Charge in free fall

[quiet]

Hi. First of all I beg to be apologized, in case of creating the thread in the wrong section. Maybe you can help in a question that causes me doubts.

- Ideal conditions, all theoretical, as simple as possible, mental experiment, everything happens in a vacuum.

- Device formed by an electrically charged spherical body and a radiation detector, rigidly connected by an insulating connection.

- Fixed to the planet there is another radiation detector.

- Condition A: Initially the device is at a distance from the ground and there is no relative movement between the device and the planet.

- Condition B: The device is in free fall towards the planet.

- The planet is electrically neutral.

- What does each detector record in condition A?

- What does each detector record in condition B?

[MigL]

This is all 'frame of reference' related...

In condition A, from the charged object frame, it is accelerating with respect to the planet ( to stay stationary ), and so its detector will detect radiation.
In condition B, the object is not accelerating in its own frame but it is in the planet's frame, so the planet's detector will detect radiation.

I know we've discussed this at length previously, but cannot recall the thread.

[quiet]

1 hour ago, MigL said:

This is all 'frame of reference' related...

Thank you MigL for providing a response. I need now to think a little.

[Markus Hanke]

15 hours ago, MigL said:

In condition A, from the charged object frame, it is accelerating with respect to the planet ( to stay stationary ), and so its detector will detect radiation.

This is not correct. There is no relative motion between detector and electric charge, so, from the perspective of the detector, there is no magnetic field. Without a magnetic field, there can of course be no radiation (which is an oscillating electromagnetic field) - despite the fact that this is a “supported” (i.e. accelerated) charge. Hence, in condition A), the radiation detector detects nothing, because there is no magnetism there to form any radiation. More generally, a “supported” observer in the same frame as a “supported” charge will never detect the charge emitting any radiation. 

The same is also true when you put the entire assembly into free fall - again there is no relative motion between detector and charge, and hence there is also no radiation detected in condition B). This preserves the equivalence principle - it’s in fact just a simple charge without any forces acting on it locally.

It is very important to understand though that, in a curved spacetime background, the answer to the question of whether or not there is radiation detected is a purely local one; if the detector were placed (e.g.) on the surface of the Earth, as opposed to in a co-moving frame with the charge, the outcome would be different. For condition A) you still would not have radiation, but for condition B) you would now detect a radiation field, because there is accelerated motion between detector and charge. Freely falling charges will always appear to radiate from the point of view of a “supported” observer.

The paradox is only an apparent one though, because observers in curved spacetimes are not related by Lorentz transformations, so they do not necessarily agree on which aspects of the electromagnetic field they detect, and which ones not. It’s not a contradiction, just a question of perspective. The other thing of course is that an observer who experiences proper (!) acceleration (a supported observer) has a horizon associated with his reference frame, so there are regions of spacetime that are inaccessible to him - simply put, the entirety of the electromagnetic field is always there, but not all observers are able to detect all aspects of it. Hence under the right conditions, some observers see radiation, whereas others do not. Since energy is also observer-dependent, there is also no contradiction in that regard.

[MigL]

The problem specified two detectors Markus, one on the charged object, and the other on the planet.

So, while you are right about condition A ( frames can get confusing, and I was probably considering a third frame ), for condition B, the detector on the charged object will not detect radiation, but the one on the planet will.

Thanks for the clarification.

Edited August 27, 2018 by MigL

[Markus Hanke]

47 minutes ago, MigL said:

and the other on the planet.

You are right, I misread this. My bad. In this case, what I had written in the third paragraph of my earlier post applies - as you correctly say, the earth-bound detector will detect radiation.

[quiet]

10 hours ago, MigL said:

The problem specified two detectors

 

9 hours ago, Markus Hanke said:

the earth-bound detector will detect radiation.

Hi. Surely it is a mistake to use classical mechanics in this problem. Please also allow me to raise a doubt in those terms.

In classical mechanics the mass of a body links acceleration with force. For example, the resultant force is equal to zero in a system of two equal and opposite forces. Is that equivalent to two equal and opposite accelerations, which give a resultant acceleration equal to zero? If that equivalence were coherent, then in condition A a force is the weight of the set formed by the charged body and the detector that accompanies it. Another force is the sustentation, opposed to the weight. In case of reasoning well, in classical terms, in condition A the net acceleration is equal to zero and the charge does not radiate.

In condition B there is only the weight and there is no sustaining force. Then the net acceleration is nonzero and the charge radiates.

