I Paradox: Electron Radiates in a Gravitational Field

  • #101
Tom Kunich said:
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.
If you mean accelerated expansion of the Universe caused by dark energy, that's "acceleration" in a different sense which is not relevant here.
 
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  • #102
stevendaryl said:
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.

What Demystifier said in post #88 and 100, and PeterDonis in post #93.

I think the problem is easier if it is assumed that the rocket engine thrust is continuously adjusted for constant acceleration, regardless the effects of the lab experiments, and regardless of the variable mass due to fuel consumption, etc. Consider the chamber of the rocket as separate from the engine that is accelerating it (similarly with the rope accelerating the chamber in the case of an accelerated elevator).
 
  • #103
Demystifier said:
The principle of equivalence is valid only locally. The concept of radiation, on the other hand, is a global concept. The claim that an accelerating charge in Minkowski spacetime "radiates" means that the flux of Poynting vector through a sphere far from the charge does not vanish. Hence you cannot apply the principle of equivalence to determine whether the static charge in gravitational field radiates. The equivalence principle is only helpful to study EM field in a vicinity of the charge, but this short-range effects do not tell much about radiation.

Can we achieve a valid EP region by making the region much larger, perhaps such as on a planet of far greater radius than the earth? Or what if we put a filter on the apparatus such that a parallel beam of radiation is emitted? Or what if we consider high energy radiation with very small wavelength?
 
  • #104
Demystifier said:
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. :biggrin: )

This is not actually clarifying anything. Let me try to explain again.

Let's pick an inertial coordinate system. Suppose that at time ##t_0##, you have a situation where there is a region of space that can be described in this coordinate system by:
  • There is a charged particle with mass ##M## and charge ##Q##.
  • The particle have 3-velocity ##\vec{v}##.
  • There is an electromagnetic field ##F^{\mu \nu}##.
I think everyone would agree that in accounting for energy conservation, you can't just use ##E = \gamma M c^2##. The total energy must also take into account electromagnetic energy.

So if you replace the charged particle by an uncharged particle, and keep the 3-velocity ##\vec{v}## the same and keep the mass ##M## the same, the total energy will not be the same, because the electromagnetic energy will be different. Does anybody disagree with this?

If you agree, then it follows that you can't just use ##E = \gamma M c^2## to calculate how much energy was needed to accelerate the mass ##M## to achieve velocity ##\vec{v}##. The charge ##Q## and the electromagnetic field ##F^{\mu \nu}## must be taken into account, as well.

If it turns out that the energy required to accelerate a particle from rest to velocity ##\vec{v}## is independent of whether it's charged, that's a pretty remarkable thing. I'm pretty sure it's not true, in general.

Now, instead of looking at the general case, let's look at the special case where ##F^{\mu \nu}## is the electromagnetic field due to the charge ##Q##. In that case, maybe you can say that some of the electromagnetic energy is already accounted for in the mass ##M##. That is, the mass ##M## already takes into account the electromagnetic self-energy. Does it then follow that the amount of energy required to accelerate a charged particle is the same as for an uncharged particle of the same mass? Maybe that's true, but it certainly isn't true by definition. It requires an argument.
 
  • #105
MikeGomez said:
What Demystifier said in post #88 and 100, and PeterDonis in post #93.

Those three posts don't answer the question, at all. There is no reason, a priori, to believe that it takes the same amount of energy to accelerate a charged particle in an electromagnetic field as it does to accelerate an uncharged particle of the same mass. You can't compute the energy by simply knowing the mass and the final velocity.
 
  • #106
stevendaryl said:
Those three posts don't answer the question, at all. There is no reason, a priori, to believe that it takes the same amount of energy to accelerate a charged particle in an electromagnetic field as it does to accelerate an uncharged particle of the same mass. You can't compute the energy by simply knowing the mass and the final velocity.

When a charged particle is at rest in some frame, then I suppose that the mass of the particle already takes into account the electromagnetic energy. But is that true when the particle is accelerating, as well?
 
  • #107
stevendaryl said:
If it turns out that the energy required to accelerate a particle from rest to velocity ##\vec{v}## is independent of whether it's charged, that's a pretty remarkable thing. I'm pretty sure it's not true, in general.
I agree with that. That's similar to the fact that a car moving with given acceleration spends more fuel when the car's lights are turned on.
 
  • #108
Demystifier said:
I agree with that. That's similar to the fact that a car moving with given acceleration spends more fuel when the car's lights are turned on.

So it isn't obviously true that the energy required to transport a cargo of mass ##M## on board a Rindler rocket is independent of whether the cargo is charged.

But presumably, if the same rocket is hovering above a black hole, the energy required is independent of whether the cargo is charged.
 
  • #109
stevendaryl said:
But presumably, if the same rocket is hovering above a black hole, the energy required is independent of whether the cargo is charged.
I think that it isn't independent.
 
  • #110
Demystifier said:
You cannot add together forces that belong to different categories.
Sure you can. You just need to make sure that the categories are mutually exclusive and collectively exhaustive.
 
  • #111
Dale said:
Sure you can. You just need to make sure that the categories are mutually exclusive and collectively exhaustive.
But in this case they are not mutually exclusive because the rocket force has an electromagnetic origin.
 
  • #112
Demystifier said:
But in this case they are not mutually exclusive because the rocket force has an electromagnetic origin.
Ok, so make that point instead. Adding forces of different categories is fine. Double counting is not.
 
  • #113
Dale said:
Ok, so make that point instead. Adding forces of different categories is fine. Double counting is not.
Technically, you are right. But when you add forces from different categories, then there is a great risk of double counting or some other type of confusion. So in practice, it's better to avoid it.
 
  • #114
Demystifier said:
But in this case they are not mutually exclusive because the rocket force has an electromagnetic origin.

But the whole point I was making is that the force supplied by the rocket need not be the same for a charged versus uncharged cargo, because the charged cargo is moving through an electromagnetic field, while the uncharged cargo is not.

@PeterDonis wrote

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.

It's certainly true that the total force must be the same (by definition), but it doesn't follow that the force that the rocket must exert on the particle is the same, since the rocket is not the only thing exerting forces on the particle.

Certainly if you weigh a ping-pong ball on a scale in the presence of an electric field, the measured weight will be different if the ball has a charge on it.
 
  • #115
I think of Einsteins elevator experiment. A guy and an electron in a box, all three are members of the same inertial system. If you accelerate the box, the guy and the electron drop to the floor, floor pushes back, and they are accelerated. If you put the box in a gravitational field, again the guy and the electron drop to the floor, floor pushes back to counter the force of gravity, and they are at rest. The principle of equivalence says that no experiment inside the box can detect the difference. If an electron is being pushed up by the floor in a gravitational field by whatever apparatus, and the guy in the box says it does not radiate, then the electron in the accelerated box being pushed up by the floor by the same apparatus will be deemed to be not radiating by the guy being accelerated along with that electron.
 
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