Room-temperature superconductivity?

[johanfprins]

Johan,

I don't think you should try to correct anything till you have attended Walter Lewi's youtube lecture and corrected your misapprehension about Doppler shift!

It did not help you out of your stupor: So why should I "attend" this lecture when I can deliver a better lecture on the Doppler effect than Walter Lewi can ever hope to do.

Now who is disrespecting MIT?

It is not disrespect if I claim that I can give a better lecture. I really have difficulty with your reasoning and logic. You state that I state what I have never stated and then you attack your own wrong conclusions.

But Johan, don't you feel this wholesale denial of standard physics because it does not fit your preconceptions is unbecoming?

Here you again demonstrate your immaturity and also the fact that you are not a REAL scientist. The supreme law of a natural scientist is to accept that there is no such thing as "standard physics". If you cannot put yourself in the frame of mind that what is considered "standard physics" today might be proved wrong tomorrow, you should get the hell out of physics. One of the supreme tasks of a real physicist is to also continually question standard dogma AND to be willing to reject or reinterpret it when an error in experiment and/or logic is found.

I am just doing my job and then you call it "the wholesale denial of standard physics". It is NOT denial when you ask: "What if that which be believe in today is wrong tomorrow; and to contemplate what could be wrong. A person like you who is not willing to think outside the official dogmatic boundaries should be barred from doing any physics; since you are the type of person who ends up burning people on the stake.

Why not take a deep breath and admit that Doppler effect does exist,

You see what I mean? I have NEVER claimed that the Doppler effect does not exist; but in order to try and score a cheap point you accuse me of something I have not stated or even reasoned in any manner. During the inquisition this was a famous technique used by tormentors to condenmn a person to be burned at the stake. "Are you a true believer? Then you know you are saved and therefore it does not matter if I wrongly burn you on the stake, does it?

So why do YOU not "take a deep breath" and at least try to apply the Lorentz transformation logically. Let me see if I can help you by at first posing a simpler problem that might be within your grasp:

A spaceship of length L within its own inertial refrence frame Kp moves with a speed v pass you (within your own inertial refrence frame K) and it switches on two lights, one in its nose and one in its tail, simultaneously.

You are at the coinciding position when the light in the nose switches on and you synchronise youir clock with a clock in the nose of the spaceship. Now the question: "When and at which position, will the tail light appear within your inertial refrence frame K?

Ah hell, you will probably get it wrong, so let me give you the answer. The position as measured from your position, where the tail light will switch on within K is given by:

x=-(gamma)*L

And the time is

t=-(gamma)*(v/c^2)*L

Note the MINUSSES. Firstly the tail end switches on when it is further away from you within K than the actual length L of the spaceship. Secondly, the tail end switches on BEFORE the nose light is seen switching on (by you) at your own position. Does this mean that the tail light anticipated (ahead of time) that the noselight will switch on at your position? Obviously this cannot be so since this will be Voodoo physics: However, if you assume that the different times at position x=0 and x=(gamma)*L are time coordinates within a 4D manifold, then you have to conclude that the light in the tail anticipated that the light in the nose is going to switch on and you will then reach a Voodoo conclusion, just as you are reaching when you conclude that one twin will age less than the other twin.

The more logical conclusion to reach is thus that the different times are NOT time coordinates within a 4D manifold and that therefore you cannot interpret the effect of the Lorentz transformation as following different paths through a 4D space-time manifold with different "proper times". The latter concept is highly "improper" when there is no gravity present.

[tomclarke]

A spaceship of length L within its own inertial refrence frame Kp moves with a speed v pass you (within your own inertial refrence frame K) and it switches on two lights, one in its nose and one in its tail, simultaneously.

I have answered every one of your arguments, and will this one. But it rests on your initial condition above. This is ambiguous. Simultaneously in which frame? The spaceship Kp or the observer K?

[johanfprins]

If they only use the data from when the aircraft are at altitude, how does the gravity change (other than the normal differences found on the planet from point to point)?

You have to use ALL the data and changes that happenned at every second during the flights. It is well known that in order to save fuel, aircraft tend to fly a near parabolic path if they can by accelerating up and then decelerating down. AS far as I remember, in the flying clock analyses they used average values for speed. This is obviously nonsensical. What happens during acceletration and deceleration? Does SR apply or does Einstein's formula for time dilation caused by acceleration apply?

Conceptually, Johan, I think the test you are arguing for is the comparison of a GEO based clock against a surface clock. Although I will have to think a moment about how they adjust for Earth Centered.

