Will a charged particle in free fall wrt to the earth radiate? As I’m sure everyone knows, an accelerated charged particle radiates.
Yes… did you just answer your own question?
An accelerated charged particle radiates… And a particle in freefall is not accelerating, so it does not radiate. The core of the General Theory of Relativity is that acceleration is indistinguishable from gravity, and if a free-falling particle radiated, that would be a way to distinguish the two.
Of course, this is completely insignificant anyway, unless you have a black hole or such.
A particle in freefall is not accelerating as measured by an inertial reference frame. What about a (non-inertial) reference frame attached to the earth? The particle is accelerating in that frame. Do charged particles only radiate when they accelerate in an inertial frame?
I’m going to assume there’s a tube through the center of the Earth (the one that people are always falling into) so the charged particle keeps oscillating back and forth. From an inertial frame far from the Earth, there’s what appears to be a classical oscillating charge, so it must radiate.
From the frame of reference of the charge, it’s in an inertial frame, not accelerating at all, so it can’t radiate, as chronos said. But from this same frame, the rest of the universe is oscillating back and forth. If the total charge of the universe is zero, this “rest of the universe” must also be charged. No way am I attempting the math, but it’s at least plausible that the electromagnetic fields generated by the ROTU act on the charged particle in such a way to accelerate the particle, bringing it into motion with the ROTU. From the charged particle’s point of view, it’s been accelerated into motion, from the ROTU’s point of view, the charged particle radiated away it’s energy. Everybody’s happy.
Huh? You lost me on that one ZenBeam.
The problem as I understand it is that the particle is definitely accelerating wrt to an observer on the Earth. But, nonetheless, the particle is in an inertial frame (following a geodesic) and it is the observer on the earth that is really in the accelerated (non inertial) frame.
I agree (for whatever that’s worth) with Chronos, but it seems there is quite a bit of literature out there that says this is not a solved problem.
I may also agree with ZenBeam but I’m not sure what he said.
Huh? These statements seem contradictory to me.
How can a particle in freefall not be accelerating wrt to the Earth? If it has no initial velocity, it will presumably fall to the Earth under the influence of gravity, accelerating all the way. If it has some tangential velocity so that it is in orbit, it is also being constantly accelerated.
Am I missing something? [Probably…]
If the total universe charge is zero, and for discussion, you seperate out one charge particle, then the remainder of the universe has a charge, opposite that of your particle so the total charge is still zero.
When Chronos said that a particle in freefall is not accelerating, I think he meant that it is not accelerating in an inertial reference frame. An inertial frame within a gravitational field must be in freefall to cancel the effects of the gravitational field, so the particle is not accelerating wrt the frame (which is itself in freefall too). Someone standing on the earth is accelerating upward at 32 ft/sec[sup]2[/sup] relative to an inertial reference frame (since an inertial frame must be in freefall).
OK, I just checked with my advisor, who knows way more relativity than I, and he confirmed the following points:
1: A charged particle in a gravitational orbit will not radiate.
2: A charged particle on the surface of the Earth, despite being accelerated, will not appear to be radiating, as viewed by a person in the same frame (i.e., also on the surface of the Earth).
3: This effect is essentially the same as the gravitational redshift of light, which has been measured and experimentally verified.
He then referred me to a couple of papers on the topic, which I have not yet had an opportunity to look up. When I do, I’ll try to post a summary here.
Here’s a couple of questions I asked at another forum - and I still don’t know the answers.
Thank you. One last question if I may - does this mean that a charged particle held stationary in Earth’s gravitational field will radiate?
Construct a thought experiment: I sit in a space ship that’s accelerating, looking at a charged particle that’s mounted in a fixed position with respect to the ship. By the equivalence principle, this should be locally equivalent to sitting on the ground looking at a charge that’s stationary with respect to the planet. So it won’t radiate, right? But the guy sitting outside the space ship in a non-accelerating frame, sees an accelerating charge go by, and it radiates. So can it really be that whether or not a charge radiates depends on frame of reference? Does it radiate if and only if it accelerates with respect to the observer? Is the distinction between a virtual and a real photon a function of the
acceleration of the reference frame?
Wow, is that really so? I thought one of the problem’s with early atomic models was that electrons would radiate away all their energy had they been charged particles in a circular orbit around the nucleus. Does the electric vs. gravitational force change the parameters in this case? Is it because gravity does strange things to space-time here?
If we’re all in happy laboratory inertial frames, than that particle will not radiate. In this case, one could argue that Earth is accelerating through space-time though.
With accelerating frames of references, you are going into General Relativity, and I have a feeling you’re invoking the equivalence principle of Special Relativity here. You’ll have to ask Chronos or his instructor on this one…
I’ve skimmed some papers on the topic. The main thing I’ve discovered is that there are lots of papers on the topic. And, every one of them begins with, “We return to the familiar problem of a charge…”
My findings thus far: people don’t agree. I have seen papers arguing these (and other) points:
No radiation. They show this by explicity transforming the EM fields into a co-accelerating frame.
No radiation. There is no source of energy for radiation for a supported charge in a gravitational field.
Radiation. The charge’s fields are not accelerated with the charge itself or (equivalently) the charge on the table does not follow its freely falling fields ==> distortion in the field ==> radiation.
Radiation. The equivalence principle doesn’t apply / energy is poorly defined.
