Charged Particle on Earth's Surface: Will It Emit Radiation?

[Amaterasu21]

TL;DR Summary

In GR, gravitational acceleration is upwards. Maxwell's equations tell us an accelerating charge produces EM radiation. If that's the case, would a charge "at rest" on Earth's surface do so? Where does it get the energy from?

General relativity tells us that an object in free-fall is actually inertial, following a geodesic through curved spacetime, and not accelerating. Instead, it's objects like us, on the surface of a large body, that are accelerating upwards.
Maxwell's equations also tell us that accelerated charges emit electromagnetic radiation.

So it seems to me that a charged particle in free fall would not emit EM radiation, while a charged particle "at rest" on Earth's surface should.
I have two questions:
1) Does this really happen and has this radiation been detected?
2) If so, where do particles get the energy from to continuously radiate? A charged particle can stay "at rest" on Earth's surface indefinitely, but it can't keep emitting energy forever... is either of those statements wrong?

 

[hutchphd]

I would ask you to look at Larmor's formula and calculate the power radiated by an electron on Earth's surface. I see no reason why it would not classically radiate, but consider the energies involved...please report your results..they will be small.
The energy will come eventually from the rotation of the Earth which somehow rotates this charge

 

[Amaterasu21]

The energy will come eventually from the rotation of the Earth which somehow rotates this charge

So what would happen to a charge sitting on the surface of a non-rotating mass?
(I got 5.5*10^-52 watts btw - okay, a very small amount of power. But the fact there's any power radiated at all confuses me since it'd need some sort of energy source, and the rotation of the Earth doesn't seem like enough because a non-rotating mass would have gravity too.)

 

[hutchphd]

It will not classically be accelerating.
I understand where you are going with this and I defer to the GR mavens...I don't even know if this is well known

 

[Dale]

Summary:: In GR, gravitational acceleration is upwards. Maxwell's equations tell us an accelerating charge produces EM radiation. If that's the case, would a charge "at rest" on Earth's surface do so? Where does it get the energy from?

So it seems to me that a charged particle in free fall would not emit EM radiation, while a charged particle "at rest" on Earth's surface should.

How are you planning to detect this radiation?

 

[Vanadium 50]

Here we go again...

(1) Please start by doing a forum search.
(2) No, it doesn't radiate by any conventional definition.
(3) I predict a bunch of people will come in here to inject A-level subtleties that will confuse more than clarify.

 

[Amaterasu21]

It will not classically be accelerating.
I understand where you are going with this and I defer to the GR mavens...I don't even know if this is well known

Here we go again...

(1) Please start by doing a forum search.
(2) No, it doesn't radiate by any conventional definition.
(3) I predict a bunch of people will come in here to inject A-level subtleties that will confuse more than clarify.

1) I can't find a query similar to this one on the forum, sorry.

2) I've found a few sources that say it will, though:
This paper: https://core.ac.uk/download/pdf/25269331.pdf

"It is found that the “naive” conclusion from the principle of equivalence - that a freely falling charge does not radiate, and a charge supported at rest in a gravitational field does radiate - is a correct conclusion, and one should look for radiation whenever a relative acceleration exists between an electric charge and its electric field. The electric field which falls freely in the gravitational field is accelerated relative to the static charge. The field is curved, and the work done in overcoming the stress force created in the curved field, is the source of the energy carried by the radiation. This work is done by the gravitational field on the electric field, and the energy carried by the radiation is created in the expense of the gravitational energy of the system."

Apparently the energy carried by the radiation comes from the gravitational energy of the system, but I'm not sure what that would imply. How could the system just lose gravitational energy if it's just sitting there?

This video from Veritasium (15.05 onwards) - not exactly a peer reviewed source, of course, but he's very well known and respected in the science communication community. I concede he could have got this bit wrong though, but if he did, I'd like to know why:


Other sources I'm seeing seem to suggest it doesn't radiate in its own rest frame, but does to a freely falling observer (which is surely a paradox - either there is a photon or there isn't?) I'm also finding suggestions Maxwell's equations should be modified in curved spacetime, is that the case?

 

[hutchphd]

I like Veritasium but if you really need to know this answer you need to embark on a journey of discovery. Goi for it.

 

[Dale]

which is surely a paradox - either there is a photon or there isn't?

No, radiation is frame variant. This is why I asked how you intended to measure the radiation. So again, how do you intend to measure the radiation?

 

[Amaterasu21]

No, radiation is frame variant. This is why I asked how you intended to measure the radiation. So again, how do you intend to measure the radiation?

Let's say we use a very high charge at the surface of a planet with very strong gravity, so the acceleration's enough to produce a photon we can detect with the photoelectric effect.

Does that mean if we allow the photosensor to fall freely, it could detect a photon, whereas if we put it on the planet's surface, it won't?

 

[A.T.]

Maxwell's equations tell us an accelerating charge produces EM radiation.

Not if the acceleration is constant, and we are in the rest frame of the charge. Then you just have a constant but distorted E-field (Figure b), as opposed to radiation for changing acceleration (Figure a).

From: https://arxiv.org/abs/1503.01150

 

[Dale]

Does that mean if we allow the photosensor to fall freely, it could detect a photon, whereas if we put it on the planet's surface, it won't?

Yes. It is easier to understand with something like an antenna, but the principle is the same.

 

[Vanadium 50]

Does that mean if we allow the photosensor to fall freely, it could detect a photon

Your original question was simple. This question is not. It requires GR, fairly advanced classical EM, and now that you have introduced photons, QM. There will not be a B-level answer. The closest will be that the sensor will detect something, but there is no unique explanation of what was detected and why. Different observers will see different things.

 

[vanhees71]

Not if the acceleration is constant, and we are in the rest frame of the charge. Then you just have a constant but distorted E-field (Figure b), as opposed to radiation for changing acceleration (Figure a).

View attachment 288684

From: https://arxiv.org/abs/1503.01150

For hyperbolic motion there is radiation after all, but a comoving observer cannot observe the radiation. There's a nice AJP paper by Franklin and Griffiths investigating this (in)famous problem very thoroughly. It underlines the importance of taking Dirac-$\delta$-like contributions into account properly and to be aware of the specialty of this situation, where the asymptotic in and out state of the particle's motion are to approaching the speed of light. In this case a naive application of the Lienard-Wiechert potentials is tricky:

https://doi.org/10.1119/1.4875195
https://doi.org/10.1119/1.4906577 (erratum)