Do inspiral charged black hole pairs radiate light?

[md65536]

I figure they would because their orbits should have an effect on the electromagnetic field detectable at a distance?

[Markus Hanke]

So long as you are far enough from the BH pair, you can consider this situation as being two charged point particles in free fall. We already know that freely falling charges do not radiate (irrespective of metric), so my educated guess would be that there is no light detected. Given that, I don’t see how the situation could be different in the near field, so my guess is that there’s no radiation anywhere.

However, this is an unusual and mathematically pretty involved scenario, so it is possible that I might be wrong. I don’t see how they could radiate though without violating the equivalence principle.

[KJW]

15 hours ago, Markus Hanke said:

We already know that freely falling charges do not radiate (irrespective of metric)

...

I don’t see how they could radiate though without violating the equivalence principle.

Why do you say that freely falling charges do not radiate? A charged object sitting on a table doesn't radiate even though it is accelerating. I think the circumstances under which a charge radiates is not straightforward. This is a topic that interests me a lot and I've even been considering this issue quite recently. Part of the difficulty as I see it is that the mathematics that says an accelerated charge radiates is limited to the Minkowskian metric and needs to be generalised to arbitrary metrics. Another difficulty as I see it is how the Lorentz force relates to Maxwell's equations. I'll discuss this in more detail in the hopefully-not-too-distant future.

 

[MigL]

We had this discussion a few years back, and @Markus Hanke participated in it.
At the time, he convincingly stated that an accelerometer attached to a free-falling charged particle shows no acceleration, so there is no reason to radiate.
I'm willing to re-open the discussion, perhaps in a split-off thread as it is off-topic here, to address your question, as well as my own ( which is at least, more closely related to the OP )

The accretion disk of a Black Hole strips atoms of their electrons, leaving charged particles spiraling towards the Event Horizon.
These charged particles are in free fall, and BHs with accretion disks emit massive amounts of radiation as polar jets.
It would seem that observational evidence supports radiation from free falling charges.

[Genady]

19 minutes ago, MigL said:

These charged particles are in free fall

Are they? Don't they collide with each other and affect each other non-gravitationally?

[KJW]

28 minutes ago, MigL said:

We had this discussion a few years back, and @Markus Hanke participated in it.
At the time, he convincingly stated that an accelerometer attached to a free-falling charged particle shows no acceleration,

I fully agree with this.

 

 

28 minutes ago, MigL said:

so there is no reason to radiate.

One possibility that I'm currently unwilling to accept is that radiation from a charge is not covariant, that it does depend on the frame of reference from which it is observed.

 

Edited February 19 by KJW

[Genady]

1 hour ago, KJW said:

A charged object sitting on a table doesn't radiate even though it is accelerating.

AFAIK, not every acceleration causes radiation from a charged particle. For example, in synchrotron radiation the acceleration is perpendicular to the particle velocity. In some other cases, a magnitude of acceleration is variable. Is it correct?

Edited February 19 by Genady

[KJW]

22 minutes ago, Genady said:

AFAIK, not every acceleration causes radiation from a charged particle. For example, in synchrotron radiation the acceleration is perpendicular to the particle velocity. In some other cases, acceleration is variable. Is it correct?

But when a charge accelerates, there has to be a back-reaction on the electromagnetic field causing it to accelerate. So, I don't think it is as straightforward as invoking derivatives of the acceleration. On the other hand, radiation from a charge can be obtained from the solution to Maxwell's equations.

 

Edited February 19 by KJW

[swansont]

41 minutes ago, Genady said:

AFAIK, not every acceleration causes radiation from a charged particle. For example, in synchrotron radiation the acceleration is perpendicular to the particle velocity. In some other cases, a magnitude of acceleration is variable. Is it correct?

But there is radiation, so how is this an example of an acceleration not resulting in radiation?

1 hour ago, KJW said:

A charged object sitting on a table doesn't radiate even though it is accelerating.

