Room-temperature superconductivity?

[johanfprins]

In order for them not to waste time, I will probably have to also reveal some proprietary information. The latter will obviously have to be covered by an NDA until the date that it is safeguarded by appropriate patents. After that date hey will also be free to publish any other data.

Have you communicated directly about the use of an NDA? Most will not be willing to go that route. I hope you have a plan B that does not include you divulging proprietary information. Even if it for some reason were to take more time, seems the simplest, safest route to testing.

I have excellent advisors, thank you.

[tomclarke]

Johan - have you given up substantive argument?

What substantive argument? The only substantive argument there is, is that for the time dilation formula you have (delta)t(T)=(gamma)*(delta)tp where (delta)tp is the "stationary" time interval of the moving clock within its own inertial reference frame within which it is stationary while (delta)t(T) is the transformed time within the inertial reference frame relative to which the clock is moving and is therefore the apparent relativistic rate at which the moving clock keeps time when it is viewed from the inertial reference frame relative to which it is moving.

agreed this is true. Substituted "calculated" for "apparent".

An identical clock which is stationary within the latter reference frame will keep time at the exact same rate within the latter reference frame than the identical moving clock keeps within its own inertial reference frame; Thus for a time interval (delta)tp of the moving clock within its own inertial reference frame, an identical clock which is stationary within the inertial reference frame relative to which it is moving, will measure a time interval (delta)t=(delta)tp: Exactly the same time interval!

Of course. It must in this case be symmetrical.

Muons within both reference frames must thus decay at exactly the same rate relative to their own inertial reference fames.

That is obvious, and must be true anyway.

From the other reference frame it will seem as the moving muons decay slower. This is s totally symmetric view

Agreed

and is not really happening within either inertial reference frame.

That is your interpretation: it is in the realm of philosophy, and not testable, so i won't argue.

You are arguing that if you consider one inertial reference frame as stationary, and bring the other reference frame back to this "stationary" reference frame, then the muons within the reference frame you brought back will be younger owing to the time-dilation formula.

Yes, but I would prefer it if we use "reference frame" only for inertial frames, as you have done above. The Lorentz formulae only apply to intertial frames. It is important that in the travelling twins case at least one twin must change frames to return. (In the symmetrical case both change frames to return).

Special relativity does not predict this at all since its formulas do NOT apply to situations where you accelerate.

We can divide the travelling twins case into two segments, outbound and inbound. Each segment can be dealt with using SR. We then need to see whether some additional argument can be used to "join up" the two segments assuming the frame chnage is quick. I'll do this in my next post if you agree. The results will surprise you.

It only unequovocally states that the clocks within the two reference frames MUST keep time at the same rate relative to their own inertial reference fames.

still agreed.

Which means that there is no difference in clock-rate with position or when changing inertial reference frames.

I need to change this to there is no change in clock rate relative to instantaneous rest frame with position or when chnaging inertial frames.

Thus, for as long as there is no gravity it is senseless to interpret Special Relativity as if it is defining time as an actual fourth coordinate which has different values at different postions, since at the same instant in time on any clock the other clocks everywhere will show exactly the same time.

So far your statements about "keeping time" have all been relative. Now you extend this to "all clocks in all frames". You have not shown this. You have not even shown how you would synchronise all clocks in all frames. (It can be done, but only relative to a given frame).

It is also worth pointing out that if, in a given single frame, all clocks are synchronised, (You agree this can be done using Einsteinian syncronistaion), the times on all clocks throughout space then do define a fourth coordinate which has different values at different points in spacetime, which we could call "time relative to the given frame". I have not shown that Einsteinian synchronisation in a different frame will result in different clock values for distant events, but conversely you have not shown that it would give the same values. I will show that it gives different values below if you like.

The transformed time into another inertial reference fame does show that separated events that occur simultaneously within an inertial reference frame will occur at different times within another reference frame. It is, however, totally daft to interpret this result as actual clocks keeping different time at different positions.

All true, I have never done this, providing you stick to the relative version of "keeping time" that you started with in this thread. Whenever your argument has sought to extend this to the global case, without saying how, I have corrected this.

What it really means is that clocks at these positions which keep the exact same time, record simultaneous events that occur within another inertial reference frame, at different successive times.

