- #316
Mike_Fontenot
- 249
- 17
GrayGhost said:[...]
Velocity is velocity. It's silly to re-define it.
Mike Fontenot
GrayGhost said:[...]
Mike_Fontenot said:Velocity is velocity. It's silly to re-define it.
What is "instantaneous slope" if not "change in position over duration". You are contradicting yourself. This is why working out the math is so important.GrayGhost said:these problems are resolved by using the instantaneous slope of the A-worldline for v in the LTs (where v is always < c), as opposed to B determining the A-velocity from change in position over duration (which can be superluminal).
When the traveler decelerates and becomes at rest in the frame in which the home-twin has always been at rest, the home-twin has no awareness of this event until long after it has occurred. You cheat when you give him knowledge from the frame of reference that we are aware of.Mike_Fontenot said:Here's something for you to think about, while you are mulling all this over:
Take the standard twin "paradox" scenario, with gamma = 2. Suppose that immediately before the turnaround, their separation according to the home-twin, is L lightyears. The traveler says their separation is L/2 lightyears then.
Half way through the turnaround (when the home-twin says their relative velocity is zero), the home twin says their separation is still L, and the traveler NOW also says their separation is L lightyears. So the traveler says that their separation has changed by L/2 lightyears, during an infinitesimal amount of his ageing, so he says that their relative velocity during that first half of the turnaround has been infinitely large.
Denote the age of the traveler at the beginning of the turnaround as t1, and the age of the traveler at the midpoint of the turnaround as t1+delta, where delta is infinitesimally small, but non-zero). Denote the MSIRF at the beginning of the turnaround as MSIRF(t1), and the MSIRF at the midpoint of the turnaround as MSIRF(t1+delta) ... they are DIFFERENT inertial frames.
Ask yourself this: what do MSIRF(t1) and MSIRF(t1+delta) say about THEIR own separation (with respect to the home-twin) during the first half of the turnaround? Do either of them agree with the traveler, that the separation changes by L/2 during the infinitesimal time delta, and thus that the velocity during the time delta is infinitely large?
Mike Fontenot
DaleSpam said:What is "instantaneous slope" if not "change in position over duration". You are contradicting yourself.
GrayGhost said:[...]
The link doesn't work.GrayGhost said:For reference ...
https://www.physicsforums.com/attachm...2&d=1299884366"
DaleSpam,
Look at the illustration in the link above.
It is not an assumption, it is a definition.GrayGhost said:If we are to assume that velocity is nothing more than the change of position over time
DaleSpam said:The link doesn't work.
Mike_Fontenot said:In the twin "paradox" example I gave previously (with gamma = 2), IF the traveler concludes that his distance to the home twin increases by L/2 lightyears during his infinitesimal ageing during the first half of his turnaround, then he MUST conclude that their relative velocity was infinite (on average) during that section of his life. Period. Non-negotiable.
Mike_Fontenot said:And IF (as in the CADO reference frame) the traveler's conclusions about those two distances (at the beginning and at the midpoint of the turnaround) AGREE with the respective MSIRFs' conclusions about those distances at those two instants, then those two distances ARE L/2 and L, respectively. So the traveler MUST conclude that the distance to his home twin increases by L/2 during the first half of the his turnaround. Period. Non-negotiable.
Mike_Fontenot said:IF the traveler ALWAYS agrees, about the instantaneous distance to his home twin, at each instant of his life, with his MSIRF at that instant, then he WILL disagree with that MSIRF about the relative velocity of the home twin at any instant during the first half of his instantaneous turnaround. Each MSIRF is an inertial frame, and NO inertial frame will EVER conclude that the home twin has an infinite relative velocity with respect to the traveler.
Mike_Fontenot said:If you want to use a reference frame for the traveler, for which the relative velocity of the home twin during the first half of the turnaround ISN'T infinite, then that reference frame CAN'T agree with the conclusion of the traveler's MSIRF about his current distance to the home twin, at each instant during that first half of the turnaround. You just CAN'T have it both ways.
Exactly. And again, it is not an assumption, it is a definition. If you wish to re-define velocity then you may, but you need to be completely specific about it.GrayGhost said:If we are to assume that velocity is nothing more than the change of position over time, then twin A moves superluminally from point 1 to point 3 thru B-space during B's own virtually-instant proper acceleration.
