Room-temperature superconductivity?

[DeltaV]

I am sorry about your stuff being taken. That is just obnoxious.

And suspicious.

I pray that you had nothing unrecoverable on the laptop.

Like superconductor details.

[tomclarke]

To make it easier I will give a choice question: Consider a moving clock. Relative to how many inertial reference frames does it move?

(i) One?
(ii) Ten?
(iii) A Thousand?
(iv) A billion?
(v) An infinite number?

An infinite number, of course.

And a clock's time dilation relative to any of them will be observed if an clock changes frame in such a way as to colocate with another clock in the other reference frame at two different times, allowing clock time comparison.

[johanfprins]

Johan,
What do you see as the reasons for the 46us and 8us adjustments on the GPS bird clocks prior to launch to give the 38us correction factor on orbit?
As I understand you are ok with gravity well correction...

I am really OK with BOTH corrections but they are done for different reasons:

The “gravity adjustment” is required because the gravity field is weaker “up there” than “down here” on earth. Light speed is INCREASED (NOT reduced: SORRY) within a smaller gravity field (this is not caused by “acceleration without gravity” but gravity itself) and it is for this reason that the clock “up there” does thus actually keep time at a faster rate, and must thus be adjusted to synchronise with ALL the clocks on earth; which ALL keep exactly the same time no matter where they are situated as long as they are all experiencing exactly the same gravity field.

The adjustment required for the relative motion (special relativity adjustment) is also required; BUT, in this case, not because the clock “up there” is actually running slower than a clock “down here” on earth, BUT because the Lorentz transformation causes the same clock “up there” to run slower “down here” than ALL the clocks that are actually down here. Thus, to synchronise it with a clock on earth one has to set the clock “up there” a bit faster. This is in contrast to the previous adjustment that has to be made purely to accommodate the lesser gravity up there; which causes the clock “up there” to actually keep faster time.

Why is the adjustment caused by special relativity at all required? Einstein started of well when he realised that the Lorentz transformation of two (or more) spatially separated, simultaneous events, can destroy simultaneity within the reference frame into which the events are being transformed. He then slipped up by forgetting about simultaneity when he derived “length-contraction” and “time-dilation”. What did he forget? He forgot that in order to have simultaneous events at different positions within an inertial reference frame the time at all positions within such a reference frame MUST be exactly the same: Just like Newton has assumed.

Let us look at two identical, perfect clocks a distance Lp apart within a passing inertial reference frame Kp: For two events to occur simultaneously within Kp at the starting point of Lp (say at xp=0 at which clock1 is situated) and the nose”-point of L at xp=L (at which clock2 is situated), the time MUST be the same and change at the same rate at these two points within Kp. Now let us synchronise a clock (clock 3) within K at position x=0, with clock1 within Kp at xp=0. ALL the clocks everywhere within K, no matter where they are situated, are at this instant synchronised with clock1. Since clock 1 is synchronised with clock2 and ALL the clocks within K must thus be synchronised with clock 1, it seems logical that they should all then also be synchronised with clock2 at position xp=L. BUT now the Lorentz transformation comes in to create a “relativistic effect” and you find that within K you cannot observe clock2 at a distance x=Lp within K but only at a distance given by:

L=(gamma)*Lp

Where L is LARGER than Lp (NOT SHORTER!!!!!!)

In addition the nose-point of L within K can only be observed at a later time t on ALL the clocks in K than at the synchronization time t=tp=0. And this time is given by:

t=(gamma)*(v/c^2)*Lp

It is this formula that causes time dilation, but it only does so in conjunction with the length “extension” given by the first formula It is a good excecise to now derive the time dilation for a passing clock by remembering when at t=tp=0 the distance between the clocks is zero. And a time (delt)tp later on the moving clock the distance between the clocks is L=v*t.

