Paradox of a Charged Particle in Gravitational Field
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- Опубликовано: 28 сен 2024
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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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Equivalance principle is valid only in general relativity. Special relativity is incompatible with equivalence principle.
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.
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.
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.
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.
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).
exactly ....
Or capitalizing on a misrepresentation on purpose. He knows he doesn't have access to competent people
@@fixgirod Exactly! Clearly, he thinks his audience is as dumb as _he_ is.
@@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
@@chrisvinciguerra4128 how is that LESS nefarious? lmao
My brain is now a paradox, I both feel smarter and dumber listening to this xD
You're already starting to see it work! 😎
The more you know, the more you realize you you don't know.
@@livingcorpse5664stupidoo
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.
But they do get accelerated via Newton force for gravity, right?
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.
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.
The equivalence principle has been proven over and over since GR requires it and we know GR is correct.
Nice try ChatGPT you failed again.
@@Saki630 not chat GPT but I'll take it as a compliment 😊
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.
@@szilardecsenyi516but an observer sitting on earth's surface is not in an inertial frame. It is in an accelerated frame.
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.
Veratasium is a click teaser.
Block his channel.
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.
@@karkaroff1617 He has a Ph.D. in physics education: _"Designing Effective Multimedia for Physics Education"._ So a Ph.D. in RUclips physics presentation.
Sure, and?
Are you aware that he doesn't do his videos alone, right? There are physicists and other professionals as consultants.
@@karkaroff1617 let him have it karkaroff , its the only thing he has about him .
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.
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.
Do you mean like the Millikin Oil Drop Experiment?
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.
where is you username?
@@weeb3277 Honestly, I don't know
when it .... what?
@@weeb3277 the username is there, but it's behind event horizon of your reference frame 😜
Quite interesting. So can we calculate the reverse force? Wouldn't the same apply for acceleration caused by any external force other than electrostatic?
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.
It’s enough to just stop saying that there are two forces and two fields. There is only one electromagnetic field.
@@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).
@@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.
@@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).
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.
What do you mean by "two opposing acceleration vectors"; Are you referring to different reference frames or equal and opposite gravitational forces, or BOTH❔ 🤔
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).
@@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.
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.
@@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.
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.
Q: If you transform the electric and magnetic field to curved spacetime it would solve the problem?
@@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.
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.
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.
@@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.
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?)
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.
10:10 Brazil mentioned🍺
Unruh was mentioned too!!! Good grief.
?
Brasileiro é foda
"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).
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
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.
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.
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.
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.
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?
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
Thank you for kind words :)
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.
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.
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.}
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?
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.
This is a good perspective.
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.
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...
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.
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?
Im confused, why we are assuming that radiation doesnt depends on the observer?
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.
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")
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.
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).
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.
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?
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?
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
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?
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? .
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?
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?
I don't buy the argument about the field about the charged particle moving, but that the way to interpret the Larmor formula is to replace the acceleration by the proper acceleration - that which is relative to an inertial (free-fall) frame. One upshot is that one could make accelerometers out of suspended charged particles.
feels like a question for Sean Carrol on his next AMA
Use a wild card - priority answer, or whatever its actual name is.
He already answered that question. He said something like "what do you mean by radiation? A changing electromagnetic field? Then yes, if you see a charge accelerate, you will see the field change accordingly. But "radiation", as in electromagnetic waves, comes from an oscillation, not uniform acceleration. And that, you won't see."
But I don't know how deeply he actually thought about that answer.
@@narfwhals7843 Xrays are produced by the decellaration of elctrons in an electron beam striking a target. No oscillatory motion required.
3:59 Maybe you’re going to hit upon this, but at relativistic speeds radiation may appear or disappear. Like how the Hawking radiation of a black hole disappears if you’re in free-fall around it (but just in your reference frame, because everyone else will still see it), or how sometimes relativistic observers can’t even agree on what particles exist.
The latter is very easily resolved: there is no such thing as a particle in quantum mechanics. There are only quanta of energy, but they are NOT atomistic constituents of the field. To be more precise, in quantum field theory the only states one can agree on are asymptotic states at infinity. This gets slightly more complicated in GR, but it's not fundamentally different. There is simple one additional reason why the only "real" states are asymptotic: they only exist in flat spacetime.