Your point (1) is clearly not true in your diagram. The slope of the A worldline clearly exceeds 0.866c, or even c. Regarding (2) "the intersection of the A-worldline and the B-line-of-simultaneity" is just a long-winded way of saying the position of A in the B frame. The velocity of A in the B frame is by definition the derivative of this. Again, if you wish to redefine velocity you will have to be very specific. More math less english. Even if B's acceleration is finite this can lead to v>c.GrayGhost said:If you imagine a uniform almost-virtually-instant twin B proper acceleration, it's quite easy to envision how the A-worldline progresses from vertical to the slope of 0.866c, never exceeding 0.866c, let alone c.
Now you assume I am contradicting myself. However, there are 2 processes occurring wrt the A-worldline (per B) as B accelerates ...
(1) The A-worldline rotates steadily from vertical to a slope indicative of 0.866c, never exceeding 0.866c.
(2) The intersection of "the A-worldine and B-line-of-simultaneity" moves thru B-space, and can be superluminal.
Yes, that is the Dolby and Gull approach, not the naive/CADO approach.GrayGhost said:Now if the velocity is to be determined by the receipt of light signals, and doppler shifts converted into their appropriate dilation equivalent, then superluminal motion is never detected.
What does this mean? What is the formula that describes this "motion of events"?GrayGhost said:The established velocity will match the current slope of the A-worldline per B. The reason this works out as such, is because events move in space and time per B during his acceleration, something that does not happen classically.
GrayGhost said:Let's see how twin B plugs that infinite twin A velocity value into the LTs, see where twin A is placed by B into twin B's own system.
[...]
GrayGhost said:[...]
Seems to me that when twin B and the MSIRF observer are colocated, twin A must exist somewhere in spacetime and both those fellows must agree (even though they will disagree as to how twin A got there over time) ...
[...]
The fact that it is more difficult for twin B to (keep track and) determine the location of twin A at any instant during his own acceleration, does not lead that he should disagree with the momentarily colocated MSIRF-observer. All observers at rest with each other agree on the location and clock readout of a moving observer.
DaleSpam said:I think perhaps this is the key thing that you need to define, then you could possibly define your new concept of velocity as some sort of motion in addition to or relative to this motion of events. But you really need to be clear and mathematically precise here if you want to ensure a self-consistent outcome.
Honestly, rather than patching up such a strange ad-hoc concept, I think you would be much better served actually learning some differential geometry. But if you do want to pursue this the place to start seems to be this concept of the motion of events. Start by expressing that mathematically.
Mike_Fontenot said:[...]
[...] The result is (for units where c has unity magnitude, so I'll leave c out of the equation, for simplicity):
V = v - (L*v*a)/gamma ,
[...]
GrayGhost said:Mike_Fontenot,
I'm curious, could you please show your derivation for how you arrived at ...
V = v - (L*v*a)/gamma
That is only true in inertial reference frames. In non-inertial coordinate systems you can easily have v>c. For example, in the rotating reference frame attached to and co-rotating with the Earth even the nearest star travels an orbit of more than 12 ly/day which is >>c.GrayGhost said:Dale Dale DaleSpam. It's quite interesting you know. The theory demands that no material body can accelerate to speed c, because of energy considerations. Add, speed c represents a cosmic speed limit.
Then write down the conceptually clear equations so that others may benefit. I suspect that in the process of doing so you will find that it only seems conceptually clear now because you have not actually thought through the details.GrayGhost said:Conceptually, it's quite clear in my mind, personally. Events move in B-space per B while B continues his own proper acceleration.
Which is the definition of velocity.GrayGhost said:A moves superluminally per B from a standpoint of the-change-in-position-over-time.
What do you mean here? Are you talking about the change in position wrt the change in proper time for A? That would be the spatial component of the four-velocity, which would indeed be useful. The nice thing about the four-velocity is that the spatial component is not limited to <c, and regardless of its value you are guaranteed to not exceed a speed of c.GrayGhost said:In this case, IMO, twin A moves superluminally (per B) only if B focuses only on the spatial contraction while ignoring the time dilation component altogether.