The important point to notice is that ALL the clocks at ALL and EVERY position within K are in actual fact keeping the exact same synchronised time, and this is also the case for ALL the clocks at ALL the positions within Kp. Furthermore, it is impossible to see the front end and back end of a passing rod (stationary within Kp) at the SAME instance of time on ALL the clocks within K: Thus one CANNOT see the length of such a passing object AT ALL within K: It does not have a simultaneous length within K: So how can it contract? In fact, as derived above, the space coordinate BECOMES longer.

PS: I am sorry about your stuff being taken. That is just obnoxious. I pray that you had nothing unrecoverable on the laptop.

Thanks. I have lost some data which means that my book will only appear on Amazon at a later date than I anticipated.

[GIThruster]

The adjustment required for the relative motion (special relativity adjustment) is also required; BUT, in this case, not because the clock “up there” is actually running slower than a clock “down here” on earth, BUT because the Lorentz transformation causes the same clock “up there” to run slower “down here” than ALL the clocks that are actually down here. Thus, to synchronise it with a clock on earth one has to set the clock “up there” a bit faster.

I'm sorry but would you please confirm that what you're saying is, that the two clocks, one stationary and one in a plane should indeed move at different rates? You're saying that the observations normally presented as evidence of time dilation due to velocity, is actually evidence of "the Lorentz transformation"?

Is that correct?

[tomclarke]

The adjustment required for the relative motion (special relativity adjustment) is also required; BUT, in this case, not because the clock “up there” is actually running slower than a clock “down here” on earth, BUT because the Lorentz transformation causes the same clock “up there” to run slower “down here” than ALL the clocks that are actually down here. Thus, to synchronise it with a clock on earth one has to set the clock “up there” a bit faster.

I'm sorry but would you please confirm that what you're saying is, that the two clocks, one stationary and one in a plane should indeed move at different rates? You're saying that the observations normally presented as evidence of time dilation due to velocity, is actually evidence of "the Lorentz transformation"?

Is that correct?

Johan,

The only definite way you can answer which clock is running slower is to compare them at two different times and see which has larger elapsed time. That means bringing clock "up there" "back here".

You are however now agreeing the elapsed time must be different, and hence that time dilation is real.

Thus:

(1) keep the clock up there for a long time, till the accumulated SR error is 10s.

(2) Bring the clock back. The displacement doing this is < 1light-second so changes in time relative to its observed time at distance are bounded by this. Thus the difference is still at least 9s.

The 9s is "real".

[johanfprins]

I'm sorry but would you please confirm that what you're saying is, that the two clocks, one stationary and one in a plane should indeed move at different rates? You're saying that the observations normally presented as evidence of time dilation due to velocity, is actually evidence of "the Lorentz transformation"?

Is that correct?

You are STILL missing it: The clock in the plane and the one on the ground keep time at exactly the same rate RELATIVE TO THE PLANE AND THE GROUND RESPECTIVELY. When you transform the time rate of the clock on the plane into the reference frame of the clock on the ground (NOTE I DO NOT CALL THIS A STATIONARY CLOCK: since there is NO uniquely staionary reference frame as far as light speed is concerned in the Universe) then the consecutive events related to conecutive ticks on the "moving" clock are observed to be longer on ALL the clocks on earth that are not moving with the "moving" clock. THIS DOES NOT MEAN THAT THE "MOVING" CLOCK IS ACTUALLY TICKING AT A SLOWER RATE AS THE "IMMOBILE CLOCKS" on the earth.

[GIThruster]

"Ticking at a slower rate" is of course, hopelessly confused verbiage. Everyone knows the whole notion of clocks is that they are measures of the passage of time in the various frames.

If you can agree, that the clocks when rejoined no longer show the same time, then you're not "correcting" Einstein. Likewise, your confused language would seem to have confused DeltaV and you owe him an explanation now too. He seems to think despite the evidence, that you are right and Einstein was wrong.

That would be now 2 people on the planet. Could be worse I suppose.