No it is not the same since you obviously DO require an SR correction for GPS satellites EVEN though the cl;ock on the sattelite DOES NOT keep dilated time ON THE satellite. There is thus no argument here.

The argument is when you let a clock go on a long fast journey and bring it back and then compare to a clock that styed behind. This is a TOTALLY DIFFERENT experiment than when the clocks are not brought together again.

But, in the proposed setup, neither clock is moved once on station. Thus both have no change in gravity (other than the GEO clock moving closer and farther from the sun than the surface clock does as it orbits). The surface clock is moving at earth surface speed as the earth turns, covering the 25K miles around the equator at about 0.289 miles per sec. The GEO bird is moving at its orbit speed to hold station with the surface clock, covering 165K miles on each rotation at about 1.91 miles/sec. So from an Earth centered frame, they are moving at different speeds, and thus would incur an SR difference?

Yes as long as they are not brought together the SR transformed time will be dilated both ways. But the actual time on the clocks will not be dilated.

Curt Renshaw argues that the effects seen are based on the rotational aspect of the motion, and that is is essentially equal to an accelleration (like gravity). He points to the spinning disk clock experiments that induce clock error with high speed centrifuge type set ups.
He also points out that the Keating experiment was done only as an after the fact comparison of clock data, vice monitering while in flight, and also was corrected to Earth Centered.

Exactly. It proves that the flying clock experiments are not clearcut evidence and we should have two clocks moving relative to one another along a linear path, without any change in gravity, and then returned along the same linear path. If there is a time difference this will not be caused by SR, but can then only be caused by the acceleration and decelleration.

http://renshaw.teleinc.com/papers/london1/london1.stm

I also liked this article: Optical Clocks and Relativity
Found here:
http://tf.boulder.nist.gov/general/pdf/2447.pdf
where they explore relativistic issues using optical clocks.

I also found this one interesting (older article) where the author discusses issues regarding the reformulation of relativistic transformation between coordinate time and atomic time:

http://ipnpr.jpl.nasa.gov/progress_repo ... 21/21E.PDF

Thank you for all the references but I really have to concentrate on my room-temperature superconducting devices.

[tomclarke]

Now let us return to my argument above. You said it was wrong because Doppler did not apply to the pulse rate (frequency).

You now agree that received pulses from pulsars have pulse frequency which varies with the relative speed between pulsar and earth according to the Doppler effect. Otherwise a lot of radio astronomers are going to argue with you. And I bet they have observed more pulsars than you!

Therefore the same must be true of your synchonising light pulses. So in the twins example my argument above shows that the asymmetry comes from the fact that the outbound/inbound journey switches round at the travelling twin end.

Received pulses (by the travelling twin) become higher frequency at that time. Transmitted pulses (from the travelling twin) are received at a higher freqency some time later, because they must travel to the stationary twin.

The received pulse frequency, transmitted 1s local, is affected by affected by both time dilation (symmetric) and Doppler (asymmetric) We can see that the total number of pulses from the travelling twin must be less than from the stationary twin, because the stationary twin's pulses are received at low (red shifted) frequency for a shorter time. And we both agree that number transmitted must equal number received throughout the journey.

[johanfprins]

A spaceship of length L within its own inertial refrence frame Kp moves with a speed v pass you (within your own inertial refrence frame K) and it switches on two lights, one in its nose and one in its tail, simultaneously.

I have answered every one of your arguments,

You have NOT! All you have done has been to spout Voodoo!

But it rests on your initial condition above. This is ambiguous. Simultaneously in which frame? The spaceship Kp or the observer K?

My God! Are you REALLY so slow? I have obviously stated that the lights are switched on simultaneously on the spaceship and the spaceship is obviously stationary within Kp. GET IT??

[johanfprins]

Now let us return to my argument above. You said it was wrong because Doppler did not apply to the pulse rate (frequency).

You are lying again: I DID NOT SAY THAT DOPPLER DOES NOT APPLY. The Doppler effect is inherent within the Lorentz transformation and is thus included within the equations which you are incapable of deriving for yourself. To again hold your hand let me give you one of the formulas:

After synchronisation of their clocks the twin in K sends out a light pulse at time t=t(E) and the time at which twin2 receives the pulse within Kp, is tp(A) where

tp(A)=(1+v/c)*gamma*t(E)

Do you know why you have to add v/c to unity in the brackets? Let me help you further: BECAUSE OF THE DOPPLER EFFECT.