It seems like when you start to mix gravity and electric fields you get to the border of our understanding of how these things actually work. Since electromagnetism is a quantum force the only way to truly reconcile the behavior of a charged particle in a gravitational field without some kind of paradoxical weirdness is to discover quantum gravity.
Im pretty sure we have a good understanding of electromagnetism in general relativity in general (it’s just a pain in the butt to calculate, 4D manifolds, ugh) it’s just the charged black holes/ naked singularities etc extremes that are broken
I have an issue with your holding up an elementary particle as if it were a golf ball. Surely we're beyond that model by now, especially as regards questions like this one?
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.
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.
Have you looked at the Abraham-Lorentz force calculations? You can derive the force an accelerating charge "feels" working against its own field and the Abraham-Lorentz derivation will give you that resistance to motion the charge experiences. It follows fairly simply from the Larmor formula. As it turns out, the force the charge experiences is not simple acceleration but the change in acceleration or the time derivative of the acceleration. In other words, work being done is proportional to the time derivative of the acceleration. Therefore, I have come to the conclusion that a uniformly accelerating charge will not rediate since the is no work being done but rather it requires a change to the acceleration in order to radiate.
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.
I suggest an easier way to handle this problem. Accelerating charged mass absorbs radiation, whereas a decelerating one emits radiation. In an antenna, the electrons absorb energy(also radiation) from the high voltage source and accelerate, then shed this radiation to space (when impedance is matched) when they decelerate on the way back. Cherenkov- radiation happen because of deceleration and synchrotron radiation also happen as electrons in a circle are de-celerating not accelerating- forced to change direction and rotate by the magnetic field.
In cathode-ray tube there is no radiation on acceleration, but a lot of it on deceleration on hit-ting the screen and stop. regards
Equivalence principle generally holds, but I agree that questions like these make a person wonder. 😊
This thing EM wave never gets old. 300 years of discussion and still no end in sight.
Can you do a video on quantized inertia?
Quite an interesting paradox. It becomes even more puzzling when you consider a "neutral" atom like eg a H atom. If you consider it to be a neutral particle, the radiation should be zero. If you consider it be a proton and an electron then the radiation of both add up.
This is another example a famous consideration. " Why does the electron not fall into the nucleus of an atom like planets fall into the sun if they loose energy to maintain the integrity of their orbit around the sun.
And you know the answer to that was that John's release energy in the form of discrete packets. And jump between positions of the electron orbitals nearby expressing their acquiring of or loss of, energy. So logically I would have to take the stand and electron falling in the gravitational field would neither accelerate towards the gravitational source nor decelerate by losing energy. It would gain energy and release it to once again find its low energy state of. The United States of the electron will fluctuate while in the state of acceleration. If the electron is accelerating at the appropriate speed to release all the energy required all at once it may release the energy in the form of a larger particle, whichever outcome it would be the cause of the principal of the electron seeking the lowest energy state.
Have you guys noticed that most of our problems in physics is always with gravity?
Look: quantum gravity, three bodies problem and this one.
The most ironic is that we see gravity in everything in universe and it was the first "force" that we thought that we understood. How ironically, isn't it?
deppends of frame of reference, if you be at same acceleration the net radiation is zero, in case of free falling charge the radiation gives from electric fields induced magnetic fields on anything that be "accelerating up" and in the case of charge at ground and observer in free falling we have a direct acceletarion and we have the radiation too , the GR picture is correct, but you need a theory of quantum gravity for full understand this
Perhaps look at the problem in GR, for gravitational waves. After all EM gets tricky cause there is no microscopic model of the electron, etc. GR is quadrupole, so that makes some things different, but there would be some lessons on the equivalence principle, etc. The production of GW can be modelled in all sorts of conditions.
The answer, my friend, is blowin' in the aether wind
The answer is blowin' in the wind
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.
The electromagnetic four-potential is invariant.
@@juliavixen176 It is _Lorentz_ invariant. But the change to an accelerated frame is not a Lorentz boost.
Wouldn't it just interact with the radiation it has previously emitted? Also technically the charge wouldn't be stationary on a rotating body. It's just relatively stationary. Technically the ground state of the electron still would still have radiation at rest because of the uncertainty of it's wave function. Whether or not that radiation is released however is what determines if an emission event occurs, but just thinking it would still technically have to be there due to energy conservation. The loss of energy due to photon emission would be internally regenerative due to self absorption interactions with the emission fields since now you would have previously emitted photons interacting with an electron.