[johanfprins]

The only definite way you can answer which clock is running slower is to compare them at two different times and see which has larger elapsed time. That means bringing clock "up there" "back here".

If you want to make this a precondition go ahead but it is not required, since ALL clocks at ALL positions within any inertial reference frame keep the same time. And this time is the same within ALL inertial reference frames. If this is not the case then Einstein's first postulate (the principle of relativity) is wrong and therefore the whole theory of Special Relativity will be wrong. WE know the latter is not the case: The only problem has been that this theory has not always been correctly interpreted: Even by Einstein himself!

You do not have to bring the clocks together to know that THIS MUST KEEP THE SAME TIME, snce if it is NOT SO, then different events at different positions within an inertial reference frame CAN NEVER OCCUR SIMULTANEOUSLY.

This then makes mockery of the reality that two events that are simultaneous within an inertial reference frame no.1, TRANSFORM into two events which are NOT simultaneous within another inertial refrence frame no. 2. The stupid mistake has been made to conclude that this implies that the two simultaneous time coordinates within FOR1, which give different time coordinates within FOR2 demands that the clocks at the transformed positions within FOR2 are out of synchronisation. THIS IS NOT THE CASE!!! It only means that one of the simultaneous events in FOR1, occurs at an earlier time within FOR 2, than the other simultaneous event in FOR 1 occurs in FOR2.

When the first simultaneous event occurs within FOR2, ALL the clocks within FOR2 are showing the exact same time no matter where they are situated within FOR2. When the second simultaneous event within FOR1 occurs at a later time, ALL the clocks in FOR 2 are showing this later time. It is thus absurd to reason that there are two clocks at the positions in FOR2 where the first and second event occur WHICH show "simultaneously" different times. Only a typical deluded mathematiocian will think that this mathematically convenient picture is what is actuall happening physically.

You are however now agreeing the elapsed time must be different, and hence that time dilation is real.

I have NEVER disagreed that the transformed time coordinate is commensurste with time dilation WITHIN THE REFERENCE FRAME INTO WHICH THE TIME IS BEING TRANSFORMED. I HAVE CONSISTENTLY DISAGREED THAT THE TIME ON THE MOVING CLOCK IS ACTUALLY SLOWER THAN THE TIME ON AN IDENTICAL CLOCK WITHIN THE REFERENCE FRAME INTO WHICH THE TIME COORDINATE OF THE RELATIVELY MOVING CLOCK HAS BEEN TRANSFORMED.

Thus:

(1) keep the clock up there for a long time, till the accumulated SR error is 10s.

(2) Bring the clock back. The displacement doing this is < 1light-second so changes in time relative to its observed time at distance are bounded by this. Thus the difference is still at least 9s.

If this actually happens, it would mean that the Lorentz transformation MUST be wrong since as I have shown time and again on this thread, within each and every inertial reference frame the time is THE SAME at EVERY position. Thus in terms of the Special Theory of Relativity, the clocks will NOT DISAGREE.

[The 9s is "real".

It is real for as long as you DO NOT bring the clocks together in a single inertial refrence frame. Buit if you bring them together they MUST show the same time, unless there is a reason not covered by the Lorentz transformation why they must disagree.

To conclude: Do you agree with the following description?
“The entire trip of 30,100 light years distance will require 30,102 years as measured on Earth; but as measured on the starship it will require only 20 years. In accordance with Einstein’s laws of special relativity, your ship’s high speed will cause time, as measured on the ship, to “dilate” and this time-dilation (or time warp) in effect, will make the starship behave like a time machine, projecting you far into the Earth’s future while you age only a modest amount.”

[tomclarke]

The only definite way you can answer which clock is running slower is to compare them at two different times and see which has larger elapsed time. That means bringing clock "up there" "back here".

If you want to make this a precondition go ahead but it is not required, since ALL clocks at ALL positions within any inertial reference frame keep the same time. WE know the latter is not the case: The only problem has been that this theory has not always been correctly interpreted: Even by Einstein himself!