SO PLEEEEAAAASE STOP SPOUTING NONSENSE and try and do the rest of the calculation.

[tomclarke]

does Einstein's formula for time dilation caused by acceleration apply?

An easy one. No. Gravitational time dilation correlates with phi (gravitational potential) not with dphi/dx (gravitational force).

There have been experiments done to verify that acceleration does not cause time dilation. So this is theory backed by experiment.

Here is somone else presenting the arguments, and summarising the experimental evidence.
http://van.physics.illinois.edu/qa/listing.php?id=1360

Johan - I like to work things out myself too. But only a fool would not stop and check again if they found other people had arguments contradicting their conclusions.

BTW - even if there were time dilation due to gravitation it is irrelevant for long plane journeys where the accelerate/decellerate time is insignificant compared with the crusiing time.

Best wishes, Tom

[tomclarke]

Now let us return to my argument above. You said it was wrong because Doppler did not apply to the pulse rate (frequency).

You are lying again: I DID NOT SAY THAT DOPPLER DOES NOT APPLY. The Doppler effect is inherent within the Lorentz transformation and is thus included within the equations which you are incapable of deriving for yourself. To again hold your hand let me give you one of the formulas:

After synchronisation of their clocks the twin in K sends out a light pulse at time t=t(E) and the time at which twin2 receives the pulse within Kp, is tp(A) where

tp(A)=(1+v/c)*gamma*t(E)

Do you know why you have to add v/c to unity in the brackets? Let me help you further: BECAUSE OF THE DOPPLER EFFECT.

SO PLEEEEAAAASE STOP SPOUTING NONSENSE and try and do the rest of the calculation.

Excellent. We agree.

Note that the LT has two components:
gamma (time dilation)
(1+ v/c) (Doppler effect).

My argument applies: the received pulses from the outbound journey are lower frequency than from the inbound journey, but outbound and inbound is defined at the travelling twin end, and so asymmetric. Hence the stationary twin receives low frequency pulses for longer than the moving twin. Otherwise all is symmetric, including the frequencies of the pulses, high and low (which must be the same at the two ends).

Specifically the high frequency is f0/[gamma(1 -v/c)], the low frequency is f0/[gamma(1+v/c)]

Best wishes, Tom

PS - this argument - Johan will approve - comes from me directly - though I am sure others have used it in the past.

[johanfprins]

Johan,

again, could you please explain which two quantities the "time rate" you are talking about is a quotient of? Without an unambiguous definition any statement using it will not be unambiguous, either.

Do you not know what time rate is? It is defined in terms of the periodic rate at which certain well known physical processes change. Initially it was defined in terms opf the total time it takes the sun to "circle" the earth. i.e. in terms of the time it takes the earth to complete a single revolution (if you are not a flat earther).

Since then we have found that the earth's revolutions are not reproducible enough and now use atomic transitions to define time rate. Thus time rate is defined in terms of the laws of physics and the laws of physics are the same within any and all inertial reference frames: Therefore an atomic clock will keep the SAME time rate within each and every inertial reference frame: Therefore when time interval (delta)t passes within an inertial refrence frame K, the clock within an inertial reference frame Kp will show that exactly the same time interval has passed. If NOT then Einstein's first postulate must be wrong. How many times must I repeat this simple logic?

I did the calculation. The result is that the observers disagree on the amount of time which elapsed during the journey, which is consistent with the travelling person experiencing a shortening of the distance during travel.

Then you have done the calculation incorrectly: Paste you calculation, which must be quite a number of pages long if you did the calculation correctly, and I will show you where you have gone wrong.

The calculation is actually very simple, and it's already in this thread:
viewtopic.php?t=2137&start=1610[/quote] Your calculation is wrong. They cannot differ since their respective clocks will measure exactly the same timespan to move from here to another position in space with a speed v.

[johanfprins]

Excellent. We agree.

Note that the LT has two components:
gamma (time dilation)
(1+ v/c) (Doppler effect).

My argument applies: the received pulses from the outbound journey are lower frequency than from the inbound journey, but outbound and inbound is defined at the travelling twin end, and so assymetric.

No it does not apply. If you calculate the number of pulses sent out and received by both twins during the whole round trip journey; also taking the changes in length into account which you are ignoring, then you will find that they are equal and that therefore the clocks MUST show the same time when they meet up again. Do the calculation; and if you cannot, then find something else to do than physics!!

