Room-temperature superconductivity?

[ScottL]

Isn't SR an effect on the observed time from the satellite clock to Earth and not on the actual time of the satellite clock it's self. Every lecture I've read says SR relates to the observed time, not the actual time of the satellite clock. We can either adjust the satellite clock daily causing it to get ridiculously ahead in it's actual time, or we can mathematically make up for it by equation on Earth (GPS receivers).

[tomclarke]

You say you are arguing the facts as you undersdtand them: Then PLEASE tell me when (delta)tes=(gamma)*(delta)ts, is the the time interval (delta)ts on the clock within the sattelite dilated or is it the transformed time interval (dela)tes of the time interval (delta)ts on the satttelite that is dilated? Do you really think that (delta)ts and (delat)tes are the same as you are claiming that they are?

So, I think the LT gives the relationship between (proper = clock) time in the two frames. In the case of two frames this is frame-dependent, because two clocks will never meet again. Thus each frame can claulate that the other clock goes slower perfectly consistently.

In the case of a twin who returns (or a periodically oscillating satellite, which is essentially the same) the clocks return and can therefore be directly compared. In this direct comparison a moving (and frame changing) clock will experience time dilation relative to a stationary (single frame) clock, as mandated by the LT which can be applied directly in the rest frame of the stationary clock.

The apparent contradiction, that LT applied similarly in the frame of the moving clock leads to opposite results, is not real. That is because the moving clock must change its frame/. The process of changing the frame alters the (LT-calculated) time of the stationary clock within the moving clock's frame. Thus the LT provides a precise linkage between time in the two clock frames subject to the following provisos:
(1) times of distant clocks are frame dependent, so that the time difference varies according to which frame is used.

(2) bringling the clocks together means that at least one clock must change frame. As the frame is changed so the LT-calculated time of the far clock also changes, in such a way as to make the final result consistent with the (much simpler) LT calculation in the constant frame of the non-accelerating clock.

The subtlety here is that the LT-calculated times do correspond to proper time, but when the clocks are separated this correspondence is frame-dependent, hence the two clock rest frames give different answers. However when the clocks are brought together the calculated time is frame-independent since proper times can be precisely and umabiguously compared.

The frame-dependence is simply because separated points in relativistic space do not have any frame-independent notion of global time or (equivalently) instantaneity.

A More simple example that comes from Newton: When an object with mass passes by with momentum p, does it also have momentum p within the inertial refrence frame within which this object is stationary?

No. We need to look then at the relativistically invariant 4-momentum:
(gamma*c, gamma*px, gamma*py, gamma*pz)

Obviously momentum (px,py,pz) varies with frame, since an object moving in one can be stationary in another. Which corresponds to a transformation from p=(px,py,pz) to relativistic mass. However the vector which describes both p & relativistic mass is covariant and (in geometric terms) constant.

[tomclarke]

Isn't SR an effect on the observed time from the satellite clock to Earth and not on the actual time of the satellite clock it's self. Every lecture I've read says SR relates to the observed time, not the actual time of the satellite clock. We can either adjust the satellite clock daily causing it to get ridiculously ahead in it's actual time, or we can mathematically make up for it by equation on Earth (GPS receivers).

Observed and actual times are linked by LT, as you say. More precisely, the proper (clock) time in the earth frame is linked to the cproper (clock) time on the sat by LT, with an additional correction for the change of frame if you consider the sat clock frame, this correction is not neded applying LT from the earth clock frame.

Proper times can be compared, because the difference between LT comparison in this way and "real" is at most a deltaT of light time from sat to earth. As time passes the dilation effect becomes much larger than this "measurement" error. Actually you can adjust for the measurement error if you are careful anyway, but for a thought experiment it is simpler to wait a long time, in which case speed of light issues are much smaller than the time dilation difference.

[johanfprins]

Johan,
With a clock that remains in orbit, it accumulates offset. So if one were to take the ground clock to the clock in orbit, and compare them side by side, the total error to the point of comparison would be that of the SR and GR components.

The GR correction is an actual correction since the clock is actually running slower on the satellite. Thus, if there were no SR correction needed, the clocks on being united will show no difference in time after this continuous correction in time rate.

But since there is an SR correction required relative to earth, even though the clock would be keeping the exact same time than a clock on earth if a gravitational correction is not required, the only difference in time when the clocks are brought together will be the adjustments that have been made for SR.