I agree with this, as long as the keeping time is relative to clock frame, and the simultaneous is relative to some specified other frame.

Thus, to interpret the Lorentz transformation as an actual coordinate transformation of a time coordinate that changes with position might be mathematically useful, but it is physics-nonsense.

You have not shown it is physically nonsense. "Keeping time" is always relative to the local frame. Obviously, things will be different in some non-local frame. That is not inconsistent.

And, BTW, since muon lifetime will be longer due to relative velocity, all you need is gamma ~ 10,000, or 1TV, for 1% return rate

I still say that one can use any radio-active material and if there is no change in gravity, the amount of decay on each material will be exactly the same when they are brought back together. If they are not the same the difference cannot be a result of the the time-dilation formula of SR, since this formula applies in both directions while the clocks within both inertial reference frames keep exactly the same time.

Remember that to conduct a complete analysis we need to join up the inbound and outbound journeys. each journey can be analysed with SR, and is symmetrical as you say. However when you join up the two journeys the argument is no longer symmetrical, and this will allow for the final assymetry in the ages of the muons.

Thus, when twins fly apart at a speed v, a twin that ages by 10 years will know that at that exact instant in time on his clock, his borother has also only aged 10 years. There is no "time warp" that projects the one twin into the future of the other. The latter conclusion is obviously paranormal metaphysics which makes interesting movies like Star Trek, but is pure physics nonsense since it is based on the assumption that mathematics IS reality while this is not the case.

This argument is based on your deduction that because each journey alone is symmetrical, when you join two journeys together the result is also symmetrical. You cannot know that without looking carefully at how the two journeys are joined together. I am happy to do that below, the result will surprise you.

Reality determines the mathematics that must be used, not the other way around.

In this case the maths and the reality are equivalent.

If you want to argue that the twins will have different ages when brought back together again, you must seek the reason for this in the time that the one twin decelerates to meet up with the other twin.

Yes

But before this deceleration starts, the twins will be the exact same age as mandated by the symmetry of the Lorentz transformation and Einstein's first postulate. Therefore such an age difference cannot be caused by the formula of time dilation that is derived from SR.

Ah! You have nearly got there. This argument is true for a twin in its instantaneous rest frame. But this does not join up the ages of the two twins. To do this we need to compare the age of a twin with the observed age of his sibling throughout the journey. At the end, when they come together, the observed ages must also be the real ages.

The answer to your statement that the situation remains symmetric is that what happens when the twin frame changes is not symmetric (because it happens to one twin and not the other). Although this obviously does not change the "keeping time" of local clocks it does change the relationship between local times of the two twins. It is this relative timing that is key to understanding why different paths through spacetime between the same two events can have different proper times. I will show you how (using only the agreed stuff above0 this can be calculated below, if you like.

[johanfprins]

Johan, I admit that I am not a physicist but as I understand it, as a particle approaches C the time of the frame of the particle slows down to infinity at C.

This is where we differ: The time on a clock travelling WITH the particle does NOT slow down. It only seems on a clock remaining behind that the clock with the particle slows down.

Therefore by my numbers, for a muon with a 2.2 microsecond clock to reach the moon (at 384402 km) and back it must have a time dilation factor of 1165.664 which is a speed of 0.9999996320207617 C.

But by that time there will be no muons left to compare with on earth. I calculated the time required before all the muons remaining on earth has decayed; which is different than the "apparent" lifetime of the muons relative to earth while they are moving relative to earth.

Have I done my math wrong?

No your math is not wrong but it is based on the physical impossibility that the muons on the trip decays at a slower rate within their own inertial refrence frame. According to Einstein's postulates and the Lorentz transformation this is not possible.

What went wrong was when Minkowski confused the issue: Even Einstein opposed Minkowski's interpretation that time is an actual fourth time coordinate that changes from one position to another. He stated at that "since the mathematicians (referring to Minkowski) involved themselves with Speial Relativity that even he (Einstein) does not understand it anymore".

Einstein only changed his view many on the validity of Minkowski space many years later when he found that in order to model gravity, he had to use time as a fourth coordinate that actually, ONLY in this case, does changes from one position to the next. But he should have realised that this is only the case when you include gravity.

