Jump to content

# Do inspiral charged black hole pairs radiate light?

## Recommended Posts

On 2/20/2024 at 3:24 PM, Markus Hanke said:
On 2/20/2024 at 7:32 AM, 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.

Actually, my question was semi-rhetorical because I knew you had to answer the way you did. But there is the possibility that radiation from a charge is a non-local phenomenon, in which case the equivalence principle doesn't apply.

On 2/20/2024 at 3:24 PM, Markus Hanke said:
On 2/20/2024 at 7:32 AM, 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.

One argument against the notion that a charged object sitting on a table doesn't radiate is the absence of a magnetic field. A locally inertial detector can be at instantaneous rest relative to the accelerated charge, so that even though they have different accelerations, there is no magnetic field detected from the charge.

On 2/20/2024 at 3:24 PM, Markus Hanke said:
On 2/20/2024 at 8:25 AM, 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.

The electromagnetic field is a tensor, but this is about the splitting of the electromagnetic field into the radiation field and the Coulomb field, and it is this splitting that might not be covariant. So far in this discussion, I've not seen any explanations that are covariant.

On 2/20/2024 at 3:24 PM, Markus Hanke said:

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.

This is the first time I've seen the Rindler horizon invoked to resolve this issue. At present, I don't accept the Rindler horizon explanation for two reasons:

1: What is happening in the spacetime region between the accelerating charge and its Rindler horizon? Note that the proper distance between a constantly accelerating charge and its Rindler horizon is $\dfrac{c^2}{\alpha}$.

2: Rindler horizons don't exist in reality. A Rindler horizon in Minkowskian spacetime occurs as a result of an object outrunning a pulse of light after being given a head start. This requires that the object asymptotically approach the speed of light, the head start given corresponding to the proper distance to the Rindler horizon. But in reality, the object will eventually stop accelerating, the pulse of light will inevitably catch up, and there was no Rindler horizon after all.

Edited by KJW
##### Share on other sites

1 minute ago, KJW said:

that radiation from a charge is a non-local phenomenon

Ahem

##### Share on other sites

On 2/23/2024 at 12:26 AM, KJW said:

But there is the possibility that radiation from a charge is a non-local phenomenon, in which case the equivalence principle doesn't apply.

I guess what you mean is that the radiation field (and the EM field in general) will always be much larger than the local free-fall frame.

There’s also the issue of the field “back-reacting” with the charge, which would make true free-fall impossible in the first place.

These are good points, and I’m not sure how they influence the analysis of this situation.

On 2/23/2024 at 12:26 AM, KJW said:

One argument against the notion that a charged object sitting on a table doesn't radiate is the absence of a magnetic field.

I’m struggling to understand this - why would the absence of a magnetic field contradict the charge not radiating?

On 2/23/2024 at 12:26 AM, KJW said:

The electromagnetic field is a tensor, but this is about the splitting of the electromagnetic field into the radiation field and the Coulomb field, and it is this splitting that might not be covariant.

I completely agree, and this insight should be all that’s needed to understand why some observers see radiation and others don’t.

On 2/23/2024 at 12:26 AM, KJW said:

At present, I don't accept the Rindler horizon explanation for two reasons:

That’s fair enough - how would you yourself evaluate and understand this situation?

##### Share on other sites

22 hours ago, Markus Hanke said:
On 2/23/2024 at 9:26 AM, KJW said:

But there is the possibility that radiation from a charge is a non-local phenomenon, in which case the equivalence principle doesn't apply.

I guess what you mean is that the radiation field (and the EM field in general) will always be much larger than the local free-fall frame.

There’s also the issue of the field “back-reacting” with the charge, which would make true free-fall impossible in the first place.

These are good points, and I’m not sure how they influence the analysis of this situation.

I should point out that I don't have a complete answer to the question. I'm just suggesting plausible possibilities as well as pointing out particular problems as I see them. That is, I'm not claiming that the radiation from a charge is a non-local phenomenon, but merely offer it as something that may nullify the equivalence principle. But it is also worth noting that equations in integral form may also have a differential form, a local form of otherwise non-local equations.

For a charge in an external electromagnetic field, the acceleration of the charge must be accompanied by a back-reaction on the field. And this is both covariant and local.

22 hours ago, Markus Hanke said:
On 2/23/2024 at 9:26 AM, KJW said:

One argument against the notion that a charged object sitting on a table doesn't radiate is the absence of a magnetic field.

I’m struggling to understand this - why would the absence of a magnetic field contradict the charge not radiating?

It would seem that I misspoke. What I meant was that an argument supporting the notion that a charged object sitting on a table doesn't radiate is the absence of a magnetic field. But this argument applies regardless of whether the observer is in the same accelerated frame of reference as the charge or is in an inertial frame of reference that is at instantaneous rest relative to the charge.

