>> The particle can start the scenario with velocity and have never accelerated. >> not in the real world. All particles with velocity are intrinsically always undergoing acceleration. All motion is orbitally referenced. This acceleration maybe 'inertial acceleration' in your thoughts inertial acceleration is always and solely in respect to the field spin (as in an orbit). What most call acceleration is over and above this... it is non inertial acceleration. But it is important to realise everything is accelerating at all times. [nothing is inertial as per Newtonian inertia]
GMontag , An awfaul lot of double talk here. I'll see if I can correct it. Thank you we now know your mentality and what to expect. The station "Does Not" accelerate . There are no forces of acceleration, no Bremsstrahlung, etc. Your arguement and James R, etc are oversimplified nonsense. More nonsense. Acceleration is not a matter of measurement. Acceleration has with it a component of F as in F = ma. You and others like to ignore this simple physics fact and pretend that if you can't feel uniform motion then you are at rest. That is frankly assinine and leads to the total corruption of physics. The error being made is to assume that equivelence in some respects means equivelent in all respects. It is not. There is nothing incorrect in my statment. Your response is assinine. And your response has nothing to do with my comment. If you don't understand that, you don't understand physics or electronics. And you have convieniently left out if an object accelerates in any frame it has with it an associated changing field which would be detectable in it's frame regardless of it's inertial status. The problem is there and your deliberate effort to ignore it or bury it show your lack of imagination and understanding of actual physics. Again with the half assed physics. You left out the crucial aspect of acceleration which is F = ma. Relative velocity of a non-accelerting body caused by one object accelerating is only "Relative" acceleration, not real acceleration. When you and other relativists eventually understand you are equating illusion with reality perhaps physics will once again begin to actually define the mechanics and why and how things occur. And where do you get off disregarding certain real physical facts because they are inconvenient to your concept of reality? Agreed. And it has nothing to do with the discussion. Only if you chose to ignore physics. Yes or no. If a charged particle accelerates it has a changing EM field? Yes or No. When you have a changing EM field is it detectable in the same frame? (i.e. - transformers) Double talk. Then transformers do not work and we must start the industrial revolution all over. You will not see radiation if the "Equivelence Principle" is incomplete or incorrect.
I am not going to enter into the confusion errors here, but the title of thread is a little strange. MacM will just agrue with me and I am tired of that, no longer willing to expend effort to convence him SRT and relativity are the best model of many experimental results. Cosmic bremstrahlung is one of the three types of experimental data that confirms relativity. It is surely the hardest to understand why this is true, but I can at least hint at why. The bending of light (gravitational lensing) and gravitational red shifts are relatively easy to see why they support / confirm relativity. Bremstrahlung, as anyone knowing German might guess, occurs when a charged particle is violently deacelerated in a collision. The intensity of the cosmic Bremstrahlung is thus directly proportional to the density of both "stopping particles" and charged particles, which can be measured. I.e. there are other means of knowing the density of charged particles (most of ordinary matter is in space is ionized.) and line radiation is the path for this independent measurement of the charged particle density. There is too much Bremstrahlung so there is some other non ordinary matter doing some of the "stopping" at this point is gets too complex for me, but this "dark matter" and relativity (I think) related to star velocities and thus get tied together, and only if relativity is correct does the measured Bremstrahlung data agree. (or something like that)
