Time Dilation in Reissner-Nordström Metric: Even or Odd?

[H_A_Landman]

TL;DR Summary

Is time dilation an even function of charge in the R-N metric?

In the Reissner–Nordström metric, the charge $Q$ of the central body enters only as its square $Q^2$. The same is true for the Kerr-Schild form. This would seem to imply that all effects are even functions of $Q$. For example, the gravitational time dilation is often written as
$$\gamma = \sqrt{|g^{tt}|} = \sqrt{\frac{r^2}{Q^2+(r-2M)r}}$$
which is an even function of $Q$ and does not depend on the charge $q$ of the test particle at all.

However, I've also seen the equation for time dilation (relative to infinity) written as
$$\gamma = \frac{qQr^3 + Er^4}{r^2(r^2-2r+Q^2)}$$
which has a term that's odd-order in $Q$ and depends on $q$.

So, I'm a little confused. Is it an even function in $Q$, or not? Does it depend on $q$, or not?

 

[PeterDonis]

I've also seen the equation for time dilation (relative to infinity) written as
$$\gamma = \frac{qQr^3 + Er^4}{r^2(r^2-2r+Q^2)}$$

Where have you seen this? Please give a reference.

 

[Ibix]

What's $E$ in the second expression? And is it correctly transcribed? As written, factors of $r$ cancel, and I wonder if there's an $M$ missing in the denominator.

Edit: I mean, if $E=1-qQ/r$ and the missing $M$ is a typo, the second expression is the square of the first.

 

[H_A_Landman]

Both are in the Wikipedia article on the R-N metric. The first is in the Gravitational Time Dilation section (https://en.wikipedia.org/wiki/Reissner–Nordström_metric#Gravitational_time_dilation), and the second in the Equations Of Motion section (https://en.wikipedia.org/wiki/Reissner–Nordström_metric#Equations_of_motion). I was uncertain what E meant too, but since $Q/r$ is the electrostatic potential and $qQ/r$ the electrostatic potential energy, your guess seems dimensionally plausible. Wouldn't it be $1+qQ/r$ though? The energy is negative if the charges $Q$ and $q$ have opposite signs.

 

[Orodruin]

Those equations are telling you completely different things.

The first is the gravitational time dilation of a stationary observer, the other the time dilation of a free-falling observer.

 

[PeterDonis]

Those equations are telling you completely different things.

The first is the gravitational time dilation of a stationary observer, the other the time dilation of a free-falling observer.

Strictly speaking, an observer with charge $q$ in Reissner-Nordstrom spacetime and no forces acting on it other than those due to the spacetime itself is not freely falling; they have a nonzero proper acceleration due to the interaction of their charge $q$ with the charge $Q$ of the spacetime. But these observers are the closest analogues for charged objects in R-N spacetime to freely falling observers in a non-charged spacetime.

 

[Orodruin]

Strictly speaking, an observer with charge q in Reissner-Nordstrom spacetime and no forces acting on it other than those due to the spacetime itself is not freely falling;

True. I was a bit fast on the mobile there.

 

[H_A_Landman]

Maybe it would help to slice this a different way. Replace the "black hole" of mass $M$ and charge $Q$ with an insulating hollow sphere of radius $R$ and mass $M$ and charge $Q$. Since $M$ and $Q$ are the same, presumably spacetime outside the sphere is still described by the same R-N metric, and spacetime inside the sphere is at constant gravitational and electrostatic potential.

(Feel free to assume that $R \gg r_s$, so that there is no singularity or event horizon.)

Now place a test particle with mass $m$ and charge $q$ and velocity $0$ inside the sphere. There is zero force on it, so it is free falling and not moving. What is the time dilation of the particle? Is it even in $Q$, or are there any odd-order terms? Does it depend on $q$ at all (other than the effective charge of the sphere going from $Q$ to $Q+q$)? If so, would particles with charges $q$ and $−q$ have identical, or different, dilations?

 

[PeterDonis]

Replace the "black hole" of mass $M$ and charge $Q$ with an insulating hollow sphere of radius $R$ and mass $M$ and charge $Q$. Since $M$ and $Q$ are the same, presumably spacetime outside the sphere is still described by the same R-N metric

Yes.

and spacetime inside the sphere is at constant gravitational and electrostatic potential.