Gravity is not included in Special Relativity and therefore time is not an actual coordinate that changes from one position to the next position. To use Minkowski's construction in this manner when there is no gravity is pure physics-nonsense. By doing this you are adding aspects to Special Relativity which violate Einstein's postulates and the Lorentz transfornation.

hus if tomclarke wants to assume aspects that contradict Einstein's postulates and the Lorentz equation, by assuming that you can follow a path through space-time when there is no gravity, he is welcome to follow this path into Alice's Wonderland, but I will not accept this mathematics as being valid when there is no gravity.

[johanfprins]

I admit I am confused. Johanfprins arguments may or may not be true and consistant. But, what I am confused with is how his interpretation can explain the survival of the Muons till they reach the surface of the Earth

I have derived this from the Lorentz transformation in one of my posts above and showed that it si not just caused by time-dilation but by a simultaneous increase in length of the path as measured relatiuve to earth (NOT a length contraction).

The math may be elegant and even true in certain situations, but unless it can explain real things (to a simpleton like me) then it is useless (at least to me)...

You are not a simpleton: All physicists, including Einstein, made mistakes when it comes to Special Relativity since it is difficult to release our minds from thinking linearly. The Lorentz transformation is not a linear transformation, AND it does not just transform time to be longer without also also transforming distances to be longer. When looking at events occurring withi a passing inertial refrence frame the Lorentz transformation magnifies time-intervals as well as length intervals. As soon as you model the result just in terms of time-dilation, without also including the concomitant length-stretching, you make the mistakes that tomclarke is making.

PS- There is a free E book written about Einstein by Lorentz. I have not read it completely, but it seems the man (author of the transformation formulas?) has no problems with Einstein's conclusions.

Lorentz did not consider his equations as relativistic coordinate transformations but as being caused by an actual length contraction of a rod. Einstein postulated that the equations are the result of the constance of the speed of light within each and every inertial reference frame. Thus actual length contraction as Lorentz had assumed, is not occurring.

[tomclarke]

hus if tomclarke wants to assume aspects that contradict Einstein's postulates and the Lorentz equation, by assuming that you can follow a path through space-time when there is no gravity, he is welcome to follow this path into Alice's Wonderland, but I will not accept this mathematics as being valid when there is no gravity.

All objects follow a path through spacetime. The function:
t -> (x,y,z) specifying their position in space at given local time t defines such a path (relative to some frame which defines x,y,z).

In the twins paradox case one twin follows a straight path. The other follows a path consisting of two straight segments joined where the twin velocity changes to return home.

Gravity is not included in Special Relativity and therefore time is not an actual coordinate that changes from one position to the next position. To use Minkowski's construction in this manner when there is no gravity is pure physics-nonsense. By doing this you are adding aspects to Special Relativity which violate Einstein's postulates and the Lorentz transfornation.

Flat spacetime (without curvature from gravity), & SR, clearly deals with both space and time. It is normal to view time as one dimension in a 4D spacetime. For computational purposes, any SR frame puts (x,y,z,t) axes onto the set of all events in (4D) spacetime. I'm not sure why Johan objects to this.

[johanfprins]

Flat spacetime (without curvature from gravity), & SR, clearly deals with both space and time. It is normal to view time as one dimension in a 4D spacetime. For computational purposes, any SR frame puts (x,y,z,t) axes onto the set of all events in (4D) spacetime. I'm not sure why Johan objects to this.

Time rate determines light speed and light speed determines time rate. This is clearly shown in Einstein's model of gravity where both time-rate and light speed slow down as gravity increases. Both become zero at the event horizon of a black hole. It also happens on the atomic scale when a photon is absorbed by an electron orbital-wave around the nucleus of an atom. Thus, on the macro-scale ONLY within a gravity-field can the rate of time be different at different positions in space. And this is so since light speed can be different at different positions in space.

Within SR the speed of light is constant at every point in space and within every inertial reference frame, thus demanding that the time-rate must also be constant at every point in space and within every inertial reference frame. Thus, to assume otherwise for special relativity is to assume physics-nonsense: Since one is assuming a gravitional effect without a field of gravity. And as I have pointed out repeatedly on this thread, it also violates Einstein's first postulate. In fact, it also violates Einstein's second postulate since the constancy of light speed everywhere demands the constancy of time-rate everywhere.