I agree completely. This is exactly what you should do if you wish to use this idea.GrayGhost said:Indeed, what needs to be done is to math-model the twin B experience as the-collection-of-momentary-colocated-inertial-frames-of-reference that B co-occupies.
No. It is quite different. See figures 5 and 9 in the link below. Note particularly in figure 9 that the inertial twin's worldline never has v>c in the non-inertial twin's frame (using the standard definition of velocity).GrayGhost said:Wrt another comment you made ... it seems to me that if the Dolby and Gull approach converts doppler shift to the appropriate spatial offset, it should be equivalent to what I've been saying. Yes?
Is this math model the same one I was asking about here?GrayGhost said:I'll leave it at that, and take my time to attempt a math model. I'll first search about to see who else has already done so, because I have little doubt many qualified folks have argued my position before me.
ghwellsjr said:Here's another thing I don't understand. You keep talking about an observer using Lorentz Transforms to solve for something involving summing but you have not made it clear what the starting inertial frame is that he is working with, nor the set of events (1 time and 3 spatial coordinates) in that frame, nor the relative speed between that first FOR and the FOR he wants to convert the events in to. And after doing that for one FOR and he does it again for the next FOR, what is it that he sums and what is the significance of the sum? I just have no idea what you are thinking. Please elaborate instead of just repeating the same general recipe.
ghwellsjr said:Is this math model the same one I was asking about here?
The Lorentz transform transforms between different inertial frames where v<c. There is no requirement that v<c in non-inertial reference frames as I have already demonstrated. Again, you can re-define velocity, but you need to be clear that you are doing so. Alternatively, you can leave velocity unchanged and define some other parameter that allows you to select the correct Lorentz transform.GrayGhost said:Just to summarize here, there is the classical velocity definition being the change in position over time, which can be superluminal (eg A per B when B is non-inertial in my illustration). Then there is the velocity as used in the LTs to map spacetime coordinates between systems, and this velocity must be v < c (ie luminal).
I think that is a good idea. You will learn a lot in the process regardless of the eventual outcome.GrayGhost said:I'll leave it at that, and take my time to attempt a math model.
Do you realize that Mike believes his approach is the only valid way? And that is the argument that we all have with him? Not that his approach is wrong, it's just not preferred, but he thinks it is in some fundamental way. I doubt that you are going to understand his approach unless you get a copy of his paper because he never explains it fully on this forum. He leaves important definitions out of his explanations because he wants everyone to buy his paper for $15. If he explained what is contained in his paper on this forum, why would anyone want to buy his paper?GrayGhost said:I'm not certain as yet, however I suspect that Mike Fontenot's approach is the same as my reasoning. Our differences are likely the result of looking at the same thing in 2 different ways. In particular, the issue as to whether twin B and the momentarily co-located MSIRF observer "agree on the twin A velocity at that instant".
DaleSpam said:The Lorentz transform transforms between different inertial frames where v<c. There is no requirement that v<c in non-inertial reference frames as I have already demonstrated.
DaleSpam said:Again, you can re-define velocity, but you need to be clear that you are doing so. Alternatively, you can leave velocity unchanged and define some other parameter that allows you to select the correct Lorentz transform.
DaleSpam said:I think that is a good idea. You will learn a lot in the process regardless of the eventual outcome.
This shows that you lack a basic understanding of Einstein's theory presented in his 1905 paper. He didn't talk about non-inertial frames and he didn't talk about each observer in a scenario being stationary in their own frame or having their own rest frame. His original version of the Twin Paradox had the traveler going in a big circle with the inertial observer located at one point on the circle, not at its center. The traveler is never inertial and yet Einstein analyzes him using a frame in which the inertial observer is stationary. Get it? One inertial frame to analyze all observers. That was Einstein's theory.GrayGhost said:The only reason I hadn't pursued a math-model in the past, is because I felt certain it had already been done by many others since 1905.
ghwellsjr said:Do you realize that Mike believes his approach is the only valid way? And that is the argument that we all have with him? Not that his approach is wrong, it's just not preferred, but he thinks it is in some fundamental way.
ghwellsjr said:I doubt that you are going to understand his approach unless you get a copy of his paper because he never explains it fully on this forum. He leaves important definitions out of his explanations because he wants everyone to buy his paper for $15. If he explained what is contained in his paper on this forum, why would anyone want to buy his paper?