Not in its rest frame, but that’s not an inertial frame. In an inertial frame (i.e. in freefall) you would detect radiation.

https://en.m.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field

(see Resolution by Rohrlich)

“The key is to realize that the laws of electrodynamics, Maxwell's equations, hold only within an inertial frame”

[KJW]

The way I see it, if one expresses the electromagnetic energy-momentum tensor in terms of the electromagnetic field tensor, then the divergence of this quadratic expression, which gives the Lorentz force, should also be the source of this quadratic expression similar to the way the charge current is the source of the electromagnetic field. That is, instead of considering the divergence of the energy-momentum tensor itself, which is a force, one considers the divergence of a particular quadratic expression of the electromagnetic field, which should be part of Maxwell's equations rather than requiring the Lorentz force explicitly.

 

 

[Genady]

22 minutes ago, swansont said:

But there is radiation, so how is this an example of an acceleration not resulting in radiation?

It is not. It is an example of not every acceleration resulting in radiation, i.e., in these examples a changing acceleration results in radiation.

[swansont]

22 minutes ago, Genady said:

It is not. It is an example of not every acceleration resulting in radiation, i.e., in these examples a changing acceleration results in radiation.

Moving in a circle is not a changing acceleration. It’s always in the radial direction

[Genady]

3 minutes ago, swansont said:

Moving in a circle is not a changing acceleration. It’s always in the radial direction

The direction of the acceleration changes. The vector rotates.

[swansont]

1 hour ago, Genady said:

The direction of the acceleration changes. The vector rotates.

But that’s not the reason for the radiation. The radiation is along the direction of motion, perpendicular to the instantaneous acceleration.

[Genady]

40 minutes ago, swansont said:

But that’s not the reason for the radiation. The radiation is along the direction of motion, perpendicular to the instantaneous acceleration.

Well, the instantaneous change of acceleration, in the case of circular motion, happens to be also "along the direction of motion, perpendicular to the instantaneous acceleration". However, I don't know how they (i.e., the time derivative of acceleration and the radiation) are related and don't claim anything in this regard. My point in this post was that AFAIK, a constant acceleration of charged particle can be insufficient to cause radiation.

Edited February 20 by Genady

[Markus Hanke]

7 hours ago, KJW said:

Why do you say that freely falling charges do not radiate?

Because this would provide a way to locally test whether you’re in free fall in a gravitational field, or just in an “ordinary” inertial frame - which is a violation of the equivalence principle.

Either way, I think the answer to this has been worked out mathematically by different authors, for example here.

7 hours ago, KJW said:

I think the circumstances under which a charge radiates is not straightforward.

I agree, we need to be very careful with reference frames and the form of the laws we apply.

7 hours ago, KJW said:

A charged object sitting on a table doesn't radiate even though it is accelerating.

Not if the detector is comoving wrt to it. However, if the detector is in a locally inertial frame (ie freely falling past the charge), then radiation is detected.

6 hours ago, KJW said:

One possibility that I'm currently unwilling to accept is that radiation from a charge is not covariant, that it does depend on the frame of reference from which it is observed.

That can’t be the case, since the electromagnetic field is a tensorial quantity.

However, we must remember that accelerated reference frames have Rindler horizons - so for a comoving detector the radiation is essentially in a region of spacetime that’s inaccessible to it.

Everyone agrees (tensor!) that there’s a radiation field, but not everyone has access to it.

[J.C.MacSwell]

14 hours ago, swansont said:

 

Not in its rest frame, but that’s not an inertial frame. In an inertial frame (i.e. in freefall) you would detect radiation.

https://en.m.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field

(see Resolution by Rohrlich)

“The key is to realize that the laws of electrodynamics, Maxwell's equations, hold only within an inertial frame”

If it's radiating in any frame it should be radiating in every frame.

Since it is being accelerated in it's instantaneos inertial frame should it not radiate?

