Coulomb force over Rindler horizon

[jartsa]

Let's say a rocket carrying a positively charged balloon starts to accelerate with a constant proper acceleration at time t.

After a long time another rocket carrying a positively charged balloon is launched. The crew of this rocket drives the rocket to a position right below the Rindler-horizon of the first rocket.

Now the crew in the first rocket should not be able to detect the second rocket in any way. But the crew in the first rocket can detect the Coulomb force between the balloons. Right?

 

[stevendaryl]

Let's say a rocket carrying a positively charged balloon starts to accelerate with a constant proper acceleration at time t.

After a long time another rocket carrying a positively charged balloon is launched. The crew of this rocket drives the rocket to a position right below the Rindler-horizon of the first rocket.

Now the crew in the first rocket should not be able to detect the second rocket in any way. But the crew in the first rocket can detect the Coulomb force between the balloons. Right?

You can't just create a positively charged balloon from nothing, you have to gather positive charges from elsewhere (or take negatives charges away from the balloon.

So let's make the scenario a little more complete, and suppose that we have some large object, say a planet, and we have neutral balloons, and we transfer charge from the planet to the balloon in order to make the balloon charged. So if this activity takes place beyond the Rindler horizon of a rocket, then nobody on the rocket will be able to detect that it's being done. Nobody on the rocket will be able to detect a positive electric field coming from the horizon.

But notice: both the balloon and the planet are below the horizon. So the total charge below the horizon is still neutral. So there is no $\frac{1}{r^2}$ field due to the balloon in the limit of large $r$. What you have instead is a field that at large distances looks like a dipole field, which I think goes like $\frac{1}{r^3}$. So the conclusion (I think) is that dipole fields can't extend past the Rindler horizon.

Monopole fields can. If the whole universe has a net positive charge below the horizon, then there will be a detectable field at the rocket.

 

[jartsa]

You can't just create a positively charged balloon from nothing, you have to gather positive charges from elsewhere (or take negatives charges away from the balloon.

Can you do such thing as "drive your rocket right below the Rindler-horizon of another accelerating rocket" as i suggested?

(The chasing rocket was supposed to get quite close to the first rocket)

Well now I think that it's probably impossible.

So if I may change the scenario a bit, the second vehicle could be carried by the first rocket, then it is dropped, and its crew drives it far below the Rindler horizon of the first rocket. So now the Coulomb force is very small according those people far below the Rindler horizon, but larger according to the people in the accelerating rocket.

 

[stevendaryl]

So if I may change the scenario a bit, the second vehicle could be carried by the first rocket, then it is dropped, and its crew drives it far below the Rindler horizon of the first rocket. So now the Coulomb force is very small according those people far below the Rindler horizon, but larger according to the people in the accelerating rocket.

I'm not understanding your scenario. Can we change it to this?

• Inside the accelerating rocket, we create two balls with equal and opposite charges.
• We keep one ball in the accelerating rocket.
• We let the other ball drop below the Rindler horizon.

The problem with this is that, from the point of view of the Rindler coordinates of the accelerating rocket, it takes an infinite amount of time for the ball to drop below the horizon. So there will always be a force between the two balls.

From the point of view of the falling observer, it does cross the horizon in a finite amount of time, but from that point of view, there is nothing special about the horizon. He will always see a force between the two balls, even after dropping below the horizon.

 

[jartsa]

I'm not understanding your scenario. Can we change it to this?

• Inside the accelerating rocket, we create two balls with equal and opposite charges.
• We keep one ball in the accelerating rocket.
• We let the other ball drop below the Rindler horizon.

The problem with this is that, from the point of view of the Rindler coordinates of the accelerating rocket, it takes an infinite amount of time for the ball to drop below the horizon. So there will always be a force between the two balls.

From the point of view of the falling observer, it does cross the horizon in a finite amount of time, but from that point of view, there is nothing special about the horizon. He will always see a force between the two balls, even after dropping below the horizon.

Yes very good change.

Now the falling observer says that the Coulomb force approaches zero as the distance approaches infinity.

What does the accelerating observer say? Maybe he says "the Coulomb force approaches zero as the speed of the falling charge approaches the local speed of light at the charge's position"?

 

[Ibix]

Can you do such thing as "drive your rocket right below the Rindler-horizon of another accelerating rocket" as i suggested?