[tomclarke]

Therefore when time interval (delta)t passes within an inertial refrence frame K, the clock within an inertial reference frame Kp will show that exactly the same time interval has passed. If NOT then Einstein's first postulate must be wrong. How many times must I repeat this simple logic?

Question: how in this case do you define "same time interval". Relative to which frame? You are making a comparison between the rates of two clocks in different frames. fair enough, but you need to make explicit how you are synchronising them (relative to which frame)?

Best wishes, Tom

[tomclarke]

Excellent. We agree.

Note that the LT has two components:
gamma (time dilation)
(1+ v/c) (Doppler effect).

My argument applies: the received pulses from the outbound journey are lower frequency than from the inbound journey, but outbound and inbound is defined at the travelling twin end, and so assymetric.

No it does not apply. If you calculate the number of pulses sent out and received by both twins during the whole round trip journey; also taking the changes in length into account which you are ignoring, then you will find that they are equal and that therefore the clocks MUST show the same time when they meet up again. Do the calculation; and if you cannot, then find something else to do than physics!!

Johan: do you agree:

(1) at each end frequency received is low first then high for outbound and inbound journey.
(2) low, high frequencies are same the two ends as stated above
(3) the number of pulses received at low frequency by the stationary twin is larger than the number of pulses received at low frequency by the moving twin. Conversely the number of high frequency pulses received is smaller for stationary compared with moving.

If not, which do you disagee with and I will give reasons again?

If you do agree with the above my result follows pretty immediately. Again if you can't see this I'll expand the reasoning.

[johanfprins]

Therefore when time interval (delta)t passes within an inertial refrence frame K, the clock within an inertial reference frame Kp will show that exactly the same time interval has passed. If NOT then Einstein's first postulate must be wrong. How many times must I repeat this simple logic?

Question: how in this case do you define "same time interval". Relative to which frame? You are making a comparison between the rates of two clocks in different frames. fair enough, but you need to make explicit how you are synchronising them (relative to which frame)?

Best wishes, Tom

Do you not know how you synchronise two passing clocks at a joint instant in time within both inertial reference frames so that they show the same time afterwards ? All elementary books on SR tell you how. After synchronisation, the two clocks must keep on showing the same time since they are keeping the same timev rate as demanded by Einstein's first postulate: Thus the clocks must stay synchronised during the whole in and out journey no matter how far they are apart. Therefore they CANNOT show different times when they get together again.

[tomclarke]

Therefore when time interval (delta)t passes within an inertial refrence frame K, the clock within an inertial reference frame Kp will show that exactly the same time interval has passed. If NOT then Einstein's first postulate must be wrong. How many times must I repeat this simple logic?

Question: how in this case do you define "same time interval". Relative to which frame? You are making a comparison between the rates of two clocks in different frames. fair enough, but you need to make explicit how you are synchronising them (relative to which frame)?

Best wishes, Tom

Do you not know how you synchronise two passing clocks at a joint instant in time within both inertial reference frames so that they show the same time afterwards ? All elementary books on SR tell you how. After synchronisation, the two clocks must keep on showing the same time since they are keeping the same timev rate as demanded by Einstein's first postulate: Thus the clocks must stay synchronised during the whole in and out journey no matter how far they are apart. Therefore they CANNOT show different times when they get together again.

Yes, you can synchronise at one instant, when the two clocks are colocated.

but to compare a time interval you also need to synchronise (compare times) at another instant, when the clocks are not colocated. That second synchronisation depends on frame.

To elaborate: "rate of time" comparison requires comparing times (same as synchronising) at two different instants.

[johanfprins]

[Johan: do you agree:

(1) at each end frequency received is low first

Wrong!

then high for outbound and inbound journey.

The received frequency at both ends is different for outward and inward journeys but the emitted frequencies stay exactly the same.

(2) low, high frequencies are same the two ends as stated above

Received frequencies are different for outward motion than for inwrad motion, but the emitted frequencies stay the same all the time.

(3) the number of pulses received at low frequency by the stationary twin is larger than the number of pulses received at low frequency by the moving twin.

There is no stationary twin since they are moving relative to one another, and since there is no uniquely stationary reference frame relative to which you can say either one of the twins is stationary.

The total number of pulses sent out by each twin is exactly the same at the end of the journey. For God's sake please do the full calculation as I have been asking you all along. Stop thinking in terms of Galileo's transformation.

Please stop wasting my time and try to THINK.

Best regards,
Johan