In case of being the previous erroneous thing, a didactic explanation would be very useful for me, that allows to easily understand the error.

[MigL]

It  is not a problem with classical mechanics, but rather a problem with frames of reference.
The radiation, like energy, is frame dependent, and so, is different in differing frames.
If you are not measuring an acceleration of the charge, in the particular frame you are measuring from, then the charge needs not  radiate. But if you are measuring an acceleration of the charge, from whatever frame you are in, you will also detect radiation.

[quiet]

Is acceleration absolute? I ask for what is observed, for example in a car that travels occupied by 3 people. We admit ideal conditions to simplify. When braking intensely, all people suffer the shaking without seeing relative acceleration between people. At the same time, between a person standing on the sidewalk and a car accelerating, there is relative acceleration. But only the occupants of the car feel the effect. That is, the relative acceleration is physically irrelevant. Only own acceleration causes effects and, judging the evidence, I am forced to understand that it is absolute. Errors in my interpretation of those examples?

[Markus Hanke]

5 hours ago, quiet said:

Is acceleration absolute?

There are two types of measuring acceleration - there is proper acceleration, which is what an accelerometer in the accelerated frame would physically record. This is “absolute” in the sense that all observers agree on it. And then there is coordinate acceleration, which is what a distant observer would calculate using his own set of clocks and rulers; this cannot be measured with any kind of accelerometer, and it is an observer-dependent quantity, and hence relative.

The two are not the same at all. Consider the case of the charge in free fall again - an accelerometer attached to the charge will read zero at all times (since it is in free fall), so there is no proper acceleration. However, a distant observer who looks on from “the outside” sees the charge fall faster and faster as it approaches the Earth; using his own clock and his own way to measure velocities, he thus calculates a non-zero coordinate acceleration. What the magnitude of that coordinate acceleration will be depends on the observer - not everyone agrees on it.

[quiet]

Thanks MigL, thanks Markus for what you have exposed.

Perfectly understood. I suppose I can think without confusing the proper acceleration with the acceleration that is detected only from some reference systems.

Radiating is not the only phenomenon caused by acceleration. Think of an object designed to break when it suffers a proper acceleration. The charge and the fragile object are on board the same vehicle. If we say that the charge is seen radiating from some reference systems and from others not, none of our claims destroys the charge. But we must be more careful in what we say about the fragile object. If a rupture is observed from a reference system can, from another reference system, be observed that the object has not been broken?

[Markus Hanke]

7 hours ago, quiet said:

If a rupture is observed from a reference system can, from another reference system, be observed that the object has not been broken?

No, because all observers agree on the value of proper acceleration of that object, no matter what reference frame they are in. This proper acceleration is calculated in a way that is independent of the coordinate system chosen.

[quiet]

I will express my conviction. The charge radiates only in the conditions that break the object. In cases that do not cause breakage, the charge not radiate. Why ? Because to break and to radiate, a contribution of energy is necessary. And only the own acceleration corresponds to an energy received by the object that is broken or by the load that radiates.

Edited August 30, 2018 by quiet

[Markus Hanke]

6 hours ago, quiet said:

The charge radiates only in the conditions that break the object.

As I said before, the electromagnetic field (and it’s associated energy-momentum tensor) is always the same, it’s just that different observers see different aspects of it, depending on their own perspective in spacetime. That is why some see radiation, whereas others do not. But breakage of a macroscopic object isn’t like that - it either happens or it doesn’t. Everyone agrees on the outcome, they just won’t agree on the where and when.

[quiet]

Thanks for point out that.

[John Ye]

On 2018/8/28 at 8:03 AM, MigL said:

It  is not a problem with classical mechanics, but rather a problem with frames of reference.
The radiation, like energy, is frame dependent, and so, is different in differing frames.
If you are not measuring an acceleration of the charge, in the particular frame you are measuring from, then the charge needs not  radiate. But if you are measuring an acceleration of the charge, from whatever frame you are in, you will also detect radiation.

If each detector is installed a radiation detonated explosive, how about the outcome?

[Markus Hanke]

6 minutes ago, John Ye said:

If each detector is installed a radiation detonated explosive, how about the outcome?

We have already established that one detector goes off, whereas the other one does not. Obviously, it will be the same with the explosives.

[John Ye]

On 2018/8/30 at 8:07 AM, quiet said:

I will express my conviction. The charge radiates only in the conditions that break the object. In cases that do not cause breakage, the charge not radiate. Why ? Because to break and to radiate, a contribution of energy is necessary. And only the own acceleration corresponds to an energy received by the object that is broken or by the load that radiates.

