I Paradox: Electron Radiates in a Gravitational Field

[Sorcerer]

After all that has gone through this thread, I would very much like to see where anyone says that a charge undergoing constant acceleration doesn't radiate. I think I will dig through Jackson on this subject, though I am not really at Jackson's level in my review of physics.

Well, I think there are good arguments for a uniformly accelerated charge not emitting observable radiation according to someone comoving with it.

I don't know how much trust you put in the University of Campinas in Brazil, or FAPESP in Brazil (their wiki page: https://en.wikipedia.org/wiki/São_Paulo_Research_Foundation), or the National Council for Scientific and Technological Deveolopment in terms of peer review (their wiki page: https://en.wikipedia.org/wiki/National_Council_for_Scientific_and_Technological_Development), but they did support the paper I linked to earlier, which, after a lengthy series of derivations and examinations, said this:

This result answers our question. A comoving observer will not detect any radiation from a uniformly accelerated charge. The comoving observer can receive signals only from regions I and IV. The field emitted by the accelerated charge does not reach region IV, and in region I, it is interpreted by the comoving observer as a static field. We note that essentially the same argument was used by Rohrlich to show that in a static homogeneous gravitational field, static observers do not detect any radiation from static charges.

https://arxiv.org/pdf/physics/0506049‎

Now, this is very math heavy, and a bit beyond me. But, as a kind of analogy, not everyone is going to see a magnetic field just because someone else does. One observer moving with respect to an electric field will see a magnetic field. Measurements are what matter. All the measurements have to agree. I believe Dale said this.

Which leads me to believe that if the paper above is correct, then observation of the radiation is frame dependent, but if I understood the paper correctly, the radiation is still happening, it just is impossible to be seen by the comoving observer (due to a what amounts to a "radiation event horizon.")
If any of you post-grad or graduate student people have time, I'd appreciate your insight on the above link, and if I've made any blatant misunderstandings of it. Thanks.

 

[stevendaryl]

There is a discussion of the equivalence principle as applied to a uniformly accelerated charge given here:

https://arxiv.org/pdf/gr-qc/9303025.pdf

Apparently, it's very complicated to untangle all the issues.

 

[Dale]

Apparently, it's very complicated to untangle all the issues.

It is only complicated if you focus on radiation instead of measurements. Hence my insistence from the beginning on defining the measurement procedure.

The EP is very specific. It says that the outcome of an experiment (a measurement) is the same if performed under uniform gravity or under uniform acceleration. It does not say that intermediate quantities, like radiation, are the same.

 

[Sorcerer]

It is only complicated if you focus on radiation instead of measurements. Hence my insistence from the beginning on defining the measurement procedure.

The EP is very specific. It says that the outcome of an experiment (a measurement) is the same if performed under uniform gravity or under uniform acceleration. It does not say that intermediate quantities, like radiation, are the same.

Is this similar to the magnet and conductor issue? I'm referring to the opening paragraph of "On the Electrodynamics of Moving Bodies," where the current is the same but explained differently by two observers. Same measurement, but frame dependent explanations?

 

[Dale]

Is this similar to the magnet and conductor issue? I'm referring to the opening paragraph of "On the Electrodynamics of Moving Bodies," where the current is the same but explained differently by two observers. Same measurement, but frame dependent explanations?

Yes. The EP is about the equivalence of measurements, not explanations.

 

[Vanadium 50]

Here (I don’t have the book, but I think it is available online)

It is not available.

Again, I ask where did Feynman say it? I want to know exactly what he said, because this is a place where context matters.

 

[stevendaryl]

It is only complicated if you focus on radiation instead of measurements. Hence my insistence from the beginning on defining the measurement procedure.

The EP is very specific. It says that the outcome of an experiment (a measurement) is the same if performed under uniform gravity or under uniform acceleration. It does not say that intermediate quantities, like radiation, are the same.

But Parrot is saying that it's not true. You have two different situations:

1. A rocket ship hovering above a massive star.
2. A rocket ship accelerating at constant proper acceleration in flat spacetime.

Parrot is claiming that careful measurements would reveal a difference in these two cases. And not because of tidal effects. He says that in case 1, there is no difference between the rocket power needed to hold up a charged particle of mass $M$ and the power needed to hold up an uncharged particle. In case 2, it requires more force to hold up the charged particle (because some of the energy that goes into accelerating the particle is lost to radiation).

