Paradox of a Charged Particle in Gravitational Field

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It might seem like we already know everything regarding the general theory of relativity and electrodynamics but certain simple questions can turn out to be extremely difficult to answer. I have been struggling with this one for a long time and the resolution is unsatisfactory to say the least.
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[@lukasrafajpps]

To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/ProblemsAndSolutions . You’ll also get 20% off an annual premium subscription.

[@classicalmechanic8914]

Equivalance principle is valid only in general relativity. Special relativity is incompatible with equivalence principle.

[@_John_P]

The equivalence principle does not exist in nature because nature does not produce uniform gravitational fields. All gravitational fields necessarily produce tidal forces, thus breaking the equivalence. With the equivalence principle out of the way, there's no paradox.

[@pwinsider007]

solution of problem is that accelarating charge will create event horizon due to unruh effect and virtual particles will fall in event horizon therefore energy lost by radiation of charge will be compensated by energy gained by absorption of virtual particles so that particle has a constant accelaration and it will not violate equivalence principle.

[@lordkalem]

This question requires a universe that does not have the same properties as our own. aka, sure in a hypothetical universe where gravity does not impart energy to objects, sure, but gravity DOES impart energy to objects... seriously man this is high school physics.... go read up on what happens to particles near black holes.

[@frankwijnans444]

The paradox is completely gone, when you realize different observers see different electric and magnetic fields. The charged particle inside the lift doesn't radiate but gets accelerated from radiation from outside. That's consistent with the fact kinetic energy is different for different observers.

[@DrDeuteron]

you're asking the wrong physicists if they all say "of course", this is a famous problem for QED in curved spacetime, with the vacuum transforming non trivially...leading to things like radiating horizons (famously: Hawking radiation, and less so: Unruh radiation).

[@quasarsupernova9643]

exactly ....

[@fixgirod]

Or capitalizing on a misrepresentation on purpose. He knows he doesn't have access to competent people

[@undercoveragent9889]

@@fixgirod Exactly! Clearly, he thinks his audience is as dumb as _he_ is.

[@chrisvinciguerra4128]

⁠​⁠@@fixgirodwhy do you make it sound so nefarious? I think he’s simplifying for the sake of the video narrative. To me, it’s kind of rude to assume he must not be talking to smart enough people, instead of assuming he is trying to make compelling content and incorporating the wrong opinions only for the sake of discussion

[@cykonot]

@@chrisvinciguerra4128 how is that LESS nefarious? lmao

[@zoetropo1]

1. Standing on the Earth, what you see is in a non-inertial frame. 2. Maxwell’s equations require modification in a non-inertial frame. 3. Special relativity was devised to make Maxwell’s equations covariant. 4. Covariance implies an inertial frame. 5. Therefore in a non-inertial frame you cannot use Maxwell’s equations to predict an absence of a time-varying magnetic field. 6. Thus there’s no paradox. 7. What we need are measurements precise enough to detect radiation from particles that are stationary in a 1g non-inertial frame.

[@at0mly]

Just came to write this but you said it better than I would have.

[@LaplacianDalembertian]

Frame of Earth is only slightly non-covariant, and Lienard formula (Larmor formula for relativity) says that it starts radiating only when Beta is close to 1. So there will never be an observable radiation for particles with insignificant speed. It will radiate only when particle falls into Black Hole. Non-inertiality of Earth does not matter, while this inertiality is with its Beta much much less then 1.

[@Frisbieinstein]

"3. Special relativity was devised to make Maxwell’s equations covariant." I'd say Maxwell's equations were already covariant. These equations were not however compatible with the Cartesian view of the world, so the latter had to change. Maxwell's equations stood unaltered. They're covariant now so they were covariant then. They were perhaps Einstein's main guide to special relativity.

[@duoko98]

EXACTLY, CORRECT COMMENT

[@ShandyTheMan]

Doesn't necessarily have to be a precise measurement to detect the effect if you can either or both increase the intensity of the radiation (multiple charged particles) or run it enough times for a chance detection or an average above background noise, if applicable. Afterall nature has provided the most precise photon measurement devices possible; electrons in their exclusion energy wells.

[@nihlify]

My brain is now a paradox, I both feel smarter and dumber listening to this xD

[@lostson1st]

You're already starting to see it work! 😎

[@livingcorpse5664]

The more you know, the more you realize you you don't know.

[@bgpcusercheater5174]

​@@livingcorpse5664stupidoo

[@I-M-2.]

Maxwell equations are not valid 'as is' to curved spacetime nor Larmur formula (nor QED). To tackle it, we need to use the curved spacetime version of the EM Lagrangian and derive the curved spacetime radiation condition. My (usually bad) intuition says the equivalence principle still holds.

[@renscience]

The equivalence principle has been proven over and over since GR requires it and we know GR is correct.

[@Saki630]

Nice try ChatGPT you failed again.

[@I-M-2.]

@@Saki630 not chat GPT but I'll take it as a compliment 😊

[@szilardecsenyi516]

It is not necessarily required to adhere to the equivalence of inertial frames either; observers in their own inertial systems can choose to always calibrate the speed of light as isotropic. This would mean that genuine Lorentz contraction and time dilation would occur relative to the 'stationary ether', and with proper synchronization of clocks, the space and time coordinates interpreted in different inertial frames could be connected by the Galilean transformation. It is just difficult to accept that in this case, the length of a stationary rod in its own inertial frame and the length of an identical rod moving relative to it in the moving inertial frame would not be the same.

[@I-M-2.]

​@@szilardecsenyi516but an observer sitting on earth's surface is not in an inertial frame. It is in an accelerated frame.

[@christophjansen646]

The question is simply to be answered: the Larmor formula relates to an accelerated electrically charged particle losing energy in an electromagnetic field which is the cause of the acceleration. So it does not relate to any such particle which is accelerated in a gravitational field or a curved space time. Therefore the Larmor formula does not tell us anything about radiation being generated in the described setting.

[@pidaras_pidarasina]

But they do get accelerated via Newton force for gravity, right?

[@perarve2463]

So there is no brehmsstralung if the acceleration is caused by the strong nuclear force or the weak force? It seems me that the QFT calculation of photon emission in a collision is very much the same independent of what force caused a 4.momentum exchange.

[@hugoballroom5510]

many good comments so far. I just want to applaud your sketch of this as a kind of "paradox" along the lines of other relativity "paradoxes."

[@karkaroff1617]

I've been thinking about this problem ever since i saw Veritasium talking about it, but your video goes in much more detail. Good job.

[@mikhailfranco]

Veratasium is a click teaser. Block his channel.

[@karkaroff1617]

No. I'm a physics grad student, I know where to look for precise information. I watch Veritasium as a entertainment activity. Sometimes he talks about things that I don't know. Gravity is not my subject of research. Besides, he also have a PhD in physics.

[@mikhailfranco]

@@karkaroff1617 He has a Ph.D. in physics education: _"Designing Effective Multimedia for Physics Education"._ So a Ph.D. in KZbin physics presentation.

[@karkaroff1617]

Sure, and? Are you aware that he doesn't do his videos alone, right? There are physicists and other professionals as consultants.

[@PluetoeInc.]

@@karkaroff1617 let him have it karkaroff , its the only thing he has about him .

[@BarriosGroupie]

Rhorlic's paper resolved this IMO: "The principle of equivalence". Annals of Physics. 22. This is described in the section _Resolution by Rohrlich_ in the Wikipedia article _Paradox of radiation of charged particles in a gravitational field._ When a charge radiates, the incoming Schott energy cancels the outgoing radiation giving a curved electric field, zero magnetic field and apparently no radiation which has tricked many experts. Here's a question: how would one levitate a charge so it's stationary on the surface of the Earth? When thinking about the issues you've raised in your video, I've found using a charged sphere as a model instead of a charged particle is very helpful in understanding what's going on physically.

[@FadkinsDiet]

You have to levitate it (counteract Earth's gravity) by either gravity or one of the other forces. If gravity, then the whole scenario is trivial. If electromagnetism, well that has to be accounted for somehow. I suspect the amount of EM force needed will change depending on the observer, and that's where the apparent discrepancy comes from. I also suspect that there's no way to keep the falling charge adjacent to the falling observer without some additional force.

[@juliavixen176]

Do you mean like the Millikin Oil Drop Experiment?

[@Saki630]

yikes another person who apparently uses ChatGPT

QED-wise, the radiation for a charged particle happens when it flips the spin and releases a photon. This can happen spontaneously (but also the recapture) OR due to an external field acting on the charge. In both cases the tree level approximation is that it flips the spin - tree level means, that is the most probable case. In gravity, there is no need to flip the spin (except diverging gravitational field may). So one problem is that the Maxwell equations and the Lorentz force does not take spin into account, the other is the lack of self-interaction (which is present even in the propagator of QED). In QED you have a vacuum full of virtual photons, representing the field of the particle, and when it accelerates the photon is not recaptured - however you need an other photon to accelerate the particle, whose remnant could also be the outgoing radiation. So in QED you have a convoluted answer, which does the radiated photon comes from, but you still need an external field that causes the acceleration.

[@weeb3277]

where is you username?

@@weeb3277 Honestly, I don't know

[@dennisbrown5313]

when it .... what?

[@lazy_ai]

@@weeb3277 the username is there, but it's behind event horizon of your reference frame 😜

[@jasdeepsinghgrover2470]

Quite interesting. So can we calculate the reverse force? Wouldn't the same apply for acceleration caused by any external force other than electrostatic?