ghwellsjr said:I have a copy of his paper but it is copywrited. I'd like to discuss his ideas but if I give away all his secrets or quote from his paper, am I violating his copywrite? Is if fair for me to be discussing his ideas on this forum when only he and I (and anyone else with his paper) will know what we are talking about? The last person that was promoting a book he wrote for $8 got instantly banned. Why is Mike still able to promote his $15 paper and not get banned? Huh?
ghwellsjr said:This shows that you lack a basic understanding of Einstein's theory presented in his 1905 paper.
ghwellsjr said:He didn't talk about non-inertial frames ...
ghwellsjr said:... and he didn't talk about each observer in a scenario being stationary in their own frame ...
ghwellsjr said:His original version of the Twin Paradox had the traveler going in a big circle with the inertial observer located at one point on the circle, not at its center. The traveler is never inertial and yet Einstein analyzes him using a frame in which the inertial observer is stationary. Get it? One inertial frame to analyze all observers. That was Einstein's theory.
Hehe, yes, I do tend to think my own ideas are good ideasGrayGhost said:It was your idea.
I have not read his paper, since I am unwilling to pay for it. However, I have no objection to what he has described of his CADO equation on this forum. My objection is limited to his occasional incorrect claims that his CADO convention is the only correct simultaneity convention for a non-inertial observer and all other conventions are wrong. He has avoided making that claim in this thread, so I am OK with what he has said here.GrayGhost said:What sounds promissing is that you and DaleSpam seem to agree his paper is valid.
Good, here is a post that Mike linked to in his first post on this thread (page 11, post #167)GrayGhost said:I think I understand Mike's approach, since it's similar to what I myself have long envisioned, if not the very same thing.
Maybe you could tell us what he means by "elementary measurements and elementary calculations", because he won't tell us:Mike_Fontenot said:I would say that there IS a valid frame for an accelerating observer.
That frame consists of the (infinite) collection of inertial frames (the MSIRFs), one for each instant of the accelerating observer's life, each of which being momentarily stationary wrt the accelerating observer at that given instant in his life.
This frame is a well-defined...there is no ambiguity or inconsistency at all. And it is NOT a "convention": there are no other reasonable alternatives, because it is the ONLY possible frame for the accelerating observer which doesn't contradict his own elementary measurements and elementary calculations.
Oh, now he calls them "elementary observations and elementary calculations". I guess "observations" is the same as "measurements". In any case, since you think you understand his approach, and it is similar, if not identical to what you have long envisioned, what exactly does he and would you mean by these terms?Mike_Fontenot said:Those elementary observations and elementary calculations are given, in detail, in my paper. I'm not willing to reproduce them here.
I wonder why you think that. Can you provide the link that gave you this idea?GrayGhost said:What sounds promissing is that you and DaleSpam seem to agree his paper is valid.
Help me? You mean help me understand his position? Yes, I believe I understand exactly what his fatal mistake is.GrayGhost said:Did buying Mike's paper help you in any respect?
I'm sure you do, but I was talking about Einstein's theory presented in his 1905 paper which you alluded to in a previous post. Special Relativity has evolved since then in all kinds of directions which I have never seen the need to investigate and I don't claim to understand these additions to what Einstein first presented.GrayGhost said:I understand the special theory as well as you do, probably better.ghwellsjr said:This shows that you lack a basic understanding of Einstein's theory presented in his 1905 paper.
(Please note, you are quoting from the end of section 4, not 3.)GrayGhost said:ghwellsjr said:He didn't talk about non-inertial frames ...OEMB section 3: It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide. If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be ½tv2/c2 seconds slow.
In a theory devoid of gravity, it seems to me that a clock moving in a continuously curved line is non-inertial. Therefore, even though Einstein's SR was a theory of uniform translatory motion, he extrapolated what the effect of acceleration would be based upon the all-inertial theory.