You do not have to bring the clocks together to know that THIS MUST KEEP THE SAME TIME, snce if it is NOT SO, then different events at different positions within an inertial reference frame CAN NEVER OCCUR SIMULTANEOUSLY.

This then makes mockery of the reality that two events that are simultaneous within an inertial reference frame no.1, TRANSFORM into two events which are NOT simultaneous within another inertial refrence frame no. 2. The stupid mistake has been made to conclude that this implies that the two simultaneous time coordinates within FOR1, which give different time coordinates within FOR2 demands that the clocks at the transformed positions within FOR2 are out of synchronisation. THIS IS NOT THE CASE!!! It only means that one of the simultaneous events in FOR1, occurs at an earlier time within FOR 2, than the other simultaneous event in FOR 1 occurs in FOR2.

When the first simultaneous event occurs within FOR2, ALL the clocks within FOR2 are showing the exact same time no matter where they are situated within FOR2. When the second simultaneous event within FOR1 occurs at a later time, ALL the clocks in FOR 2 are showing this later time. It is thus absurd to reason that there are two clocks at the positions in FOR2 where the first and second event occur WHICH show "simultaneously" different times. Only a typical deluded mathematiocian will think that this mathematically convenient picture is what is actuall happening physically.

You are however now agreeing the elapsed time must be different, and hence that time dilation is real.

I have NEVER disagreed that the transformed time coordinate is commensurste with time dilation WITHIN THE REFERENCE FRAME INTO WHICH THE TIME IS BEING TRANSFORMED. I HAVE CONSISTENTLY DISAGREED THAT THE TIME ON THE MOVING CLOCK IS ACTUALLY SLOWER THAN THE TIME ON AN IDENTICAL CLOCK WITHIN THE REFERENCE FRAME INTO WHICH THE TIME COORDINATE OF THE RELATIVELY MOVING CLOCK HAS BEEN TRANSFORMED.

Thus:

(1) keep the clock up there for a long time, till the accumulated SR error is 10s.

(2) Bring the clock back. The displacement doing this is < 1light-second so changes in time relative to its observed time at distance are bounded by this. Thus the difference is still at least 9s.

If this actually happens, it would mean that the Lorentz transformation MUST be wrong since as I have shown time and again on this thread, within each and every inertial reference frame the time is THE SAME at EVERY position. Thus in terms of the Special Theory of Relativity, the clocks will NOT DISAGREE.

[The 9s is "real".

It is real for as long as you DO NOT bring the clocks together in a single inertial refrence frame. Buit if you bring them together they MUST show the same time, unless there is a reason not covered by the Lorentz transformation why they must disagree.

To conclude: Do you agree with the following description?
“The entire trip of 30,100 light years distance will require 30,102 years as measured on Earth; but as measured on the starship it will require only 20 years. In accordance with Einstein’s laws of special relativity, your ship’s high speed will cause time, as measured on the ship, to “dilate” and this time-dilation (or time warp) in effect, will make the starship behave like a time machine, projecting you far into the Earth’s future while you age only a modest amount.”

Johan, I defy anyone, reading the above, to understand what you mean.

Are you saying that the travelling clock, brought back to its twin, will never show different time? In which case you cannot explain the GPS satellite consistent error/day which over time will make an error much larger than any "observational" correction (limited to light speed between clocks) could account for.

Your argument here is suspiciously similar to to your justification for superconductivity.

You assert strongly an incorrect theoretical premise which implies your result, and then argue that your result must be true, saying that experiment is unnecessary to verify this!

In the superconductivity case your premise is that field in your diamond must be zero.

In this case your premise is that:

And this time is the same within ALL inertial reference frames. If this is not the case then Einstein's first postulate (the principle of relativity) is wrong and therefore the whole theory of Special Relativity will be wrong.