[@AndrewBrownK]

I really like this question. I've always suggested that the equivalence principle should go both ways. Acceleration by electromagnetic fields should be indistinguishable from curved spacetime (gravity). Here's a thought experiment someone smarter than me could try playing with. Take yourself a universe. It has a charged black hole (let's say negatively charged). This black hole is in an especially clean corner of space, where it cannot easily find positive charge to balance out to a neutral charge. Now take an electron, and shoot it gently towards the black hole. Hypothetically, the repulsion of negative charges will allow the electron to "hover" inside the black hole's gravitational field (without needing to orbit) (at the point where the gravitational acceleration inward is equal to the electromagnetic acceleration outward). So with this setup, what does both the gravitational field and electromagnetic field look like?

[@thrall1342]

Nice one. Not sure though, I'd expect the particle to radiate, since it is accelerated in the space-time picture. Just like a particle in an ion trap on earth, actually ... should be measureable this way ^^

[@nicolasreinaldet732]

The equivalence principle can't go both ways because the electromagnetic field differs between particles. The equivalence principle only holds because gravity behaves like a curvature, In the sense that It affects EVERYTHING the same.

[@undercoveragent9889]

"Hypothetically, the repulsion of negative charges will allow the electron to "hover" inside the black hole's..." Nonsense. You are claiming that the black hole can't 'see' the electron until it passes the event horizon at which point, the hole develops a sudden allergy.

[@michaelcarey26]

@@nicolasreinaldet732 you dismiss this perspective too easily. it is possible to imagine that different particles experience different space-time curvature.

[@zemm9003]

​@@michaelcarey26no it isn't without creating a completely new theory.

[@debrainwasher]

In my humble opinion, there is a quite simple solution for this problem without QED: The base of this thought experiment is the fact, gravity is really not a force, but is created by warping spacetime e.g. in the vicinity of heavy masses - or the warping time to be exact. Since force is F=ma, we can concentrate on acceleration. Since there are two absolutely (!) opposing acceleration vectors created by gravitation, they simply cancel in classical and relativistic electrodynamics. Hence free charges within gravitational fields do not radiate. Things however change completely within a rocket, since there or no longer opposing acceleration vectors but only one, that goes with the velocity vector, since a=dv/dt. Therefore, neither Einstein's strong equivalence principle is violated, nor Maxwell's laws, nor Larmor's formula. The only thing, that really happens, is things slow down within a gravitational field relative to a flat Minkovsky-spacetime. Every energy «lost» in a gravitational field (e.g. gravitational redshift) is gained, when a photon escapes the field. This saves even conservation of energy and momentum.

[@The_Green_Man_OAP]

What do you mean by "two opposing acceleration vectors"; Are you referring to different reference frames or equal and opposite gravitational forces, or BOTH❔ 🤔

[@The_Green_Man_OAP]

In relativity, force is not F=ma, it is F=d(mv)/dt and m is not constant. F would then have 2 components, one in the "a" direction and one in the "v" direction (or if you do the derivative from 1st principles & reinterpret as a difference equation you get the "v+dv" direction).

[@debrainwasher]

@@The_Green_Man_OAP This is correct, as far as relativistic velocities or strong gravitational forces are involved. In this case, masses must be multiplied by the Lorentz-Factor γ. Since this is not the case, we can use Newtonian dynamic. Furthermore, it is true, that on large scales - very week gravitational fields - strange things happen, that some people attempt to overcome with MOND (MOdified Newton Dynamics), while another hypothesis goes toward WIMPs. However, in case of elementary charges, we know, these are quantized, Therefore, they can't give up a tiny piece of their ground state energy. Furthermore, we would see a self-discharge of every capacitor on earth, since E=CU²/2=QU/2 because C=Q/U - but not in deep space or orbit. No such things have been observed.

[@tonibat59]

A dumb question to which I don't know the answer: Does a mass in free fall increase its mass by gamma? I suspect the answer is no, but I'd happy to listen to arguments.

[@debrainwasher]

@@tonibat59 Yes, it does, since gamma is a function of velocity, while gained energy is a function of relativistic momentum - the product of velocity times rest mass times gamma: Einstein told it as follows: E²=(mc²)²+(pc)². This is his complete equation.

[@Mahesh_Shenoy]

I loved the video. It's incredible to think that we understand so much complexities of how the universe behaves. From the orbitals of the quantum world to the hawking radiations of black holes and yet scientists disagree on what a plastic comb rubbed with my hair will do! Btw, I loved the presentation and the story telling. The way it starts with, 'of course free falling charges must radiate', and how it leads to 'falling charges having to fall slower' leading to the 'breaking the equivalence principe' leading to a need for a deeper investigation was brilliant (no pun intended). Also, congrats on getting Brilliant sponsorship :) Cheers.

[@lukasrafajpps]

Thank you very much :) I am very glad you liked the video :) :)

[@MichalPlichta]

I love both you guys.... I guess it would be incredible awesome and extremely funny when you get collaborate.... really looking forward to it! Cheers!

[@pwinsider007]

​@@lukasrafajpps solution of problem is that accelarating charge will create event horizon due to unruh effect and virtual particles will fall in event horizon therefore energy lost by radiation of charge will be compensated by energy gained by absorption of virtual particles so that particle has a constant accelaration and it will not violate equivalence principle.

[@kylelochlann5053]

Note: Measuring a changing electric field does NOT necessarily mean magnetic field intensity is non-zero, only that there's a non-trivial curl of the magnetic field.

[@АндрейМирон-х2н]

It’s enough to just stop saying that there are two forces and two fields. There is only one electromagnetic field.

[@kylelochlann5053]

​@@АндрейМирон-х2н No one is saying that there are two electromagnetic fields, but given the two Maxwell equations (one of which couples to the matter fields and the other which describes the free-electromagnetic field) we have 4 independent field components (the two excitation fields (D,H) and two that define a pair of field intensities (E,B) which can be related via the constitutive relations). What I am saying has nothing to do with this, rather, given some time-like curve what does the observer measure of the electromagnetic field components if the electric field intensity is a function of time (a non-trivial curl of the magnetic field intensity, and not necessarily a non-zero value of the field itself).

[@АндрейМирон-х2н]

@@kylelochlann5053 But the author of the video clearly divides this into components. First, he needs to disassemble the physical meaning without confusing himself with a distorting perception of dividing formulas. That in different reference systems the electrical and magnetic calculations are different, but their resulting influence is identical. And after that it is more correct to express the meaning of the situations under consideration.

[@terrapin323]

​​@@АндрейМирон-х2нThis is just semantics. Including the idea of their being a single electromagnetic field rather than two fields that induce eachother. What's crazy is you can't derive the magnetic field from the coloumb field, even given the fact the circular shape of the magnetic field around a moving electron is just the cross product of its velocity vector and the coloumb field. even given special relativity, physicists always presuppose the magnetic field already exists in order to separate them in lorentz force formula, and the em field tensor, etc it just makes the math 10x more elegant to represent it as two fields which separately induce one another. Theres a significantly more complicated double cross product representation where you treat the B field as only a lorentz boosted E field and B's dissapear but is it really worth it? Its like its only simple as two fields. Even the word electromagnetic implies this duality, if you were really capable of viewing it as a single 4d vector manifold thing youd just call it the electric field, because magnetic fields dont exist if you view its just "one" thing (the force transforms under lorentz boost).

[@DavidGoldgruber]

How do we know today for sure that linearly accelerated charges radiate, and jerk isn't needed fx.? I mean, what has happened since Pauli, Born and Feynman? At 8:50 : In this form, Maxwell's equations are only valid in flat spacetime, in an inertial frame, bc. those are not covariant derivatives. At 10:05 : This kind of Rindler horizon explanation only works in flat space time, but around the Earth, not neccessarily. In this case, the workings of such a horizon is not as trivial as in Minkowski, neither as in Schwarzschild, where every Schwarzschild observers Rindler horizon is just the event horizon.

[@moisessalazar4432]

Q: If you transform the electric and magnetic field to curved spacetime it would solve the problem?

[@DavidGoldgruber]

​@@moisessalazar4432 I don't know. Maxwell's equations in curved spacetime are nonlinear, so people can't/don't like working with them. Wikipédia says that Fritz Rohrlich examined the flat version of this problem in 1965 (a Rindler charged particle and an inertial observer). Based on how weak the curvature is around Earth, that approximation may be sufficient in our case.

[@dialectphilosophy]

Hey, another great video! We once pondered over this exact question, and you know, there’s an EXTREMELY simple resolution - but you’re not going to like it. An illuminating paper we found on the topic was Rohrlich’s 1963 “The Principle of Equivalence”. The conclusion of the paper was that a co-accelerating observer does not observe any radiation - and that this was a direct consequence of the covariance of Maxwell’s equations. What does this tell us? Why - simply that acceleration is relative. An observer who constructs their coordinates via accelerating matter will not observe physical effects of that acceleration. Thus, much like it’s impossible to know one’s true velocity, it’s impossible to know one’s true acceleration. Hmmm, sounds an awful lot like being stuck in a sort of Matrix 🤔 Almost like there’s a theory about that… 🙃 At any rate, you’ve reminded us to do a video on this topic sometime soon. Great work again!

[@aniksamiurrahman6365]

Co-accelerating observers will see each other steady (as we see while driving), but both will agree about feeling a gravitation-like force opposite to acceleration. Every observer irrespective of their state will agree on one of the two - acceleration or gravitation. But, even if we forget all that, how does relative acceleration prove your matrix theory?

[@zemm9003]

Acceleration just means a deviation from the geodesic it doesn't mean anything else.