So he's saying that local measurements can in principle tell the difference.

[edit] His conclusion is about measurements. From the abstract:

We argue that purely local experiments can distinguish a stationary charged particle in a static gravitational field from an accelerated particle in (gravity-free) Minkowski space.

 

[stevendaryl]

But Parrot is saying that it's not true. You have two different situations:

1. A rocket ship hovering above a massive star.
2. A rocket ship accelerating at constant proper acceleration in flat spacetime.

Parrot is claiming that careful measurements would reveal a difference in these two cases. And not because of tidal effects. He says that in case 1, there is no difference between the rocket power needed to hold up a charged particle of mass $M$ and the power needed to hold up an uncharged particle. In case 2, it requires more force to hold up the charged particle (because some of the energy that goes into accelerating the particle is lost to radiation).

So he's saying that local measurements can in principle tell the difference.

I'm not saying that Parrot is correct in his conclusions, but only to say that if he's right, it's a violation of the equivalence principle, according to the formulation "local measurements cannot distinguish between blah and blah".

 

[stevendaryl]

It is not available.

Again, I ask where did Feynman say it? I want to know exactly what he said, because this is a place where context matters.

[edit]Kevin Brown discusses Feynman's argument here:

http://www.mathpages.com/home/kmath528/kmath528.htm

 

[Dale]

So he's saying that local measurements can in principle tell the difference.

[edit] His conclusion is about measurements. From the abstract:

Hmm, I don’t think this is an accepted view, but I haven’t finished the article yet

 

[stevendaryl]

John Baez discusses Feynman's argument here:

http://www.mathpages.com/home/kmath528/kmath528.htm

An excerpt from Feynman is found here:

http://www.applet-magic.com/feynmanEM.htm

 

[MikeGomez]

John Baez discusses Feynman's argument here:

http://www.mathpages.com/home/kmath528/kmath528.htm

That link looks oddly familiar. Hmm, just like one that I provided in post #44. Why do you think it is John Baez?
Or did you post the wrong link by accident?

 

[stevendaryl]

That link looks oddly familiar. Hmm, just like one that I provided in post #44. Why do you think it is John Baez?
Or did you post the wrong link by accident?

My mistake. I thought I remembered him writing on that topic, but that particular article doesn't have an obvious author.

[edit] Apparently, it is KEVIN S. Brown.

 

[MikeGomez]

It is not available.

Again, I ask where did Feynman say it? I want to know exactly what he said, because this is a place where context matters.

I actually don't know exactly what he said, and you're right, context matters.

If you wish I can retract my statement that Feynman said that accelerated particles don't radiate.

Edit: I meant accelerated charged particle.

 

[stevendaryl]

I actually don't know exactly what he said, and you're right, context matters.

If you wish I can retract my statement that Feynman said that accelerated particles don't radiate.

Edit: I meant accelerated charged particle.

He definitely did not say that accelerated charged particles never radiate. He was specifically talking about the case of constant proper acceleration.

 

[girts]

can't we use the example of steady DC electrical current, excluding the parts where it is turned on or off otherwise it is constant proper acceleration of electrons which doesn't lead to any radiation as is the reason why DC cannot be used in transformers since the field is static and it can't transfer energy aka radiate.

 

[Nugatory]

can't we use the example of steady DC electrical current, excluding the parts where it is turned on or off otherwise it is constant proper acceleration of electrons which doesn't lead to any radiation as is the reason why DC cannot be used in transformers since the field is static and it can't transfer energy aka radiate.

There is no acceleration of charge in a constant DC current if you exclude the startup and shutdown periods.

 

[girts]

Exactly but that was my point, there is also no acceleration or de-acceleration for a charged particle of constant velocity which was my point. Pardon my misunderstanding but then why we are considering radiation from the examples said previously of a constant accelerated charge? Unless ut somehow changes its speed relative to an observer?

 

[stevendaryl]

can't we use the example of steady DC electrical current, excluding the parts where it is turned on or off otherwise it is constant proper acceleration of electrons which doesn't lead to any radiation as is the reason why DC cannot be used in transformers since the field is static and it can't transfer energy aka radiate.

Are you mixing up constant acceleration and constant velocity?

Constant velocity means zero acceleration. The issue is whether nonzero acceleration necessarily leads to radiation.

