Room-temperature superconductivity?

[happyjack27]

say you tie a rock to each end of a string, then you spin it and release it. the string will stay taught because the two ends are constantly acting on each other's inertia. now say you have each rock emit light - or lets say sound - since each one's inertia is constantly changing relative to the other, and the sound (or light) they emit is relative to their own inertial reference fame, they will each hear a constant doppler shift. or were it light, red shift.

at the end of the day, rock 1 will say to rock 2: you emitted at a constant frequency of 3.7 herz, for what i count to be 8 seconds, so you emitted a total of 3.7*8=29.6 pressure waves. and rock 2 will say: but no, i emitted at 4hz! i emitted exactly 32 pressure waves! YOU emitted only 29.6! and rock 1 will of course retort: well your clock must be off.

[johanfprins]

So if you are on board the GPS Satelite as a rider, and look at the clock output, it will look slow, just as it did on the surface of the earth prior to launch (due to the embedded adjustment)? Or will it tell you that one second is one second, and all is good?

As I have explained before: If there were no gravity difference, a person on the sattelite and a person on the earth will measure the same time rates on their respective clocks on the sattelite and on the earth. The person on the sattelite will however measure that the clock on earth has a slower clock rate within the sdattelite, and the person on earth will measure the clock on the sattelite to be slower clock rate opn erath than the clock on earth. Thus, to measure positions on earth, the person on earth will have to increase the time rate of the clock on the sattelite as if it is running slower although it is in reality NOT running slower than the clock on earth.

This is also the crux of the Twins argument. Is the rate of Cesium Decay different in orbit on the GPS bird different from the rate of Cesium Decay in the clock on the ground?

ONLY because of gravity. If there were no gravity difference the rate of Cesium Decay will be exactly the same on the sattelite than it is on earth.

To the point, if the clock rates were not different, then the corrections SR and GR would not be required.

Wrong! Ignoring GR, the clock rates will still have to be adjusted to calculate distances on earth since the transformed time rate on earth is slower, EVEN tough the real time rate on the sattelite is NOT slower than the clock rate on earth!!!!!

[happyjack27]

So if you are on board the GPS Satelite as a rider, and look at the clock output, it will look slow, just as it did on the surface of the earth prior to launch (due to the embedded adjustment)? Or will it tell you that one second is one second, and all is good?

As I have explained before: If there were no gravity difference, a person on the sattelite and a person on the earth will measure the same time rates on their respective clocks on the sattelite and on the earth. The person on the sattelite will however measure that the clock on earth has a slower clock rate within the sdattelite, and the person on earth will measure the clock on the sattelite to be slower clock rate opn erath than the clock on earth. Thus, to measure positions on earth, the person on earth will have to increase the time rate of the clock on the sattelite as if it is running slower although it is in reality NOT running slower than the clock on earth.

This is also the crux of the Twins argument. Is the rate of Cesium Decay different in orbit on the GPS bird different from the rate of Cesium Decay in the clock on the ground?

ONLY because of gravity. If there were no gravity difference the rate of Cesium Decay will be exactly the same on the sattelite than it is on earth.

To the point, if the clock rates were not different, then the corrections SR and GR would not be required.

Wrong! Ignoring GR, the clock rates will still have to be adjusted to calculate distances on earth since the transformed time rate on earth is slower, EVEN tough the real time rate on the sattelite is NOT slower than the clock rate on earth!!!!!

i have to agree w/johanfprins, here.

[johanfprins]

happyjack, that is not the question. We all agree that time rate in rest frame of clock can be so defined, in fact this is tautology.

So you agree that the time rate within different inertial reference frames MUST be the same? This is so obvious that it is not really an achievement to realise this!

Johan is saying it is also true for time in different frames. If he is not making a statement about comparison of different frames, but restating the local frame argument that is fine, but he then cannot use this to compare clocks in different frames. He does this, e.g. to argue sthe two twins must have the same age.

it cannot be otherwise since the clocks of both twins keep the same time rate!

You say Johan's asserts that Einstein's postulate proves his point. But it says nothing, nada, zilch about what happens when you compare time in different frames. That BTW is exactly what Johan's hated LT does.

It clearly states that the laws of physics MUST be the same within ALL inertial reference frames and this is NOT possible unless the time rate on all clocks are the same within their respective inertial reference frames within which each one of them is stationary!

Further, Johan will not when asked define how he compares time in different frames -

Why do you keep on blatantly lying? I have told you over and over and over and over again how to compare the times on the two clocks after thyey have been synchronised so that t=tp=0

so even if Eunstein's postulate applied across frames, Johan cannot use it till he tells us how physically he defines the comparison.