[@lukasrafajpps]

Thank you for the comment. About the relativity of acceleration, there is still not consensus whether the General Relativity truly satisfy general principle of relativity maybe just qualitatively. What we know that it is not relative in the same sense as velocity because to transform among reference frames of different velocities you always use Lorentz transformations. So tranforming from stationary to moving you use LT and if you want to transform back then you also use LT but with opposite sign but you are still in the same group of transformations. If you want to transform from inertial frame to accelerated frame you use different transformation than when you transforming from accelerated frame to inertial frame. These transformations are different and therefore these frames are not equivalent quantitatively. (If they were the twin paradox would not be solvable) Anyway I can see a nice symmetry there particle on the surface + observer on the surface = no radiation particle falling + observer falling = no radiation particle on the surface + observer falling = radiation observer on the surface + particle falling = radiation!?!? would be nicely symmetric. I would love to see your video about this. Please make one :D

[@dialectphilosophy]

@@lukasrafajppsThanks for replying! We really appreciate that you’ve been choosing to tackle these interesting and more involved physics topics, rather than doing the usual circuit of broad-appeal subjects - we always learn something new watching your videos. You’re right to point out the transformations are different between two inertial frames versus an inertial and non-inertial frame - this is the distinction that separates the General Principle of Relativity (in which any frame of reference could be subsumed under a single set of transformation-invariant laws of physics) and the Principle of General Covariance (in which the inertial laws of physics are covariantly extended to non-inertial frames). So we wouldn’t disagree with you there - our point was slightly subtler. If you as observer B adopted as your inertial frame what was a non-inertial frame from the perspective of observer A, then to go between your frame and observer A’s frame, you would need covariant transformations (in each of you would describe the other’s laws as involving fictitious forces, etc.). However, if we consider a third observer C who shares A’s definition of the inertial and a fourth observer D who share’s B’s definition of the inertial frame, then to go from A to C or from B to D we need merely only use the Lorentz transformations, while going from C to D or A to D or vice versa we again need the covariant transformations. Thus, we see the nature of the transformations are untouched - it’s simply our starting point in choosing what constitutes the inertial frame that is subject to a certain arbitrariness. This is a point we’ve been trying (perhaps unsuccessfully) to communicate across various videos. Meanwhile, General Relativity asserts that all observers will unambiguously agree on what constitutes an inertial frame via use of an accelerometer. A rest frame on the surface of the earth is non-inertial, so GR definitively tells us a particle at rest on earth’s surface is “really” radiating. However, it appears Maxwell’s equations tells us we can’t actually measure this radiation if we are co-accelerating. This lands us, as you quite rightly assert at the end of the video, in a paradoxical situation, indicating something in our thinking needs revision. Hopefully we will get around to a video on this topic soon, and if we do so we will definitely cite and utilize the research you’ve done here!

[@dialectphilosophy]

@@aniksamiurrahman6365They would only “feel” a gravitational acceleration if they were using accelerometers that had been previously calibrated in some inertial frame - otherwise the accelerometers would register no change in dimension, and the observers would be unable to measure that force at all. We discuss this concept more thoroughly in our video “Why Relativity Doesn’t Add Up”. Interestingly, we actually first developed Matrix Theory to explain the conundrum of relative acceleration - how could acceleration be relative and yet still be responsible for causal propagation of information in our universe? The answer we finally hit upon was that it was our knowledge of acceleration which was relative to our coordinate constructions, and that we could still posit absolute acceleration, but we would have to concede it was ultimately unknowable. Later we realized we could say the same thing for absolute velocity (and absolute space), and then in turn realized that could be applied to resolve the puzzles of relativity.

[@ZenithWest169]

Good video but I think you should of included orbiting charge analysis too. For example, what's the difference between a charge moving in a circle in a magnetic field versus a gravitational field? (Like how you animated the electric field not being accelerated in the magnetic field, wouldn't that apply to an orbiting charge?)

[@mauricegold9377]

Questions: could it be that a charged particle is always radiating (and perhaps borrowing and returning energy from the vacuum), unless it is sitting on the surface of some body or other? And even if that were largely true, could it not be that few bodies are exactly at rest with respect to one another because of thermal motion? In free space, surely the particle's field is stretching out, and therefore at some point, some part of it may eventually encounter the weak gravitational field of a yet distant body, whereas the particle itself, further away, now moves relative to at least part of the field it produces. It could be I am havering, and/or maybe the effects are too minuscule to ever measure (or contemplate).

[@김영철-p7v]

This problem is older than general relativity. When Einstein was devising the equivalence principle before the announcement of general relativity in 1909, this problem was pointed out and refuted by Max Born. Nevertheless, Einstein had to stick to the equivalence principle to create general relativity. My answer to this problem is “General relativity itself is wrong.” I succeeded in producing the same result as the three early evidences of general relativity, using special relativity and Maxwell gravity, and in particular, I could remove any doubt by showing it through simulation rather than mathematical tricks. In that method, the equivalence principle had to be denied. If you want to know more about it, look for the book "Relativistic Universe and Forces" or related posts or site "金睛塔". I will not post a link because KZbin prohibits links.

[@carparkmartian2193]

I'm a physicist and I think this is a great video in that it is seeking to move problem solving to a broader domain than the hyper conservative confines of academia and the scientific paper manufacturing industrial complex. Which is something that Hossenfelder aligns with. Now what I get out of this, and agree with, is that the electric field of a charge is an extrinsic property of that charge. This is an unavoidable conclusion from my own analysis that comes from a very different fundamental model that derives the formation of the electron and positron. In this model the electric field is a result of the capacitive electrical energy storage of the electrical permittivity of "empty" space. That capacitive energy stored in that "empty" space is due to the spin interaction of the electron ( a charge). So the fundamental intrinsic property of the electron is its spin and its magnetic dipole moment. The electric field, mass and gravitational field of the electron are all extrinsic properties of the electron. They are literally all "off particle" properties. ( by way of simple proof - the higgs field is an extrinsic field so the electron's mass and associated gravity field are all "off- particle " properties. The non- obvious realisation here is that the spin of the electron must extend out as far as the electric field it induces in the surrounding space - that generates that electric field. Quantum theory tells you that particle spin is non-physical. Not because it is proven not to be physical, but because no one can make sense of how it could be physical given the strangeness of spin half and g=2 for the magnetic moment. But that is an assumption and an assumption is a very very long way from a proof that these properties are not for example the result on an emergent physical property of a simple underlying but dynamic physical system. Not forgetting that we need to question the fundamentals of quantum theory - because it allows us to loose track of 96% of the universe. Anyway back to the problem. It is the extrinsic electric field of the charge that performs the function of ( electric and magnetic )energy storage of the electron. The electron at rest on the surface of the earth has an extrinsic electric field also at rest relative to the gravitational field it is in. The extrinsic electric field of the electron is not changing so there is no net energy exchange or emission from the electron. The proof of this is that all the electrons of the earth would be radiating and that would be detectable and wind down the existence of electrons pretty quickly due to persistent energy loss. This contradiction is pretty much the same as the strange non-radiative loss of an orbiting electron that is constantly changing direction ( accelerating). There is a very good simple reason for this behaviour. Which involves understanding the role of superfluids in this dynamic emergent behaviour. ...more on that very soon...

[@giammarcociaccioni469]

I really find interesting what you said, but i can not deal with the sentence “ the electric field is separered from the change”. Also the fact that electric field is attracted from the gravitational pole. If you check in “Purcell or Griffiths” (2 well know books on elecromagneticity) nothing about what i mentioned is written

[@annoloki]

The "field" is only a mathematical description of the pattern of the effect that you would get in different places, for example, if there was an electron there... it makes no sense to talk about the field as if it is a thing in its own right, it isn't, it's a way of modelling potential outcomes of interactions that could occur, but it is only a model... for example, you don't need to model balanced forces that cancel out, like the charges in a neutron, but that doesn't mean that the universe doesn't "do" them.

[@raycar1165]

Gravity is a byproduct of the electromagnetic force. The greater the mass on a planet the larger the gravity field. And mass is dependent on the accumulation of charge. So the total amount of charge determines the gravity of a given object. The standard model architects lifted their noses at observational data and chose to use conceptually based mathematical models. And today there are millions of intelligent people fighting to find out who will have the next impossible answer. There will be no explanation in any text book after the 1920’s. We’ve been duped. There is no room in today’s science for antigravity, transmutation, cold fusion. All key words we’re taught to dismiss. They are all required in an electric universe. The Theory of Everything is just another way to describe Natural Philosophy.

[@kevinvanhorn2193]

First things first: leaving out GR or QM, in flat space and using the Lorentz transformation to convert coordinates between inertial frames, what do Maxwell's equations say about about a uniformly accelerating charge and a non-accelerating observer vs. a non-accelerating charge and an accelerating observer?

[@account0199]

The most Zen of knowledge: "A charged particle does not radiate across space, you radiate across the charged particle's space", I wanted to add (ᵃⁿᵈ ᶦᵗ ˢᵀᴱᴬᴸˢ ʸᴼᵁᴿ ᴱᴺᴱᴿᴳʸ ᵀᴼ ᴰᴼ ᴵᵀᵎ), But I don't wanna ruin it.

[@justintime9714]

1. Free-Falling charges don't radiate in their rest frame because they follow their geodetics. 2. Charges on the surface of the earth radiate because they are accelerated. We can't measure it in the rest frame of the charges because it's behind rindler horizon.

[@pinocleen]

KZbinrs famous words "there are some experts in the comments" :D Great topic and video btw.

[@tw5718]

Interesting question. Does whether a charge radiates or not have to be frame independant? I don't think so. Neither the Electric nor Magnetic fields are tensors, meaning they can be 0 in one frame but not in another. In one frame you can see an electric current and therefore a magnetic field, but not in another.

[@4lumi]

Bah… you completely missed the point … for the equivalence principle to hold the sistem must be isolated, and this also implies that it is electrically and magnetically isolated. The equivalence hold only if no radiation and no information is exchanced from inside to outside and viceversa.

