# Is there measureable spacetime distortion from EM fields

1. Jul 10, 2015

### bwana

Please forgive me if I am posting in the wrong forum. Also the board limits the length of the title. I wanted to ask about Effect of high intensity electric and magnetic fields on spacetime.
To begin, 15 years ago this was published:

http://www.scientificamerican.com/article/do-electric-charges-and-m/

They refer to the "Reissner-Nordstrom" solution to Einstein's gravitational field equations. Basically what gravity would do near the surface of a very strong VanDe Graaf generator. They also mention the solution describing the special case in which the net electric charge is zero is the famous "Schwarzschild solution" to the gravitational field equations.

and in this forum 11 years ago we read:

But I am more interested in what progress has been made in measuring the effect of a strong electric or magnetic field on gravity (and hence spacetime). Is the prediction that it would increase gravity? or decrease it? or change the curvature of space so that gravity changes differently than the square of the radius between two masses?

Physical examples are not cited as they would be hard to find. But arent quasars a logical source of high intensity EM emissions? So the obvious question is:
Does light bend around a quasar differently than it bends around an equivalently massive but non radiating object?

2. Jul 10, 2015

### Staff: Mentor

It depends on what you mean by "gravity" and how you do the comparison.

If you mean, what proper acceleration would a neutral object need to have in order to "hover" over a charged black hole at some finite radius $r$, compared to a neutral black hole of the same mass (meaning the same $M$ appearing in the metric), it would be smaller. The reason is that the "mass" $M$ that appears in the metric for a charged black hole is really the "mass measured at infinity". Since $M$ is, strictly speaking, measured at infinity, it includes the equivalent mass of all the energy stored in the electric field of the hole. At any finite radius, however, some of the energy stored in the electric field is above you, not below you, and only the energy below you contributes to the effective "mass" that you feel, which is what determines the proper acceleration you need to hover.

In short, the "vacuum" region around a charged black hole, unlike that around a neutral (Schwarzschild) black hole, is not really vacuum; it contains energy stored in the electric field. That makes a difference in what is observed.

The Reissner-Nordstrom solution describes a static, charged black hole; it doesn't describe a black hole with EM radiation around it. (The hole in the R-N solution has a static electric field, but no magnetic field.) So this question requires a different model from the one we used above. The usual model for a quasar is a spinning black hole which may or may not be charged (usually it's assumed it isn't), but which has charged plasma around it that gets spun up by the hole's spin. The spinning of the plasma is what creates the magnetic field and the EM radiation.

As for light bending around a quasar, first you would have to try to separate out the effect of the plasma on light. Supposing you could do that, you would find that the remaining light bending would be the same as for any other spinning black hole with the same mass and angular momentum (assuming the hole itself was not charged).

3. Jul 10, 2015

### bwana

thank you. I appreciate your insights from first principles. Getting back to your first answer, I was hoping if we could 'flesh it out' a bit. You said:

>>If you mean, what proper acceleration would a neutral object need to have in order to "hover" over a charged black hole at some finite radius r, compared to a neutral black hole of the same mass (meaning the same M appearing in the metric), it would be smaller.<<

Does this imply if that the force of gravity in this example would decrease faster than r2?

4. Jul 10, 2015

### Staff: Mentor

Gravity is not a force in GR. That's why I specified the proper acceleration required to "hover". In Newtonian terms, the proper acceleration of a "hovering" object is numerically the same as what we normally think of as the "acceleration due to gravity" (the acceleration that a free-falling object will have if it is dropped from rest). But in GR, even for an uncharged black hole (in fact, even for an ordinary gravitating object), the proper acceleration does not decrease quite as r^2. The formula for proper acceleration for an uncharged black hole is:

$$a = \frac{M}{r^2} \frac{1}{\sqrt{1 - \frac{2M}{r}}}$$

Note the extra factor in the denominator (and note also that I am using "geometric units", in which $G = c = 1$). For a charged black hole, this formula changes to

$$a = \left( \frac{M}{r^2} - \frac{Q^2}{r^3} \right) \frac{1}{\sqrt{1 - \frac{2M}{r} + \frac{Q^2}{r^2}}}$$

where $Q$ is the charge of the black hole in "geometric units".