I don't think so. A Rindler horizon is a null surface - whatever distance below it you are now, you must be further below it later since it "travels" at the speed of light.

I can approach and cross your Rindler horizon from above. From below, I can only get further away.

 

[Orodruin]

Now the crew in the first rocket should not be able to detect the second rocket in any way. But the crew in the first rocket can detect the Coulomb force between the balloons. Right?

The entire form of the Coulomb force rests on the assumption that you have a charge at rest for an infinite time. Assuming that your rocket carrying the charge is not accelerating, it will have crossed the Rindler horizon at some point. Any effects you feel on the accelerating ship must be due to the part of the world line of the other rocket that is before crossing the Rindler horizon.

 

[Ibix]

What does the accelerating observer say? Maybe he says "the Coulomb force approaches zero as the speed of the falling charge approaches the local speed of light at the charge's position"?

Work in the inertial rest frame of the released charge. You can easily write down the field at the ship's position as a function of coordinate time $t$. You can also write down the velocity of the ship at $t$, which will let you calculate the EM field measured by the ship. Then you write down the coordinate time as a function of the ship's proper time and substitute in.

 

[PAllen]

I think there is a scenario that captures what the OP is intending, and it is a case of unusual initial conditions for a hypothetical universe.

There are two rockets that have eternally been accelerating at 1 g, and per some inertial frame, have always had a separation of 1.5 light years. Then, one has always been behind the Rindler horizon of the other. The rockets have also eternally had charged balloons.

Then, I think it is true that one would detect the charge of the other (after a particular time on the rearward world line), but not vice versa.

 

[Orodruin]

Then, one has always been behind the Rindler horizon of the other.

I disagree. The "lagging" rocket at some point passes through the future Rindler horizon of the "leading" rocket and the "leading" rocket at some point emerges from the past Rindler horizon of the "lagging" rocket.

The future Rindler horizon in the standard setup is the line $x = t$ and the past the line $x = -t$. It is true that the lagging rocket will never be in the coordinate patch of the leading rocket's Rindler coordinates, but this does not mean that the lagging rocket is out of sight. The leading rocket will be able to (at some point) see everything that happens to the lagging rocket before it passes the future Rindler horizon.

Conversely, the lagging rocket will also at some point emerge from the past Rindler horizon and until that point it cannot be affected by anything going on at the leading rocket.

 

[jartsa]

Work in the inertial rest frame of the released charge. You can easily write down the field at the ship's position as a function of coordinate time $t$. You can also write down the velocity of the ship at $t$, which will let you calculate the EM field measured by the ship. Then you write down the coordinate time as a function of the ship's proper time and substitute in.

Okay, now I have checked out some formulas. This one is very convenient for me: When the direction of the velocity is along a field line of the electric field, then E'=E.

The released charge says that the electric field of itself at distance r from itself is: $E=\frac{kQ}{r^2}$

And the distance to the ship at time t is: $r=\frac{ c^2}{a} (\sqrt{1+(at/c)^2}-1)$

Now if we substitute that latter thing to the formula for E we get:

$E=\frac{kQ}{(\frac{ c^2}{a} (\sqrt{1+(at/c)^2}-1))^2}$That formula above tells how the E at the position of the ship changes as time passes, according to the released charge.The formula that tells how the E at the position of the ship changes as time passes, according to the ship, would be that formula with t substituted by ... ... I'm not sure by what.

Hmm.. maybe the ship crew reads their clock which says that the time is T, then they calculate corresponding time in the frame of the released charge like this:

$t= \frac{c}{a} \operatorname{sinh}( \frac{aT}{c})$

That value of t above changes at increasing rate according to the ship crew. So we can conclude that if we asked the released charge and the ship crew the question: "at what rate is the field of the released charge changing at the position of the ship" , the ship crew would report a higher rate. Is that correct?I got the distance formula and the time formula from here:
http://www.math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

 

[sweet springs]

Now the crew in the first rocket should not be able to detect the second rocket in any way. But the crew in the first rocket can detect the Coulomb force between the balloons. Right?

Does charged BH (ref. https://en.wikipedia.org/wiki/Reissner–Nordström_metric ) suggest coulomb force works beyond event horizon? Teaching of anyone should be appreciated.