Would it not? And vice-versa as has been done with clocks that returned from orbit?

The flown clocks were not adjusted and should thus only show an offset caused by gravity and not owing to SR, since this is what Einstein's first postulate demands. If there is an offset that can be ascribed to SR then the predictions by Einstein's Special Theory of Relativity will be wrong.

Then from that point on, the clocks would track together, but with an offset (GR+SR) incurred during the initial period of seperation, yes?

Only the offset required for SR since only this offset is required as measured from earth. It is NOT required because the clock on the sattelite itself is losing time; as in the case of gravitation.

I am still unsure how you think that the SR component is not an actual accumulation when observed functioning of exisiting orbital clocks says that it is.

The time dilation formula tells you clearly that a clock within its own inertial reference frame (within which it is stationary)t is keeping the exact same time as any other identical perfect clock that is stationary within any other reference frame. ONLY the transformed time into another inertial reference frame relative to which the clock is moving becomes dilated: Not the time on the clock itself. The formula is simple, straightforward and crystal clear on this.

I think that this is also possible to demonstrate as with two opposing orbits at the same altitude. The clocks would have experienced the same gravity history, but would accumulate error based on the closing velocity as they orbit against each other.

Sound feasible but expensive.

I also agree that the idea of a truly gravity free test also has merit. Although finding a space volume with no gravity is obviously a challenge while we are deep in the solar well.

Yopu do not need "no gravity" but only no-change in gravity for the whole experiment from beginning to end. The latter is not a challenge. I suggested a simple experiment of sending a clock at high speed from one end of a tunnel (within which the gravity does not change) to the other en and back, and repeating this until SR tells us that the time dilation must have grown to a measureable value; and then comparing it with a clock that stayed at home. My prediction derived directly from Einstein's Special Theory of Relativity is that there will be no time difference wharsoever between the clocks that can be ascribed to SR.

I also posit that something of the proposed nature of my experiment has already been done, if not directly, then as a data analysis of existing flown clocks. I will look around for it.

If the data involved the mixing of gravitational changes and SR, forget it.

[krenshala]

Proper times can be compared, because the difference between LT comparison in this way and "real" is at most a deltaT of light time from sat to earth. As time passes the dilation effect becomes much larger than this "measurement" error. Actually you can adjust for the measurement error if you are careful anyway, but for a thought experiment it is simpler to wait a long time, in which case speed of light issues are much smaller than the time dilation difference.

Out of curiosity, you have mentioned before that the time change due to the Lorenz Transformation cannot exceed the light-time from the observer (on earth) to the observed (clock on the sat in orbit). On what do you base this? (not saying you are wrong, just curious as to the justification)

[johanfprins]

Observed and actual times are linked by LT,

There is only one time-formula that applies when a clock is moving relative to another clock and that is the time dilation formula and it has nothing to do with a special notion of local time derived from Minkowski Space.

And the time dilation formula is symmetric in that the standard clock rates on both clocks within their respective reference frames within which they are stationary are the same since they are not moving relative to these reference fames and therefore their time rates as measured relative to these reference frames are not dilated.

Dilated time only occurs when a clock is moving relative to an inertial reference frame and it is ONLY observed within the reference frame relative to which the clock is moving. Dilated time CANNOT be observed within the moving clock's own reference frame within which it is stationary SINCE THE TIME ON A STATIONARY CLOCK DOES NOT DILATE.

Get IT?

[ladajo]

Thanks for the feedback.

There is some data that exists with the type of orbits we are speaking of. The question is finding if someone has done the analysis of it to look at SR. I do know for example that their has been a analysis of an earth equatorial orbit clock in relation to the surging of clock speed around the solar orbit plane as the clock moves forward with the earth and then agaisnt the earth. This then resulting in a net velocity oscillation when compared to the sun. This analysis has shown that there is a velocity impact on the clock rate in regard to the solar orbital track. Obviously the gravity component change is negligiable given the solar distances involved for the clock as it orbits the earth.

I also know that one oculd compare the ISS clock or Shuttle clock data against existing GPS constellation clocks and subtract the predicted (and observed gravity components. I guess that one could also try to compare GPS orbit to GEO orbit clocks and again subtract the gravty component.
I am sure at some point someone has looked at this as it is simple readily available data to crunch. I just need to find it. I did find some studies previously that looked at GEO to GEO and GEO to GPS clock synch issues. And they did speak to a need to compensate for velocity differences.