I always wonder why anyone would want to go from one inertial frame to another inertial frame, I can't image why you would want to try to go from an inertial frame to a non-inertial frame. What's the point? Suppose you can find someone who has done this somewhere during the last century or suppose you figure out how to do it on your own. What do you learn by doing this?GrayGhost said:All I've been doing here in this thread ghwells, is hypothesizing by extrapolation (of the special theory) how twin B might apply the LTs to accurately transform his spacetime coordinates into the twin A system. Clearly, twin B cannot apply the LTs as easily as one would in the all-inertial scenario. That point was made way back yonder, and it's not as though anyone didn't already know it.
Here's the whole quote:GrayGhost said:ghwellsjr said:... and he didn't talk about each observer in a scenario being stationary in their own frame ...OEMB section 1: We might, of course, content ourselves with time values determined by an observer stationed together with the watch at the origin of the co-ordinates, and ...
and that statement allows us to imagine the same anywhere else in the OEMB paper, including section 3 where the LTs are derived. Although it does not have to be, said coordinate system may well be assigned by the observer to himself, as his own frame of reference.
My motive is to help you learn, and I hope you get it this time.GrayGhost said:Get it? You're kidding, yes?ghwellsjr said:His original version of the Twin Paradox had the traveler going in a big circle with the inertial observer located at one point on the circle, not at its center. The traveler is never inertial and yet Einstein analyzes him using a frame in which the inertial observer is stationary. Get it? One inertial frame to analyze all observers. That was Einstein's theory.
So now I must ask you, why did you feel the need to tell me all this in the first place? What's your motive here?
GrayGhost
ghwellsjr said:Good, here is a post that Mike linked to in his first post on this thread (page 11, post #167)
Maybe you could tell us what he means by "elementary measurements and elementary calculations", because he won't tell us:
Oh, now he calls them "elementary observations and elementary calculations". I guess "observations" is the same as "measurements". In any case, since you think you understand his approach, and it is similar, if not identical to what you have long envisioned, what exactly does he and would you mean by these terms?
ghwellsjr said:I wonder why you think that. Can you provide the link that gave you this idea?
ghwellsjr said:Help me? You mean help me understand his position? Yes, I believe I understand exactly what his fatal mistake is.
ghwellsjr said:I'm sure you do, but I was talking about Einstein's theory presented in his 1905 paper which you alluded to in a previous post. Special Relativity has evolved since then in all kinds of directions which I have never seen the need to investigate and I don't claim to understand these additions to what Einstein first presented.
ghwellsjr said:Please note, you are quoting from the end of section 4, not 3.
ghwellsjr said:Your comments show that you don't understand the difference between a non-inertial object/observer and a non-inertial frame of reference...
ghwellsjr said:This indicates to me that you have this erroneous concept that Special Relativity requires you to assign each object/observer to its own frame. This is completely wrong...
ghwellsjr said:Einstein's SR is a theory about a single inertial frame of reference in which all objects/observers are described and analyzed, and each object/observer can have its own velocities and/or accelerations but still described by that one single frame.
ghwellsjr said:In this example, he (Einstein) talks about two clocks, one at rest at location A and the other traveling in a circle starting at A, moving away from A, and then returning to A, accelerating all the time. In other words, this clock is non-inertial. But he doesn't assign a non-inertial frame of reference to it in which it is continuously at rest nor does he assign a series of inertial frames to it in which the clock is at rest in all of them. He wasn't extrapolating SR from an all-inertial theory to include accleration.
ghwellsjr said:From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by ½tv²/c²(up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.
Now, after you describe and analyze all the stationary, moving, and accelerating objects and observers in a scenario according to one inertial frame of reference, you can switch to a different inertial frame of reference which is described as having a motion with respect to the first frame of reference. And then by looking at the space-time coordinates of different events in the first frame, you can use the Lorentz Transform to see what the space-time coordinates are in the second inertial frame. That's what SR is all about.
ghwellsjr said:I always wonder why anyone would want to go from one inertial frame to another inertial frame, ...
ghwellsjr said:... I can't image why you would want to try to go from an inertial frame to a non-inertial frame. What's the point? Suppose you can find someone who has done this somewhere during the last century or suppose you figure out how to do it on your own. What do you learn by doing this?
ghwellsjr said:... My motive is to help you learn, and I hope you get it this time.
Sure he can. See the Dolby and Gull figure 9 that I pointed out earlier.GrayGhost said:One thing's for certain, while non-inertial, B cannot assume A sits at half the EM's roundtrip length.