[@threeMetreJim]

Did you mix up falling charge (constant speed) with accelerating charge (changing speed) at 3:38? To me it sounds almost like it should be similar to dropping a magnet down a copper tube; acceleration slowed in a similar way, also involving charges and photons.

[@lukasrafajpps]

isn't a falling charge also accelerating?

[@threeMetreJim]

@@lukasrafajpps I think what happened is I commented too early without watching the rest of the video. It might be confusing trying thought experiments with a single charge. Maybe a large group of charges would be easier to analyse - you could then confine them to a conductive object for experiments; an experiment along those lines is the one where length contraction of moving charges causes the same effect but by different mechanisms depending on your point of view. From my point of view is the question is: "does a static charge radiate (photons)" - The largest collection of static charge I can think of, that is easily testable, is a thunderstorm. I have no idea how much radiation there should be, but I can imagine it would have been detected by now if it were present.

[@threeMetreJim]

@@lukasrafajpps I've been thinking about accelerating charges and wondered if surface charging of satellites causes them to slow down due to radiation, and whether that could be detected. Also ionised particles orbiting a black hole probably emit some radiation due to the huge acceleration they must experience, not just the friction with other particles.

[@makeitreality457]

Given the Lamour formula at 15:50, and the subsequent solutions, the amount of radiation emitted by charged particles in a gravity field would indeed be an undetectable fluctuation against the enormous electrostatic attraction any experimental setup would introduce. Such a small difference reminds me of gravity. Like if you turned the equation around, the gravitational constant would fall out.

[@makeitreality457]

The implication is, let me spell it out for you tinfoil hat guys, (IF true) bombarding a large mass of charged particles with a tiny amount of a specific type, wave shape, and polarization of non-ionizing radiation, should produce an acceleration not unlike gravity.

[@stephanelem822]

I though about that problem when I wanted to apply the rain cycle we usually know on Earth, but for a hypothetical charged matter on the surface of a star. The basic hypothesis are a very strong gravity, a very hot heart and charged matter. The cycle of the charged matter would be: - from the heart, the very hot matter rises the gravity field (and an electric field may also) to gain altitude in the star. - at high altitude into the star, the charged slowed down by the gravity field, is at much lower temperature. - over that surface of charged electric matter at high altitude but at low temperature that creates near by an ocean of electric charged matter, there is a hole, and the charged electric matter falls into that hole to get back into the heart of the star. Of course, when the electric charged falls into the hole, it creates a huge beam of light. The is the image I've of a pulsar star. The beam we can see from a pulsar star is the hole where the charged electric matter falls into.

[@PiercingSight]

"Whether something radiates or not should be an objective event." Not in relativity. Also, your visualization of radiation is implying an undulating wave-like radiation of multiple photons, which isn't the radiation that would be experienced from an observer viewing the electron accelerate uniformly only in one direction. It would function more like the gravitational influence felt by someone accelerating past a planet. It wouldn't be a bunch of photons, it would be a single electromagnetic wavecrest. Both the electron on the ground/person falling, and the person on the ground/electron falling, situations would function identically. They would actually both create wavecrests from the perspective of the other, and influence each other as they accelerate by (assuming the observe can interact with the EM field).

[@vanikaghajanyan7760]

15:33 The energy of the particle (or charged particle) is equal to the derivative of the action in time x(0)/c: and defined in world and proper time: E(0)=с∂S/∂x(0), E=∂S/∂t [momentum p(k)=∂S/∂x(k)]. Then E(0)=E√g(00)=const and E(0) is preserved, and E is not preserved. E=mc^2/√1-v^2/c^2, where in the static case v=dl/dt=-dl/dt√g(00). Thus, when a particle moves in a gravitational field, the energy E(0)=mc^2(√g(00)/√1-v^2/c^2 is preserved*. This formula remains valid in the case of a stationary field if, when determining the velocity v, one uses the proper time measured by the clock synchronized along the trajectory of the particle. P.S. "Due to the dependence of the speed of light and the speed of the clock on the gravitational potential, after the introduction of the gravitational potential, the laws of nature must be understood as the relationship between the gravitational potential and other physical quantities." (Pauli, RT). ---------- *) - So, it should always be remembered that according to GR, it is possible to make a general statement that: the free movement of the test body occurs along geodesic lines. And this is the most general formulation of Newton's first law. That is, in a Riemannian manifold, where formally there is no gravity and the masses move freely, free fall turns out to be not accelerated, but uniformly rectilinear motion. {Unless, of course, GR obeys a strong equivalence principle, that is, if in fact a locally Lorentzian system = a globally Lorentzian system.}

[@martinmelhus7324]

The charge does not radiate in its frame of reference, but does radiate in a 'rest' frame. Consider the falling charge as a current, which will generate a B field. As the charge accelerates, the current increases, which increases the B field. Couple that with the charge's natural E field (which is moving), and there's radiation. However, in reality, the effects of gravity are tiny relative to electromagnetic effects for charged particles. For macroscopic objects with appropriate charges, these forces can be comparable.

[@shadow15kryans23]

This is a long comment, But gets in depth on the details so you can rest easier knowing it: I think it makes the most sense that a charge falling from a stationary observer on the ground would radiate just over time randomly. As in, Even if the Local reference frame didn't radiate given its gravity moves along with the electromagnetic field of the charge. Even then the electromagnetic field lines extend beyond the local scale of the particle to global scales. This entails it'll have a higher probability of creating radiation when the electromagnetic lines intersect the gravitational fields geodesic lines (aka where they move in different directions and aren't perfectly parallel even if on a extremely small scale). Similarly… The closer this intersection occurs, The higher chance it'll ACTUALLY release radiation. This would start to occur on molecular length scales just above the particles local scales, but would be so weak that it would be primarily virtual radiation and not legit radiation. This Virtual Radiation also is called “Virtual Photons”. Remember! Particles Technically are just localized Wavefunctions aka waves in disguise. This localization occurs because of the Quantum Decoherence as a result of the High Energies created by Strong Nuclear Forces Gluons. This high energy is akin to the energy created by acceleration inwards of the protons and neutrons (caused by strong nuclear force confinement battling against random wiggles of particles) in all directions and so the wave becomes compressed in all directions. Hence where the “wavefunction collapse” term mathematically comes in. You can even see this localization in the large hadron collider, Where Quarks that break away from the atom can be traced as if their particle-like due to the high velocities relative to the wavefunction. This pretty much compresses the quarks wavefunction like how you might compress a spring, reducing uncertainty in position. Now imagine this same compression but in all directions, THAT is what a atomic nucleus would be like! Yet even then. It can still interact non-locally at a distance just like any other wavefunction, just in this case via these electromagnetic field lines. What this would suggest, Is as the particles electromagnetic field lines intersect gravitational field lines at a distance the particle would release some virtual photons in bursts. With more virtual photons releasing the closer this intersection to the particles local reference frame is. It's just these virtual particles in the form of radiation would be so weak there isn't enough stuff to quantize it, and likely is quickly absorbed by the surrounding virtual particles on miniscule scales. In this sense it's kinda like the Weak Nuclear force. Just instead of rolling a dice to split a piece of the atom off over time, We are rolling the dice to release a bit of actual radiation over time. Such timescales are gigantic though, hence why evaluating a larger area around the particle is the better bet although more challenging. _________ Summary: It's almost guaranteed that particles falling relative to you on the ground releases virtual photon radiation. It's just that it doesn't become actual radiation unless you get lucky and one of the electromagnetic field lines intersects the gravitational field line at a close distance to the particle (instead of going around as it usually does). This means the radiation likely will occur in a very weak form (aka at a lower energy scale) further away from the particle. It's just correlating such low energy radiation further away, to the particle itself is extremely difficult. Soooo, We are running into the same situation we ran into with the Weak Nuclear force. Just now with virtual radiation becoming actual radiation. How do we induce this actual radiation though? Well… Perhaps accelerating the particle through a high density substance will work (to force more field intersections and closer to the particle). But then we need to find a dense substance which doesn't absorb the particle and radiation. And we'll also need to be capable of capturing what is going on at high FPS from multiple angles to accurately correlate the radiation to the particle. Needless to say this is one of those “Easier Said than done” type of things. No wonder a successful and repeatable experiment has yet to be done ooof.

[@Macialao]

Can you uniformly accelerate charge, without acting with force on it? And there are 4 basic forces, electron will first interact with EM, and only EM force. Thus i would say entire discussion is about hypotetic object as accelerated electron without external field (force). Such object cannot exist under physical laws. Acceleration must comes from fundamental force, which has it's relative field, which means there must be external field to accelerate electron, thus it's moving in it's and something elses field.

[@hugmynutus]

One of those questions where you need to break out both the GR & EM tensors to answer correctly. The correct answer is "depends, relative to what observer?" As acceleration is observer dependent as is mode of radiated energy.

[@annoloki]

Right... you need a second charged particle for it to be interacting with, otherwise there is no force generated by the presence of the charge to act on it to cause it to be slowed down.

[@hugmynutus]

@@annoloki Not exactly. Photons can still be radiated & absorbed by non-charged observer. But you still need an observer & reference frame to describe the interaction.

[@Hack3r91]

10:55 I don't understand, shouldn't the Unruh effect be experienced by the person on the ground? The presence of the charged particle and the other, free falling observer shouldn't have any infuence on the observation of the grounded person, at least I don't think it should...

