, particles that only feel gravity, move on geodesics determined by the metric, so all particles "pervceive" the curvature. Furthermore, the energy-momentum tensor on the right-hand side of your first equation contains contributions from all fields and/or particles in your theory, charged an uncharged.

On the other hand, electromagnetism is not universal: The path of a particle in an electromagnetic field depends on the charge. Hence, you shouldn't have an energy-momentum tensor of uncharged matter influnce the electromagnetic force on a chraged particle.

Second, gravity is always attractive -- for example,there are no "repulsive" geodesics in Schwarzschild spacetime.

Third, upon quantisation, the electormagnetic (also weak, strong) force become spin-1 fields, while the metric is (presumably) described by a spin-2 field (graviton). Also, you would expect that there can be only one graviton field in a consistent theory.

Finally, note that there is a formulation of electordynamics in a manner very similar to gravity in a certain way via bundles -- here, the correspondence is between, e.g., the electromagnetic field strength and the Riemann tensor, not between photon and graviton field directly.

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    $\begingroup$ The equations for the E and Metric equatuon have q and m terms so the perceived curvature for each particle is different. Why cannot this be the case. This makes intuitive sense because light has no charge so it should not "perceive" the curvature. And with this equation, in the weak field limit, we can derive electromagnetism. So why does it matter if it is universal? $\endgroup$ Jun 18, 2020 at 12:16
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    $\begingroup$ It's not the energy momentum tensor, Qμν is the tensor describing the electromagnetic properties (e.g. charge and current). The pertubation is analogous to the gravitational pertibation, it's how much it differs from flat space due to the electromagnetic interaction. $\endgroup$ Jun 18, 2020 at 13:10
  • $\begingroup$ "Second, gravity is always attractive -- for example,there are no "repulsive" geodesics in Schwarzschild spacetime." Maybe experimentally correct, I think this is "technically" not true. Einstein Field Equation don't make any statement on attractive/repulsive by itself and a common example for repulsive gravity is the curvature from the vacuum-energy of a quantum field. $\endgroup$
    – hagebutte
    Jun 19, 2020 at 6:59
  • $\begingroup$ @hagebutte I think it's misleading to call a cosmolgical constant "repulsive gravity". But whatever the name, it is not an example of two bodies repulsing each other, as you have with like charges. $\endgroup$
    – Toffomat
    Jun 19, 2020 at 7:44
  • $\begingroup$ you can put it into into $\Lambda$ if you like. You don't have to and theres no reason you really have to. In the first place its on the Righthandside in $T$ and thats the statement. $\endgroup$
    – hagebutte
    Jun 19, 2020 at 8:15
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No, this doesn't work. There are several reasons - first, electric charge is a Lorentz scalar, which means that electric charge density transforms like the time component of a 4-vector, namely $J^\mu = (c\rho,\mathbf J)$. Contrast this with energy density; since energy is not a Lorentz scalar, energy density transforms like the time-time component of a rank 2 tensor (the stress-energy tensor on the right hand side of Einstein's equations).

Furthermore, electromagnetism is not universal in the same way that gravity is. The motion of a test mass in the presence of a gravitational field is independent of its mass, while the motion of a test charge in the presence of an electromagnetic field is not at all independent of its charge.


There have been more exotic attempts to unify electromagnetism and general relativity. The Kaluza-Klein theory postulates the existence of a compact $4^{th}$ spatial dimension; under suitable assumptions (in particular the assumption that $\partial_4g_{\mu\nu}=0$, the so-called cylindrical condition), one can show that this predicts the existence of two additional fields - one 4-component field $B_\mu$ and one scalar field $\Phi$. If the latter is ignored (i.e. set equal to a constant) and the compact extra dimension is integrated over, then the remaining $B_\mu$ field and the 4-metric $g_{\mu\nu}$ obey the laws of electromagnetism and general relativity, respectively.

This isn't good enough, however. By coupling this 5-dimensional version of GR to a Dirac field (like an electron field), we obtain a relationship between the electric charge and mass of the electron which is off by something like 30 orders of magnitude. Kaluza-Klein is a fairly elegant idea but it falls apart under closer inspection. Other theories of this type have been (and continue to be) investigated, but as yet all have been problematic.

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  • $\begingroup$ Would the electromagnetic "stress-charge" tensor not also be Lorentz invariant? $\endgroup$ Jun 19, 2020 at 10:00
  • $\begingroup$ Would electric charge not transform like the time-time component of the tensor? $\endgroup$ Jun 19, 2020 at 10:01
  • $\begingroup$ @JoshuaPasa No. As I said, charge density transforms like the time component of a 4-vector, not the time-time component of a tensor. If you are referring to the object you wrote down, it is not a tensor because it does not have the correct transformation properties. $\endgroup$
    – J. Murray
    Jun 19, 2020 at 10:54
  • $\begingroup$ @JoshuaPasa See this related answer of mine $\endgroup$
    – J. Murray
    Jun 19, 2020 at 11:01
  • $\begingroup$ Why does it not transform as a tensor? + Why does the energy not transform as the 0 component of the four momentum $\endgroup$ Jun 20, 2020 at 17:35

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