 

[sweet springs]

After a long time another rocket carrying a positively charged balloon is launched. The crew of this rocket drives the rocket to a position right below the Rindler-horizon of the first rocket.

Say, the universe is neutral in charge, at Rindler horizon a pair of electron and positron is created, they apart so the electron is within the horizon and the positron is outside the horizon. Earth people observe that universe is neutral with an electron and a positron. Which do rocket people observe:
#1 the universe is negative charged by the electron for the world beyond the horizon does not interfere with them nor exist, or
#2 the universe is still neutral with the electron and the positive charged horizon ?
#1 cause a contradiction that electric force line do not close, so I assume #2.

 

[jartsa]

Does charged BH (ref. https://en.wikipedia.org/wiki/Reissner–Nordström_metric ) suggest coulomb force works beyond event horizon? Teaching of anyone should be appreciated.

Does this answer that question:

http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

 

[PAllen]

I disagree. The "lagging" rocket at some point passes through the future Rindler horizon of the "leading" rocket and the "leading" rocket at some point emerges from the past Rindler horizon of the "lagging" rocket.

The future Rindler horizon in the standard setup is the line $x = t$ and the past the line $x = -t$. It is true that the lagging rocket will never be in the coordinate patch of the leading rocket's Rindler coordinates, but this does not mean that the lagging rocket is out of sight. The leading rocket will be able to (at some point) see everything that happens to the lagging rocket before it passes the future Rindler horizon.

Conversely, the lagging rocket will also at some point emerge from the past Rindler horizon and until that point it cannot be affected by anything going on at the leading rocket.

While what I said was wrong, this is also not correct. The leading rocket is forever outside both the past and future horizons of the lagging rocket. The lagging rockets history may be divided into early, middle, and late periods. During the early period it is behind past/white horizon of the lead rocket, and outside the future/black horizon of the lead rocket. At the beginning of the middle period, it enters the future/black horizon, and thus it is behind both horizons. The late period of the rear rocket exits the past/white horizon, while remaining forever behind the future/black horizon.

The lead rocket can always detect charge from the lagging rocket, but only from its early period. That is, if the lag rocket starts wiggling it’s charge after its early period, the lead rocket can never detect this.

The most anomalous behavior, to me, is that before its late period, the lag rocket cannot detect anything about the lead rocket charge, not even whether it is positive or negative. However, during its late period, it can ultimately detect all of the the history of the forward rocket charge.

So there is a clear asymmetry, but not the one I originally described.

 

[Orodruin]

The leading rocket is forever outside both the past and future horizons of the lagging rocket

Yes, my intention was to say that the lagging rocket at some point emerges from the leading rocket’s past horizon (as clarified in the end of my post) but somehow I got things messed up in the post (probably too much editing pre-post ...)

I maintain that most of the point is correct apart from the second part of the first paragraph.

 

[Ibix]

Is that correct

Seems to be.

 

[stevendaryl]

Does charged BH (ref. https://en.wikipedia.org/wiki/Reissner–Nordström_metric ) suggest coulomb force works beyond event horizon? Teaching of anyone should be appreciated.

Yes. The rule of thumb is that for the long-range forces of gravity and the electric field, there is no difference between a charged point-mass and a spherically symmetric mass with the same total mass and charge. That continues to be the case even when the mass collapses into a black hole. From outside, the electric field and the metric looks just like a spherically symmetric object that has not yet collapsed.

 

[jartsa]

Okay, now I have checked out some formulas. This one is very convenient for me: When the direction of the velocity is along a field line of the electric field, then E'=E.

The released charge says that the electric field of itself at distance r from itself is: $E=\frac{kQ}{r^2}$

And the distance to the ship at time t is: $r=\frac{ c^2}{a} (\sqrt{1+(at/c)^2}-1)$

Now if we substitute that latter thing to the formula for E we get:

$E=\frac{kQ}{(\frac{ c^2}{a} (\sqrt{1+(at/c)^2}-1))^2}$That formula above tells how the E at the position of the ship changes as time passes, according to the released charge.The formula that tells how the E at the position of the ship changes as time passes, according to the ship, would be that formula with t substituted by ... ... I'm not sure by what.