In relativity, the frame of the observer matters.
If the observer is falling and accelerating with the test charge, BOTH are radiating. Both are made of charges. Both become altered in such a way that there is no net measurement.
There must be some other way to make an experiment except piling the nonrealistic number of elementary charges needed to radiate 1 W. How about letting a DC excited superconducting coil fall while being kept cool inside a huge vacuum chamber or in a Faraday cage tube (to be able to account for eddy induced braking)? Not my filed, plus I moved from a PhD in Magnetohydrodynamics for material science to electrotechnics (where everyone used a slightly different semiempirical equation, to Maxwell's dispear 😅), but I remember endless discussions from the colleagues in the Theoretical Department about Unhruh radiation and particles in de Sitter metrics (curved spacetime). Thus at one point the discussion in this video starts sounding like Aristotelian attempts of science from philosophical principles.
Moving charges produce magnetic fields and EMR. Acceleration is not needed for it to be radiating and losing energy it just meeds to be above absolute 0 and not receiving and radiation from some other source.
The following publication could be of some interest: G. D'Abramo "On the ‘Rigidity’ of the Force Field of a Moving Source." The Physics Educator Vol. 02, No. 03, 2050011 (2020).
As electric field and magnetic are irrespective of frame of reference and changing electric field causes origin of magnetic field and vise versa so whether in space (gravity free) or free fall a charge would radiate EMW if its is accelerated...
Sometimes i wonder whether gravity is a result of electromagnetic theory.
I do have a few comments:
- The tidal effects shouldn't intervene in the problem. You can imagine a similar problem without these, theoretically with a constant gravitational field, or practically with a supermassive black hole the size of the solar system, with an electron falling a few mm. You would have the same questions, and most likely very similar answers.
- The mathematical tools we use should not be treated so much as "reality". We know that E or B depends on the frame of reference, so their "reality" looks dubious. At the end of the day, all the mathematical symbols we use are interpretations. The equivalence principle on the other hand is a law which doesn't depend too much on the labels and symbols used for electromagnetism. To me, that makes it more fundamental.
- Radiation, as pointed out later in the video, is not a fully defined concept.
- Even in the current framework, the reality of force-bearing "particles" is unclear. Photons in QED are, when "force" is looked at, virtual particles.
- QED is a quantum theory. Relativity is not. Thus your emissions are quantized, but your acceleration is not, in the current theories.
Is it possible to settle this by measuring the energy of cosmic rays travelling via different trajectories through gravitational lensing?
Loved when you called SIngal, Signal xD
The problem is that people don’t really understand what a ‘particle’ is and they don’t understand what gravity is. Gravity is a voltage potential in the Aether. A black hole is nothing more than a hole in space where the Aether is exposed and it has near infinite voltage potential. That’s why gravity bends light. So what is a voltage potential? It’s a longitudinal or scalar standing wave in the Aether. Where as light is a transverse oscillation of the Aether which produces both dialectricity and magnetism. Only transverse waves produce dialectricity and magnetism.
I feel like whenever there's a scalar acceleration term, that will always refer to the proper acceleration, the acceleration invariant, the one that you feel, that you can measure with an accelerometer. That is zero in freefall, so I would've automatically said "no, it doesn't radiate"
Fascinating - I would be interested to know if Albert Einstein had anything to say regarding this paradox.
I have struggled with this question for well over forty years.
From a far observer point of view on a Schwarzschild solution where charge is static on Earth; the electric field is static. That is why it cannot radiate although it is accelerated because it is not following a geodesic curve.
The question seems somehow equal to: if object infinitely rotating about another massive, and it is really constantly accelerating toward center of mass - why then it doesn't gain infinite speed?
15:28 so are you suggesting that the EM field is direction in the case of a falling charge? The overall flux is 0, but pick a directionally there are different flux?
em field is caused by moving charges, motion is relative to the observer therefore one observer could measure this field to be electric but the other one could measure the very same foeld to be magnetic
Equivalency principle isn't exactly what it says. If you're in a gravity well the force of acceleration at the bottom of the box will be greater than at the top. If you're just accelerating it will be the same. Therefore you COULD do a test and realize if you're accelerating or in a gravity well.
That is why we often say the word "locally" meaning the box is small enough so that tidal forces are negligible
If every point continuously along the height of the elevator was each experiencing proper acceleration, rather than being pushed from the bottom and the rigid material compressing, or pulled from the top, and the rigid material of the elevator streached... _then_ it would be indistinguishable from gravity. (Gravity where the gradient is uniformly parallel (towards a line or extremely distant point, not radial towards a single nearby point.)