I agree with you. Both detectors will not detect any radiation in either case.

At the moment that the ball crashes to ground, both detectors will if they are still functioning.

The key is this: accelerating charge driven only by one force will not emit radiation. Decelerating charge will.

For example, electrons in linear accelerator will not emit energy while being accelerated. They absorb energy at this stage.

Btw, a charge producing magnetic field does not necessarily mean it emits radiation. Magnetic Fe is producing magnetic field all the time, by no energy is emitted. The same is as a superconductor wire current, which can keep flowing for years, generating magnetic field but not radiation.

 

 

Edited September 4, 2018 by John Ye

[swansont]

1 hour ago, John Ye said:

I agree with you. Both detectors will not detect any radiation in either case.

At the moment that the ball crashes to ground, both detectors will if they are still functioning.

The key is this: accelerating charge driven only by one force will not emit radiation. Decelerating charge will.

Um, no. How do antennas work, if accelerating a charge does not radiate?

1 hour ago, John Ye said:

For example, electrons in linear accelerator will not emit energy while being accelerated. They absorb energy at this stage.

The radiation is small, but not zero

http://farside.ph.utexas.edu/teaching/em/lectures/node131.html

1 hour ago, John Ye said:

Btw, a charge producing magnetic field does not necessarily mean it emits radiation. Magnetic Fe is producing magnetic field all the time, by no energy is emitted. The same is as a superconductor wire current, which can keep flowing for years, generating magnetic field but not radiation.

Quantum systems won't radiate when there is no lower energy state for them to go to.

[John Ye]

12 minutes ago, swansont said:

Um, no. How do antennas work, if accelerating a charge does not radiate?

Electrons in antenna are not keeping accelerating. They are accelerating then decelerating, that is oscillation. 

When accelerating, they are absorbing energy from power supply, when decelerating, releasing the energy to air in electromagnetic mode. 

 

23 minutes ago, swansont said:

Quantum systems won't radiate when there is no lower energy state for them to go to.

Yes. An electron goes from infinite far (highest energy level) to a lower energy level position, it must be suddenly stopped, and emits its extra energy. Decelerated electrons emit energy, accelerating ones don't.

This process is transition, I described it in my that script we have discussed few days ago.

[swansont]

10 minutes ago, John Ye said:

Electrons in antenna are not keeping accelerating. They are accelerating then decelerating, that is oscillation. 

Deceleration is lay terminology. Acceleration is acceleration. The direction is a mere detail. 

Radiation from antennas is continuous, meaning it radiates while the charges speed up, and while they slow down.

Quote

When accelerating, they are absorbing energy from power supply, when decelerating, releasing the energy to air in electromagnetic mode. 

Which is obviously not true. See comment above.

And of course we also have cyclotron radiation, in which the radiation is perpendicular to the velocity.

[John Ye]

3 minutes ago, swansont said:

Quote

When accelerating, they are absorbing energy from power supply, when decelerating, releasing the energy to air in electromagnetic mode. 

Which is obviously not true. See comment above.

It is true. Textbook gives wrong explanation. This can be tested and proved by detecting the radiation in a linear accelerator, and a linear decelerator. 

 

[swansont]

26 minutes ago, John Ye said:

It is true. Textbook gives wrong explanation. This can be tested and proved by detecting the radiation in a linear accelerator, and a linear decelerator. 

I already provided the link that shows that the radiation in a linac is small but nonzero. The stopping distance is much smaller than the acceleration distance, which is why you get all that radiation.

[Phi for All]

37 minutes ago, John Ye said:

It is true. Textbook gives wrong explanation.

!

Moderator Note

Then open a thread in Speculations and support that argument, but don't do it in someone else's mainstream thread. You've been warned about hijacking before.

 

[John Ye]

6 minutes ago, swansont said:

I already provided the link that shows that the radiation in a linac is small but nonzero. The stopping distance is much smaller than the acceleration distance, which is why you get all that radiation.

You mean that radiation is small (non zero) when accelerating and  large when decelerating. They are not equal in quantity? That might be true due to non ideal experiment environment. If absolutely ideal, the radiation should be zero at accelerating stage.

We can make real experiments. Detect radiation in each stage, radiation in   accelerating stage will not be detectable.

3 minutes ago, Phi for All said:

!

Moderator Note

Then open a thread in Speculations and support that argument, but don't do it in someone else's mainstream thread. You've been warned about hijacking before.

 

OK. I will obey your order.