Last edited: Jul 10, 2015
5. Jul 11, 2015

### bcrowell

Staff Emeritus
Gravity in the early universe was dominated by electromagnetic fields, not by matter: https://en.wikipedia.org/wiki/Radiation-dominated_era . This is well tested empirically, e.g., by testing models of big bang nucleosynthesis.

6. Jul 11, 2015

### bwana

forgive my sloppy statement- yes i realize that gravity is not a force. But I also thank you for that wonderful equation. As I interpret it, it seems that for a very large Q possessed by a central large black hole in a galaxy, then the gravitational field of the central large hole may decrease more slowly than r2. In such a galaxy, the outer solar systems would be revolving around their 'charged black hole' faster than a system at an equivalent distance in our galaxy (which has a presumably has a neutral black hole). The galaxy rotation problem observed by Vera rubin and others may reflect this.

But obviously I am missing something because the great minds of physics would have thought of this instead of dreaming up dark matter. What is the argument against charged black holes as a cause of the galaxy rotation problem?

Still the idea that gravity can be modulated by Q is something not often discussed in physics.

7. Jul 11, 2015

### Staff: Mentor

Why do you think black holes in the centers of galaxies have a large Q? Our best evidence is that they are most likely rotating, but not charged. Any EM phenomena are due to plasma in the region around the hole, not any charge on the hole itself.

You're looking at it backwards. What the equations I posted show is that, if we have two black holes with the same $M$ (i.e., mass measured at infinity), and we make measurements at the same $r$ with respect to both of them, the one with a nonzero $Q$ will have less "acceleration due to gravity". Both of the terms in $Q$ act to reduce the acceleration at a given $r$ for the same $M$.

No, they would be revolving slower. See above.

No, it can't. See above.

8. Jul 12, 2015

### bwana

thank you
I don't think that all galaxies have such a charged hole, just that some of them might and that we might be able to detect that as an 'aberration' of the gravitational law in those systems.

I was imagining that charge imbalance might have occurred during galaxy formation. Some might have a significant net positive Q, others an equally negative Q. All the solar systems in one of those galaxies might therefore have net positive Q for example. And the gravitational law for people evolving in that system would be different. Of course they would not know that they have net positive Q, just as most fish do not know they are in water.

Of course there is a technical problem here. We do not have an independent way of measuring the mass of a celestial body. We infer the mass of a celestial body from its gravitational effect on nearby bodies. Here on earth, besides weighing an object to infer the mass (since we know the acceleration), we can also measure a body's inertia and arrive at an independent calculation of a body's mass. Q does not enter into the calculation for acceleration due to motion so we get a 'pure' calculation of mass. (I love that equation you showed me) So, The two calculated masses (gravitational and inertial)are the same here on earth. In a 'charged' galaxy, they might not.

Yes That's what I thought. But I did get my orbital mechanics backward. Thanks again for catching me. Similar to the problem of a spaceship orbiting the moon or earth at the same altitude. The one above the moon goes slower. I really should plot that nice equation you gave me before I open my mouth again.

9. Jul 12, 2015

### Staff: Mentor

It's possible, of course, but we haven't detected any evidence for it.

If you mean the "acceleration due to motion" of a black hole, we have no way of making such a calculation, because a black hole is not an "object". It's pure spacetime curvature.

If you mean the "acceleration due to motion" of an object moving in the gravitational field of a black hole, the hole's charge certainly does affect that motion; that's obvious from the formulas I posted earlier.

10. Jul 12, 2015

### bwana

well i tried to graph some of this stuff on mathematica
here is a simple a= M/r^2
http://www.wolframalpha.com/input/?i=graph+a=((M/r^2)+)where++r=0+to+10+,+M=100
here is a= M/r^2 (1/ (1-2*M/r)^.5)
http://www.wolframalpha.com/input/?i=graph+a=((M/r^2)+)*(1/(1-2*M/r)^.5)+where++r=0+to+10+,+M=100
In the first case , acceleration decreases with radius squared as expected.
In the second case, acceleration remains zero - the real part happens to overly the x axis.
I guess this is a problem with how wolfram draws the graph.
I cannot even get it to graph the third equation with Q.
http://www.wolframalpha.com/input/?...+(Q^2/r^2))^.5)+where+r=0+to+10,+M=100,+Q=100

11. Jul 12, 2015

### Staff: Mentor

That's because you're graphing it inside the horizon. To get a nonzero real part, you need to have a minimum value of $r$ that is greater than $2M$. For $M = 100$, try $r = 201$ to $r = 1000$ or something like that.