 

[girts]

Thanks for clarification , okay so if we say nonzero acceleration then that basically means a charge is accelerating so we are talking about a charge that changes velocity and with respect to an observer at rest with respect to the charge it should see EM radiation , well hasnt this been known for a long time and accepted as the norm or am I still missing something?

 

[Nugatory]

Thanks for clarification , okay so if we say nonzero acceleration then that basically means a charge is accelerating so we are talking about a charge that changes velocity and with respect to an observer at rest with respect to the charge it should see EM radiation , well hasnt this been known for a long time and accepted as the norm or am I still missing something?

We are considered an object at rest in a gravitational field, such as one sitting on the surface of the earth. Intuitively we expect it not to radiate, but by the equivalence principle it should because this situation is indistinguishable from putting the same object in a spaceship accelerating at 1g in empty space.

 

[Sorcerer]

We are considered an object at rest in a gravitational field, such as one sitting on the surface of the earth. Intuitively we expect it not to radiate, but by the equivalence principle it should because this situation is indistinguishable from putting the same object in a spaceship accelerating at 1g in empty space.

What if the comoving observers (those at rest in the 1g accelerating spacecraft ) have no access to the radiation? I hate to keep harping on this given that I'm an undergrad and this is clearly a graduate level topic, but again, these people seem to have presented a good argument to me. Could someone here dismantle it if it is not valid?

Nevertheless, comoving observers, that is, accelerated observers with respect to whom the charge is at rest, will not detect any radiation because the radiation field is confined to a spacetime region beyond a horizon that they cannot access.

Published in The American Journal of Physics
https://aapt.scitation.org/doi/10.1119/1.2162548

If you don't want a paywall:
https://arxiv.org/pdf/physics/0506049‎
If I understand the set up, tidal forces should have no bearing, and yet their derivation results in a charge accelerating according to an inertial observer radiating, while according to the observer comoving with the charge never having access to the radiation. Again, me = undergrad, topic = grad or upper division at lowest, but why doesn't that solve the issue?

I mean, if the people in the 1 g rocket cannot access the radiation (for whetever reason), the equivalence principle is completely saved: someone in a uniform gravitational field will not see radiation, nor will someone in then 1 g rocket.

 

[Nugatory]

What if the comoving observers (those at rest in the 1g accelerating spacecraft ) have no access to the radiation?

That is indeed one of the possible resolutions to the apparent paradox. Another possible resolution (not generally accepted, but with a respectable pedigree from Pauli to Feynman) is that constant proper acceleration does not produce radiation (that is, the third derivative of position must be non-zero if there is to be radiation).

 

[Sorcerer]

That is indeed one of the possible resolutions to the apparent paradox. Another possible resolution (not generally accepted, but with a respectable pedigree from Pauli to Feynman) is that constant proper acceleration does not produce radiation (that is, the third derivative of position must be non-zero if there is to be radiation).

So... what you're saying is that radiation can be a real jerk?

 

[Sorcerer]

That is indeed one of the possible resolutions to the apparent paradox. Another possible resolution (not generally accepted, but with a respectable pedigree from Pauli to Feynman) is that constant proper acceleration does not produce radiation (that is, the third derivative of position must be non-zero if there is to be radiation).

So... what you're saying is that radiation can be a real jerk?

Of course, if you push him too hard to change, he might snap.

*ba dum tss*

Sorry I can't stop.

 

[Demystifier]

As I pointed out in https://arxiv.org/abs/gr-qc/9909035 , the EM field measured by a local observer depends only on the instantaneous velocity of the observer, not on the acceleration (or higher derivatives). In that sense, the radiation does not depend on acceleration of the observer.

 

[stevendaryl]

As I pointed out in https://arxiv.org/abs/gr-qc/9909035 , the EM field measured by a local observer depends only on the instantaneous velocity of the observer, not on the acceleration (or higher derivatives). In that sense, the radiation does not depend on acceleration of the observer.

I was looking for the definitive answer to the question: Does a Rindler observer see radiation from a "stationary" charge ("stationary" meaning zero spatial coordinate velocity in Rindler coordinates)? Did you give the answer and I missed it?

You seem to be saying, though, that for an accelerating charge the self-force is not isotropic, which I would think would manifest itself as an apparent change to the particle's weight compared to an uncharged particle of the same mass? So, if you have two particles of mass $M$, one has charge $Q$ and one is neutral, then their "weight" on board a Rindler spaceship (the force needed to keep them at constant Rindler spatial location) would be different?