Thus, without gravity, time becomes redundant as a coordinate, just as the z-coordinate becomes redundant within a flat x-y plane even if the x-y plane moves along the z-axis so that the z-coordinate changes, the instantaneous z-coordinate remains the same value at all positions within the x-y plane. Only on a curved surface does z again become active as a coordinate that changes from one position to the next. Similarly, although time plays a role in SR, it is a redundant coordinate which does not actually change from one position to the next. Also in this case time only becomes active as a coordinate within curved space-time: And for this to manifest there must be a gravity-field.

[johanfprins]

Remember that to conduct a complete analysis we need to join up the inbound and outbound journeys. each journey can be analysed with SR, and is symmetrical as you say. However when you join up the two journeys the argument is no longer symmetrical, and this will allow for the final assymetry in the ages of the muons.

This cannot be deduced from the Lorentz transformation since these equations do, at present, not model accelerating reference frames.

This argument is based on your deduction that because each journey alone is symmetrical, when you join two journeys together the result is also symmetrical. You cannot know that without looking carefully at how the two journeys are joined together. I am happy to do that below, the result will surprise you.

As long as you do not invoke impossible paths within Minkowski "space-time", be my guest.

Reality determines the mathematics that must be used, not the other way around.

In this case the maths and the reality are equivalent.

No they are not since you use arguments of paths through a Minkowski space-time which do not exist in gravity-free space.

If you want to argue that the twins will have different ages when brought back together again, you must seek the reason for this in the time that the one twin decelerates to meet up with the other twin.

Yes

Good! The only problem is that your arguments on how this must be done are all wrong since they are based on Minkowski space without gravity.

But before this deceleration starts, the twins will be the exact same age as mandated by the symmetry of the Lorentz transformation and Einstein's first postulate. Therefore such an age difference cannot be caused by the formula of time dilation that is derived from SR.

Ah! You have nearly got there. This argument is true for a twin in its instantaneous rest frame. But this does not join up the ages of the two twins.

You cannot join up the two ages of the two twins by using the Lorentz transformation. What you are doing is to fudge this result by assuming that there are paths within a space-time when there is no gravity. This is physically meaningless.

I believe that when one of the twins accelerates he will observe the other twin accelerating away from him and that in terms of SR this is still a symmetric condition. The only loophole you have is to say that when a twin accelerates by switching on his engines, he will experience a force which Einstein claimed is equivalent to gravity, and that for this reason the twin ages slower during acceleration. But this has nothing to do with the time dilation formula of SR which supposedly requires a twin to "change inertial reference frames" for its age to be compared.

But I think that even the acceleration argument is wrong, and that non-gravitational acceleration will one day be proved not to have any effect on clock rates: Only the strength of a gravity-field lowers the clock-rate as well as light-speed simultaneously. I believe the latter can be proved directly from Maxwell's equations: And I would probably have had the answer by now if I had rather concentrated on my research instead of arguing with you on this thread.

Best wishes,
Johan

[tomclarke]

Code: Select all

B
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A

Vertical axis is time in initial rest frame. Horizontal axis is x coord of space in the same. Twin P stays stationary, his frame is always this rest frame F0.
P: A -> B

Twin Q travels away with velocity V to event X, and then back to meet twin P at event B.
Q: A -> X -> B

Twin Q spends his time in either frame F1 (outbound) or frame F2 (inbound).

F1 and F2 have relative speed v with F0, and gamma factor 0.5.

To analyse this note that:

At A and B clocks on P & Q can be compared directly.

At A,B,X (when Q changes frame F0 ->F1 -> F2 -> F0) the change can be arbitrarily short and hence Q clock time during change can be taken as 0.

On AX, Q clock measures F1 time. On XB Q clock measures F2 time. P clock always measures F0 time.

P & Q clock both read zero at A. What do they read at B?

Gamma relates F1 & F2 time to F0 time in frame F0. We suppose V is chosen so that gamma = 0.5.

Notation:
T0, T1, T2, time in F0, F1, F2
deltaT0 - time difference in F0
T0(A->B) - time difference between A and B in F0, etc.

From SR, we know that:
gamma*deltaT0 = delatT1 = deltaT2



so 0.5*T0(A->B) = T1(A->X)+T2(X->B) or

reading in P clock (A-> B) is 2 * reading in Q clock(A->B).