I have done this over and over and over and over and over and over and over: The distance between the clocks at any instance in time on either clock, after having set t=tp=0, is D=v*t=v*tp and therefore, for any time after t=tp=0 you mUST have that t=tp. How do YOU physically compare the clocks to conclude that they DO NOT show the same time.

I agree, Johan's arguments seem quite convincing till you connect them.

Connect them to paranormal physics I presume!

[johanfprins]

Can you show where the calculation is wrong, or do you just claim it's wrong because the results don't meet your expectations?

I must admit that I do not have the time to go through your calculation. It is unnecessarily complicated because you involve more than just the two twins and you conclu0ison is wrong.

If I had the time I would have argued this case in more detail. But it is a waste of time since we ONLY have to involve the two twins and their relative speed v when they move away from one another and their relative speed MINUS v when they move towards one another: And the Lorentz transformation when applied correctly forb this return journey proves that there will be no time difference on the clocks.

And as I have stated over and over and over ande over and over, if there is a time difference this time difference violates Einstein's postulates from which he derived the Lorentz transformation: And this must then prove that the Lorentz transformation is WRONG!! At this stage I am not yet willing to accept the latter conclusion and I thus suspect that the flying clock experiments must be flawed.

[tomclarke]

two cesium particles start off in the same inertial reference from
then they go off and do whatever
then they come back to the same inertial reference from
each one would have experienced the same accelerations, decelerations, and velocities, relative to each other. after they return to the same inertial reference frame, they will see each other as having the same age; as having the same probability of having decayed.
meanwhile while they are accelerating and decelerating and what not, they will see "redshift" or "blue shift" in their electromagnetic fields. this will be interpreted as changes in energy levels or clocks relative to them going faster or slower. and the quantum events; the quantum wave collapses observed may be different in the different reference frames.
but in the end when they meet they will be in sync. and future quantum probability wave collapses will be consistent w/the laws of probability

a plane orbiting at high altitude will not be in sync when it lands, as it has experienced less total acceleration due to gravity than a clock on the ground.

likewise a geo-sycnhronous satellite will have to compensate for the reduced gravitational acceleration it experiences relative to a clock on the surface of the earth.

i remember arguing w/my high school teacher that gravity was an acceleration, not a force, and if it was a force, why was the gravitational constant in m/s^2 and not kgm/s^2? i think einstein would've taken my side.

Happyjack, gravity is indeed an issue. the experiments quoted checking SR use two planes flying at identical height. The GPS corrections are both SR and GR, since low sats have larger GR, lower SR, high sats have reverse, the two effects can be separated.

More serious is the argument that acceleration is equivalent to gravity. This is true, in the sense that an accelerated ref frame cannot be distincguished from gravity.

HOWEVER, the time dilation effect of gravity is a property of change in gravitational potential well (phi) not of the local gravitational field. So it turns out to have no effect on radioactive particles in centrifudge, or muons in a astorage ring, where the experimental results exactly correspond to SR.

[happyjack27]

What you describe is time, not time rate (a rate being a quotient of two quantities).

a rate, more precisely, is a derivative. the rate of change of one thing with respect to another. in this case, the rate of change of "time" with respect to an observable, e.g. expected cesium decay rate.

the phrase "time rate" is used to distinguish from "time", as in e.g. "8pm"

"time" is the integral of the time rate over a region of an observable, e.g. cumulative cesium decay.

when we say "time rate", we're talking dt, as opposed to t. so when someone says, something is traveling 8 m/s. (meters per second), they are giving a rate, dx/dt. (where x is position, t is time). we are talking about the "dt" part of all such statements made with respect to an inertial reference frame - the rate at which time passes. or more concisely, "time rate".


we're dealing with what in calculus is known as "related rates" problems: http://en.wikipedia.org/wiki/Related_rates

[ladajo]

So if you are on board the GPS Satelite as a rider, and look at the clock output, it will look slow, just as it did on the surface of the earth prior to launch (due to the embedded adjustment)? Or will it tell you that one second is one second, and all is good?

As I have explained before: If there were no gravity difference, a person on the sattelite and a person on the earth will measure the same time rates on their respective clocks on the sattelite and on the earth. The person on the sattelite will however measure that the clock on earth has a slower clock rate within the sdattelite, and the person on earth will measure the clock on the sattelite to be slower clock rate opn erath than the clock on earth. Thus, to measure positions on earth, the person on earth will have to increase the time rate of the clock on the sattelite as if it is running slower although it is in reality NOT running slower than the clock on earth.

