Room-temperature superconductivity?

[johanfprins]

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.

You do not need to synchronise them again since according to Einstein's first postulate they must remain synchronised after the first colocated synchronisation.

That second synchronisation depends on frame.

Why do you want to synchronise again when Einstein's first postulate demands that they must stay synchronised no matter how far they move from one another? Don't you believe in Einstein's postulates from whioch he derived teh Lorentz transformation?

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

Yes, within the same reference frame. Whether the clocks are synchronised or not, they must thus keep the same time rate within their respective inertial reference frames as demanded by Einstein's first postulate.

Let me repeat again, and again, and again: The laws of physics are the same within both inertial reference frames and the same periodic physical process that is used to define time rate within one of the inertial reference frames will give the exact same time rate within the other inertial reference frame; even if the clocks have not been synchronised at all. Thus one synchronisation is enough to keep them synchronised afterwards.

[tomclarke]

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

Yes, within the same reference frame. Whether the clocks are synchronised or not, they must thus keep the same time rate within their respective inertial reference frames as demanded by Einstein's first postulate.

Sorry Johan, which is "the same" reference frame? We are comparing two frames. The initial synchronisation was in either, with same results because the two clocks were colocated. The second one, needed to compare "rate of time" must be in one or other, or some different frame. The results will depend on which frame is chosen, because the clocks have moved away from each other.

Why do you want to synchronise again when Einstein's first postulate demands that they must stay synchronised no matter how far they move from one another? Don't you believe in Einstein's postulates from whioch he derived teh Lorentz transformation?

I don't, but you do, when you talk about two clocks having "the same rate of time" and being in different frames. The phrase only has meaning, physically, if you do these two synchronisations. Thus its physical meaning depends on the frame in which the synchronisations are done.

[johanfprins]

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

Yes, within the same reference frame. Whether the clocks are synchronised or not, they must thus keep the same time rate within their respective inertial reference frames as demanded by Einstein's first postulate.

Sorry Johan, which is "the same" reference frame? We are comparing two frames.

Can you not understand simple mathematics and what it means to choose an origin? We do not HAVE to compare time rate within two reference frames after we have chosen the origin for time and position measurements simultaneously within both reference frames. If, after this synchronisation you want to caompare the time rate on clocks within the two inertial refrence frames, YOU HAVE TO RE_SITE THE ORIGIN AND THIS REQUIRES THAT THE LORENTZ FORMULAE THEMSELVES MUST BE TRANSFORMED TO BE VALID FOR THIS NEW ORIGIN!!!!

The initial synchronisation was in either, with same results because the two clocks were colocated.

Correct, since you simultaneously chose an origin from where you measure distance and time; since this is what has been done when the Lorentz transformation equations were derived. If you want to compare the same two clocks again after their first synchronisation, you are forced to choose another origin for position and time. The Lorentz formulas themselves must then be transformed to be valid for this new origin. This is a long and lengthy calculation. It is easier to just note that the distance between the two origins demands that the clocks at the two origins MUST keep the same time after synchronisation within their respective inertial reference frames: i.e that D=v*t=v*tp and therefore t MUST be equal to tp all the time after synchronisation.

The second one, needed to compare "rate of time"

Such a measurement is not necessary since Einstein's postulates tell you that the rate of time on ALL clocks in the universe, as measured within the inertial reference frames wiohin which each clock is stationary, must be exactly the same for all these clocks. And the fact that there can only be single distance D between the clocks at any time on either one of the clocks confirms from D=v*t=v*tp the time rates are teh same also after synchronisation and that Einstein's first postulate is thus correct. This is a precondition for Special Relativity.

The results will depend on which frame is chosen, because the clocks have moved away from each other.

No it does not! Only if you do it in a stupid manner and do not remember that with a second synchronisation of two clocks within the two reference frames you are changing the origins used for both position and time within the two inertial refrence frames; and must thus recalculate all the transformations in terms of these new origins. If you ignore the latter transformation of the Lorentz formulas in terms of two new origins, you spout Voodoo as you have done all along on this thread.

Why do you want to synchronise again when Einstein's first postulate demands that they must stay synchronised no matter how far they move from one another? Don't you believe in Einstein's postulates from whioch he derived teh Lorentz transformation?

I don't, but you do, when you talk about two clocks having "the same rate of time" and being in different frames.

NO that is a given that it is so: Synchronisation is only required so that the two clocks show the same time afterwards NOT to ensure that they keep the same time-rate as they must according to Einstein's first postulate. Even if you do not synchronise them, they will still keep the same time rate. Synchronisation HAS nothiong to do with equalising time rates within the two inertial reference fames. There is no reason to equalise time rates because they ARE always equal: This is a given that it must be the same owing to Einstein's first postulate.

The phrase only has meaning, physically, if you do these two synchronisations.

BS! What has synchronisation of two clocks to show the same time have to do with time-rate?

Thus its physical meaning depends on the frame in which the synchronisations are done.