It’s always fun to read papers that flat-out contradict each other. Later I will read these papers more closely with hopes of reconciling the differences. In the meantime, my vote goes to “no radiation”. (I rather liked the explicit transformation of the fields, and I rather like the equivalence principle.)
I have a VERY strong suspicion that the resolution to this apparant paradox comes from the fact that the gravitational field is not uniform (1/r-squared dependance and all that).
The equivilence principle only truly applies if the gravitational field is uniform. If you have a uniform field (say g=10m/s^2 in Z direction), then the charge and it’s field out to infinity accelerate the same and you have no radiation (equivilence principle applies). Of course, any observer is then subjected to the exact same acceleration, and we could never detect that the gravitational field exists to begin with!
If the gravitational field is not uniform, then the equivilence principle goes out the window. “But wait”, you say " on the scale of an electron the gravity is approximately uniform!". Sorry, your looking at the wrong scale; the field of the electron is infinite in extent.
If you imagine the field lines from the electron, they are clearly being bent as it accelerates under the earths gravity (clearly since the charge appears to be accelerating to an observer in a truly inertial reference frame far from the earth).
It’s tempting to claim that the degree of bending of the field lines (and hence the amount of radiation) will be proportional to the local gravity gradient; but i don’t think that is quite true.
If you imagine a bizarro gravitational potential well with a small region of flat gravity (no gradient), the electron would still be radiating while in that region.
The radiation is clearly related to the gradient gravity, but not just locally.
My best guess would be something along the lines of:
radient power = integral of e-field dot gradient(gravitaional acceleration) over all volume. but that is just a guess.
non-standard disclaimer: I am in fact a physicist, but this is not my field by ANY streach. barely pased E&M and never took a class dedicated to relativity; much less a class on quantum chromodywhatsit. I am merely doing a little pondering on an interesting gedanken experiment. any similarity between the above and a coherent unified field theory is purely coincidental.
This is the one which seems so clear to me. If you have a charge on the surface of a non-rotating mass, and a stationary observer far away from the mass, nothing is varying with time. How can it possibly radiate?
But how do you reconcile this with Chronos’s advisor’s statement “1: A charged particle in a gravitational orbit will not radiate.”? Consider the same mass, except it has a tube through the middle. The same charge is there, but this time it’s in free-fall, oscillating back and forth through the tube. When it’s at the top of its (linear) orbit, it’s in the same location as the stationary charge. But the stationary charge and the orbiting charge are accelerating with respect to each other. I would think this would mean at least one of them must be radiating.
Chronos, is your statment 1 true for all observers? While I’m at it, in statement 2, you write “2: A charged particle on the surface of the Earth, despite being accelerated, will not appear to be radiating, as viewed by a person in the same frame” (italics mine). What about an observer in a different frame? Does it matter whether they’re higher or lower in the gravitational well?
Finally, statement 3 is curious: “This effect is essentially the same as the gravitational redshift of light…” I’ll confess, I don’t understand this. Does the radiation “occur”, but get redshifted to nothing as the observer moves farther out the gravitational well? (“occur” is in quotes because I don’t know what I mean by that.)
But what if radiation is relative to the observer’s motion? In other words, an inertial observer in free fall would see the supported particle as accelerating.
It appears at least in the vicinity of a black hole that radiation is relative. To put the following quote in perspective - real photons mean radiation; virtual photons just follow along with the particle and according to “Black Holes and Time Warps” (Box 12.5, p. 444) the distinction between virtual and real, at least in the vicinity of EV, does depend on the relative acceleration.
In the example, everything is stationary. The observer is far away, and built a large geodesic sphere around the mass, and is dangling from it, with a suitable counterweight on the opposite side of the sphere. No motion at all. If the charge is radiating, you have a perpetual motion machine.
This is kind of why I was asking Chronos if his statement 1 was true for all observers. If different observers see the stationary charge radiating or not, depending on their frame, then the same could be true for an orbiting charge.
Ok, I agree, but since neither observer is following a geodesic they are both in non-inertial accelerated frames. And in this case you would not expect to see any radiation since they are not accelerating relative to each other.
There are still a lot of unanswered questions. (Don’t leave us hanging, Chronos :().
I was thinking about this some more since my bump, and in particular about Luckie’s post. I believe he’s correct that we have to look carefully at the field of the charge, and not just the charge itself.
Consider a charge moving in a circle at some frequency, but not in orbit about a mass. Obviously, we are applying a force to keep the particle moving in a circle, and, also obviously, the charge is radiating.
Now look at it from the non-inertial frame of reference moving with the charge. In this frame, the electron is motionless, and there is a time-varying, spatially uniform gravitational field. This field is the field needed so that the force applied to the particle exactly counteracts the gravitational field. In our original system, energy is being radiated away, so in this system, there must also be energy radiated away. Since the electron is motionless, it can’t be radiation due to acceleration of the charge. The only thing left is interaction between the time-varying gravitational field and the electric field. This must be the source of the radiation.
Now consider the case where the charge is in the same motion, but due to being in orbit about a mass. The electron is now in an inertial frame, and still is not accelerating, so the charge can not radiate. As Luckie pointed out, the electromagnetic field is not in an inertial frame. In particular, away from the mass, the time-varying gravitational field due to choosing the electron’s frame of reference will dominate over the gravitational field of the mass. Since we’ve established above that electromagnetic fields in such a time-varying gravitational field will radiate, there will also be radiation in this case.