Hmm.. maybe the ship crew reads their clock which says that the time is T, then they calculate corresponding time in the frame of the released charge like this:

$t= \frac{c}{a} \operatorname{sinh}( \frac{aT}{c})$

That value of t above changes at increasing rate according to the ship crew. So we can conclude that if we asked the released charge and the ship crew the question: "at what rate is the field of the released charge changing at the position of the ship" , the ship crew would report a higher rate. Is that correct?I got the distance formula and the time formula from here:
http://www.math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

A much simpler version of that:

As the electric field strength measured by the crew of the accelerating spaceship does not depend on the instantaneous velocity or acceleration of the ship, we can make things simple for us by saying that just before the measurement of the electric field the ship stops accelerating and moving relative to the dropped charge.

Then the crew will say that they have managed to travel across a very wide electric field in a very short time, because the field was length-contracted. That's why the field strength at the current position of the ship is very small.

And an observer standing next to dropped charge says the travel took a small amount of ships's proper time, because ship's clock was time dilated.

 

[Ibix]

Then the crew will say that they have managed to travel across a very wide electric field in a very short time, because the field was length-contracted. That's why the field strength at the current position of the ship is very small.

I'd be very careful saying that because you're mixing reference frames. So I'm not completely sure that what you are saying is consistent. I think your individual statements about interpretation in the various inertial frames is correct, but it's not completely clear which reference frame you are using for what statement, so I may be wrong.

Avoiding this whole issue is one of the reasons I was advocating the use of the electric field as a function of the ship's proper time. No ambiguities.

 

[sweet springs]

Which do rocket people observe:

I am afraid my scenario was in failure.

Say rockets of various constant acceleration a>0 approach from X=+infinity, stop at T=0 and X=c^2/a in an IFR and leave toward X=+infinity. The plane X=0 is their Rindler event horizon.
Say there is a charged star at X=-d<0 from enough past to future in the IFR.

The ships cannot observe events in the space time region X > T where c=1. The world line of the star lies beyond the line X=T so some events on the line is not observed by the ship.

The ships, however, do not observe the star and generated em field enter into the horizon. The star stay frozen touching the horizon forever and do not go beyond the horizon. So the horizon itself do not have to undertake charge.

Say a pair creation of charged particles event take place in the space time region X>T. The ships cannot observe at all the charges including generated em fields, e.g. dipole one. Again horizon itself has nothing to do with charge.

 

[PeterDonis]

The plane X=0 is their Rindler event horizon

No, their Rindler horizon is the two null lines $X = T$ (the future horizon) and $X = -T$ (the past horizon).

 

[sweet springs]

Thanks so much, PeterDonis.
I do correction.

Say rockets of various constant acceleration a>0 approach from X=+infinity, stop at T=0 and X=c^2/a in an IFR and leave toward X=+infinity. The plane X=T,-T, where c=1,0 is their Rindler event horizon.
Say there is a pair of positive and negative charges in the IFR. They are at rest and far apart and come closer by Coulomb force. At T=0 space coordinate of positive charge is d>0, that of negative charge is -d. They will collide some time later.

The ships observe motion of positive charge only. Past frozen particle at X=0 comes up, stops, goes down to X=0 and frozen again. The ships also observe Coulomb force applying on the positive charge due to untisymmetry motion of coming up and going down. The ships people would assume that there is negative charge behind the horizon.

 

[jartsa]

Say rockets of various constant acceleration a>0 approach from X=+infinity, stop at T=0 and X=c^2/a in an IFR and leave toward X=+infinity. The plane X=T,-T, where c=1,0 is their Rindler event horizon.
Say there is a pair of positive and negative charges in the IFR. They are at rest and far apart and come closer by Coulomb force. At T=0 space coordinate of positive charge is d>0, that of negative charge is -d. They will collide some time later.

The ships observe motion of positive charge only. Past frozen particle at X=0 comes up, stops, goes down to X=0 and frozen again. The ships also observe Coulomb force applying on the positive charge due to untisymmetry motion of coming up and going down. The ships people would assume that there is negative charge behind the horizon.

At first the rocket's acceleration and velocity vectors point to opposite directions. That causes the rocket to move so that its "Rindler horizon" is leading. And that will result in all kinds of stuff going through the "Rindler horizon". Right?

I put quote marks around "Rindler horizon", because that thing doesn't seem to be a proper Rindler horizon.