You have absolutely no justification for this statement. Your reason for accepting it (I think) is your confused notion of "keeping the same time" which has no clear physical definition. This lack of clarity carries over to all your otehr argument.

Best wishes, Tom

[johanfprins]

"Ticking at a slower rate" is of course, hopelessly confused verbiage.

Not confused verbiage but proof that your brain is muddled. If two clocks do not keep time at the same rate the "seconds" on one clock will tick pass slower than the "seconds" on the other clock.

Everyone knows the whole notion of clocks is that they are measures of the passage of time in the various frames.

Really? Then why do you not know that this is so and that therefore two identical perfect clocks moving with a speed v relative to one another MUST according to the Lorentz transformation keep exactly the same time within their respective inertial reference frames? This is also demanded by Einstein's "principle of relativity".

If you can agree, that the clocks when rejoined no longer show the same time,

If they do show different times it will violate the Lorentz transformation. Then either the Lorentz transformation must be wrong, or the times are different for another reason which has nothing to do with the Special Theory of Relativity.

then you're not "correcting" Einstein.

I am correcting Einstein's interpretation of time dilation.

Likewise, your confused language

The confusion is in your head since you just do not have enough synapses that can fire simultaneously

He seems to think despite the evidence, that you are right and Einstein was wrong.

I did not convince DeltaV since he has enough grey matter to have seen the actual physics involved long ago.

That would be now 2 people on the planet. Could be worse I suppose.

There are many more physicists of good standing that disagree with the mainstream interpretation of the Special Theory of Relativity. Most of these superior physicists have, however, learned the hard way to keep quiet or else they are branded as "crackpots" by the real crackpot-morons like you.

[tomclarke]

Not confused verbiage but proof that your brain is muddled. If two clocks do not keep time at the same rate the "seconds" on one clock will tick pass slower than the "seconds" on the other clock.

This is circular. Define "pass slower" in a way which is in principle measurable, and unique. I bet you cannot do it!

[johanfprins]

This is circular. Define "pass slower" in a way which is in principle measurable, and unique. I bet you cannot do it!

Oh Tom! I thought that in contrast to GIThruster you have some gey matter. To read time on any clock you must have an interface. The simplest since clocks were first designed and built is a watch-arm that turns around through an angle of 2*pi after the arm 12 hours has passed. If the one clock keeps slower time than the other, its arm will only complete 2*pi after the faster clock has already done so.

Niow you are probably arguing that one cannot "compare" two clocks when they are moving relative to one another. In fact you can since you can, for example, send a light beam from one clock to the other and calculate the times on the clocks by means of the Lorentz transformation to prove that they are keeping the same time within their respective reference frames. This is, however, superfluous since Einstein's "principle of relativity" and the Lorentz transformation demand that the clocks MUST keep the same time within their respective reference frames. If this is not the case, then the Theory of Special Relativity MUST be wrong. And we have enough evidence that it is not wrong. So there is no reason in terms of the Lorentz transformation that the clocks will ever show different times, as long as they did not experience different gravitational forces.

[tomclarke]

This is circular. Define "pass slower" in a way which is in principle measurable, and unique. I bet you cannot do it!

Oh Tom! I thought that in contrast to GIThruster you have some gey matter. To read time on any clock you must have an interface. The simplest since clocks were first designed and built is a watch-arm that turns around through an angle of 2*pi after the arm 12 hours has passed. If the one clock keeps slower time than the other, its arm will only complete 2*pi after the faster clock has already done so.

Niow you are probably arguing that one cannot "compare" two clocks when they are moving relative to one another. In fact you can since you can, for example, send a light beam from one clock to the other and calculate the times on the clocks by means of the Lorentz transformation to prove that they are keeping the same time within their respective reference frames. This is, however, superfluous since Einstein's "principle of relativity" and the Lorentz transformation demand that the clocks MUST keep the same time within their respective reference frames. If this is not the case, then the Theory of Special Relativity MUST be wrong. And we have enough evidence that it is not wrong. So there is no reason in terms of the Lorentz transformation that the clocks will ever show different times, as long as they did not experience different gravitational forces.