22 hours ago, Markus Hanke said:
On 2/23/2024 at 9:26 AM, KJW said:

The electromagnetic field is a tensor, but this is about the splitting of the electromagnetic field into the radiation field and the Coulomb field, and it is this splitting that might not be covariant. So far in this discussion, I've not seen any explanations that are covariant.

I completely agree, and this insight should be all that’s needed to understand why some observers see radiation and others don’t.

That's true. Give up covariance and the whole problem goes away. Seems like a Faustian deal to me.

22 hours ago, Markus Hanke said:
On 2/23/2024 at 9:26 AM, KJW said:

This is the first time I've seen the Rindler horizon invoked to resolve this issue. At present, I don't accept the Rindler horizon explanation for two reasons:

1: What is happening in the spacetime region between the accelerating charge and its Rindler horizon? Note that the proper distance between a constantly accelerating charge and its Rindler horizon is c2α .

2: Rindler horizons don't exist in reality. A Rindler horizon in Minkowskian spacetime occurs as a result of an object outrunning a pulse of light after being given a head start. This requires that the object asymptotically approach the speed of light, the head start given corresponding to the proper distance to the Rindler horizon. But in reality, the object will eventually stop accelerating, the pulse of light will inevitably catch up, and there was no Rindler horizon after all.

That’s fair enough - how would you yourself evaluate and understand this situation?

As I said above, I don't have a complete answer. But I do see two possibilities that are not necessarily consistent:

(1): Radiation only occurs if an accelerated charge is in an external electromagnetic field. That is, there is no radiation purely from an accelerated charge's own electromagnetic field. Such a hypothesis is not invalidated by cyclotron radiation, bremsstrahlung, nor Cherenkov radiation, but would explain in a covariant manner why some accelerated charges do not radiate.

(2): The divergence of the quadratic expression in terms of the electromagnetic field tensor of the energy-momentum tensor gives not only the Lorentz force, but also some expression of the source in terms of an accelerated charge. This may actually be inconsistent with (1), but I think it is fair to assume that the Lorentz force law (along with its equivalent alternative expressions) is consistent with Maxwell's equations.

Edited by KJW
##### Share on other sites

52 minutes ago, KJW said:

That's true. Give up covariance and the whole problem goes away.

But we’re not giving up covariance, are we? We’re simply considering how the EM field - a covariant object - decomposes in a particular frame.

59 minutes ago, KJW said:

1): Radiation only occurs if an accelerated charge is in an external electromagnetic field.

Is this not a form of non-locality? Basically you’re saying that whether or not a charge radiates in some local region depends on the existence of potentially distant sources (=external field).

1 hour ago, KJW said:

(2): The divergence of the quadratic expression in terms of the electromagnetic field tensor of the energy-momentum tensor gives not only the Lorentz force, but also some expression of the source in terms of an accelerated charge.

I need to think about this one first

##### Share on other sites

23 hours ago, Markus Hanke said:

But we’re not giving up covariance, are we? We’re simply considering how the EM field - a covariant object - decomposes in a particular frame.

But if the decomposition depends on the particular frame, that is giving up covariance.

23 hours ago, Markus Hanke said:
Quote

1): Radiation only occurs if an accelerated charge is in an external electromagnetic field.

Is this not a form of non-locality? Basically you’re saying that whether or not a charge radiates in some local region depends on the existence of potentially distant sources (=external field).

The response of the charge is to the electromagnetic field at the location of the charge, and is therefore local to the charge. We are not interested in the source of the external electromagnetic field. The problem then becomes decomposing the electromagnetic field at the charge into the field from the charge and the field not from the charge. But, the response of the charge is to the total electromagnetic field, so the hypothesis is that the field from a point charge is zero at the charge regardless of its motion. However, if a charge can be accelerated non-electromagnetically, then one can test if it radiates in the absence of an external electromagnetic field.

##### Share on other sites

On 2/25/2024 at 3:21 PM, Markus Hanke said:
Quote

(2): The divergence of the quadratic expression in terms of the electromagnetic field tensor of the energy-momentum tensor gives not only the Lorentz force, but also some expression of the source in terms of an accelerated charge.

I need to think about this one first

Have you ever derived the relativistic Lorentz force law from the divergence of the electromagnetic energy-momentum tensor? If you haven't, it may be difficult to appreciate where I'm coming from. I won't do it here, but basically one starts with the energy-momentum tensor on the left side of the equality, and a particular quadratic expression in terms of the electromagnetic field tensor on the right side of the equality. Perform the divergence of both sides. On the left, one has a force, and on the right after a number of steps, the product of the electromagnetic field tensor with the charge current vector (or charge times velocity). If one is willing to accept without question the electromagnetic energy-momentum tensor, then the force on the left becomes mass times acceleration, and we have an expression that says that the source of the particular quadratic expression in terms of the electromagnetic field tensor, divided by the mass of the charge, is the acceleration of the charge. This is analogous to the expression that says that the source of the electromagnetic field tensor, divided by the charge, is the velocity of the charge.