 

[stevendaryl]

What if the comoving observers (those at rest in the 1g accelerating spacecraft ) have no access to the radiation? I hate to keep harping on this given that I'm an undergrad and this is clearly a graduate level topic, but again, these people seem to have presented a good argument to me. Could someone here dismantle it if it is not valid?

Published in The American Journal of Physics
https://aapt.scitation.org/doi/10.1119/1.2162548

If you don't want a paywall:
https://arxiv.org/pdf/physics/0506049‎

There is still a mystery to clear up, in my mind. If a charged particle radiates, according to an inertial observer, then that means that some of the energy used to accelerate the particle goes into radiation, by energy conservation. This means that it requires more energy to accelerate a charged particle than an uncharged particle of the same mass. This in turn means that for a Rindler rocket carrying a charged particle, more fuel is required than if it were carrying an uncharged particle of the same mass. So the charged particle would appear to weigh more.

But then is the same true for a charged particle at rest in a gravitational field?

 

[Sorcerer]

There is still a mystery to clear up, in my mind. If a charged particle radiates, according to an inertial observer, then that means that some of the energy used to accelerate the particle goes into radiation, by energy conservation. This means that it requires more energy to accelerate a charged particle than an uncharged particle of the same mass. This in turn means that for a Rindler rocket carrying a charged particle, more fuel is required than if it were carrying an uncharged particle of the same mass. So the charged particle would appear to weigh more.

But then is the same true for a charged particle at rest in a gravitational field?

Complete shot in the dark here (from a very unqualified person, myself), but: how much energy? And is it comparable to the effect the charged particle would have on the stress-energy tensor by virtue of being charged, and would that be relevant in any way?

EDIT: wait, a charged particle must move to created an EM field, correct? Still, could there be any effect?

Does the presence of a charged particle in some way affect the other particles around it in a way that would create any kind of pressure that could be relevant? Etc, etc.

 

[girts]

Now I do understand that my questions here are more like brakes on a high speed intellectual debate but please bear with me as I too want to understand this topic deeper.I read through the article @Sorcerer posted, and I think I kind of understand the reason why two charged particles traveling at the same speed which is then uniformly increasing due to uniform acceleration (gravitational or electrostatic doesn't matter in this case I assume) can't see each other's radiation but instead see a static field much like two charged particles would see while being at rest with both a third observer and also with one another.
So If I understand correctly the reason for this is that as the particles move along they move with some speed which is less than c, but changes in the EM field travel at c in vacuum, so as the two charges enter every new frame both traveling at the same velocity the changes in their EM field "lines" as perceived by an inertial observer (aka the third observer at rest with respect to the charges) are radiating outwards in all directions with speed "c" but since the charged particle speed itself can never exceed or even come close to c, they themselves (the traveling particles) cannot reach their own created changes in the static field lines.
Nor can they see each other's field lines being distorted because their speed even under acceleration is the same so the field line changes happen simultaneously and appear as not happening at all for the two particles?But still even if this all is so, the question stated by stevendaryl in post#77 and before is valid and interesting, aka where does the energy come from?
I mean gravitational field is static and so is the electric field of a charged particle,so how come two such static fields can produce radiation which is a dynamic property that can radiate (transfer) energy outside these two systems (two fields) without there being any apparent energy input?
Just a quick though experiment along the way, if this is true then in theory we could say somehow have a place here on Earth where the gravitational acceleration constant is higher then nearby and then we could put a bunch of charged particles at that higher gravity and stand next to them and receive radiation, while no energy input would be done, doesn't this somehow violate energy conservation and perpetual motion?

 

[Sorcerer]

Just a quick though experiment along the way, if this is true then in theory we could say somehow have a place here on Earth where the gravitational acceleration constant is higher then nearby and then we could put a bunch of charged particles at that higher gravity and stand next to them and receive radiation, while no energy input would be done, doesn't this somehow violate energy conservation and perpetual motion?

Going to throw more shots in the dark:

I don't think this one would work because the distance for a noticeable gravitational difference is probably too large for the equivalence principle to remain valid, but if it did, presumably there would be different pressure up there, and pressure is part of the stress-energy tensor, right? Maybe the lower pressure can correspond to the "loss in energy?"

 

[Demystifier]

I was looking for the definitive answer to the question: Does a Rindler observer see radiation from a "stationary" charge ("stationary" meaning zero spatial coordinate velocity in Rindler coordinates)? Did you give the answer and I missed it?