7 hours ago, Markus Hanke said:

 

Not if the detector is comoving wrt to it. However, if the detector is in a locally inertial frame (ie freely falling past the charge), then radiation is detected.

 

Same event should take place in each frame, even if seen differently, correct?

What am I missing?

Other than inside neutral atoms, shouldn't any charged particle constrained in a gravitational field (so not in free fall) radiate, even if negligibly/undetectably?

 

Special conditions (that I don't understand) aside:

 

"According to the Larmor formula in classical electromagnetism, a single point charge under acceleration will emit electromagnetic radiation. In some classical electron models a dmitistribution of charges can however be accelerated so that no radiation is emittedhttps://en.wikipedia.org/wiki/Nonradiation_condition#:~:text=According to the Larmor formula,that no radiation is emitted."

 

https://en.wikipedia.org/wiki/Nonradiation_condition#cite_note-Pearle_1978-1

Edited February 20 by J.C.MacSwell

[MigL]

16 hours ago, Genady said:

Don't they collide with each other and affect each other non-gravitationally?

Yes they do.
But such interactions tend to make the accretion disk planar; see Saturn's rings for an example.
Further, such interactions would be random, and would produce random radiation.

The inspiralling accretion disk produces a very specific radiation pattern, polar jets, of the type associated with orbiting charges.
All 'active' Black Holes display such jets; some around early massive galactic center BHs produce jets that outshine the parent galaxy, and are known as Quasars.

Since the infalling plasma of the accretion disk is in free fall, why is the radiation apparent/accessible to an outside observer ?

[Genady]

9 minutes ago, MigL said:

Yes they do.
But such interactions tend to make the accretion disk planar; see Saturn's rings for an example.
Further, such interactions would be random, and would produce random radiation.

The inspiralling accretion disk produces a very specific radiation pattern, polar jets, of the type associated with orbiting charges.
All 'active' Black Holes display such jets; some around early massive galactic center BHs produce jets that outshine the parent galaxy, and are known as Quasars.

Since the infalling plasma of the accretion disk is in free fall, why is the radiation apparent/accessible to an outside observer ?

I remember that charged particles moving in rotating magnetic fields are involved in the description of this, but it was too long ago (and not in my main line of study) to recall details. Hope to see some expert answers.

[Markus Hanke]

15 hours ago, J.C.MacSwell said:

If it's radiating in any frame it should be radiating in every frame.

15 hours ago, J.C.MacSwell said:

Same event should take place in each frame, even if seen differently, correct?

What am I missing?

What is present in every frame and for all observers is the electromagnetic field due to the presence of the charge (which is what you refer to as “event” above). This is a tensorial quantity, so all observers agree on it being non-zero. However, observers do not necessarily agree on the value of the individual components of the tensor, since these will be functions of space and time, which are observer-dependent concepts. 

Physically speaking this means that everyone agrees there’s an electromagnetic field, but not everyone agrees what this field “looks like” in terms of its decomposition into E and B components in a given frame, since this decomposition is again observer-dependent. For the field to look like radiation, E and B must be periodic functions of space and time of a specific form, and they must be related in specific ways; this may not be the case in all reference frames.

When you do the maths, what you find is that the radiation emitted by a charge supported in a gravitational field is in fact present even for a comoving (=accelerated) observer, but it is located in a region of spacetime that is inaccessible to him (it is beyond the Rindler horizon). On the other hand, the freely falling observer is locally inertial, so there’s no Rindler horizon, and he can detect the radiation. There’s no contradiction, it’s just that one must be careful about frames and their particular conceptions of space and time.

[MigL]

21 minutes ago, Markus Hanke said:

When you do the maths, what you find is that the radiation emitted by a charge supported in a gravitational field is in fact present even for a comoving (=accelerated) observer, but it is located in a region of spacetime that is inaccessible to him (it is beyond the Rindler horizon). On the other hand, the freely falling observer is locally inertial, so there’s no Rindler horizon, and he can detect the radiation.