Here is a link that from section 7. forward includes a good development of the LR verses SR argument. You may find it interesting. Although, to accpet some of the points, you need to accept the referenced experiments in context and validity (Keating for example).

http://metaresearch.org/cosmology/gps-relativity.asp

[tomclarke]

Observed and actual times are linked by LT,

There is only one time-formula that applies when a clock is moving relative to another clock and that is the time dilation formula and it has nothing to do with a special notion of local time derived from Minkowski Space.

And the time dilation formula is symmetric in that the standard clock rates on both clocks within their respective reference frames within which they are stationary are the same since they are not moving relative to these reference fames and therefore their time rates as measured relative to these reference frames are not dilated.

Dilated time only occurs when a clock is moving relative to an inertial reference frame and it is ONLY observed within the reference frame relative to which the clock is moving. Dilated time CANNOT be observed within the moving clock's own reference frame within which it is stationary SINCE THE TIME ON A STATIONARY CLOCK DOES NOT DILATE.

Get IT?

Johan, something is preventing you from understanding what I write.

"time dilation" is a relative phenomena. When applied to two different inertial frames, where clocks can only ever be directly compared at one point in time, it shows how a clock in one frame ticks in the coordinate system of the other frame. And of course, therefore, time dilation does not change rate of clock in its own rest frame.

I specifically said that LT relates proper (clock) time in one frame to proper (clock) time in another: specifically, it gives the rate at which proper time in one frame pases as measured in another frame. That is simple, and it is indeed counterintuitive, but correct, that it is symmetric so that in two different frames the other frame clock will be time dilated. There is no contradiction.

You are thinking that because time dilation does not affect proper time directly, it cannot affect clock relative times when clocks are brought together again.

That is because you do not understand the relationship between LT measured time and proper time. When the two clocks are colocated the two must be identical. Therefore LT measured time (with time dilation) can be used to determine the time shown by a frame-changing (moving) clock relative to a frame-constant (stationary) clock providing the frame chosen for the measurements is constant. And when the clocks are colocated again this calculated time difference is also the proper clock time difference.

That naturally rules out doing the same thing from the POV of the the frame-changing twin, but we can do this, as long as we are careful and add in a time-shift correction caused by the chnage of frame.

So:

time dilation is symmetric

twins paradox is not symmetric when one twin stays in frame, and other twin chnages frame because we must add time-shift on frame change to time dilation. Time shift is always more important than time dilation, and has opposite effect. Time shift is clearly asymetric since the stationary twin does not shift frames.

[ladajo]

I found this paper by Clifford M. Will instructive and interesting.

http://arxiv.org/PS_cache/gr-qc/pdf/9504/9504017v1.pdf

He defers to Ashby's 1993 writeup on GPS and relativity.

[pbelter]

This thread is neither news nor about room-temperature superconductivity.
Can you move to General forum?

[Betruger]

It is about room temp SC. Till someone, a proper skeptic like Tom Clarke or MSimon, or one of the loud mouths like GIT manage to show Dr Prins has nothing but vaporware with his "RTSC" materials, this thread is pertinent to Polywell. RTSC is a big deal for e.g. WB-D.

[johanfprins]

Johan, something is preventing you from understanding what I write.

I understanf perfectly well what you are writing and I also understand perfectly well that it is utter nonsense.

"time dilation" is a relative phenomena. When applied to two different inertial frames, where clocks can only ever be directly compared at one point in time,

Exactly here you are talking utter nonsense. You can compare the clock rates all the time when both clocks send out light pulses every second (as measured on each clock) and the other clock records the arrival of the light pulses. It is a simple excercise to derive from the Lorentz transformation the time difference on both clocks between the pulse a this clock sends, and the time when the pulse arrives from the other clock. If this time difference has a certain value on a clock, it proves that the other clock is totally in sync all the time, and vice versa. And it is then easilty proved that not to have an absurd situation the two clocks must keep time ALL the time at exactly the same rate within their respective inertial reference frames.

You can do this calculation for the clocks on their way out and on their way back and count the number of pulses each clock sends out and receives and you will find that they are exactly the same. Thus when the clocks meet up they must both show that exactly the same time has elapsed.

I specifically said that LT relates proper (clock) time in one frame to proper (clock) time in another: specifically, it gives the rate at which proper time in one frame pases as measured in another frame. That is simple, and it is indeed counterintuitive, but correct, that it is symmetric so that in two different frames the other frame clock will be time dilated. There is no contradiction.