Let's consider something more basic:

The Coulomb field around a charge drops with $\dfrac{1}{r^2}$. Enclose the charge in a sphere and perform a surface integral of the field to obtain the charge as the source of the field, noting that the surface area increases with $r^2$, cancelling the $\dfrac{1}{r^2}$ drop of the field. But the radiation field drops with $\dfrac{1}{r}$, so that one has to square the radiation field to obtain a field that drops with $\dfrac{1}{r^2}$ and whose surface integral obtains the source of this radiation field by cancelling with the $r^2$ increase of the surface area.

For me personally, I don't accept without question the electromagnetic energy-momentum tensor. Thus, instead of obtaining acceleration of the charge from the left side of the equality as above, I would like to obtain it from the right side of the equality, thereby disconnecting the acceleration of the charge from the energy-momentum tensor. It is worth noting that due to the antisymmetry of the electromagnetic field tensor, the product of the electromagnetic field tensor with the charge current vector is orthogonal to the charge current vector, just as the acceleration vector is orthogonal to the charge current vector. But that doesn't imply that these two vectors that are orthogonal to the charge current vector are parallel to each other. Nevertheless, the orthogonality of the product of the electromagnetic field tensor with the charge current vector to the charge current vector is a right side of the equality result.

##### Share on other sites

On 2/26/2024 at 6:20 AM, KJW said:

But if the decomposition depends on the particular frame

Sure - isn’t that already a suitable model for the situation at hand? The charge is seen to radiate in some frames but not in others.

On 2/26/2024 at 6:20 AM, KJW said:

However, if a charge can be accelerated non-electromagnetically, then one can test if it radiates in the absence of an external electromagnetic field.

Ok - this doesn’t sound like too hard of a test to perform, I wonder if this has been done?

13 hours ago, KJW said:

Have you ever derived the relativistic Lorentz force law from the divergence of the electromagnetic energy-momentum tensor?

I remember having seen this done, so I’m aware of the concept.

13 hours ago, KJW said:

For me personally, I don't accept without question the electromagnetic energy-momentum tensor.

What problem do you see with this?

##### Share on other sites

On 2/27/2024 at 3:44 PM, Markus Hanke said:

Sure - isn’t that already a suitable model for the situation at hand? The charge is seen to radiate in some frames but not in others.

I'm not aware of any experiment where charge is seen to radiate in some frames but not others.

On 2/27/2024 at 3:44 PM, Markus Hanke said:

Ok - this doesn’t sound like too hard of a test to perform, I wonder if this has been done?

I don't think it requires the absence of electromagnetic acceleration, only the presence of an acceleration due to a non-electromagnetic force. Then one can examine if the radiation is accounted for by the total force or just the electromagnetic force. A proton-proton interaction experiment might be a suitable if the energy is large enough to overcome the electromagnetic repulsion so that the strong force is felt, but not too large to cause a transmutation.

On 2/27/2024 at 3:44 PM, Markus Hanke said:

I remember having seen this done, so I’m aware of the concept.

👍

On 2/27/2024 at 3:44 PM, Markus Hanke said:

What problem do you see with this?

The problem I see is not a disagreement but an inability to see the principle behind the electromagnetic energy-momentum. It seems to come out of thin air... AFAICT, it does not come from Maxwell's equations. Actually, it has occurred to me that it cannot come from Maxwell's equations for dimensional reasons (no mass), but if it can be shown that the quadratic expression in terms of the electromagnetic tensor has energy-momentum-like properties, then this may satisfy me.

##### Share on other sites

11 hours ago, KJW said:

The problem I see is not a disagreement but an inability to see the principle behind the electromagnetic energy-momentum. It seems to come out of thin air... AFAICT, it does not come from Maxwell's equations.

AFAIK (and can remember) it comes basically from the general definition of the Hamiltonian, with the potential field $$A_{\mu}$$ plugged in. I’m a bit pressed for time these days (involved in a big project here at the monastery), so I wasn’t able to immediately find a proper textbook reference; but the second-to-last answer on this PSE thread outlines what I mean:

I’ll have to dig through my notebooks when I get a chance, I know I’ve got a proper reference on this somewhere.

##### Share on other sites

On 2/26/2024 at 10:44 PM, Markus Hanke said:

Sure - isn’t that already a suitable model for the situation at hand? The charge is seen to radiate in some frames but not in others.

What does this mean? Wouldn't it radiate as light? If so, what frame wouldn't it radiate in?

On 2/26/2024 at 10:44 PM, Markus Hanke said:

Ok - this doesn’t sound like too hard of a test to perform, I wonder if this has been done?