I'm sure your attempt will be a laugh riot. Of course it does, as I already showed you. Acceleration is simply the second derivative of displacement. It is non-zero for the station with respect to the train, therefore it is accelerating. Irrelevant. If the station is charged, the train certainly will see the station radiate. Honestly, Mac, have you ever even taken a Physics course? This is Mechanics 101. Acceleration is nothing more and nothing less than the second derivative of displacement. Just because a force *can* be a cause of acceleration does not mean that every acceleration must have a force causing it. There was quite a bit wrong with your statement. You were implying that you would see radiation in a non-inertial frame from a particle that is at rest in that frame, which is simply wrong. I understand how transformers work perfectly fine. They still are completely irrelevant to the topic at hand. Once again, you simply have the physical facts utterly wrong. You will *never* see radiation due to an accelerating charge in a frame that is accelerating along with that charge. And btw, frames aren't associated with particles, they are associated with frames. The fact you don't seem to get is that electric and magnetic fields are not frame-invariant. Your continued imagination of a problem has no effect on reality. F=ma is not a part of the definition of acceleration at all. Seriously, Mac. Go pick up a vector based physics text and read through the first chapter. *All* acceleration is relative. You have to specify some other point for displacement (and its derivatives) to have meaning. Since there is no way to differentiate this "illusion" from reality, it is silly to talk about it as if there is a difference. And just what physical fact am I disregarding? The one you just made up off your ass? Particles don't start at rest just because you want them to, Mac. If you didn't observe it accelerate, directly or indirectly, you have absolutely no reason to believe it ever accelerated. This question is meaningless for two reasons. First, you didn't specify what the particle is accelerating relative to. The frame? Some other particle? Santa Claus? Second, particles don't have fields. Frames have fields. Particles have charges. To answer what I think what you would have asked if you actually had some idea of what you were talking about: If a charged particle is accelerating in a frame, then yes, that frame's field will be changing. If you are in the rest frame of a charged particle, inertial or not, then you will never see that frame's field change without another charged particle moving in that frame. Another unparseable question. Detectable in the same frame as opposed to what? Certainly you can detect if a frame's field is changing at any given point in that frame by placing a detector at that point at rest in the frame. But a detector will only ever detect the field of its rest frame, not some other frame. Translation: I didn't understand it. Do you even think about what you post before you type it, or do you just enjoy talking out your ass? Here's a hint, Mac. The electrons in the transformer windings aren't at rest relative to the transformer.
I guess you skipped my introductory comments about most jpeoiple believing Bremsstrahlung being only breaking energy. I guess you also skipped clicking on the link I provided which gives a more full description regarding Bremstrahlung. Where relativity is valid it is quite accurate. However, that does not alter nor excuse the fact that it is completely invalid in its overall presentation which includes reciprocity. Once that is understood it becomes clear that emperical data only demonstrates an absolute view and invalidates the relative velocity view. You have some catching up to do.
I guess you skipped my introductory comments about most people believing Bremsstrahlung being only breaking energy. I guess you also skipped clicking on the link I provided which gives a more full description regarding Bremstrahlung. Where relativity is valid it is quite accurate. However, that does not alter nor excuse the fact that it is completely invalid in its overall presentation which includes reciprocity of time dilation and mixes frames to claim length contraction. (i.e. - disregards physical conclusion of one frame and relables the affect in another frame). Once that is understood it becomes clear that emperical data only demonstrates an absolute view and invalidates the relative velocity view. You have some catching up to do.