Here I'm not sure. The usual model of a metallic hollow spherical shell with a surface charge is an approximation that neglects gravity. I'm not sure it works if the only thing holding the shell static is hydrostatic equilibrium--i.e., the shell's self-gravity exactly balancing the repulsion of the charges on it. I'd have to take a look at the detailed math.

There is also the question of what the metric would be inside the shell. In the absence of charge, it would be the flat Minkowski metric. But I'm not aware of any known solution for "constant gravitational potential" inside a shell with nonzero charge, so there is a "constant electrostatic potential" inside that is not zero. Again, I'd have to take a look at the detailed math and see if there's a way to figure out what the metric would be. I'm not aware of any papers in the literature that study this case.

(Feel free to assume that $R \gg r_s$, so that there is no singularity or event horizon.)

That of course makes things simpler, but by itself it doesn't address the concerns I described above.

Now place a test particle with mass $m$ and charge $q$ and velocity $0$ inside the sphere. There is zero force on it, so it is free falling and not moving. What is the time dilation of the particle?

To know that we would have to know the metric inside the shell, and we don't. See above.

 

[H_A_Landman]

In the Newtonian limit, both $1/r^2$ forces from opposite parts of the shell would exactly cancel, so we'd only be left with the potentials. And so for $Q=0$ the interior metric would be flat Minkowski plus the gravitational time dilation, which would be constant inside the sphere. I think that limiting metric still counts as flat (constant time dilation doesn't curve any geodesics). The real answer is probably close to that, but maybe not exactly that.

All observers have to agree on the relative time dilation between the particle inside the sphere and another observer far from the sphere but at rest with respect to it. That's just a dimensionless number. It can't be gauge-transformed away.

I'm not worried about the structural strength of the sphere. It's a thought experiment, we can postulate very high structural strength and approximately zero thickness. It's not metal though, I specified insulating so that the sphere charge can't flow around if the particle moves.

Even if the metric inside the sphere isn't flat everywhere, it has to be flat near the center (by symmetry). Maybe it would be easier to solve just for the center point first.

 

[PeterDonis]

for $Q=0$ the interior metric would be flat Minkowski

Yes.

plus the gravitational time dilation, which would be constant inside the sphere.

The time dilation doesn't make any difference to the flatness of the metric. It only comes into play if you have to compare timelike worldlines inside and outside the shell.

However, all this is only known to be true for the $Q = 0$ case. As I said before, I am not aware of any known solution for the interior of a hollow shell for the $Q \neq 0$ case.

All observers have to agree on the relative time dilation between the particle inside the sphere and another observer far from the sphere but at rest with respect to it. That's just a dimensionless number. It can't be gauge-transformed away.

Yes, agreed, but again, this doesn't change the fact that we don't have a known solution for the $Q \neq 0$ case. We might wave our hands and say that heuristically the time dilation ought to behave the same, but that's just hand-waving, not solving.

I'm not worried about the structural strength of the sphere. It's a thought experiment, we can postulate very high structural strength

But not infinite structural strength. Relativity places a finite limit on the strength of materials. The question is whether that is sufficient.

Also, remember that the problem is not just to support the shell but to contain it--to keep it from flying apart due to the electromagnetic repulsion between all the charges on its surface.

It's not metal though, I specified insulating so that the sphere charge can't flow around if the particle moves.

Under a high enough potential, any material becomes a conductor. There is no such thing as infinite insulating capacity in relativity, any more than there is infinite structural strength.

Even if the metric inside the sphere isn't flat everywhere, it has to be flat near the center (by symmetry). Maybe it would be easier to solve just for the center point first.

You can't solve for just a single point. The EFE is a differential equation; it can only be solved on an open region.

 

[H_A_Landman]

The time dilation doesn't make any difference to the flatness of the metric.

Constant time dilation doesn't. Time dilation that varies by location does; it's equivalent to a gravitational field, as was noted well before GR. (See e.g. Jun Ishiwara, “Zur Theorie der Gravitation.” Physikalische Zeitschrift 13: 1189–1193 (1912).) So the question of whether the interior is force-free and the question of whether the time dilation is constant over the interior are closely linked.

We might wave our hands and say that heuristically the time dilation ought to behave the same, but that's just hand-waving, not solving.

Agreed, but that doesn't prevent us from asking what would happen if it DID behave the same.