This is also the crux of the Twins argument. Is the rate of Cesium Decay different in orbit on the GPS bird different from the rate of Cesium Decay in the clock on the ground?

ONLY because of gravity. If there were no gravity difference the rate of Cesium Decay will be exactly the same on the sattelite than it is on earth.

To the point, if the clock rates were not different, then the corrections SR and GR would not be required.

Wrong! Ignoring GR, the clock rates will still have to be adjusted to calculate distances on earth since the transformed time rate on earth is slower, EVEN tough the real time rate on the sattelite is NOT slower than the clock rate on earth!!!!!

I think you misunderstood me here, or I was not clear. I fully agree that if you magically remove the GR part, there remains an SR component to correct for. As seen in the below chart. But I also re-state, if there was not a rate difference regarding the SR component, then there would be no SR correction required. If there is a rate difference, then the "age" must accumulate differently. A count is a count.
The below chart clearly shows the required SR component correction factor which is a function of speed, not gravity. I do see some merit to the argument that the speed component is actually an induced secondary "gravity" component given the Keplarian accelleration. But on that, I am not yet convinced.
If I delete or "zero out" the GR function in the Kalman filter, the SR part remains and still adjusts clock count. The clock will be correcting for that part. The count rate will still require it. That means that the clock is "aging" differently. I am not sure if we are talking past each other or not. But the rate is different. That is a fact. The "why" may be up for debate, but the "is" is not. The error is "Net", SR and GR.

I am thinking on your repeated comment that the earth clock will be seen as slower due to SR by the orbiting clock. I do not see that as so at this point. I think it would invalidate the math currenlty being used to correct between clocks not only on the surface, but also in various orbits, and non-orbits. Although, you could also argue that everything orbits everything, and orbits are either stable (repeating) or changing. After all, a moon of mars does orbit the earth in a highly eliptical path.

As I ponder this, I wonder if this is the point in the math that matters. For example, do we go around the sun, or does the sun go around us? Mathematically I can do either. It is a function of the choice of whether I assign motion to the sun, or the the earth. Ie. which I choose to keep centered in my computer screen.

[ladajo]

This link shows how the "orbit" of Juptier looks centered on Earth. There is a relativistic component to Juptiers position in Earth's gravity well as well as a speed component regarding Jupiters overall progression "around" Earth.

Scroll to the bottom to see the animation.

http://astronomy.nmsu.edu/geas/lectures ... ide01.html

[happyjack27]

in a world without gravity, the two twins DO have the same age. they have traveled the same distance, at the same speed, and have had the same accelerations and decelerations. they are mirror images of each other; the only difference between the two is one's coordinate system is orientated clockwise, and the other's counterclockwise. so the question breaks down to: which is older, clockwise or counter-clockwise? left or right? which is, of course, absurd. they are fictituous distinctions.

i think the paradox can be demonstrated with 2 cesium particles. the question being whether or not each one has decayed. and it's a shrodinger's box. while in separate reference frames, they each see each other's time going faster than they see their own. so one particle sees the other particle as having PROBABLY decayed, while itself as having PROBABLY not decayed. and the other particle sees the same thing, only reversed. so now they come back to the same inertial reference frame, what happens? schrodinger's box is opened.

and the quantum uncertainty inherent in our clock is related to the neccessarily limited information we have of the particles' spatial histories, even in a system of only two paritcles.

on that note, it would be interesting to recreate SR experiments using materials that have been quantum entangled.

[ladajo]

This is a good link to show earth centered orbits: Unfortunately it does show tracks like the first one I put up.

I think there is something to the math in regard to non-retrograde motion and determining relative relationships (who is fast or who is slow).
I would posit that maybe the sun is our local non-retrograde motion reference point. And if something is in orbit around earth, it is non-retro, thus the earth remains "ground". However that said, there is no reason one can not calculate the relativistic impact of Jupiter's relative orbit around Earth.

http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=1 ... ;cad=0#orb

[johanfprins]

on that note, it would be interesting to recreate SR experiments using materials that have been quantum entangled.

Very good! At this point I am going to let another cat loose amongst the pigeons by claiming that time does not exist within an entangled wave. Entanglement and gravity are related more than gravity is related to Special Relativity. Einstein's arguments based on Special Relativity to conclude that curved space-time is required to model gravity are fortuitous.

