# Gravitational field from virtual photons; does $\rho c^2 = \frac{1}{2} \epsilon_0 |E|^2$?

Answers to Is energy localised in space? got me thinking about answer(s) to Sun constantly converts mass into energy, will this cause its gravity to decrease? and the contribution to the Sun's gravity by the energy of the real photons slowly making their way out of the Sun (small but interesting).

As a gedankenexperiment, I make a strong electric field $$E$$ (of virtual photons) of a particular shape by bringing two or more charges close together, and then map the apparent gravitational field (distortion of space-time) around it and then calculate a hypothetical "mass distribution" $$\rho$$ that would produce such a field.

Question: Would the calculated "mass density" reflect the actual energy density as $$\rho c^2 = \frac{1}{2} \epsilon_0 |E|^2?$$

• virtual photons, photons with gravity, electric field from photons, gravity from space time distortion, an equivalent $\rho$ as the source of gravity-a lot of these terms juxtaposed don't gel into a coherent whole Commented Apr 16, 2020 at 7:54
• @lineage I think that it's clear enough to be answered.
– uhoh
Commented Apr 16, 2020 at 7:56
• i get that, just an aside Commented Apr 16, 2020 at 7:56
• Imo, you do not have to go through the virtual photon representation of static fields. General Relativity is a classical theory and I think just the energy in the electric field is enough to make an energy-momentum tensor contribution. Commented Apr 16, 2020 at 8:03
• the equivalent stress energy tensor, when compared to that of the em field would indeed have its $00$ component as the relation you mention. Commented Apr 16, 2020 at 8:08

## 2 Answers

Answer: yes, so long as you use the word "reflect" broadly enough.

Einstein's equation says that the curvature of the metric is related to the stress-energy tensor by $$G_{\mu \nu} = 8 \pi G T_{\mu \nu},$$ where $$T_{\mu \nu}$$ is the stress-energy tensor and $$G$$ is Newton's constant. For an electromagnetic field in flat (or weakly curved) spacetime, the stress-energy tensor is $$T_{\mu \nu} = \frac{1}{2} \epsilon_0 \begin{Bmatrix} |E|^2 & 0 & 0 & 0 \\ 0 & E_y^2 + E_z^2 - E_x^2 & 0 & 0 \\ 0 & 0 & E_z^2 + E_x^2 - E_y^2 & 0 \\ 0 & 0 & 0 & E_x^2 + E_y^2 - E_z^2 \end{Bmatrix}$$ From top to bottom, these components represent the energy density, followed by the pressures normal to surfaces in the $$x$$-direction, $$y$$-direction, and $$z$$-direction respectively.

The important point is this: the spacetime metric doesn't just respond to energy density, it also responds to pressures and fluxes. For most conventional matter, the energy density is much much much much bigger than any of the other components of the stress-energy density, and so we just focus on the energy density and ignore the rest of the stress-energy tensor as negligible. But for an electromagnetic field, all of these components are the same order of magnitude as each other, and so we will have to take them all into account.

Or, to put it another way: Suppose you created your very strong electric field. Suppose I then measured the electromagnetic energy density $$\frac{1}{2} \epsilon_0 |E|^2$$ at all points in space, and painstakingly put dust around a pair of uncharged spheres so the dust had exactly this energy density everywhere. Since my dust would have negligible amounts of pressure, you and I would have different stress-energy tensors, and we would end up with different spacetime metrics.

Generally, the interplay of general relativity and electromagnetism is known as gravitoelectromagnetism. I thought I had looked this up one before and found some good information, but skimming through the top couple google results seems to only turn up confusion now.

What I would like to do here is comment on how GR and E&M will generically interact as classical field theories. In particular, there is no need to think about "virtual" particles of any kind. Virtual particles themselves are nothing more than a suggestive name given to a particular feature of a graphical representation of a Taylor series (Feynman diagrams).