Although, there is also a way to measure the mass of the elevator vs. the mass of the Earth from inside the elevator, by jumping up and measuring how much the elevator's proper acceleration decreased. The ratio of your mass to the masses of the elevator and the Earth will give you the answer.
But... this principle is about coordinate systems and not physical elevators.
If there is a magnetobremsstrahlung radiation why wouldn't there be a gravitotobremsstrahlung one as well?
Nice thought experiment. Sounds like a equivalent principle of a different kind.
Regarding the co-moving observer in free-fall ... isn't that similar to the falling particle ? The observer falls slightly differently than the particle, since they both fall to the same gravitational centre, but at slightlx different angles ...
5:20 - hence the resolution that "gravity is not a force" - gravity is space itself moving - see veritasium, pbs space-time, or others who talk about this
So is the photon observed real? If gravity is just space, this observed photon should be an illusion. Yet I'm positive that the radiation measured via experiments show that it is real.
2:36 this is the point where i completely lost it lol
Why does a particle emiting energy shoud deacelerate? Is it bc some energy is being radiated in the opposite direction of the aceleration?
I have a strong feeling that im wrong lol
ρ = m/V, ρ₀ = m₀/V₀ ,sg = ρ/ρ₀
Den-sity is defined as mass per unit volume. It has the SI unit kg m-3 or kg/m3 and is an absolute quantity. Specific gravity is the ratio of a material's density with that of water at 4 °C (where it is most dense and is taken to have the value 999.974 kg m-3). It is therefore a relative quantity with no units.
Objects that are denser have a greater concentration of mass, thus creating a larger gravitational pull than a same-sized object with lower density-i.e., a cubic foot of solid rock will exert a larger pull on its surroundings than a cubic foot of ocean.
No it will not radiate anything, there is no reference point in your question to have an influence over the particle, a photon travels as a wave until it hits your retina. The only loss/gain of a particle would be wavelength being pulled by gravity or ejected from a neutron star. Otherwise a perfect vacuum and the lack of an observer would make a particle retain its energy until it will be force by its wavelength to transition into a different state eg. From xray to UV.
In theory if the particle accelerates towards Earth with g it cannot radiate energy exactly based on g because that would make it loose energy and therefore add a small deceleration to g, so the actual value of the acceleration should be a bit smaller than g so that this kind of EM friction is at equilibrium with the force of gravity
16:14 Wouldn't it be just sqrt( 5*10^52)= 2.24 *10^26 charges, because the charge in the formular is squared?
Here is my interpretation: Any body free falling in a gravitational field is not changing in energy. Then why would it radiate? The charge radiation equation should be seen as an approximation. The acceleration should be understood in terms of energy change, not absolute acceleration. In most practical situations it works well because the acceleration is accompanied by a change in energy, but it would not be the case for a free falling particle, or a particle lying on the earth's surface. A particle that falls in free fall does not radiate and a particle lying on the earth's surface doesn't radiate, but it does radiate when it hits the earth's surface as it changes from free fall to stationary on the earth's surface. This is not unique to charged particles. A neutral space rock also emits radiation when it hits the surface of the moon at high speed.
How do you produce a freely falling charge?
What does the experiment says? None of these seems to require tought to do experiment. After all, we got MEMS sensors and superconductors now.
Our measurement tools are just not sensitive enough for it to happen.
@@pinocleen Don't we hv MEMS sensor capable of sensing nanowatt?
@@aniksamiurrahman6365 Nanowatt isn't even close. Nanowatt is 10^-9. This is 10^-52. Literally 10000000000000000000000000000000000000000000x less than a nanowatt.
But if your charged + particle is resting on the Earth's surface it will become neutral anyway by charge transfer. So there is no problem or paradox. In free fall for an accelarating charged mass the effect is the same as loosing mass during the fall and will not affect the gravitational accelaration of the Earth g=9.81 m/s^2 since it is independent of the mass value. of the falling object.
Fantastic analysis!
I think this just shows that the Larmor formula, although very accurate, is just an approximation of a more fundamental mechanism in action...
Already using today's theories we can rule out its accuracy in quantum mechanical contexts.
First, this formula doesn't take into account quantum mechanical effects at all...