Hence the expected result.

So far there is nothing surprising, except to Johan who does not accept the F0 to F1 or F2 time relationship.

Now let us do the same thing from Q POV to see why we do not get contradiction.

In F1 frame, true, Q clock will be 2X faster than P clock. Equally in F2 frame. This looks like a contradiction with previous result!

We are going to work out P's clock times in Q clock's frame throughout Q's journey. The problem is the frame changes: F0->F1->F2->F0. The previous calculation did not have this problem because P never changed frame.

We can shift from F1 to F0 frame times at A since P & Q are at same postion, so there is no difference. Similarly at B we can shift from F2 to F0 frames with no difference, because P&Q are at same position. So:
The total time change on P clock, as measured in Q instantaneous frame, is:

half of Q clock time (for A->X and X->B)
+ change in P clock due to change from F1 to F2 at X. Since P is far away for Q at this point we cannot work out the time directly. We have to calculate it. But how we do this depends on the relative velocity of P & Q. When Q is moving towards P all the calculated times for P get increased, when away they all get decreased. The shift in time due to frame velocity is proportional to the distance away and goes to zero when there is no separation.

This time shift when Q changes from from F1 to F2 thus has the effect of making the calculated time of P clock suddenly much larger. So that when P & Q finally meet the time on P clock will be double that on the Q clock, even through the contribution of the A->X and X->B segments gives only half of Q clock.

Befor Johan gets apoplexy let us wee why the calculated time, with time shift, is all we have. There is no frame-independent way to compare P & Q times when they are far apart. The three frames in which we could do the comparison (F0, F1 & F2) all give different answers.

So when Q changes velocity, its rest frame changes and the time on P which is instantaneous with now on Q therefore also changes. The P clock remains unchanged.

This time shift phenomena is a fact of relativistic life. I guess I will have to prove it a bit more carefully using the standard light ray arguments that SR loves.

The time shift needs to be factored in with time dilation when comparing twin ages. Time shift is not so often talked about. It is more complex as a concept than time dilation, and only comes into play when resolving the apparent paradox of symmetry between twins time dilation and assymmetry between their ages.

A much longer but clearer presentation of time slip and time dilation in SR can be found in Tom van Flandern's paper:
http://www.volta.alessandria.it/epistem ... -vanfl.htm

Ironically, TvF actually advocates Lorentzian relativity, a slight variant of SR in which absolute time exists! Johan would like this. However as TvF shows, LR and SR give identical results in the twins paradox case, equal to the ones conventionally shown. The difference between LR and SR is very subtle and outside the arguments here.

Enjoy!

[johanfprins]

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Enjoy!

You see how arrogant you are. I know this wrong reasoning. So why are you posting it as if you want to teach me something. Your knowledge of physics is so rudimentary compared o mine that you are only making an ------- of yourself.

As I told you time and again that this time-space diagram is BS since it is not possible that the timeline for any one of the two twins can be anything but parallel to the time axis. This is so since both inertial reference frames are keeping the exact same time and since both twins are stationary within their respective reference frames. Prove to me by using the Lorentz transformation that you can bring the twins together by not using a Voodoo space time diagram within a gravity free space.

Stop boring me with your dogmatic chants: PLEASE!

The fact is that SR cannot model the return line.

I do not know how old you are, but it seems that you are still quite young. I suspect that it may be possible that within your lifetime it will be proved that this so-called derivation of yours is Voodoo. I then wish you a terrible conscience, if it is possible for you to ever have one? Only saints and psychopaths have no regrets. You just like me is not a saint. I have regrets, since in my youth I also argued illogically to defend the mainstream and thus did terrible injustices to people whio did actually know better.

[tomclarke]

Johan,

I expect that the concept of time-slip, as well as time dilation, being important in comparing times between two reference frames is new to you. It certainly answers your claims that SR as others know it must lead to paradox.

Anyway, old or young, I don't think there is more I can say to help you. I've notice dthat whenever te arguments get detailed you return with generalities that do not address the specifics, as here.

Best wishes, Tom

[tomclarke]

perhaps...