I'm awaiting your (thought) experiment to compare these clocks, Johan.

[johanfprins]

I'm awaiting your (thought) experiment to compare these clocks, Johan.

You can just send a light pulse from the one clock to the other where it is reflected back, and then apply the Lorentz transformation to derive the exact times on both clocks WITHIN THEIR RESPECTIVE REFERENCE FRAMES, at the instant that the pulse was sent from the one clock.

It involves quite a bit of algebra which you will probably not be able to follow. I am stating the latter since you have not yet been able to follow the much simpler algebra that I have already posted above and which proves that the two clocks keep the same time WITHIN THEIR RESPECTIVE REFERENCE FRAMES since their transformed times do not show ANY time dilation. Thus, I think you must first try to master the simpler case, which I have already posted (if you are capable of doing so) before I present more complicated situations.

So let us start again with two clocks passing each other with a relative speed v. Let us choose the the inertial reference frame Ks as the one that is "stationary" and in which we are the observers, and the inertial reference frame Kp as the one moving past and within which the "moving" clock is stationary. When the origins of Ks and Kp, at which the two clocks are situated respectively, pass each other we synchronise the clocks so that at this instance ts=tp=0.

Now the first question: How far will the "moving" clock be from us when our "stationary clock reads a time ts>0.
Do you agree that within our reference frame (Ks) the "moving" clock will be a distance xs from our stationary clock where xs=v*ts? Yes or No?

Do you agree that both clocks still keep time after they have been synchronised? Yes or No?

Now consider any time tp>0 on the moving clock within its own inertial reference frame Kp, within which it is situated at xp=0.

Do you agree that there must be transformed coordinates tps and xps of the coordinates tp>0 and xp=0 of this clock within Ks? Yes or NO?

Do you agree that these transformed coordinates follow from the Lorentz transformation as:

xps=(gamma)*(0+v*tp) and tps=(gamma)*(tp+0) Yes or No?

Do you agree that for ANY VALUE of the distance xps, there will be a value for the transformecd time tps within K?

So let us choose to calculate the transformed time tps when xps=xs. Then according to the Lorentz transformation above, one has that:

xps=v*tps=xs: which according to the equation above means that tps=ts. There is no time dilation of the transformed time at all.

You can now use the reverse Lorentz transformation to transform the time on the clock ts into the time tsp it has within Kp, and then you will find that tsp=tp. Again there is NO TIME DILATION. From the symmetry it is clear that ts=tp. Thus the two clocks must keep the same time within their respective reference frame.

Time dilation only occurs when the time on a clock within Kp, which IS NOT situated at xp=0, is transformed into Ks AND THIS IS ALWAYS ACCOMPANIED BY A LENGTHENING (never A CONTRACTION) OF THE POSITION COORDINATE OF THE "MOVING" CLOCK WITHIN Ks. If you ignore the position coordinate when doing a time transform, as Einstein has done, you obtain rubbish. Similarly if you neglect the time coordinate when transforming the length of a rod YOU ALSO OBTAIN RUBBISH. The latter rubbish is the conclusion that the length of a rod that is stationary within Kp contracts within Ks. So when applying "time-dilation" as if one should not also consider the position coordinate at the same time, you derive NONSENSICAL physics like the TWIN PARADOX.

Added comment: You do not even have to do the Lorentz transformation to conclude that ts=tp since the same distance D between the clocks is given by v*ts as well as v*tp. This madates that ts MUST be exactly the same as tp. The clocks carried by the two twins show the same time ALL the time.

[GIThruster]

Just curious, Johan; what was the other online forum you caused a stir and were "teaching physics" as regards the Twins Paradox?