Isn't the problem that if there was radiation, the expected energy would be undetectably low?

##### Share on other sites

5 hours ago, md65536 said:

What does this mean? Wouldn't it radiate as light? If so, what frame wouldn't it radiate in?

The kind of wavelengths you get would depend on the specifics of the setup - it’s conceivably possible to get visible light too.

For a stationary charge supported in a gravitational field, the result I am familiar with from the literature (see link further up in the thread) would indicate that a comoving detector would not detect any radiation, but another detector freely falling past the charge, would.

5 hours ago, md65536 said:

Isn't the problem that if there was radiation, the expected energy would be undetectably low?

Good point. But I think given enough charge and enough acceleration, it should be detectable. I must admit I’m not sure what the actual numbers are like, I never looked at this in that much detail.

##### Share on other sites

11 hours ago, Markus Hanke said:

For a stationary charge supported in a gravitational field, the result I am familiar with from the literature (see link further up in the thread) would indicate that a comoving detector would not detect any radiation, but another detector freely falling past the charge, would.

Oh, right. I see that was already resolved earlier. The frame where no light (of any wavelength) is radiated is an accelerating frame.

##### Share on other sites

Posted (edited)

I read https://arxiv.org/abs/physics/0506049 [1] from swansont's earlier link, and it clears up some of my misconceptions.

The basic conclusion, in the case of a comoving observer and a uniformly accelerated charge, is that there is only a certain region of spacetime that "would allow us to detect unambiguously the radiation emitted by the charge," and that region is outside the light cone of any event on the particle's world line, meaning that no radiation is detectable at all. Within the light cones, "the detection of radiation has no absolute meaning because the detection depends both on the radiation field and the state of motion of the observer." So I guess if you wanted to argue that any radiation could be detectable, you'd have to be really creative with definitions in order support that conclusion.

Also it is not electromagnetic radiation as I'd assumed. "The radiation content can be extracted by separating the components that drop off as 1/R from the usual Coulomb 1/R2 fields."

1. The radiation of a uniformly accelerated charge is beyond the horizon: a simple derivation
Camila de Almeida, Alberto Saa

Edited by md65536
##### Share on other sites

5 hours ago, md65536 said:

Also it is not electromagnetic radiation as I'd assumed. "The radiation content can be extracted by separating the components that drop off as 1/R from the usual Coulomb 1/R2 fields."

How does your conclusion follow from the quote? What kind of radiation is it?

##### Share on other sites

16 hours ago, swansont said:

How does your conclusion follow from the quote? What kind of radiation is it?

I don't know! Is it all/only electromagnetic radiation, ie. photons?

I see references to "radiation field", but it's described separately from the electromagnetic field? I assume that if electromagnetic radiation is detected, that means a photon is emitted at one event and absorbed at another event. That seems at odds with the quote from the paper, "the detection of radiation has no absolute meaning", so I'd already concluded my assumption of EMR was wrong. Then I (mis?)interpreted the quote as a description of what the radiation is, as something that remains when things like EMR are separated out. Doesn't EMR drop off as 1/R2?

It's all very confusing...

##### Share on other sites

2 hours ago, md65536 said:

I don't know! Is it all/only electromagnetic radiation, ie. photons?

I see references to "radiation field", but it's described separately from the electromagnetic field? I assume that if electromagnetic radiation is detected, that means a photon is emitted at one event and absorbed at another event. That seems at odds with the quote from the paper, "the detection of radiation has no absolute meaning", so I'd already concluded my assumption of EMR was wrong. Then I (mis?)interpreted the quote as a description of what the radiation is, as something that remains when things like EMR are separated out. Doesn't EMR drop off as 1/R2?

It's all very confusing...

The Coulomb field is static; that’s the 1/r^2 field. EMR intensity drops off as 1/r^2 from a point source, but intensity is the square of the field strength.

##### Share on other sites

1 hour ago, swansont said:

The Coulomb field is static; that’s the 1/r^2 field. EMR intensity drops off as 1/r^2 from a point source, but intensity is the square of the field strength.

Then the radiation field of an accelerating charged particle drops off as 1/r because it propagates perpendicular to the acceleration of the charge, the field lines distributed over a circle for a given r rather than a sphere?

An oscillating charge radiates EMR with a frequency equal to that of the oscillation. Apparently, Maxwell's equations imply that even a charge with a constant acceleration must also radiate. However, the frequency and energy would be zero, or at least approach zero as time approaches +/- infinity. So one could say that a charge at rest on the surface of Earth does not radiate energy, or that it radiates light with infinite wavelength, which is not physically detectable nor has an absolute meaning, but is consistent with all physical laws.

I hope this is right instead of me just getting more confused.

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account

## Sign in

Already have an account? Sign in here.

Sign In Now
×

• #### Activity

• Leaderboard
×
• Create New...

## Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.