Hi MacM, I read the reference you posted. It was quite interesting, but I don't see the conflict with (my limited understanding of) relativity. I certainly didn't get the impression that the authors of the reference thought they had something as groundbreaking as a proof against relativity would be. One really interesting thing about the reference you provided was the formula for the cutoff frequency of the Bremstrahlung radiation, this was calculated as the inverse of the "interaction time". In the case of a massive charged particle accelerated by free-fall on earth the interaction time would be quite long and so the cutoff frequency would be very low. I think a careful experiment could probably detect it, but I don't know if any such experiment has been done. In any case I don't think that the Bremstrahlung radiation has much to do with relativity. I think relativity just says that it doesn't matter if the detector is at rest on the groud and you drop the charged particle or if you drop the detector and the particle is on the ground. You should observe the same radiation in each case. You should also observe the same radiation if you are in orbit and accelerating (e.g. with a string) the detector relative to the particle with a=g or vice versa. -Dale
Although this thread doesn't seem to me to be a proof against relativity it does bring up a point that I had not considered and that I find very confusing. If the Bremstrahlung radiation is frame-dependent then it would seem that the existense of particles is also frame-dependent. In other words, lets say that I perform an experiment in my rest frame. I have an accelerating charged particle and a co-accelerating detector. I also have a stationary detector. The stationary detector will detect radiated energy and will therefore say that Bremstrahlung photons were produced while the accelerating detector will not detect any radiated energy and will therefore say that Bremstrahlung photons were not produced. I knew that different frames could disagree on time, distance, wavelength, and frequency, but I thought that all frames agreed on the existence of particles. I think that all inertial frames in SR will agree if a given particle is accelerating and they will agree on the existence of the photons. They will disagree on the magnitude of acceleration, the rate of photon production, the wavelength, etc., but not the existence of the photons. In GR, however, it seems to me that not all inertial frames will agree if a given particle is accelerating since some inertial frames are mutually accelerating due to the spacetime curvature. This then should lead to a disagreement on the existence of the photons in addition to the other disagreements from SR. Could someone with GR experience please explain what I am missing according to GR? Are these somehow "frame photons" similar to "frame forces" like the centripetal force? Or do you need to introduce "frame anti-photons" in non-inertial frames? How do mutually accelerating inertial frames affect the situation? Do the missing photons just get redshifted to zero in some frames and can be therefore considered to exist in all frames? -Thanks Dale
To DaleSpam: Perhaps ths will help: Imagine a long line of uniformly-spaced parallel bar magnets, but the "N up" ones alternate with the "S up" ones. Now cosider two observers, one, A, is at rest WRT the bar magnets and B traveling just above the tops of the magnets at such a high speed that the electric field (due to the dB/dt term in maxwell's equations) has the same energy content as the magnetic field for B. I.e. for B a photon exists and for A only a static magnet field. The classical particle nature/concept that photon can display is creating in your mind (I think) the confict: either the particle does or does not exist. Truth is (I think) the if you could travel with a photon, it cease to exist. All you can find are orthogonal static E and B fields - no photon. Now slow down a little -see the radio wave passing you? etc. Slow down greatly - see the photon zip by? etc. Note it die and eject an electron by Photo electric effect. I don't have time now to clearly think thru your questions but hope these comments may hit some of the problem.
I did not intend to suggest that the authors of this in any way were applying it to relativity. My point had to do with the physics of acceleration affect on a charge. Acceleration of a charged particle causes a changing EM field. A changing EM field is detectable in a rest frame of the particle (the basis for the referance to transformers where the field is changing and couples across the primary to secondary winding which are at rest to each other.) It follows therefore that if you claim the station accelerated away from the train (an SRT view) then Bremsstrahlung must occur and be measuraeable since it would have a changing field. Since it doesn't then it suggests the view is flawed and "Relative Acceleration" and "Relative Velocity" are illusionary and not actual motion. Physics should conclude the only actual acceleration and hence actual velocity is held by the object which under went the F = ma force. If anything is ficticious (referance relativity and centrifugal force) it is relativity's claims of velocity for an object which has had no F = ma associated with relative velocity to some other object.