As I told you time and again that this time-space diagram is BS since it is not possible that the timeline for any one of the two twins can be anything but parallel to the time axis. This is so since both inertial reference frames are keeping the exact same time and since both twins are stationary within their respective reference frames. Prove to me by using the Lorentz transformation that you can bring the twins together by not using a Voodoo space time diagram within a gravity free space.

Johan, the diagram is a picture of position versus time which I use to illustrate what is going on, pictures being better than words. I have no idea what is your objection with it. Clearly if one of the twins has no spatial movement the other must heva spatial movement, since their separation changes!

The argument below this (uncontentious) diagram shows how the P & Q clocks relate to each other via Lorentz transforms AKA time dilation and "time-slip" of the distant twin's time when viewed in a different frame.

It is the true answer to your erroneous "the twins must be symmetrical" argument. It does not relate to acceleration altering time rate (which it does not).

It is very neat. It could be presented without any diagram, of course, if that is your preference.

You might wish to read the longer version in Tom Van Flandern's paper. TvF did not agree with conventional SR theory, so he was hardly a conformist. But he was a clear thinker.

Best wishes, Tom

[johanfprins]

You might wish to read the longer version in Tom Van Flandern's paper. TvF did not agree with conventional SR theory, so he was hardly a conformist. But he was a clear thinker.

I am nearly 70 years old and have taught physics for 50 years, also the dogmatic nonsense that you still believe in. There is nothing you can recommend on SR that I have not yet read. So please stop being so arrogant.

At least we both agree that the clocks of the two twins must keep time at the same rate within their respective inertial reference frames while they move at a relative speed v. You claim that one can only compare the clocks by bringing then together, I claim it is not necessary, and even if it is necessary the Lorentz transformation does not give a recipe how it must be done; and that the arguments that are used when modelling the "bringing" together based on Minkowski diagrams are all flawed: Also those of Flanders.

I would like to spend time arguing this case further, but I just do not have the time anymore since I have to urgently develop prototype superconducting devices. The money for this is in the pipeline. It will thus be stupid of me to spend my time arguing about SR while neglecting room temperature superconducting prototype devices.

In either case, what we differ on shoudl be determined by a clear unequovical experiment, which does not mix in time dilation caused by gravity, as is the case for the flying clock experiments. I have proposed a simple experiment that should be done. Maybe somebody will pick this up and do this experiment. If you are proved correct I will apologise: I hope that if I am proved correct you will do the same.

Unfortunately, I have to now say farewell until, if it happens, I again have free time.

Best wishes,
Johan

[D Tibbets]

I admit I am confused. Johanfprins arguments may or may not be true and consistant. But, what I am confused with is how his interpretation can explain the survival of the Muons till they reach the surface of the Earth

I have derived this from the Lorentz transformation in one of my posts above and showed that it si not just caused by time-dilation but by a simultaneous increase in length of the path as measured relatiuve to earth (NOT a length contraction).

The math may be elegant and even true in certain situations, but unless it can explain real things (to a simpleton like me) then it is useless (at least to me)...

You are not a simpleton: All physicists, including Einstein, made mistakes when it comes to Special Relativity since it is difficult to release our minds from thinking linearly. The Lorentz transformation is not a linear transformation, AND it does not just transform time to be longer without also also transforming distances to be longer. When looking at events occurring withi a passing inertial refrence frame the Lorentz transformation magnifies time-intervals as well as length intervals. As soon as you model the result just in terms of time-dilation, without also including the concomitant length-stretching, you make the mistakes that tomclarke is making.
...