You can't solve for just a single point. The EFE is a differential equation; it can only be solved on an open region.

We can choose a tiny open ball centered on the origin, and (if desired) take the limit as $r\rightarrow 0^+$.

I'm wondering whether we can't attack this with simple algebra. Define $Q' \equiv Q + q$ to be the total charge of the sphere plus particle. We know that the R-N metric outside the sphere is a function of $Q'^2 = Q^2 + 2qQ + q^2$, and since $Q\gg q$ we can sometimes ignore the last term. That leaves us a small but non-zero term that is first order in $qQ$. Is that enough to answer the original question?

A counter argument would be to suppose we take $q$ from the sphere itself. Then the total charge would be $Q' = (Q-q) + q = Q$, the metric outside the sphere doesn't change, and there's no first order term. But this is (almost) physically identical to the previous case, and should have (almost) the same answer.

 

[PeterDonis]

Constant time dilation doesn't. Time dilation that varies by location does; it's equivalent to a gravitational field, as was noted well before GR.

This is not correct. There is time dilation that varies by location for a family of accelerated observers in flat spacetime (Rindler observers), and those observers also observe a "gravitational field" (they observe objects released into free fall to accelerate downward). So it is not the case that the presence of time dilation that varies by location implies a curved spacetime.

that doesn't prevent us from asking what would happen if it DID behave the same.

You can ask all you like, but without a solution to the EFE that has the desired properties, you can't answer any of the questions you want to ask.

We can choose a tiny open ball centered on the origin, and (if desired) take the limit as $r\rightarrow 0^+$.

You can take the open ball, but the limit is useless. There is no such thing as a "solution" to the EFE (a metric) that only covers one point. A metric by definition tells you about distances between distinct points in an open region.

I'm wondering whether we can't attack this with simple algebra.

If all you're concerned about is time dilation outside the sphere, then we know the metric is R-N there, and the key point is the one @Orodruin gave in post #5. If the only force involved is the electromagnetic force between the sphere with charge $Q$ and the test object with charge $q$, then the time dilation of the test object will depend on how it moves, which obviously will depend on the sign of $q$ relative to $Q$. You will have two different motions, one for like charges (test object gets repelled) and one for unlike charges (test object gets attracted), and the time dilation (which, btw, you now have to define very carefully since the test object is not stationary) will be different for those two cases--and both of those cases will be different from the case of a stationary object, whose motion is independent of the relative signs of $q$ and $Q$, and whose time dilation only depends on $Q^2$ (and $M$ and $r$, of course).

If you're concerned about time dilation inside the sphere, then, as I've said, we can't answer any questions about that unless and until we find a solution for the metric there.

 

[H_A_Landman]

it is not the case that the presence of time dilation that varies by location implies a curved spacetime.

Yeah, I had a brainfart there. Agree. A time dilation field with constant gradient $\approx$ a uniform gravtiational field, which has zero curvature. I should have said any change in the gradient by location implies curvature.

 

[PeterDonis]

A time dilation field with constant gradient $\approx$ a uniform gravtiational field, which has zero curvature.

Actually, the gradient of the "time dilation field" is not constant for Rindler observers in flat spacetime (which is the closest we can come to a "uniform gravitational field"--note that the "acceleration due to gravity" for such observers is not constant either). So your criterion for curvature is incorrect. The correct criterion for curvature has nothing to do with time dilation or its gradient; it is the presence of geodesic deviation (geodesics which are initially parallel either converge or diverge).

 

[H_A_Landman]

Nod. Curvature is defined in such a way that a cylinder has zero curvature and is not considered to be curved at all. However the forward implication ("A time dilation field with constant gradient ... has zero curvature.") is still correct, it's just the other direction that doesn't work.

 

[PeterDonis]

the forward implication ("A time dilation field with constant gradient ... has zero curvature.") is still correct

Not really. In flat spacetime, AFAIK, there is no congruence of worldlines that fits the description "a time dilation field with zero gradient". But there are curved spacetimes that do fit that description, although they're not physically reasonable (IIRC we had a thread some time ago here on "infinite flat plate" solutions that are the closest thing in GR to "uniform gravitational field" solutions).

Once again, the correct criterion for curvature has nothing to do with time dilation; it's the presence of geodesic deviation. You will be much better off just sticking to that.