You are mixing too many metaphors for me to follow here. Is the train the particle or the detector? Similarly, is the transformer supposed to be the detector? Or perhaps the primary winding is the particle and the secondary winding is the detector? It would be helpful (to me) if you could dispense with the metaphors for a moment and just clearly describe the experiment you are proposing and the results you expect. By the way, with all the metaphors I am not 100% sure if this is a point of confusion for you or not, but the train station and the train example is not symmetrical according to SR. In other words, any two inertial frames are symmetrical, but the train's rest frame is not inertial while it is accelerating away from the station. While the train is accelerating a dropped ball will appear to be pulled to the back by a "frame force". The same effect will not occur in the station's rest frame which is inertial. That said, even though the situation is not symmetrical it is perfectly valid to talk about the frame force accelerating the station in the train's rest frame. In fact, it is necessary in order to properly describe the motion of the station in the train's rest frame. The symmetry in GR is that a train in flat space accelerating at g is symmetrical with a person standing on the earth. Both are in a non-inertial frame which is accelerating at g and so dropped balls will appear to be pulled by a frame force. In other words, any two frames with identical frame forces are symmetrical and gravity is considered a frame force. The accelerating train and the inertial station do not have identical frame forces and are therefore not symmetrical according to GR either (at least according to my fairly weak understanding). -Dale
Correct. We agree and this is not a point of confusion for me. It is however, for those that want to claim that the station accelerated away from the train - which is what James R and others have argued saying its changing relative velocity means it is accelerating. My position is it is neither accelerating nor has velocity as a consequence of any other body's motion, only it's own and F = ma is required to have acceleration and consequently actual velocity change. Here we disagree. It is where physics begins to go wrong. It is best to recognize the station status as a consequence of the trains motion. The inability to sense the trains inertial motion does not and has not altered the fact that it was the train that actually accelerated and under went F = ma. I have no dispute with this. But given the mandated reciprocity one MUST expect Bremsstrahlung of any charged particle under acceleration, be it uniform linear or gravitational (if you want to claim equivelence). I do not.
The field will only change in frames where the particle actually accelerated. In the particle's rest frame, the field will not change. And btw, any motion causes a changing field, not just acceleration. Wow, are you this dense naturally, or do you have to work at it? As I already pointed out to you, the electrons in the primary winding are not at rest relative to the secondary winding, or the electrons in it. The electrons in the primary windings are accelerating, moving back and forth, due to the AC current. In the frame where the station is accelerating relative to the frame, the station *is* radiating.
Ok, maybe this will help. Lets consider an inertial frame X and an accelerated frame X* where x* = -1/2 g t² + x. In other words the frames are co-incident at t=0 and X* is accelerating relative to X in the positive direction at an acceleration of g. Now, if we have particle A (aka the station) at rest on the orgin of X then, since X is an inertial frame, there are no forces acting on the particle. We can also express the motion of A in the X* frame. In particular, x<sub>A</sub>*(t) = -1/2 g t² + 0. Because A is accelerating in X*, with a<sub>A</sub>* = -g, and there are no real forces acting on A we must postulate a "frame force" of f<sub>A</sub>* = -m<sub>A</sub> g in order to explain its motion according to Newtons laws in X*: a<sub>A</sub>* = f<sub>A</sub>*/m<sub>A</sub> = -g. Now if we have a second particle, B (aka the train), starting at rest on the origin and we apply a constant force, f<sub>B</sub> = m<sub>B</sub> g, which is sufficient to accelerate it with an acceleration of g then we will have x<sub>B</sub>(t) = 1/2 g t². We can translate x<sub>B</sub> into the accelerating frame by x<sub>B</sub>*(t) = -1/2 g t² + x<sub>B</sub>(t) = 0. So B is at rest on the origin of the non-inertial frame X*. Because B is at rest in X*, so a<sub>B</sub>* = 0, and B is experiencing an applied force of f<sub>B</sub> we must postulate a frame force of f<sub>B</sub>* = -f<sub>B</sub> = -m<sub>B</sub> g in order to explain its motion according to Newton's laws in X*: a<sub>B</sub>* = (f<sub>B</sub> + f<sub>B</sub>*)/m<sub>B</sub> = 0. So in summary, in the reference frame X we see that A is stationary and B is accelerating at a<sub>B</sub> = g. In the reference frame X* we see that B is stationary and A accelerates at a<sub>A</sub> = -g. Since the choice of coordinate system is completely arbitrary the description of the motion is equally valid in each case. So it is just as valid to talk about A accelerating in X* as it is to talk about B accelerating in X. Despite this, the situation is not symmetric because the frame forces only exist in X*. In X only a single force is needed to describe the motions, while in X* three forces are needed. Also, note that the presence of the frame forces dramatically changes the results of even simple physics experiments like throwing a ball, again highlighting the lack of symmetry. I don't know the original context of the train-station example, but I suspect that the knowledegable posters on this forum did not mean that the situation was symmetric simply because the acceleration can be validly described in either frame. -Dale
Bremsstrahlung is associated with acceleration of charged particles. Accelerated charged particles produce radiation. Take an electron travelling around a synchroton ring, for example. In the ground frame, the electron is accelerated, and so radiates. MacM seems to be asking - what happens if we go to the rest frame of the electron? Does it radiate or not? Because in its own rest frame, the electron isn't accelerating. The answer is: in the electron's frame, it still radiates. Obviously, this must be true, since the radiated photons are detected in every frame. They can't exist in one frame and not exist in another. Next question: WHY does the apparently unaccelerated electron radiate, in its own rest frame?