[@3zdayz]

Acceleration does not cause time dilation (other than the resulting change in velocity which makes the dilation have a delta) the time dilation in gravity is constant. The time dilation of gravity is equivalent to moving at the escape.velocity

[@3zdayz]

Just saying equivalence principal is good for thought experiments,but there are flaws that can be found in reality

[@The_Green_Man_OAP]

The Lamour formula, named after physicist Louis de Launay Lamour, is a mathematical expression that relates the energy of a charged particle to its momentum and charge. The formula is given by: P = 2q²a²/3c³(4πεo) where: * P is the energy of the particle (W/m²) * q is the charge of the particle * a is the acceleration of the particle * c is the speed of light * εo is the vacuum permittivity constant This formula is a simplification of the more general Lorentz force equation, which describes the force experienced by a charged particle in a magnetic field. The Lamour formula is often used in situations where the magnetic field is strong and the particle's motion is relativistic, meaning that its speed is close to the speed of light. The Lamour formula is a useful tool for understanding the behavior of charged particles in high-energy environments, such as those found in particle accelerators and space plasmas. It is also a fundamental equation in the study of plasma physics and electromagnetic wave propagation.

[@lion1o21]

I am glad you touched upon this question. I have the feeling that when analysing this process we cannot view the electron as a classical object but as a quantum system which radiates only upon change of quantum state. Some people mention QED in the comments but from the little I have read it seemed that there is still debate when using the theory. There is though another way. What if you used the electron's zitterbewegung? I have worked a bit on this and it appears that the electron will radiate only if there is a change in the zitter frequency of the electron as viewed from the lab frame. Does this statement apply to all the cases you have shown? I will think about it..

[@markattila9835]

Radiation and E&M fields are not absolute, they depend on reference frames. You notice Lorentz force law has velocity vector component. So in one ref frame you may have magnetic field whereas in the other you don't for the same moving charge. Furthermore that formula for radiation is incomplete as it's only the scalar component. The acceleration is also a vector, so radiation will also depend on if you're in an accelerating ref frame. Gravity's effect is already an accelerating ref frame. So a "falling charge" may experience zero net acceleration in that ref frame. If so it will not radiate when observed from such ref frame. As for the equivalence, if I remember correctly it's un-proven in the same way as inertial mass and and gravitational mass is considered equal.

[@lukasrafajpps]

Hi, yes E&M fields depend on the observer but you can't transform away an electromagnetic wave using lorentz transformations. So if there is an EM wave in one frame there is also in other.

[@markattila9835]

@@lukasrafajpps The EM wave would be different frequency depending on the ref frame, but it would be non-vanishing in all ref frames. So that would mean the the falling charge is either radiating in all ref frames or it isn't. The question is, does a perfectly uniform gravitational potential (doesn't exist) cause a charge falling in it to radiate due to acceleration? I would say no because the acceleration is effectively zero due to the equivalence principle. The charge does not experience acceleration it is moving along a geodesic. A charge standing on the "ground" would observe the "falling" charge as accelerating. I'd argue that the charge on the ground is radiating not the falling charge.

[@narfwhals7843]

​@lukasrafajpps the transformation to an accelerated frame is not a Lorentz transformation.

[@lukasrafajpps]

@@narfwhals7843 I know I wasn't talking about such transformations. As I understood the author was talking about the mixing of the electric and magnetic field of moving charges since he was talking about the velocity component of the Lorentz force.

[@lukasrafajpps]

@@markattila9835 That is something I would be fine with but there are good arguments that a charge in freefall would radiate. Although Lorentz transformations can't create or destroy EM wave, the transformations between inertial and accelerating observers are not LT.

[@johnkeck]

At last! I've been asking this question for decades. I appreciate your presentation, which brought out aspects of the problem I hadn't considered so clearly.

[@karkaroff1617]

10:10 Brazil mentioned🍺

[@SusanSousa-no1km]

Unruh was mentioned too!!! Good grief.

[@karkaroff1617]

?

[@JustCJson]

Brasileiro é foda

[@mrslave41]

your videos are so amazing. if only because of how humble you come across. your pace is just so perfectly slow. it gives me just enough time to digest each word. your animations are also slow and crisp. an absolute joy. 🥹 ❤❤❤

[@davidhand9721]

I'm not so sure on the idea that a field is falling gravitationally, and I have a feeling that the resolution will come from a better understanding of fields under GR. The EM math you're using throughout the video is not Lorentz invariant, which is well known to cause apparent paradoxes.

[@benspratling9041]

I think the solution is in that "radiation" as we have thought of it is only a coupling effect, not an objective reality. It is not the case that charged particles do not broadcast anything when not accelerated, but it is the case that when they are accelerated, they produce a coupling effect we call EM radiation. Energy is not a real quantity of something that gets "transferred" from one charged particle to another by this "Radiation", instead it is just a convenience calculation of totals. what's really going on is every charged particle is always "Radiating" the capability to change momentum per unit charge in all directions at speed c. that's happening regardless of whether the particle is accelerating or not. To test whether there is "radiation", we hold a second charge at a fixed distance away from a charged particle, and therefore there must be many many many other charged particles holding the second charge in place, held in position by EM oscillatory systems we call "atoms". Specific derivatives in the charge field due to the acceleration of the first charge particle will influence the second within the context of the oscillatory system's they're trapped in. if the derivatives don't happen in a way that resonates with the trap, then the second charged particle gets a net 0 acceleration because the change cancels out with it's oscillatory motion half the time, on average. only if the derivative of the first particle has a rate that resonates with the second will it produce enough change in the second to register a change in our instruments.

[@Aworldof1]

This is a good perspective.

[@SeanCMonahan]

Is it possible that the accelerating charge appears to emit photons in the direction of its acceleration when viewed from the ground, but that the charge sees the ground emitting photons at it (or vice versa)? Or would that just make the paradox worse?

[@arctic_haze]

My first answer was "Of course not". The surrounding of a free falling particle is actually an inertial frame of reference in General Relativity.

Mine too. But then the conclusion that a sitting charged particle radiates is super weird. 🤷‍♂️

[@arctic_haze]

Not so weird if one really understands general relativity. Especially as well as Rohrlich who solved this paradox in 1965. The "sitting" particle does radiate but only if observed from a free falling frame of reference. This is described in Section 8.3 of his handbook.

[@lukasrafajpps]

but what is weird that as far as I know Rohrlich also claims that observer on the surface would see radiation from freelly falling charge.

[@arctic_haze]

@@lukasrafajpps His point is that the observer on the surface is not inertial one so Maxwell's equations do not work correctly in this reference frame. The one we discovered them in!

@@arctic_haze I guess, I'm just an amateur. 🙂

[@imsorryyourewelcome]

Correct me if I'm wrong (which it looks like I might be) but... isn't the particle in free-fall experiencing zero acceleration since it is simply following the curve of spacetime? And isn't the observer on the ground the one being accelerated by the surface of the earth by 9.2m/s^2 away from Earth's barycenter? So, wouldn't the person on the ground be the one radiating from the perspective of the charged particle? Doesn't the same thing apply to a charge sitting "still" on the surface of the Earth, as in, the particle is only accelerating if something is preventing it from following its free-fall trajectory?

[@jakob3270]

Can you make a video about Amanda Gefter's book "Trespassing on Einstein's Lawn"? Can't stop thinking about it.

[@brickie9816]

I'm trying to learn basic electricity/magnetism physics, I need not to enter this rabbit hole of madness. Still, interesting af, hopefully I will be able to understand the problem at some point. Good luck

[@Soniesfera]

Im confused, why we are assuming that radiation doesnt depends on the observer?

[@IsZomg]

1:40 ok I haven't watched the rest of the video nor read any comments but my gut feeling immediately is that gravity is a curvature in spacetime and does not accelerate the particle, the particle keeps moving in a straight line through the curved spacetime. So I would expect no radiation coming from it. Watching the rest of the video now...

[@ericastross4177]

I guess larmor precession works in bulk where there is an energy source and a circuit. You can then track where the work is done. Scenarios for one individual particle doesn't complete a circuit, ie we don't have before and after measured states. suspect, the gedanken scenarios out of GR are masking assumptions.

[@DarrenReynoldsBly]

Request for another video! People often say you can't travel faster than light, but it seems to me that relative to the speed of light, you can't travel at all. No matter how fast you go, any light you emit always travels away at the same relative speed. Since in common parlance we have no trouble moving, this demonstrates that there is no fundamental limit to the rate at which we can get closer to a destination because our speed relative to light is always zero. It is merely that one can never *observe* something moving faster than light, including the launch pad you left behind you, or your destination. Instead what happens the faster you try to go is that distances start to contract. It is perfectly plausible to reach a four light-year destination in less than four years. It would be great to get a proper explanation of that. And, what does length contraction mean for the size of the observable universe if you can bring the unobervable parts closer by moving fast towards them?

[@alwayscurious413]

I think this is a very interesting subject. I would be grateful if you could review the context of a transverse EM field (light wave) being emitted from an RF antenna. The emission arises from the oscillation of the electrons back and forth along the antenna- they are accelerating back and forth as the drive voltage is cycled. So far so good. The EM intensity is torus shaped being zero along the axis of the antenna - so there is a moving electric field back and forth along the axis but it is not a propagating light wave. I would argue the accelerating charge does not radiate energy along its direction of movement- in fact it cannot do so. Does this tally in any way with this discussion? The next question is whether we need the electron to reverse its path (at least once) to propagate the EM field. In other words simple harmonic motion. Apologies in advance if I’ve slipped a gear on this - just wanting to get to the very bottom of EM radiation at the fundamental level.

[@discoveringthegardenofeden7882]

Simple: what do meteoroids do when they fall through the atmosphere? They explode. Not because of heat but because of charge equilibration. Overcharging and explosions happen before heat should become a problem. They (try to) radiate.