With all due respect this reasoning is unadulterated claptrap. The other clock is not dilated at all. The simple fact is that:

(dilated time)=(stationary time) divided by SQRT(1-v^2/c^2)

The "other" clock is stationary within its inertial refrence frame: THIS MEANS THAT v=0 for this clock within its own inertial refrence frame. THUS IT'S TIME cannot be dilated within its own inertial reference frame. It is only slower than the statonary clock in the other inertial refrence frame which also show (stationary time) since its speed relative to its own inertial reference frame is also v=0. The reasonwhy the time for the moving clock is slower than the stationary time on the stationary clock relative to which it is moving is because the speed of this clock relative to the other stationary clock IS NOT ZERO.

You are thinking that because time dilation does not affect proper time directly, it cannot affect clock relative times when clocks are brought together again.

I do not have to "think that": The Lorentz transformation DEMANDS that.

That is because you do not understand the relationship between LT measured time and proper time.

A relationship you derive from a misinterpretation of Minkowski space.

That naturally rules out doing the same thing from the POV of the the frame-changing twin, but we can do this, as long as we are careful and add in a time-shift correction caused by the chnage of frame.

Utter nonsense.

So:

time dilation is symmetric

twins paradox is not symmetric when one twin stays in frame, and other twin chnages frame because we must add time-shift on frame change to time dilation. Time shift is always more important than time dilation, and has opposite effect. Time shift is clearly asymetric since the stationary twin does not shift frames.

The latter conclusion is in TOTAL violation of Einstein's first postulate AS WELL AS the Lorentz transformation; and must therefore be wrong.

I also think that your harping away from the Lorentz transformation to reach conclusions that violate this transformation is becoming boring. I really have better things to do on my superconducting materials and would thus prefer to answer questions on superconduction.

[tomclarke]

Exactly here you are talking utter nonsense. You can compare the clock rates all the time when both clocks send out light pulses every second (as measured on each clock) and the other clock records the arrival of the light pulses. It is a simple excercise to derive from the Lorentz transformation the time difference on both clocks between the pulse a this clock sends, and the time when the pulse arrives from the other clock.

You need to correct for the fact that the frames are moving, so two successive light pulses will travel different distances. The number of pulses received will be larger when the two clocks are moving together than when they are moving away from each other. There are just two effects here, time dilation and doppler shift.

If this time difference has a certain value on a clock, it proves that the other clock is totally in sync all the time, and vice versa.

Because of the frame movement this synchronisation is not absolute. You can synchronise the two clocks relative to either of their frames - and get different answers for each frame. You are implying here that the light pulses provide "absolute" synchronisation, when because of the doppler shift they do not. Further you are implying that the light pulse receive rate gives LT time dilation. But of course we must also take into account doppler shift.

But the frames are moving wrt each other. So the synch you get from light going one way is not the same as the synch you get from light going the otherway.

Absolutely.

Indeed
And it is then easilty proved that not to have an absurd situation the two clocks must keep time ALL the time at exactly the same rate within their respective inertial reference frames.

I have never denied this. In fact I have stated it many times. But it is not "proved", it is the definition of a clock, and so tautologous.

You can do this calculation for the clocks on their way out and on their way back and count the number of pulses each clock sends out and receives and you will find that they are exactly the same. Thus when the clocks meet up they must both show that exactly the same time has elapsed.

OK. This is your substantive argument, and where we disagree. Glad we have got here.


Call the stationary (always in one frame) clock P, the moving (frame-changing) clock Q. Call the events when they are initially & finally together A and B respectively.

Suppose P emits Np pulses of light between A & B. I agree that Q will receive Np pulses of light between A & B

Equally I agree that if Q emits Nq pulses of light, P will receive Nq pulses of light.

Also, I agree that Np & Nq represent the clock time difference from A to B on the P & Q clocks respectively.

I can't see how you get from this to saying that Np must equal Nq?

The received pulses are viewed relative to receiving clock at a rate which is altered from the transmit rate by two effects:
(1) LT time dilation
(2) doppler shift, caused by the ever-increasing or ever-decreasing separation between the clocks, which gives a doppler shift effect.

Note that doppler shift would be present in a Newtonian world. Time dilation is only present in a relativistic world.

The rate of these two effects combined is different on inbound and outbound journeys. Now, from the viewpoint of the travelling clock Q an equal number of pulses are emitted inbound and outbound (Nq/2 each way). This must be so because the two journeys take the same (Q clock) time.

