# Electromagnetic spacetime curvature effects

1. Dec 2, 2008

### Naty1

I believe it was correctly concluded "yes" because an electromagnetic (EM) field has energy and pressure so it curves spacetime; it does have a gravitational effect. Although it was not discussed, this effect is much, much smaller than the electromagnetic effects because we know gravity is really weak under most conditions.

Ok, so along comes an EM field and bends space. We would normally think that because there is only an electromagnetic field present, that only charged particles would be deflected by the normal interactions.

But if spacetime is curved, then ALL particles, charged or not, should be also be deflected since spacetime is curved. I never heard of such a thing! Maxwell's equations don't say anything about this, do they? Comments??? Corrections?

2. Dec 2, 2008

### StatusX

That's true, but the deflection would be tiny in most circumstances. For an example where it's not, think about the Reissner-Nordström solution for a charged black hole. The spacetime curvature is qualitatively different than that of a Schwarzchild black hole, and so an observer falling in would notice the difference due to the field even if he was not charged. It's the gravitational field associated to the EM field, not the EM field itself, that's affecting him.

3. Dec 3, 2008

### PhilDSP

It might be worth keeping in mind that the meaning of "uncharged particle" is that the total charge within it is zero. It doesn't follow that there are no charges or subcharges inside. There may even be something which is more primitive in EM than what we consider positive and negative charge.

For example, an uncharged particle (considered in classic terms), the photon, can deflect another photon. That phenomenon is called Delbruck Scattering or Scattering of Light by Light.

4. Dec 3, 2008

### Naty1

Well if the idea I posted seems rational, how does the effect of EM curvature of space time get calculated?

I looked at Wikipedia

http://en.wikipedia.org/wiki/Electromagnetic_tensor

where the EM tensor formulation is discussed.....is there a gravitational component buried in the EM tensor??? I don't know what all the symbols represent...it would be nice if Wiki sometimes explained in words a bit more...

Last edited: Dec 3, 2008
5. Dec 3, 2008

### StatusX

The EM stress energy tensor is one of the contributions to the total stress energy tensor that appears on one side of Einstein's equation:

$$G_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$

For example, if the only matter is a point charge of charge q and mass m, the stress energy tensor consists of a delta function at the particle due to its mass, plus the stress energy of its field, which is smeared out over space. In this case, since there is no magnetic field, the total stress energy tensor would be:

$${T}_{\mu \nu} = \frac{1}{2} \epsilon_0 \left( \begin{array}{cccc} \frac{2}{\epsilon_0} m \delta^3(x) + E^2 & 0 & 0 & 0 \\ 0 & {E_x}^2 - {E_y}^2 - {E_z}^2 & 2 E_x E_y & 2 E_x E_z \\ 0 & 2 E_x E_y & {E_y}^2 - {E_x}^2 - {E_z}^2 & 2 E_y E_z \\ 0 & 2 E_x E_z & 2 E_y E_z & {E_z}^2 - {E_x}^2 - {E_y}^2 \end{array} \right)$$

where:

$$\vec E(x,y,z) = \frac{q}{4 \pi \epsilon_0 r^2} \hat r$$

So if you can solve for a metric whose Einstein tensor is equal to $8 \pi G/c^4$ times this matrix, you've found the spacetime curvature due to a point charge. But this is just the Reissner Nordstrom solution I mentioned before. (You need to be a little careful, because the $\vec E$ I gave is the solution in flat spacetime, not curved spacetime. But this is the basic idea, you just need to make sure your final field also satisfies Maxwells equations in your final spacetime)

Last edited: Dec 3, 2008