 

[Ibix]

At first the rocket's acceleration and velocity vectors point to opposite directions.

Depends on your frame of reference. One can always choose a Rindler observer's velocity to be parallel or anti-parallel to its acceleration.

Note that there are two Rindler horizons. There is a past horizon ($x=-ct$ in inertial coordinates centred on the crossing point of the two horizons), which is analogous to the horizon of a white hole, from which things can rise and be seen to rise. There is also a future horizon ($x=ct$) which is analogous to the horizon of a black hole.

Whether or not something passes through a given Rindler observer's horizon(s) is coordinate independent. The Rindler observer's simultaneity planes are such that he always regards the horizon intersection ($x=ct=0$) as "now". So in this Rindler frame, crossing the past horizon always happens "before now" and crossing the future horizon never quite happens.

 

[sweet springs]

Past frozen particle at X=0 comes up, stops, goes down to X=0 and frozen again.

Correction: Not X=0 of the IFR but x=0 for the ships or in Rindler coordinate.

 

[Ibix]

Correction: Not X=0 of the IFR but x=0 for the ships or Rindler coordinate.

I don't think anything is frozen at the past horizon. That's the boundary beyond which the Rindler observer could never have affected anything. But he can receive signals from beyond it in finite time just fine (as long as it's not also beyond the future horizon). Things freeze at the future horizon because it's always in the future for a Rindler observer and light from the horizon never quite catches up.

 

[sweet springs]

I don't think anything is frozen at the past horizon.

Aren't past and future symmetric ? I mean for the case of ten year future and the case of a million year future, the positions of free falling body are almost x=0 and they show only a little difference. For the cases of ten year past and for a million year past, the positions of free popping up body are almost x=0 and they show only a little difference. I think applicability of expression frozen are to be shared among past and future.

 

[Ibix]

Aren't past and future symmetric ?

Depends what you mean. Here's a quick sketch of a Minkowski diagram showing a black Rindler observer with a purple inertial observer crossing both the (red) Rindler horizons.

You can see that the blue signals the inertial observer sends arrive at the Rindler observer in finite time for all of history until it crosses the future horizon, so there's no freezing at the past horizon. You can imagine flipping time, but if you do so the meaning of the blue signals changes - now the Rindler observer is signalling the inertial one.

The past horizon is actually the boundary between events the Rindler observer could ever have influenced and events it could not have influenced - it's the limit as $t\rightarrow -\infty$ of its future lightcone. Similarly the future horizon is the limit as $t\rightarrow \infty$ of its past lightcone. Flipping time flips the meaning of the boundaries, and hence whether the Rindler observer can observe things crossing it or not.

 

[sweet springs]

Thanks Ibix for your nice drawing and clear explanation.

I would restate the points. The IFR space time is composed of 4 area
#1 Future cone of Origin
#2 Past cone of Origin
#3 Rindler coordinate area
#4 Opposite to Rindler coordinate area

As for causal relation
Events in #2 and #3 can effect on events in #3 Rindler area.
Events in #1 and #4 can not effect on events in #3 Rindler area.

May I add that, as for information coming from events in #2, observer in #3 i.e. the ship passengers cannot trace back that information beyond past horizon. The information have been staying close to past horizon in infinite past though the information arrive to the events in #3 as usual in due time. With regard to OP's interest, part of charged particle world line in #2 and #3 effect the ship passenger, including the case that they cannot touch charged particle in #2 area but can see its image by arrival of reflected light and em field generated at their own position and time.

 

[sweet springs]

No, their Rindler horizon is the two null lines X=TX=TX = T (the future horizon) and X=−TX=−TX = -T (the past horizon).

I would like do some clarification.
https://en.wikipedia.org/wiki/File:Rindler_chart.svg says horizons are half line starting from origin of the IFR. Rindler coordinate has nothing to do with other half part.

In the sense that

Events in #2 and #3 can effect on events in #3 Rindler area.
Events in #1 and #4 can not effect on events in #3 Rindler area.

not half lines but full lines of X=T,-T seem to have meanings in IFR. Is there definition on half or full ?

 

[PeterDonis]

I would like do some clarification.

Wikipedia is not a valid source.

Is there definition half or full ?

Both meanings could be used; you have to determine which is intended from context.