[@Achrononmaster]

This was really awesome thanks. But it falls in the category of one of those unphysical classical EM puzzles. (From the "there is no spoon" department:) There is no electromagnetic or Faraday field, there are only electron-photon vertex interactions, so that'd be the proper setting for a definitive answer (if that's what you want). For the classical EM puzzle (as a nice puzzle in its own right) don't you want to conduct a fully covariant analysis, hence use the Faraday field, not separate (E, B, ρ, J). Justathought. Unruh effect seems to be the crux, no?

[@johntravolta3235]

Maybe it has to do with the curvature of spacetime? If the observer and the charge particle were to be in a same reference frame, they don’t experience any differences between their spacetime so they don’t feel any radiation or any other form of resistance. But if this were not to be the case, then the charged particle and the observer A don’t agree in the length of the spacetime and the differences within these lengths may lead to some form of resistance: in the case of the observer A, it would be radiation. If the observer B with the same reference frame as the charged particle sees the observer A, then observer B would sense gravitational waves, which would lead to similar effects. The energy, which causes all of this, is from the decrease in the potential energy. Does the acceleration change? No, because it’s only spacetime contraction and expansion. One can measure something similar. Of course, if you push it a bit hard, I’m willing to yield.

[@simonbusman2633]

Thank you for this great video. As the electron emits very low energy when falling towards earth the wavelength must be long. Synchrotron radiation has wavelengths in the nano meter range, close to the size of the electron, and it's real. From a radio-engineering standpoint that seems logical as size of an antenna must not be to far off from the wavelength, at least not many orders of magnitude. If the antenna is way to small compared to the wavelength, it is impossible to get the energy out of the antenna because there is no way to match the transmitter impedance with the impedance of vacuum. All the energy you push into the antenna bounces back to the transmitter. It becomes very unlikely to get a single photon out.

[@geoperry]

Having no useful comment on this programme's content, I offer only my thanks for Lukas' stand-out presentation: it was my pleasure to have been confounded by his clearly-articulated conundrum, better than confused by so many presenters' verbal slurry, particularly when speaking critical points in sotto voce as though sharing a secret. On Internet. - Thanks, Lukas; I'm off to browse your channel. -g

[@lukasrafajpps]

Thank you for kind words :)

[@kylelochlann5053]

Note 2: Einstein Equivalence is a statement that the gravitational field is a metric field. The back reaction of the electromagnetic field would have nothing to do with the EEP.

[@jamspandex4973]

If you have a moving charged particle in free space, you can *always* boost to it's rest frame, and in it's rest frame it can NOT radiate photons because an electron does not have any excited states in free space. The ONLY way you can have the particle radiate, is if there is some additional interaction. To describe the interactions of particles with gravity,. we would really need a theory of quantum gravity - electrons and their behaviour can most simply be described (almost correctly) by QED. So what you need is to embed the QED lagrangian into a curved space time. A Curved space time has energy itself, so you would need to solve everything in terms of the equations of the interaction of the electron with the energy of the gravitational fiedl somehow, if that were even what was going on. We can;t write down a feyman diagram for that, so we can't really calculate it properly. The point is, that an electron radiating a photon is a QED process, but movement through a curved spacetime is a classical behaviour. In tfield theory, an electron is NOT just an electron, it is an electron surrounded by an cloud of an infinite number of virtual photons and virtual electron-positron pairs. The bare charge is infinite but you can never "see" bare charge, you can only "see" the charge of the electron shielded by all these virtual electron positron pairs, which gives it the usual "effective" charge that we actually observe. We can not switch off the quantum field, as that is what actually creates the electron - electreons are excitations of the electron field. If the electron were at rest, these virtual photons and e+e- pairs are continually being created and destroyed all the time. If it is moving through some other (gravitationsal ? ) field with which it is interacting (how does it interact with gravity at all in a field theoretic way ? Who knows, maybe via virtiual Higgs bosons ? do Higgs bosons couple to gravitons ? What *are* gravitons ? ) but lets say is *can* interact with the field, then perhaps what would happen is that some of these *virtual* photons will interact with the other field, and will then be able to take some energy from this other field enough to make themselves into real photons, which can then radiate the energy away. In the resst frame of the electron, it is *still* surrounded by these virtual photons, but it has some additional field moving past it, with which it can excahange a photon. Because we are in the rest frame of the photon, then this "radiated" photon might have a negative virtuality. Technically the electron would accelerate so maybe the excahnge of the virtual photon would be to *increase* the electron energy in the "lab" frame. In the electron rest frame the related photon could have any energy, or any momentum so long as the invariant mass of all the particles was the same. It would not really be any different at at from the physics of electron scattering at HERA (an electron-proton collider at DESY) where eg we had 27 GeV electrons or positrons, scattering off of quark within the colliding protons, wher ethe protons had 920 GeV. The kinematic peak of the scattered electrons had the electrons scattering through some angle but *without* gaining, or losing any energy. They changed direction, by exchange of.a negative invariant mass (virtual) photon with the scattering quark. This is ok, since it is avirtual photon that is absolbed by the other system, so all the invariants are conserved, energy and momentum is all conserved fine. So if the electron is interacting with the gravitational field in some way, then there would be some Feynman diagram where the electron was interacting with some other thing by the exchange of the "radiated" photon and everything would have to be Lorentz invariant. Sadly we don't have Feyman diagrams for gravity, and we don't really have a quantum field theory that explains how gravitons would couple to a charged particle. Photons have no mass, and so no (direct) coupling to the Higgs, but they still move in geodeics through a gravitational field. Perhaps we could consider the movement through a gravitational field to be an infinite sequence of movements through regions of space that were all locally flat, but where all these infinite flate space times are all orinetaed in different "directions",. Maybe that is how photons move through a gravitational field, I can't say, I don't work on general relativity, but if this is the case, then electrons moving along a geodesic would not radiate either, then as they would be constantlly be moving in a "straight line" locally. This would sort of make sense, because really the "accelerating" electroms in a gravotational filed, would nedd to *absorb* photons, not radiate them at all, in order to gain eneregy. But that would be fine, given an electron interacting with some other particle, one can always create an photon with an invariant mass, where the electron continues to move in the exact same direction, but with a larger energy and momentum, you just need to properly work out what happens at the other end of the virtual photon, because that is what would be happeneing to the photons if that was how they were interacting. If the electrons where coupling to the field by way of interactions directly with gravitons, or via a higgs boson, the kinematics at the electron end would still be the same, soe the invariant mass of the propogattor would be the same also, but what happened at the other end would be dependent on the copupling of the parfticle. None of these even included that effect of the electron spin - if the electron spin were not affected, then it would have to be interaction with a virtaual scalar, but perhaps the electron spin would not be conserved. Quantities of individual particles are only conserved when there are no interactions that can affect that quantum number. I have said enough, so for anyone who has managed top make it this far, sorry I couldn't answer it., we'll just have to wait for quantum gravity I think, but the important thing to remeber is to conserve energy and momentum: If the electron is speeding up, then treating gravity as a "basic" force, not a curved space time, in the overall rest frame of the Earth and electron then the Earth will be microspopically changing its energy and momentum, so this means that the electrom will have to be exchanging *something* (gravitons?) with the earth, and you can easily calculate the kinematics at the electron end. Although finally, the idea of moving charges radiating, is really a classical picture, when really, when dealing with individual particles, we need to consider things in a Quantum Field Theory way, at least it it is not a simple case of simple movement. In particle physics experiments, we measure charged particle trajectories by travel through silicon detectors, but when the particles move through the detectors, they interact electromagnetically with the material (ie mostly electrons in the atoms) in the silicon in well understood semi-classical ways - we know enough about QED etc, that we could probably calculate it in a field theory way if we had to, but it would be *very* tedious and *very* complicated because these other electrons would be bound in atoms, in some high voltage electric field in a doped semiconductor etc, so solid state physics is a good enough approximation for all that. With gravity, we don't have the underlying field theory to know what the underlying interactions of an electron with gravity are, so we would have to stick with some partial semi-classical approcimation, which moght well be incomplete, or just wrong, and any one of the semi-classical assumptions might actually be incorrect, and the only thing we could really count on is that energy and momentum should be conserved, including the individual particle invariant masses.

[@robsosno]

You've forget that non-charged matter consist of charged particles. First time when you compare charge particle falling from tower. In this case both charged particle and neutral atom will radiate or not in the same way. The difference could be only in case of truly neutral particles like fotons, neutrinos or graviton. Second time in the last experiment: 10^50 atoms times 10^-52W = 0.01W . But you shuld multiply this by average number of electrons. So free falling Moon to the Earth should radiate weakly. This is still very subtle effect but at least there is some hope that in future it could be detectable.

[@projectarduino2295]

Questions that arise: What does local mean in a practical sense? What physical distance is local? Why is an electric field, or any field of force affected by gravity? Could we accelerate the particle with laser pushing and differentiate between emissions from acceleration and emissions absorption? I would love to hear some answers if you could provide them.

[@Barteks2x]

My answer is that it *has to* radiate because as far as I can understand, there is no real physical difference between falling down and being in orbit around an object. And there is no way an electron being in orbit around something (gravitationally) wouldn't create any radiation. That's happens even with gravitational waves and those are definitely confirmed to exist. If this violates the equivalence proinciple, so do gravitational waves (or, it doesn't in the same way as gravitational waves don't, regardless of how that works, most likely something like "you can't locally observe radiation you emit yourself")

[@fredlamycernvm]

DId you consult Bill Unruh himself? I once met him, and we spoke about the effect he discovered, and I could see he had a clear vision of the implications such as the one you pose the question about!