The REAL reason why curved space time is required, is because matter waves have complex wave amplitudes. But as long as tomclarke wants to argue against all common sense that two clocks which keep the same time rates can measure different time intervals for a path through gravity free space, we will never be able to reconcile Einstein's gravity with the wave mechanics of matter. In fact, they have been effectively reconciled since 1926 when Schroedinger formulated his differential wave equation for an electron; bar the illogical arguments that tomcklarke has been promoting with his life on this thread.

[Teahive]

In a world without gravity there would be no twins, no matter, no acceleration, no spacetime.

What you describe is time, not time rate (a rate being a quotient of two quantities).

a rate, more precisely, is a derivative. the rate of change of one thing with respect to another. in this case, the rate of change of "time" with respect to an observable, e.g. expected cesium decay rate.

the phrase "time rate" is used to distinguish from "time", as in e.g. "8pm"

"time" is the integral of the time rate over a region of an observable, e.g. cumulative cesium decay.

when we say "time rate", we're talking dt, as opposed to t. so when someone says, something is traveling 8 m/s. (meters per second), they are giving a rate, dx/dt. (where x is position, t is time). we are talking about the "dt" part of all such statements made with respect to an inertial reference frame - the rate at which time passes. or more concisely, "time rate".

Great, thanks. Of course the issue, as I said all along, is that units of time themselves are defined as observable processes, e.g. 1 s being the duration of 9,192,631,770 periods of some specific radiation. So while you count 9,192,631,770 periods of the radiation of a cesium atom at rest, the crystal in a typical quartz clock wil have oscillated 32,768 times, and the distance that light will have travelled in vacuum relative to you is 299,792,458 m by definition.

That is all fine and allows us to say that different processes always happen at specific rates relative to each other within any given inertial reference frame. However, just like the frequency of the radiation received from a moving cesium atom isn't the same as that from a stationary cesium atom, we cannot conclude that someone moving relative to us will have counted the same number of periods after a roundtrip.

Can you show where the calculation is wrong, or do you just claim it's wrong because the results don't meet your expectations?

I must admit that I do not have the time to go through your calculation. It is unnecessarily complicated because you involve more than just the two twins and you conclu0ison is wrong.

Complicated? Surely you jest. You can even drop one of the observers, A or B, completely and the result is still the same. Furthermore I'm only looking at the problem from a single frame of reference (using doppler shift to look into the other frame), and only at one leg of the journey.

If you agree that the doppler factor for an emitter approaching at v = 0.6 c is 2, i.e. the receiver measures a frequency twice as high as the frequency an observer travelling with the emitter measures, there can be no other solution than the observer changing frames measuring a shorter elapsed time.

[ben m]

Hello all,

Newbie here. I'm a regular on the James Randi Foundation forums (forums.randi.org). I just wanted to point out (and I apologize if this is inappropriate) that Johan F. Prins spent most of the past few months conducting exactly this argument on a forum over there---there was a thread on "room temperature superconductivity"; Johan showed up, and in the course of trying to explain his ideas, he had distinct trouble with the basics of energy conservation, acceleration, Gauss's Law, the definition of voltage, the Schrodinger equation, radiation from moving charges, kinetic energy, and of course relativity. He ended up accusing Hafele and Keating of fraud.

The relativity argument is uncannily familiar; I see a handful of patient and knowledgable posters giving Johan spacetime diagrams, detailed Lorentz transformations, and so on, and getting insults in return.

Over on JREF, about a week ago, Prins bailed out of the thread announcing that he had "better things to do". I'm greatly amused that his "better things to do" apparently includes doing exactly the same thing with a different audience.

Anyway: if anyone was sitting here thinking "Boy, Johan's error is so straightforward, I will be able to clear it up with one explanation and one more spacetime diagram"---well, take a look at the JREF thread (Google for it) to see how many such explanations have scrolled across Johan's browser.

Cheers!

[kcdodd]

All this argumentation has lead me to a new question for myself though. Why does vacuum energy not break the principle of relativity. I am definitely not familiar with quantum field theory. However, it seems to me that if one assigns an energy density to a space itself that seems to imply a preferred reference frame. If you write that energy density as a four-vector then there is a "rest vacuum energy" to be a Lorentz invariant, and singles out a particular frame as being special.

EDIT:

I think I answered this for myself. If the vacuum energy was actually only one component of a vacuum stress energy tensor, it could be invariant under lorentz transformations. The Minkowski metric is invariant by definition, and is a rank two tensor. So perhaps the vacuum stress-energy has the same form? Which would mean the energy would have to be accompanied by an isotropic pressure? Strange.