My answer is that he sees radiation, provided that one accepts a certain local definition of "radiation". But one does not necessarily need to agree with such a definition. In general there is no "obvious" definition of radiation, so various ad hoc definitions are possible.

 

[stevendaryl]

My answer is that he sees radiation, provided that one accepts a certain local definition of "radiation". But one does not necessarily need to agree with such a definition. In general there is no "obvious" definition of radiation, so various ad hoc definitions are possible.

I thought that others had come to the opposite conclusion (about the same local definition of "radiation").

 

[stevendaryl]

I thought that others had come to the opposite conclusion (about the same local definition of "radiation").

We can operationalize it as: You have a charge Q that is suspended in the middle of an accelerating rocket. Will a hydrogen atom in the wall or ceiling or floor of the rocket be knocked into an excited state? Will a photographic plate in the wall or ceiling or floor be darkened?

 

[jartsa]

We can operationalize it as: You have a charge Q that is suspended in the middle of an accelerating rocket. Will a hydrogen atom in the wall or ceiling or floor of the rocket be knocked into an excited state? Will a photographic plate in the wall or ceiling or floor be darkened?

The rocket is a charged accelerating object, so it must radiate, so its radiation must not be absorbed by itself. I mean the rocket should radiate according to an observer far from the rocket.

If there is also charge -Q somewhere in the rocket ... then the rocket must still radiate - like an accelerating dipole radiates.... I mean if your spaceship is charged the enemies can detect your position from the E-field. If the position velocity changes, information about the new position velocity is transmitted by EM-waves. And Faraday cage or any other wall that moves with the spaceship does not help.

If a charged rocket travels back and forth like a sine wave, what is the radiation like? Not a sine wave, as the frequency of radiation depends on acceleration?? So antennas are (almost) linear devices, while charged rockets aren't?

 

[Demystifier]

I thought that others had come to the opposite conclusion (about the same local definition of "radiation").

Are you sure that it was the same local definition?

 

[stevendaryl]

Are you sure that it was the same local definition?

Parrot, summarizing Singal says:

He shows that the Poynting vector vanishes in the rest frames of certain co-accelerating observers and concludes from this that “in the accelerated frame, there is no energy flux, ... , and no radiation”.

 

[Demystifier]

There is still a mystery to clear up, in my mind. If a charged particle radiates, according to an inertial observer, then that means that some of the energy used to accelerate the particle goes into radiation, by energy conservation. This means that it requires more energy to accelerate a charged particle than an uncharged particle of the same mass. This in turn means that for a Rindler rocket carrying a charged particle, more fuel is required than if it were carrying an uncharged particle of the same mass.

So far I agree.

So the charged particle would appear to weigh more.

Does a light bulb in a rocket weigh more just because it radiates? I wouldnt's say so. The same for the charged particle.

But then is the same true for a charged particle at rest in a gravitational field?

If a charged particle at rest on Earth emits energy, where does it take energy from?

 

[Demystifier]

Parrot, summarizing Singal says:

That doesn't coincide with the definition of radiation I use in the paper.

 

[stevendaryl]

Does a light bulb in a rocket weigh more just because it radiates? I wouldnt's say so.

The "weight" of an object is operationally the force needed to keep it in place. It's just a word. What I meant was: does it require more force to keep a charged particle of mass $M$ at "rest" in a Rindler spaceship than it does to keep an uncharged particle? And is that true in a gravitational field.

If a charged particle at rest on Earth emits energy, where does it take energy from?

You argue in your paper that one can't assume that the self-force of a charged particle in a gravitational field is isotropic. If it's not isotropic, and has a radial component, then that would manifest itself as an additional force needed to keep the particle in place, whether or not it manifested itself as radiation.

 

[Tom Kunich]

You have to be careful here. What specific measurement supports that claim that it doesn’t radiate in a gravitational field, and what exact measurement is the equivalent one for an accelerating charge? Is there actually a different prediction for the two?

Where are you suggesting an electron is getting the energy to radiate?

 

[Dale]

Where are you suggesting an electron is getting the energy to radiate?

I’m not. Please read the many subsequent posts for clarification.

 

[PeterDonis]

does it require more force to keep a charged particle of mass $M$ at "rest" in a Rindler spaceship than it does to keep an uncharged particle?