Now, the good starting place for writing down a field theory which has some pre-determined set of symmetries is to think about what the Lagrangian should be. Here we need a gauge invariant, coordinate invariant action. I'll remind you that the Einstein-Hilbert action whose variation produces the Einstein field equations is $$S_{EH}=\frac{1}{8\pi G}\int d^4x\sqrt{-g}R$$ where I am using units $$c=1$$, $$g$$ is the determinant of the metric tensor, and $$R$$ is the Ricci scalar. For general information on this action, I recommend Carroll's GR book. In Minkowski space with metric $$\eta_{\mu\nu}$$, the Maxwell action takes the form $$S_M=\int d^4x\left(-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}\right)=\int d^4x\sqrt{-\eta}\left(-\frac{1}{4}\eta^{\mu\alpha}\eta^{\nu\beta}F_{\mu\nu}F_{\alpha\beta}\right)$$ where $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$ is the field strength. In the second equality I have explicitly put in the Minkowski metric factors (note $$\eta=\det(\eta)=-1$$ for Minkowski space). It may also be helpful to look up what the components of $$F_{\mu\nu}$$ are in terms of the $$E$$ and $$B$$ fields. Briefly, the $$F_{0i}$$ components are essentially the components $$E_i$$ of the electric field (up to minus signs) while the $$F_{ij}$$ components (for $$i,j=1,2,3$$) give the components of the magnetic field (up to minus signs).

My claim will be that the correct way to put these two actions together will be to leave the Einstein-Hilbert action as it is and change the $$\eta_{\mu\nu}$$'s in the Maxwell action to general metric factors. It may seem a little contrived to stick in the $$\sqrt{-\eta}$$ factor, but this is actually needed to make sure the action is invariant under coordinate transformations at the end of the day.

The full action for GR coupled to E&M would then be $$S=\int d^4x\sqrt{-g}\left(R-\frac{1}{4}g^{\mu\alpha}g^{\nu\beta}F_{\mu\nu}F_{\alpha\beta}\right).$$ From here, we can think about the equations of motion. There are two fields to vary here, the metric and the vector potential. Varying the metric, we find that the contribution due to the Einstein-Hilbert term gives the usual Einstein tensor, so in the end we find exactly Einstein's field equation, $$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=8\pi GT_{\mu\nu}$$ where $$T_{\mu\nu}$$ is the energy-momentum tensor, but now because we have computed this from an action, we can also compute exactly the energy momentum tensor by the variation with respect to the metric hitting the Maxwell term: $$T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu\nu}}.$$ Now, it happens that I computed this the other day and the result turns out to be $$T_{\mu\nu}=g^{\alpha\beta}F_{\alpha\mu}F_{\beta\nu}-\frac{1}{4}g_{\mu\nu}F^{\mu\nu}F_{\mu\nu}.$$

Now, looking at this stress tensor may be a little hard on the eyes, and I think it might be hard to do much analysis on this in the current form. What we can do, however, is think about the special case where the metric is actually the Minkowski metric, just to get a feel for what the components of this tensor should look like. Rather than write it out, I'll refer you to this wiki page, where someone has very helpfully written out the components of this tensor in terms of the Poynting vector and electromagnetic fields.

A very interesting consequence we can note right away is that, because the Poynting vector tells us about the angular momentum in the electromagnetic fields, the Einstein equation tells us that not only does the energy density of the electromagnetic fields force the metric to change, but the linear and angular momentum also change the metric. I always find this to be an absolutely wild and completely underrated effect of general relativity which goes by the name frame-dragging.

There's one last thing I'd like to point out in all of this: not only does the electromagnetic field influence the metric, but the metric also influences the electromagnetic field in the sense that the components of the metric will also appear in the equations of motion for the electric and magnetic fields. This can be seen by varying our action with respect to the vector potential. Clearly the Einstein-Hilbert term won't contribute, but we see that there are still metric factors in the Maxwell term now, so those will follow us into the equations of motion. Hence the influence goes both ways.