Same issue here, plus to have an ordinary event horizon, you need $Q < M$; for example, try $Q = 50$ with $M = 100$. Here the event horizon is at $r = M + \sqrt{M^2 - Q^2}$, so the minimum value of $r$ that you graph should be a little larger than that.

12. Jul 12, 2015

### bwana

:sheepishly rolls eyes

WTF is the matter w me? plotting w/o thinking about ranges.

Score another point for using pencil and paper. I wouldn't have made that mistake if I wrote those eqs down.

13. Jul 12, 2015

### bwana

so here is the completed plot for
M=100, Q=100, x=201 to 1000
http://i.imgur.com/CmOi2g0.png
the first equation is represented by the blue '+'
the second equation is plotted in red dots
the third equation is the solid orange line

Interestingly when Q = 100 is added to the mix (the orange line), acceleration of gravity decreases very slowly with distance rom the hole.
This is consistent with orbital velocities of solar systems not changing much as distance from the charged black hole increases.

14. Jul 12, 2015

### Staff: Mentor

$Q = M$ is an edge case; you should try $Q < M$ (for example, as I suggested before, try $Q = 50$).

But it's always smaller than for the $Q = 0$ case.

But it's not consistent with the observed velocities being larger than what would be estimated from the visible mass.

15. Jul 12, 2015

### bwana

ok, here are more graphs that are actually pretty. M=100
first graph is two equations. a=M/r^2 in green and also a=M/r^2 * 1/(1-2M/r)^.5 in blue
http://i.imgur.com/w4n1C30.png
second graph is the complicated equation that includes Q for Q = 20,50,100,200
http://i.imgur.com/grpXVML.png
here is a blow up that also includes Q=101
http://i.imgur.com/LUeh5yK.png
Notice that as Q increases, acceleration becomes weaker at a given distance

fourth graph is everything.
http://i.imgur.com/M9YeTS7.png
Notice that when Q=M=100 the curve is identical to the simple a=M/r^2

Now i have to translate this stuff to cgs units and plug in some real numbers to see what is real. Thanks again.

16. Jul 12, 2015

### Staff: Mentor

I don't see the square root in the second equation here (or in the fourth graph including everything).

You must be mis-labeling or mis-identifying a curve. If you plug $Q = M$ into the equation I posted earlier, you will see that the factor under the square root sign blows up at $r = M$, which corresponds to $x = 100$ in your graph. None of the curves I see graphed do that.

Last edited: Jul 13, 2015
17. Jul 13, 2015

### bwana

well after some research, it seems that the Grapher application on a mac behaves weirdly. look at these three graphs for Q= 9999,10000,10001
http://i.imgur.com/KBNDSFa.png
I removed the abscissa so the weirdness can be clearly seen. But the action centers at x=100. Obviously the code runs into rounding errors. The reason that it 'does not blow up' when it should on the graph is a software error. And I was just getting to like grapher. So now I have to find something else. I didn't like matlab or mathematica but I guess I'll just have to buck up.

BTW you said " Q is the charge of the black hole in "geometric units"." What does that mean? Dimensional analysis shows that the quantity Q should have a constant,k, multiplying it with units of (grams/coulomb^2) so the product can have the same units as mass so it can be combined with the other terms in the equation, no?

Last edited: Jul 13, 2015
18. Jul 13, 2015

### bwana

19. Jul 13, 2015

### bwana

Correction : Q^2, not Q in three graphs for Q^2= 9999,10000,10001 mentioned above

20. Jul 13, 2015

### Staff: Mentor

It means that $Q$ in these units has units of length, just like $M$. Sometimes, to avoid ambiguity, the mass and charge in "geometric units" are denoted as $r_s$ and $r_Q$, as on this Wikipedia page:

https://en.wikipedia.org/wiki/Reissner–Nordström_metric

The formulas for these in terms of the mass and charge in conventional units are also given there. Basically, "geometric units" for mass means $G = c = 1$, and for charge it means, in addition, that $1 / 4 \pi \varepsilon_0 = 1$ (i.e., the Coulomb force constant is 1).