[@lukasrafajpps]

Sadly no but I would love to hear his opinion on this matter :)

[@roberttarquinio1288]

Gravitation acts on the photon of starlight as starlight passes through a gravitational field I think the photon accelerates as it passes through and is deflected by the gravitational field Gravitation pushes; it does not pull Gravitation is a manifestation of space time curvature which distorts (bends, warps) the fabric of space and time also referred to as space-time Charged particles are deflected as they pass through a gravitational fired Gravitational fields are comprised of gravitational field waves As gravitational fields interact they produce gravitational field waves Charged particles are also deflected by magnetic field; they are also influenced by an electric field Negative electric charged particles such as electron are attracted to positive electric charge particles such as proton The proton positive charge is a net charge produced by interacting quarks (2 up and 1 down) comprising the proton An electron is attracted and becomes captured by proton via coulomb charge (also referred to as coulomb electric charge)

[@MichalPlichta]

It is extremely disturbing with current knowledge and awesome progress there are still that simple question we do not know answer for them!

[@alwayscurious413]

Excellent remark - to my mind physics is falling over because of the simple things we haven’t truly nailed down rather than the complicated things we’ve maybe over utilised. So the reason there could be a crisis in cosmology etc is because the fundamental basis is still locked down and we might have built a tall castle on sand. It is these kind of ‘break it right down’ videos that could help us unpick the tangle we have made.

[@mrslave41]

1:14 “…due to this acceleration it would radiate…”. no. a satellite 🛰 is in free fall. it’s not experiencing “acceleration”. [after listening to 12:41 … i’m now sooo confused 🫤]

[@lukasrafajpps]

That is also what I think but there are physicists that claim otherwise. On wiki you can find a resolution by Rohlich claiming it would radiate en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field

[@greggregory8311]

Good to see I'm not the only one confused 😮

[@JohnUrbanic-m3q]

This is where Wheeler-Feynman absorber theory really comes into its own. It clarifies that radiation only makes sense in the context of an interaction with another particle, and one can then dismiss the fields and self-interaction entirely, and rely on simpler coulomb-like advanced and retarded fields. However in this case one still needs to do the advanced and retarded calculations over an actual metric, which is where it takes some effort (and I am too lazy/rusty). I hope someone takes this approach, because I personally can't see which answer should pop out. Maybe the "forces" involved save the equivalence principle, because the charge is no longer free falling. But, the advanced and retarded forces might cancel in some fashion. But, they are not symmetric to each other. But... At least it seems much more obvious that the various cases discussed will be self-consistent.

[@ailblentyn]

Didn’t Feynman say that charged particles only radiate when their acceleration changes? Like oscillations, deflections.

[@dvoiceotruth]

acceleration in gravitational field is not constant, strictly speaking.

[@ailblentyn]

@@dvoiceotruth isn’t it? If you have a charged metal sphere sitting on your table, isn’t it undergoing constant acceleration?

[@geoffgjof]

So this is probably unrelated to the video, but... Now that we have better measuring equipment, have we verified that objects fall at EXACTLY the same speed? I get that the gravitational constant was discovered because, if there is a difference, it's negligible in most normal applications; but are we sure that there isn't a very micro-miniscule difference between how fast objects fall? The reason I ask is that if gravity were an elctro-magnetic phenomenon, due to electron distribution in an object, then the more mass an object has, the more "positive" that object would be, and the more it would pull other objects with a relative "negative charge" towards it at a constant that is proportional to the density distribution of electrons in the larger object. But. There would still be a teeny tiny difference between different objects. And if that force is like most things in life, then it should be exponential rather than linear. So perhaps it would be easier to measure discrepancies between the gravity of objects in space because being farther away from a large object would mean that the larger force isn't just creating "noise" in the difference of signal we're trying to evaluate?

[@jeffpkamp]

Are there quantum effects that stop a charged particle from radiating photons with power levels as low as one accelerating at 9.8m/s2, or the even slower tidal accelerations talked about in the video?

[@Tata-ps4gy]

This seems like a huge dialect W for me. Solipsism just cant describe Reality. Thank you so much for sharing this mystery with us. It was a very interesting video

[@aniksamiurrahman6365]

I got a question. Let's imagine a thought experiment. Let's say we put a ccd beside a charged something and accelerate the whole setup to, say 60g using a solid rocket motor. Let's say we got another ccd, which is not moving with respect to the lab. What will our two CCDs observe? We can do the opposite too. Accelerate (at high g) the ccd towards a charged thing and see if the charge srarts to glow. Earth's gravitation might not be intense enough to cause detectable radiation, but we can perform equivalent experiments wiyh accelerating frame, accelerate them as much as we wish and see if we see anything.

[@meenki347]

Ok, consider this experiment. Rather than just dropping a metal ball off the tower of Pisa. We first heat it up and then drop it. That way, long wave heat radiation will be radiating as it falls. But we also immediately drop a second heated ball. The first heated ball would be radiating down as well as up towards the second ball too. But the heat of the second ball would be radiating both up and down. Would this long wave radiation have any effect on the speed of the balls?

[@renscience]

If a charge moving in a curved space-time radiates energy, and assuming radiates photons (aka energy and assuming it has mass) 1st…would it not require to meet certain frequencies as in hf so it would only occur if allowed, and 2nd, if it did radiate, would it not evaporate eventually due to E/c^2= m? Also, why would it slow falling from the tower? Regardless of mass, even if lost due to radiation, all mass accelerates the same in the same curved space-time?

[@smizmar8]

So, if you dropped a charged particle on a copper plate, the copper plate would be the observer. Possibly you would need a more exotic setup, but for the point I'm making, wouldn't you be able to measure a difference in potential across different local points of the copper plate as the charged particle approached? Presumably one would want to model or predict the fluctuations in potential around the area where the particle is going to land. Does this make sense?

[@cosi321]

Could it be "harder" to accelerate, i.e. one has to expend more energy to uniformly accelerate through an non-uniform electric field? Thinking about standing on the earth is like being on a rocket accelerating relative to the charged particle, I would not expect the particle to gain momentum from moving through its field. Although, the radiative energy received could be linked to greater power consumed for the same acceleration in the variable electric field?

[@rexauer9896]

I like the idea of the electron being a single atomic point (2D space; mean all and null orientation at all times) that produces an “antimatter” field or electric field based off quantum spin, and when with a nucleus of protons and neutrons, it creates the electron field (now 3D space, all orientations and positions around the nucleus within the field limit and required energy levels) and it produces the electron field; electron=field which is why we cannot observe an electron around an atom cause the field is the electron(s). When there are many electrons or atoms collected together it creates a blanket field where the N and S poles of the quantum spin self align thus producing a 3D object. Gravity in my eyes is antimatter as it can produce force without physical interaction (mass on mass) and gravitons and positrons and gamma rays etc. may interact with the electron/field. Positron interaction with electrons produce gamma rays (cancer scanning) it goes from 2D to 3D, a sin wave is 2D and will represent positrons, and the rotational vector (the electron) will be what allows for spin(min required points for rotation=2). So based on orientation of the electromagnetic field gravity could decide the quantum spin (so could all other antimatter’s), knowing this we could induce different styles of energy, frequency, antimatter’s, and matters to detect radiation and energy levels of the electron in any reference point. Possibly able to move the field around the electron (push it left or right but can never pass outside of the electron or it would not exist and the electron would have to radiate another field around itself, gravity could be stripping away a layer of field and its perceived and radiation. with the right kind of induced process I’m sure something could be observed… OBSERVATION! Particles act differently when being observed as they have to reproduce the action for every observation (1 for its perspective and X for X many observations.) making the action of a particle 3D and materialistically observable. There’s a cool photon experiment done where they found that using a camera and itd act differently. Just my theory though lol

[@LeopoldoGhielmetti]

I'm not an expert at all but I'm questioning if maybe we should consider the time dilation in a gravitational field. A particle sitting on the surface is accelerated up and it's field is falling down, but the part of the field that is under the particle is on a slower time frame and the part of the field that is over the particle is in a faster time frame. So the down field is receding slower from the particle than the upper field restoring the field shape to a perfect sphere, so all actions combined, you have no energy liberation. I can imagine that for a falling particle it should be something similar, but the other way around. If you consider an observer that is falling in the same reference frame as the particle, for that observer the time dilation/contraction should compensate for the radiation too (but in that case you should consider the differences in the time frame not only for the particle but also for the observer if it is in a gravitational field), so both particles (ground and free falling) don't radiate. For a particle and an observer that are in space (not subject to a gravitational field) there is no time frame dilation/contraction, so there is no need to correct the shape of the electric field. Where the energy comes from the opposite observer (the one that is on the ground looking at the falling particle or the one that is falling looking at the particle on the ground) ? The particles are sitting "still" with a perfect spherical field around, the one that is moving through the field is the observer, so the energy of the photons simply comes from the kinetic energy of the observer itself, not from the particle.

[@juliavixen176]

The electromagnetic four-potential is invariant.

[@narfwhals7843]

@@juliavixen176 It is _Lorentz_ invariant. But the change to an accelerated frame is not a Lorentz boost.

[@ricardojsgw]

Newton's second law says that acceleration is equal to force over mass. So, according to the given formula a charged particle will only radiate energy when a force is applied to it, like in a particle accelerator. A charged particle does not accelerate on its own. It needs a force to accelerate. Now, when you substitute “a” for “g” (earth’s gravity) you’re taking a Newtonian view of gravity, where a force brings the particle down, which is incorrect. In Einstein’s view of gravity there is no force acting on the particle. It is curved spacetime that makes the particle fall to the ground. So, my conclusion is that a falling charged particle should not radiate energy because no force is being applied to it. But the question remains whether it should radiate energy standing on the ground, where a gravity field is present, or on the floor of a firing rocket ship in space, because in both cases a real force is involved. I think it should. But I guess you could not harness its energy because you will be also standing on the ground or on the floor of the rocket ship. Like potential energy from water which you cannot harness if you’re at the top of a dam.