Should I then, assume that, if I was a far -away observer of accretion disc plasma falling into a Black Hole, I see radiation because I am also in ( far away ) free fall ?

Yet, If I'm close to the BH, blasting my rocket  to stay a 'supported' distance away, I would detect no radiation as it is beyond a horizon ?

[J.C.MacSwell]

8 hours ago, Markus Hanke said:

What is present in every frame and for all observers is the electromagnetic field due to the presence of the charge (which is what you refer to as “event” above). This is a tensorial quantity, so all observers agree on it being non-zero. However, observers do not necessarily agree on the value of the individual components of the tensor, since these will be functions of space and time, which are observer-dependent concepts. 

Physically speaking this means that everyone agrees there’s an electromagnetic field, but not everyone agrees what this field “looks like” in terms of its decomposition into E and B components in a given frame, since this decomposition is again observer-dependent. For the field to look like radiation, E and B must be periodic functions of space and time of a specific form, and they must be related in specific ways; this may not be the case in all reference frames.

When you do the maths, what you find is that the radiation emitted by a charge supported in a gravitational field is in fact present even for a comoving (=accelerated) observer, but it is located in a region of spacetime that is inaccessible to him (it is beyond the Rindler horizon). On the other hand, the freely falling observer is locally inertial, so there’s no Rindler horizon, and he can detect the radiation. There’s no contradiction, it’s just that one must be careful about frames and their particular conceptions of space and time.

Thanks Marcus I'll check out Rindler horizon and see if I can make sense of it.

[J.C.MacSwell]

So if I have a charged particle on my desk emitting a photon due to constraint against gravity alone, that I can detect with my photon detector with the detector in free fall, there will be no detection whatsoever if it's not in freefall? Or no detection only if the detector is comoving and similarly constrained against gravity?

 

Or I guess detectable or not in other frames dependant on how there movement is with regard to the Rindler horizon? (Does the wavelength of the photon approach infinite with respect to a frame coincident (somehow?) with the Rindler horizon?

[Genady]

On 2/19/2024 at 10:32 PM, Genady said:

Well, the instantaneous change of acceleration, in the case of circular motion, happens to be also "along the direction of motion, perpendicular to the instantaneous acceleration". However, I don't know how they (i.e., the time derivative of acceleration and the radiation) are related and don't claim anything in this regard. My point in this post was that AFAIK, a constant acceleration of charged particle can be insufficient to cause radiation.

After a further contemplation. It appears that I was wrong anyway.

[Markus Hanke]

23 hours ago, MigL said:

Should I then, assume that, if I was a far -away observer of accretion disc plasma falling into a Black Hole, I see radiation because I am also in ( far away ) free fall ?

You would see radiation regardless of how you move, since this arises overwhelmingly from the interactions between particles within the plasma. This is kind of a different scenario.

But if it was only a single isolated charge falling into the BH, then you would not see any radiation from it due to its free fall motion.

23 hours ago, MigL said:

Yet, If I'm close to the BH, blasting my rocket  to stay a 'supported' distance away, I would detect no radiation as it is beyond a horizon ?

Again assuming an isolated charge in free fall. You would not see radiation from the particle, but you would see radiation all around you, since the vacuum has now become a “thermal bath” for you (Unruh effect).

13 hours ago, J.C.MacSwell said:

that I can detect with my photon detector with the detector in free fall, there will be no detection whatsoever if it's not in freefall? Or no detection only if the detector is comoving and similarly constrained against gravity?

The detector needs to undergo accelerated motion for there to be a Rindler horizon, so it needs to be supported against gravity. I don’t think detector and emitter need to be perfectly comoving (ie at relative rest), but to be honest I’m not 100% sure on this.

13 hours ago, J.C.MacSwell said:

Or I guess detectable or not in other frames dependant on how there movement is with regard to the Rindler horizon?

The Rindler horizon is a result of accelerated motion, and thus observer-dependent. But yes, essentially.