Are you stating that as time dilation occurs at relativistic speeds, the length traveled also increases in proportion? It seams that would be an argument that relativistic changes are non existant as the two effects cancel each other out, at least from a stationary lab frame of reference (?).
Wouldn't this result in the impossibility of Muons reaching the Earth's surface. It seems (to me) that the sign of the time or distance effects must be reversed for any possibility of matching observations. This is what I think is stated in the Wikipedia article on Mouns. One or the other aspect can be appreciated (time dilation or distance Contraction) depending on which frame of reference you are in. In the Muon's frame there is no time dilation, the clock runs normally and the Muon decay rate is as expected, but the Muon still manages to plow into the Earth, because from the Muon's reference frame, the Earth surface is closer (distance contraction). From the Earth reference, the Muon is time dilated. I'm not sure why one effect would not serve from either perspective in an isolated system and perhaps this is what you are arguing. But, distance is relative(?) in comparison to all the surrounding objects. The Muon may be created at 100 miles altitude, but satellites measured at the same time may be at 200 miles altitude above the Earth. At least in the altitude vector the satallite and Earth are motionless, but the Muon is moving towards one and away from the other. It is at relativistic speeds so there is Muon time dilation from the Earth/ satellite frame of reference. If there is distance elongation also, then the Muon would never reach a destination, at least there would not be any relativistic contribution, which means that the observation of Muon detection at the Earth's surface is impossible. From the Moun's perspective, any of it's relativistic speed would be cancelled out by increasing distances in all directions . The Muon would be motionless in relation to everything around it. I think this might result in a completely static universe (or not). There must be a reversal of distance effect or time effect (minus sign in there somewhere).


[EDIT]
I think this is equivalent to the Doppler argument. If relativistic time dilation and length/ distance expansion march together, it is the magnitude of the relativistic speed that determines the effect. the vectors are irrelevant (towards or away). How then would you explain the Red shift, and the Blue shift which includes a vector consideration.
And consider a photon in the vacuum of space. As it is traveling at the speed of light it's time dilation is infinite from the stationary lab reference, and if distance lengthening occurs at the same time from the photon's frame of reference, then the photon never moves relative to the rest of the universe, and conversely nothing reaches it- it is entirely isolated. Interaction can only occur if + time dilation is matched by - distance dilation. In isolation either time or distance changes might explain what is experianced by a single entity, but this is only part of the story. For interaction (existance) it must interact with something that also has it's own frame of reference. Assigned positive values must be matched by corresponding negative values, otherwise never the twain shall meet.

Dan Tibbets

[ladajo]

Johan,
As I continue to think and follow on this thread as one of the most interesting fo ra while now. I would like to revisit earlier discussion on how you think the observed and used mechanism regarding orbital clocks addresses your concerns:

But I think that even the acceleration argument is wrong, and that non-gravitational acceleration will one day be proved not to have any effect on clock rates: Only the strength of a gravity-field lowers the clock-rate as well as light-speed simultaneously. I believe the latter can be proved directly from Maxwell's equations: And I would probably have had the answer by now if I had rather concentrated on my research instead of arguing with you on this thread.

Best wishes,
Johan

I know previously that you have proposed a gravity free clock test. And I am on board with that idea. But I also have to ask how you consider the current applied theory regarding the two clock component accounting functions regarding SR and GR. Given that there aer now and have been many clocks in low orbit and high orbit, and that the compensation mechanism in place seperates the SR bit from the GR bit as a function of velocity and gravity seperately, and that this compensation mechanism seems to function quite accurately for all the devices in use, How do you account for the velocity correction requirement as a seperate entity from the gravity correction?
We know from real observation that Geo orbit devices verses LEO devices switch emphasis between the driving correction from gravity to speed and the opposite. We also know that clocks orbiting the earth in a low angle plane experience velocity impacts from the sun orbit they ride in as they speed up and slow down relative to the path around the sun while making the Equatorial orbit around the earth. These would seem to argue one of two things, that velocity does matter, or that what we think is velocity induced is a function of some other phenomena or subset of existing things that we do not yet grasp.
I will revisit the previous posts along this regard, but as I sit here and frame this post, I do not clearly recall you directly addressing these points of physical observation, if you have, my apologies, and I hope to find it in the previous entries during my review.

[johanfprins]

Are you stating that as time dilation occurs at relativistic speeds, the length traveled also increases in proportion?

The two are obviously related since the Lorentz transformation is derived from a comparison of (delta)x=c*(delta)t and (delta)xp=c*(delta)tp (see for example Einstein's book "Relativity: The Special and General Theory" Appendix 1).

What is pointed out there is that when (delta)tp/(delta)t is mu>1, then (delta)xp/(delta)x must also be equal to mu for c to be the same within both inertial reference frames. Thus, if you just use time dilation at its own without also including the concomittant length stretching, you end up making the mistakes that tomclarke is making.

This discussion is extremely interesting and I would like to proceed with it in more detail; but unfortunately I will have to postpone such participation until a later time. My sincere apologies.