When you synchronise two clocks within two inertial reference frames you set the time on the two clocks to be the same within BOTH inertial reference frames. You do not only have to do it when they co-locate: When they do not co-locate when you synchronise the Lorentz formulas change to take the absence of co-location of the time origin with the position origin into account. This complicates the mathematics but does not change the physics-fact that all clocks keep exactly the same time rate within a gravity-free space: No matter how far they are from one another and no matter with what speed they are moving relative to one another. So how is it possible to synchronise two clocks, which are stationary within their respective reference frames, to be only synchronised within one of these reference frames? Stop spouting paranormal metaphysics!

[tomclarke]

Two comments Johan:

(1) You have not replied to my proof above that the two twins clock's must read different times. I was expecting you to say which link in the reasoning you disagree with. I can only suppose you now accept this.

Thankyou.

(2) The debate about clock synchronisation was because teahive asked you how you could calculate "time rate". He said this required comparison of two clocks. I pointed out (I'm sure he would too) that needs two synchonisations.

Now you are saying that time rate is always the same by definition, so there is no need to measure it physically!

That makes your whole argument circular. You have agreement that time rate of a local clock relative to local frame physical processes is constant. That is the definition of a clock, and that is what is implied by Einstein's postulate.


You can't extend Einstein's postulate to a comparison of time in different frames without defining how you compare time rates in different frames. But when I ask you, you tell me no comparison is needed because the time rates by definition are the same.

This argument of yours gets 0 respect from me, and I guess little from anyone else here. Please if anyone can find any merit in it please will they post the reasons.

[johanfprins]

Two comments Johan:

(1) You have not replied to my proof above that the two twins clock's must read different times. I was expecting you to say which link in the reasoning you disagree with. I can only suppose you now accept this.

Thankyou.

Which proof? I did not see a valid proof. Maybe what you think is a proof I did not see as a proof. I then apologise: Would you mind to prove to me why two clocks which keep the exact same same time rate all the time during the whole journety of the twins can measure two different time durations for the same journey? I just cannot see it.

(2) The debate about clock synchronisation was because teahive asked you how you could calculate "time rate". He said this required comparison of two clocks. I pointed out (I'm sure he would too) that needs two synchonisations.

No it does not, since two synchronisations only reset the time origin twice in succession. What you have to do is to prove that after you have synchronised the two clocks they are actually keeping the same time at any instance in time on any one of the clocks. And that I have done over and over and over on this thread. As I pointed out the x-coordinate of the clock at the origin of Kp, within K is x=v*t; while the xp coordinate of the clock at the origin of K, within kp is v*tp. And at Any instance in time on either of the two clocks clocks there can only be single distance between them: So that v*t MUST be equal to v*tp and this means that t=tp as it must be according to Einstein's first postulate.

What you see when you transform events from Kp to K (for example) is not what actually occurs within Kp but a relativistic "distortion" of what occurs within Kp. Thus, when your transformation gives you a slow-down in time rate, this slow-down IS NOT actually occurrring on the clock within Kp. Just as the the path of an object within Kp is distorted when transformed into K. This does not mean that this distorted path within K is the path that is followed within Kp. To thus conclude that one twin is remaining younger sine "his time is dilated" within the other twin's reference frame is pure poppycock. This situation is symmetric and remains symmetric under the Lorentz transformation: And this symmetry is also demanded by Einstein's postulates. Since the clocks of both twins keep the exact same time rate, both twins age at exactly the same rate and will thus be exactly the same age when they meet up again.

Now you are saying that time rate is always the same by definition, so there is no need to measure it physically!

Not by definition but by the postulates that Einstein used to derive the Lorebntz transformation. If the time rtae is not always the same Einstein's postulates must be wrong, and if the Lorentz transformation is determined by these postulates and you use this transformation to prove the postulates wrong, then i suspect that your logic must be flawed.

That makes your whole argument circular. You have agreement that time rate of a local clock relative to local frame physical processes is constant. That is the definition of a clock, and that is what is implied by Einstein's postulate.

This is not so when there is no gravity.

You can't extend Einstein's postulate to a comparison of time in different frames without defining how you compare time rates in different frames.

Are the laws of physics the same within any inertial refrence frame? yes or no? What determines the clock rate of an atomic clock? The laws of physics? yes or no. This demands that for Einstein's Special Theory of Relativity to be valid, clock rates must be exactly the same on ALL clocks when there is no gravity.

But when I ask you, you tell me no comparison is needed because the time rates by definition are the same.

I did not just do that I also showed by by a direct impeccable proof that v*t=v*tp and therefore afer synchronisation when the clock within K shows a time t, the clock within Kp must at that same instant in time within K show a time tp=t within Kp. Is the algebra too complicated for you?

This argument of yours gets 0 respect from me,

Thus when v*t=v*tp one cannot conclude that t=tp? Shall we call this tomclarke algebra?

[happyjack27]

You can't extend Einstein's postulate to a comparison of time in different frames without defining how you compare time rates in different frames.

well duh. that's just basic mathematics.