[@dustinsoodak8954]

That seems to imply that if you are initially floating next to a charged particle in interstellar space, then start (extremely) rapidly accelerating past it, you are going to detect em radiation and it will experience a force in your direction of acceleration, even if your ship doesn't have a charge or magnetic field or anything else that would normally cause it to interact with the charged particle at a distance.

[@moisessalazar4432]

These are my insights: The golden rule is: accelerated charges will emit (even in the centripetal acceleration case) hence the charge should emit. Appling the conservation of energy may not apply (photons loose energy after redshift, energy conservation is not universal!) Electric and magnetic fields are affected by relative velocity and curved spacetime. I think that these will affect the energy and frequency of radiation but not if the particle would or not radiate.

[@drdca8263]

3:37 : must whether or not a particle radiates, be independent of frame of reference? Whether a detector detects the radiation has to be, I suppose, but isn’t Hawking radiation dependent on frame or reference?

[@JH-le4sd]

Another problem is there is no "detector" term in any of Maxwell's equations.

[@GaussianEntity]

​@@JH-le4sd That's because it's hidden from plain sight. Recall that deriving all of the equations required using a point charge to determine the magnitude and direction of the magnetic and electric field strength.

[@JH-le4sd]

@@GaussianEntity But a hypothetical magnitude (field point) and a physical detector aren't the same thing--e.g. in the examples, Maxwells equations should be valid for *all observers'* frames of reference, but in reality, the radiated photon will only be measured (or not) by the physical detector in its frame, but this still produces a physically observable fact which all will see (the count on the detector). Maxwell's equations have no built-in solution to this--no detector term, and that ambiguity of how you settle it is really what drives the overall problem.

[@chrischiesa609]

I have some difficulty accepting the derivations of the theory of relativity in the first place. I first learned of it as a child in a time-life book which gave the classic thought experiment of a train traveling near the speed of light through a station where a stationary observer watched a clock on the train ticking. A simple geometrical argument indicates that the clocks pendulum has to move further on a moving train than on a stationary one, and while this is no problem for a physical pendulum since material objects can move at differing speeds, upon replacing a physical pendulum with a photon ping ponging back and forth between a pair of mirrors, a paradox seems to arise because the speed of light must appear to be constant to all observers. My problem however is this: this argument assumes that the stationary observer can observe a photon bouncing back and forth between mirrors, in the same way that he could observe a clock pendulum swinging. The argument further assumes that such observations are instantaneous. I believe or rather it seems to me, that one must first take into account the fact that observations are themselves mediated by photons; and that an observer could not watch a photon bounce back and forth between two mirrors unless at some point that photon or scattered by an intervening medium and reflected toward the viewer's eye, in which case it would never reach the mirror and could not be observed to bounce from it. One gets a little better traction if one presumes a packet of photons, traveling together, of which only some are scattered to the viewer, leaving the rest to presumably reflect off the mirror, but at this point I start to lose track. Then, two, one must take into account the Doppler shift to be expected of observations carried out via wave phenomena such as light waves, when viewing phenomena occurring on a moving platform such as the train in question: surely those observations would appear to be slowed down if viewing the train pulling away, and to be sped up if doing the train approaching? Instead of watching a photon bounce between mirrors, let us say that the train is carrying a radio transmitter, broadcasting a 1 kilohertz audio beacon at a frequency of 5 KHz. Basically, every 5th peak of the carrier wave will coincide with a peak of the modulation. So as the train moves toward you, Doppler ship will drive the perceived frequency of the carrier up, and as the train moves away from you Doppler shift will move the frequency of the carrier down, and since the carrier is carrying the wave peaks that make up the modulation, the modulation itself will appear to speed up or slow down, and so if we go entirely by appearances, it looks like time on the train has either sped up or slowed down respectively. Doesn't that account for the perception of time dilation, sufficiently that we don't need to invent relativity? .

[@astro_male]

I also think a lot about this problem, although it lies outside my area of ​​research. I think I'll spend a couple more years understanding and expanding my knowledge :) I suggest keeping in mind one more non-obvious thing related to relativism. The distances between energy levels of atoms vary depending on the height. This does not answer the questions, but it is also worth keeping in mind when analyzing.

[@The_Green_Man_OAP]

The Earth's atmosphere protects us from the harmful effects of radiation from the Sun and other stars by absorbing and scattering charged particles. However, some of these particles do make it through and reach the surface. The amount of radiation that these particles cause depends on their energy and the density of the atmosphere. The energy of the charged particles from the Sun and other stars ranges from a few MeV (million electron volts) to several GeV (gigaelectron volts). The higher the energy of the particle, the more damage it can cause to living cells and materials. However, the majority of the particles that reach the Earth's surface have energies below 10 MeV, and most of these are absorbed or scattered by the atmosphere before they reach the surface. To estimate the amount of radiation caused by charged particles from the Sun and other stars, we need to consider the flux of these particles and their energy distribution. The flux of charged particles from the Sun and other stars is typically measured in units of particles per square meter per second (cm^-2 s^-1). The energy distribution of these particles is typically described by a power-law spectrum, with a few particles at high energies and many more at lower energies. Using the observed flux and energy distribution of charged particles from the Sun and other stars, we can calculate the expected dose rate at the Earth's surface. The dose rate is a measure of the amount of radiation that reaches a given location per unit time. The dose rate at the Earth's surface due to charged particles from the Sun and other stars is typically in the range of 10^-4 to 10^-3 sieverts (Sv) per year. For comparison, the average dose rate at sea level on Earth is about 2.4 Sv per year, due to natural sources such as cosmic rays and radon. The dose rate can be higher in areas with high levels of radon or other radioactive materials. In conclusion, the radiation caused by charged particles from the Sun and other stars is a small but measurable amount, and it is largely absorbed or scattered by the Earth's atmosphere before it reaches the surface. However, the dose rate due to these particles can be higher in areas with low atmospheric density or in periods of high solar activity.

[@rodschmidt8952]

So it's a second-order effect, because lines of gravity are not quite parallel? Similarly, an accelerating elevator in outer space can be distinguished from an elevator car sitting on the ground, because the lines of gravity are not quite parallel

[@truelyfine]

would symmetry considerations imply that for a down accelerating particle the radiation would be going down, carrying momentum down, slowing the charge from accelerating as fast as gravity alone would specify? Sort of touched on by the magnetic field diagram with the field below.

[@anonimoculto]

Gravity affects light, hence also electric and magnetic fields. So, there is electromagnetic emission only when a charge is not following its "default" geodesic path in spacetime - when it is being forced to change its geodesic. In both cases, the charge static with respect to the ground (the charge belongs to earth) and the charge in free fall, the charges (and fields) are following the disponible (by laws of physics) default geodesic path. Same when a charged object is in orbit. No radiation, in neither the cited cases at all.

[@LuisAldamiz]

6:28 - the field, which is made up of photons, doesn't generally fall through the ground, unless it is transparent to some frequencies (mostly it's opaque, if not totally, but guess you can try experimenting on a glass bridge and see if there's any difference).

[@JohnVKaravitis]

Shouldn't you be including magnetic fields? If I'm standing still and an electron flies by, I measure a magnetic field. If I am riding with the electron, I measure no magnetic field. At the extreme, if I am riding a photon, I will always measure both an electric and magnetic field.

[@mykrahmaan3408]

Far more important is the fact that neither the concept of CHARGE nor of FORCE nor of ENERGY is particle physically defined. Mass of a particle could be imagined as contained in the volume (size) of the particle. There is absolutely NOTHING else physical in any particle to account for the other 3 above mentioned properties, not to mention still others, like SPEED, VELOCITY, MOMENTUM, HEAT,.......

[@robinwang6399]

There are multiple things I need to correct: when ever the charge deviates from its original trajectory, there will be a change in the vector potential, which induces a change in the field. This change in the field when decomposed into electric and magnetic field, and have multi pole decomposition done onto it, we see that part of the change remains even at long distances. This is what Lamour radiation is, not the field lagging behind the particle, but the change in the particle itself propagating through the field. Now if we look at this in a semi classical manner, and interpret the particle as a spread out wave of field excitations, we see the geodesic deviation of a gravitational source does cause tidal effects that compresses and elongate the wave in various directions, partially accelerating the charge. This effect should still happen on the ground if we purely look in this perspective. This is obviously wrong, hence the need for a more quantum theory. Since the Lamour formula comes from any change in 4 velocity. This generalizes to any deviation away from a geodesic, even uniform acceleration should cause radiation in an external frame. Instantaneous comoving frames sees nothing. This should be consistent with universal expansion, and what you found.

[@I3ushmasteR]

I've a side quest-ion: looking at the Lorentz force, a charged particle in a magnetic field drifts thus gaining an acceleration (linear momentum becomes angular). It is correct that velocity isn't changing and the particle doesn't radiate even if it's accelerating?

[@I3ushmasteR]

well.. anyway, the free falling particle does not radiates because it is not accelerating. The same happens if the particle stays still and the observer is in free fall. To me the question can be: "Can I see a radiation from a particle (even if it's not radiating) just because I'm accelerating towards its?"

[@atheistaetherist2747]

The answer, my friend, is blowin' in the aether wind The answer is blowin' in the wind

[@davidusa47]

Really enjoyed this video and greatly appreciated the math done at the end for some perspective

[@Pencil0fDoom]

I was under the impression that gravity only acted upon *objects* (things composed of hadrons with the property of mass arising from the Higgs field). Now you’re asserting that the electromagnetic field is subject to gravity, which upon reflection is obviously the case with light (gravitational lensing) but this has to do with the warping of spacetime due to the presence of objects with mass. So is gravity causal or emergent? How far upstream is it? To pose the question a different way, would “gravity” or… no… do the fields themselves interact in the absence of matter?