You two disagree. This is related to my post above (Today 06:46 PM EST). I don't know the answer and would appreciate some insight. -Thanks Dale
I'll pass on the opportunity to return the negative innuendo in that James R seems to be hedging his bet slightly and I prefer to respond to him since he is sticking to physics. You pointed out to me that electrons are moving in the primary? Nice. Since you are fairly new here let me educate you just a tad. My background education and experience is mechanical, electrical and nuclear engineering, specialized in Process Control Intrumentation Systems. i.e. - Electronics, Pneumatics, Hydraulics, etc. My point is the moving electrons provide a changing EM field, the same as an accelerating charged particle. The primary winding simulates an accelerating charged particle. The secondary winding which happens to in the same frame of referance, hence at relative rest with the charged particle (winding) is a detector. Get it now?. Or should I schedule more classes for you. Some people are naturally slower than others.
I appreciate all the effort you just put forth, however, it was not necessary. My only point is I dispute the view where acceleration is defined in terms of a changing velocity in absence of the physical requirement for their to be the F = ma relationship. I dispute the view which attribute an object to having velocity affects due to relative velocity caused by another object having accelerated.
Well I'll be damned. Please Register or Log in to view the hidden image! You were the last one I expected to address the actual physics issue. However, my question isn't really why it radiates in it's own rest frame. A particle in accelerated motion radiates. While in inertial motion it also radiates but that radiation is static and undectectable in it's frame. But while accelerating the EM is changing and during acceleration the EM becomes detectable in it's own frame. My concern is the absence of detectable radiation of static charges under the acceleration of a gravity field "IF we are to accept the Equivelence Principle". Or in your prior example, why one does not measure radiation from static charges in the train station, if indeed it physically accelerates away from the train as you have previously claimed. My contention is simple. There is physics differances between acceleration via F = ma and apparent acceleration due to changing relative velocity by another object actually undergoing physical acceleration. That physics makes a fatal error by not recognizing the differance. They are not equivelent and should not be treated equivelent as they are in relativity.
Yeah, that's right. Here's where you went wrong. Physics does recognize the difference. In particular, special relativity recognizes the difference. Any good undergrad physics book discusses exactly this issue in the context of the twin paradox. From the Feynman Lectures on Physics, Vol. I, Section 16-2, "The Twin Paradox": "[...]consider a famous so-called "paradox" of Peter and Paul, who are supposed to be twins, born at the same time. When they are old enough to drive a space ship, Paul flies away at very high speed. Because Peter, who is left on the ground, sees Paul going so fast, all of Paul's clocks appear to go slower, his heart beats go slower, his thoughts go slower, everything goes slower, from Peter's point of view. Of course, Paul notices nothing unusual, but if he travels around and about for a while and then comes back, he will be younger than Peter, the man on the ground! That is actually right; it is one of the consequences of the theory of relativity which has been clearly demonstrated. Just as the mu-mesons last longer when they are moving, so also will Paul last longer when he is moving. This is called a "paradox" only by the people who believe that the principle of relativity means that all motion is relative; they say, "Heh, heh, heh, from the point of view of Paul, can't we say that Peter was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that both should be the same age when they meet." But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship - the rockets went off, things jammed up against one wall, and so on - while Peter felt nothing. So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an "absolute" sense, and it is certainly correct. "