From the viewpoint of stationary P the outbound journey appears to take much longer than the inbound journey, because it observes outbound pulses red-shifted (at a lower received frequency) and inbound pulses blue-shifted (at a higher received frequency). There are an equal number of outbound and inbound pulses (since an equal number were emitted) hence the outbound pulses must be received over a longer (P clock) time than the inbound pulses.

So when we do the calculation we get:

Time dilation (LT) is symmetric and each clock sees the other clock's pulses slower on both inbound and outbound journeys

Doppler shift is symmetric, on inbound journey both clocks receive pulses faster and on outbound journey both clocks receive pulses slower.

The EFFECT of doppler shift is asymmetric. Because by Q clock inbound and outbound journeys take the same time. By P clock inbound and outbound journeys take different times.

[johanfprins]

Time dilation (LT) is symmetric and each clock sees the other clock's pulses slower on both inbound and outbound journeys

Doppler shift is symmetric, on inbound journey both clocks receive pulses faster and on outbound journey both clocks receive pulses slower.

The EFFECT of doppler shift is asymmetric. Because by Q clock inbound and outbound journeys take the same time. By P clock inbound and outbound journeys take different times.

I am sorry but you are running in circles and grabbing at straws.

Let us say the two twins pass each other and set both their clocks to zero, then after a time t on the clock of twin1 the other twin2 is a distance D=v*t away from twin1 and according to twin2 measuring tp at that instant in time the other twin1 is at a distance Dp=v*tp aweay. But the distance between the twins MUST alawys be the same value so that at any instant in time on both clocks you MUST have that D=Dp, and thus that t=tp.

Now assume that before the twins parted, they decided jointly that each one of them will send a light pulse to the other after the same length of time has elapsed on their clocks. When twin1 sends his light pulse at a time t(E) the two twins are a distance D=v*t(E) from ecah other: In other words D/t(e)=v as it must be when the speed is v. But what does twin2 see?

To obtain what twin2 sees, you must Lorentz-transform the coordinates x=0 and t=t(E) when the light pulse appears: Twin2 thus sees the light pulse appearing at a distance

Dp(E)=(gamma)*vt(E)=(gamma)*D

Even though the actual distance between the twins is exactly D when the light pulse is born, twin2 sees the light pulse apperaing at a much larger distance away from him than twin1 actually is. In addition twin2 also sees the light pulse at a later time than the actual time t(E) twin1 sends it: i.e. at

tp(E)=(gamma)*t(E):

Thus, to repeat, much later than the light pulse is actually born in the reference frame of twin1. This does not mean that the cliock of twin2 is keeping slower time!!! It only means hat on his clock which is keeping the exact time than twin1's clock, twin 2 sees that light pulse appearing at a later time.

What do you get when you divide Dp(E) by tp(E)? You get that Dp(E)/tp(E)=D/t(E)=v. Thus what twin2 sees OWING to the speed v is a simultaneous magnification of the time lenght AND the distance length. This is also what twin 1 sees when observing a light pulse being emitted by twin2 at excatyly the same instant as twin1 is emitting.

This does not mean that at the instant when both twins send out a light pulse simultaneously after an agreed time t=tp=t(E) has elapsed, that they are actully further from one another than D=vt(E)=vtp(E). Thus to assume that time changes from position to position as is done when drawing a so called Minkowski diagram with time along one axis and space along a perpendicular axis, you ignore the concomitant length stretching (or dilation) and then you reach the paranormal conclusions that you are advocating on this thread.

[tomclarke]

So:

time dilation is symmetric

twins paradox is not symmetric when one twin stays in frame, and other twin chnages frame because we must add time-shift on frame change to time dilation. Time shift is always more important than time dilation, and has opposite effect. Time shift is clearly asymmetric since the stationary twin does not shift frames.

The latter conclusion is in TOTAL violation of Einstein's first postulate AS WELL AS the Lorentz transformation; and must therefore be wrong.

Would you care to substantiate this argument?

The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion.

In this case the travelling twin is shifting between frames. NOT uniform translatory motion. Other than this shift all is symmetric. So we are consistent with this

Re LT. The LT establishes a transformation between two frames, you are not applying it to relate stationary twin time in the travelling twin outbound and inbound frames at the point where it turns round. This application is asymmetric, becaue the stationary twin never alters frame.