But when I ask you, you tell me no comparison is needed because the time rates by definition are the same.

well yes, we are defining our unit of time rate as being a certain proportion to, say, ten half lives of a cesium particle at rest relative to our local inertial reference frame.

thus, the time rate in each inertial reference frame is the same: ten half lives of a cesium particle at rest relative to that reference frame.

there's no logic necessary here, this is just how we are defining measurement.

the idea is that there are an infinite number of clocks whose time they show are all equally correct/valid with respect to their own inertial reference frame. and in each one we are using the same unit to measure time so its nonsensical to say that their time rates are different.

[johanfprins]

well yes, we are defining our unit of time rate as being a certain proportion to, say, ten half lives of a cesium particle at rest relative to our local inertial reference frame.

thus, the time rate in each inertial reference frame is the same: ten half lives of a cesium particle at rest relative to that reference frame.

there's no logic necessary here, this is just how we are defining measurement.

the idea is that there are an infinite number of clocks whose time they show are all equally correct/valid with respect to their own inertial reference frame. and in each one we are using the same unit to measure time so its nonsensical to say that their time rates are different.

Thank you happyjack. It is so obvious that it must be so.

[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?

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? The fact that the SR and GR corrections are both required indicates to me that the rate is different.
However, as I searched around space travelling isotopes (batteries and clocks), I have not found any conclusive data in the lifespan tracking data that indicates so far the relitivistic impact is occuring. I will be interesting to see how the expected lifespan of the new Mars Rover nuclear battery tracks when compared to a like battery here on earth, when adjusted for a Sun Centered Frame and the difference in Mars solar orbital speed is compared to earth's solar orbital speed.

Back to the point, if the clock rates were not different, then the corrections SR and GR would not be required. And, given that it is, it says that the twins would age differently, because the isotopes are "aging" differently as shown by the decay rate change. Granted, Renshaw's argument is that the rate change is a function of "artificial gravity" in effect brought on by the position on the "spinning disk". Ie: you can't tell the difference between gravity and accelleration.

[tomclarke]

You can't extend Einstein's postulate to a comparison of time in different frames without defining how you compare time rates in different frames.

well duh. that's just basic mathematics.

But when I ask you, you tell me no comparison is needed because the time rates by definition are the same.

well yes, we are defining our unit of time rate as being a certain proportion to, say, ten half lives of a cesium particle at rest relative to our local inertial reference frame.

thus, the time rate in each inertial reference frame is the same: ten half lives of a cesium particle at rest relative to that reference frame.

there's no logic necessary here, this is just how we are defining measurement.

the idea is that there are an infinite number of clocks whose time they show are all equally correct/valid with respect to their own inertial reference frame. and in each one we are using the same unit to measure time so its nonsensical to say that their time rates are different.

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.

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.

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.

Further, Johan will not when asked define how he compares time in different frames - so even if Eunstein's postulate applied across frames, Johan cannot use it till he tells us how physically he defines the comparison.

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

[happyjack27]

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.

[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?

it will look slow, because it is slow. the clock is set to show time in earth seconds, not satellite seconds. things on the earth are constantly accelerating towards the satellite due to the electromagnetic forces of atoms on the earth* pushing back against "gravity" (which is really just the fact that it sits in a curved space). so the satelite has to compensate for the fact that clocks on the earth are constantly accelerating relative to it.


*really due to the quantum uncertainty principle which keeps electrons from entering the nucleus

[happyjack27]

gravity is like centrifugal force: there's no such thing. ( http://en.wikipedia.org/wiki/Centrifuga ... e_frame%29 ) this was einstein's great insight in GR.

the force that we experience beneath our feat is from the electromagnetic forces of atoms beneath our feet, just like the outward force we experience when swinging a rope around is just the muscular exertion required to apply force against the rope's inertia.

[ladajo]

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.

It is what causes the accelleration that matters.

In regard to the Cesium decays, I think the idea of a gravity constant experiment is interesting as Johan has proposed. However, I consider that finding a stretch of gravity constant test track, that does not involve a circular path is going to be challenging.

The SR component for speed in GPS is a real effect on clock rate. If it impacts rate, it must also impact "age". I am not saying that when brought back together that the rates will differ at that point. I am saying that a count is a count. And, if one clock is winding off more counts than the other, it will "age" more. It can not be otherwise.

As you can see from this chart the orbital speed does slow down the clock at higher speeds.

[happyjack27]

well yes, but the earth will have rotated the satellite exactly the same number of times that the satellite had rotated the earth. and at exactly the same velocity!

[Teahive]

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?

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

Yes, the laws of physics are the same within every inertial reference frame. No, that does not mean that two clocks, when separated and brought together again, need to show the same elapsed time.

A working clock will show proper time. This is tautologically true. As a thought experiment I could run a physically accurate simulation of a clock on a computer, and this simulated clock could show different times than the clock on my wall, depending on how fast the computer is. I could even pause the simulation for a while, and the simulated clock wouldn't notice a difference. Yet they both obey the same laws of physics within their own reference frames. Einstein's first postulate is not violated.

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

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.

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