In addition to that, QFT and QED give us a deeper and more comprehensive explanations on what is really going on when a charged particle emmits electric and magnetic radiation - which this formula just aproximats
This alone shows that this formula is no more than an approximation, and not really is the deepest truth
Now, there's an even better way to describe magnetism using the framework of relativity, by considering the fact that: A charge moving relative to another charge would measure length contraction for the other charge.
In simplistic terms this just makes the moving charge measure a "bigger charge density" for the other charge relative to which it moves - which makes the electric field it experiences from that other charge even bigger in a direction perpendicular to its own movement the faster it moves.
This property alone creates the effect we know as magnatism and is described purely by special relativity.
So again, the Larmor formula is an aproximation, and we know it is just by considering ideas from modern physics.. (which we know are more fundamental and accurate)
The problem showcased in the video just shows where this formula fails...
This even makes more sense when considering where it fails as seen in the video. It fails when considering it and questioning its behavior under relativistic effects, and again, this makes alot of sence becuase this formula was not constructed to be accurate with relativity (nor QM and QFT)!
So let's not lose our minds again thinking that Einstain was wrong, everytime we thought he was, we were proved wrong ourselves.
Where does the energy come from is always my question. If it can be shown that an energy gradient exists in relation to the particle, then it's safe to say the particle can radiate the difference.
Maybe related: If I'm moving towards a red light and it appears blue due to doppler shift, where is the extra energy coming from since blue photons have higher energy?
Good question. I’d argue that you are flying towards an oncoming wave of EM energy then the Doppler effect increases the number of bumps of peak electric field you encounter per second. Hence that defines an increase in frequency and thereby the increase in energy as defined by Planck. In other words your relative movement is a part of the system and thereby by energy. Maybe it’s a bit like listening to someone play a recorder with variable pitch control - if they hold the pitch and you speed towards then you will hear the pitch go up even though they are holding the same note. If you stay still and they start varying the pitch of the note you are playing then you would assume they are adjusting the pitch. In light the analogy would be for a tuneable laser. I’m using a crude analogy which is always risky of course!
The equivenence principle is already broken ……
If I have photon in flat space it will remain the same
If I have photon in a gravity well going inward it gains energy going out it loses energy
So blue and red shifts
9:29: it's not objective, the magnetic field is generated by relativistic movement, unlike the electric field, which is indeed objective. So the guy falling entices a magnetic field on the charged particle and this one begins objectively radiating to all, there's a guy falling from the Tower of Pisa and he's having an effect on the observer: the particle, which then goes around trying to tell everyone about the guy falling with its tiny electric field... but fails because it's such a tiny-tiny particle with such a tiny-tiny radioactive voice...
Try it with a much larger charge maybe, something in the line of those ray emitting balls of the old days of electric experimentation maybe?
Would we measure radiation, locally, If the radiation's wavelengths is of the order magnitude of Earth's diameter? So the answer: "it radiates" or "it won't radiate" could depend on the magnitude of the region considered.
My opinion : mechanically, radiation is poorly defined. What does it mean for a particle to “radiate.” What is it radiating? If we just say it’s energy then it must be accelerating. Then you have to define mechanistically what energy is…is it a bunch of other particles escaping this particle? What does that even mean? Where do those smaller particles come from? If it’s from empty space then empty space is non trivial…not simply “empty” so there are holes in everything.
The observer being part of the explanation is also relevant, or at least must be treated far more seriously because the observer plays an important role in practically every problem and every explanation for physics. For some this is rather obvious: you are part of the system so to form any kind of bijective equivalence statement (in order to preserve symmetries) involves including that observer (and in extension every other system in a super determinate fashion)
So…if space is non trivial…and there’s a relativistic construct, then we simply do not have a proper explanation for space. In some theories this is nicely wrapped up by just saying space is ether-like where a deterministic chain of causality is defined, and in the limit of the entire graph, one can form a bijections onto itself and preserve structure therefor validating equivalences such as relativity, and point out how non equivalences show up (observer effect) as a result of surjective operations done by a local partition of that graph in relation to the whole graph.
Basically : the problem is geometric…almost completely and the observer plays a crucial role in how this geometry is perceived. Equations in aren’t enough here…one has to define a construct in order to understand this problem.
A charge should only radiate if it, itself, experiences acceleration. If it is in free fall, either in electric or gravitational field, there should be no radiation.