The proper acceleration required to be "at rest" in a particular Rindler spaceship is fixed by the worldline of the spaceship; it's the path curvature of the worldline. Saying that it takes more force to keep a charged particle of mass $M$ at rest in the same spaceship as an uncharged particle of mass $M$ seems like saying that two objects following the same worldline can have different proper accelerations, which is impossible.

 

[Tom Kunich]

So far I agree.Does a light bulb in a rocket weigh more just because it radiates? I wouldnt's say so. The same for the charged particle.If a charged particle at rest on Earth emits energy, where does it take energy from?

I don't think we need to think of acceleration in a rocket. This entire universe is accelerating one way or the other. Therefore any motion other than that it is presently in is an acceleration of one sort or another.

 

[stevendaryl]

The proper acceleration required to be "at rest" in a particular Rindler spaceship is fixed by the worldline of the spaceship; it's the path curvature of the worldline. Saying that it takes more force to keep a charged particle of mass $M$ at rest in the same spaceship as an uncharged particle of mass $M$ seems like saying that two objects following the same worldline can have different proper accelerations, which is impossible.

[edit]

The total force must be the same for charged and uncharged particles, but in the case of the uncharged particle, the only force acting is the upward force of the rocket, while for the charged particle, there is also an electromagnetic force.

To accelerate a charged particle you need: $F^\mu_{rocket} + F^\mu_{field} = M A^\mu$, while for an uncharged particle, you only need $F^\mu_{rocket} = M A^\mu$

 

[Ian J Miller]

At the risk of being seen as missing something, has anybody ever observed radiation emitted by an electron, or any charged particle, falling in a gravitational field? By that, I mean there must be no change of electromagnetic potential during the charge falling. Also excluded must be heat generated radiation, e..g. radiation from mass falling into an accreting star.

 

[eachus]

At the risk of being seen as missing something, has anybody ever observed radiation emitted by an electron, or any charged particle, falling in a gravitational field? By that, I mean there must be no change of electromagnetic potential during the charge falling. Also excluded must be heat generated radiation, e..g. radiation from mass falling into an accreting star.

This is easy. Use a van de Graaff generator to charge a ball, let's say you cover a tennis ball with aluminum foil. Now drop it through a coil of wire. You will see a pulse of current. If you want to get rid of the effect of the coil, use two coils connected together and wound in opposite directions. (Or wind two strands in a coil and connect the far ends together.) Don't connect the other ends of the wires to anything. Now measure the voltage from one (free) end of a coil to where the coils are joined. Drop the ball and you will still see a spike.

 

[romsofia]

This topic has been explored in the 60s by DeWitt, and is not a paradox. Here is the paper: http://inspirehep.net/record/44837/files/vol1p3-20_001.pdf

 

[jartsa]

This is easy. Use a van de Graaff generator to charge a ball, let's say you cover a tennis ball with aluminum foil. Now drop it through a coil of wire. You will see a pulse of current. If you want to get rid of the effect of the coil, use two coils connected together and wound in opposite directions. (Or wind two strands in a coil and connect the far ends together.) Don't connect the other ends of the wires to anything. Now measure the voltage from one (free) end of a coil to where the coils are joined. Drop the ball and you will still see a spike.

The ball should not collide with anything except gravity field. Otherwise we are studying collision of charge with something else than gravity field. So:

Use a van de Graaff generator to charge a ball, let's say you cover a tennis ball with aluminum foil. Bore a hole through the earth. Now drop the ball into the hole, where it will move back and forth. Measure the generated radio-wave.Oh yes, no detectors near the hole. They would measure "near field effects". In other words the ball would collide into the electric fields of the devices.

https://en.wikipedia.org/wiki/Near_and_far_field

 

[Demystifier]

To accelerate a charged particle you need: $F^\mu_{rocket} + F^\mu_{field} = M A^\mu$, while for an uncharged particle, you only need $F^\mu_{rocket} = M A^\mu$

I think you are doing a category mistake here. You cannot add together forces that belong to different categories. The force of electric field belongs to the category of fundamental microscopic forces, together with gravitational, weak and strong force. The force of rocket belongs to the category of forces caused by various macroscopic objects, such as rocket, Earth, hammer, etc. The force of rocket on the charge is of the electromagnetic origin, so you are doing a kind of double counting. (Which reminds me of the "proof" that horse has 8 legs; 2 left legs, 2 right legs, 2 forward legs and 2 backward legs. )