Is time universal? NO (and its proof)

[2inquisitive]

Mac, there is no region in the observable universe that is free of microgravity. Gravity can only 'balance' in all directions at some specific interstellar location, but it is still there and, technically, there are still 'tidal forces' due to all distant gravitating bodies.
What does this have to do with the LISA mission or the Sagnac effect?

Edit: And before you bring it up, yes, I know gravity 'balances' at some of the LaGrange points within the solar system. But it has nothing to do with the LISA mission.

 

[geistkiesel]

Sorry if already answered (In two days absence, I am 5 pages behind, but want to answer.)
(1)Yes, unqualified.
(2)No, qualified as below:

The ground observers (especially if they accept SRT) can compute that the explosions were simultaneous in the train frame, but by their own clocks, which is what they judge simultaneity by, the rear explosion precedes the front one.

[Below you say you designed this scnario not using any theory! This seems contradictory here.]

I assume here, Billy T, that the clocks on the embankment frame are synchronized wrt each other. Meaning that instantaneously the embankment clocks read the same time.

Here is some physics that the statement above ignores. When the flash of lights occur the source of the flash has no effect on the speed of the flashes of light. As the frame is moving wrt the embankment, the "vacua", the light will surely reach the on coming light from the rear of the train before the forward moving light will arrive at the forward clock.

One cannot use a relative motion for measuring the absolute motion of light. The light knows not of the moving bombs and clocks, one approaching, the other receding both wrt the motion of the lights.

The clocks on the moving frame are also synchronized within the moving frame. There is no rational manipulation of the independence postulate that will allow the lights to arrive at the two clocks simultaneously in the moving frame.

Whatever is the perception of an observer at the midpoint of the two clocks, the light is moving at a speed c wrt to the vacua, and seen from the embankment, Vce - Vte = Vlt, which say the speed of light wrt to the embankment, minus the speed of the train, also wrt the embankment, provides the realtive velocity of frame aand photon, as seen from the embankment.

The expression above was that derived by Einstein in "Relativity", Section 7. However, AE took the expression for the relative motion of frame and photon and intended the reader to believe that this expression was using the train as the inertial coordniate system, and then he complained that this measured the speed of light less than c.

It could not be clearer that the expression is merely describing the measured relative motion of frame and photon, and is not an expression for a " measured speed of light". The speed of light is always c wrt the vacua.

Do you not use the clocks of your own frame to judge simultaneity?

The clocks on the embankment will measure sequential explosions that occur due to the arrival of the two light pulses at the points the bombs are located on the moving frame, as will the clocks on the moving frame.

Billy T, the lights do not move in some special manner to satisfy notions of the SRT observers. When the flashes arew emitted, it is the same as if the lights were emitted in the stationary frame as the speed of light is independent of the speed of the source of the lights. If there is mome time dilation and/or Lorentz contraction in the sapce covereed by the foprward moving nphoton,m then what kind of time/space manipulation occurs in the space traveled by the rear moving light?

When you impose the independence postulate on the moving lights all becomes crystal clear.

When the rearward moving photon has covered a distance ct,m the frame has moved a distance vt toward the rear moving photon. The forward moving photon has also moved ct, and is locted a distance 2vt from the forward clock. This photon covers the 2vt distance plus the small distance the frame moves in the interim, vt', in time t',or ct' = 2vt + vt', or t' = t(2v)/(c - v), the time the two lights would be moving in the same direction had the left photon reflected at the left clock. In any event, t' is time difference in the explosions measured on the moving frame, by the moving frame clocks.

Recall that there is some frame in which my birth is simultaneous with that of Christ. (I think Christ's birth proceded mine by about 2000 years - but I stupidly compare simultaneity using only my clocks, without computation for all other frames.)

Thus, to consider other frames when judging if two events are simultaneous (for you) or not is ridiculess.

Billy T, We need not compare clocks in this experiment. We need only determine if the clocks on the moving frame receive the flashes simultaneously or sequentially. (See above, the lights arrived sequentially on the moving frame).

I designed this scenario to avoid using any theory.* Only use your mind and fact speed of light is independent of direction of travel. (I do not even require that it is the same in all frames!) It could be 3x10^8 on train and 2x10^8 on the ground and my thought experiment still shows:

Events Simultaneous in one frame are NOT Simultaneous in any other.

You assumed that the explosions would occur simultaneously on the moving frame, which I showed the oppisite. That the explosions were sequential will be reflected by the clock reading in the stationary frame as well as the moving frame. It is only in the condition where the frame is actually stationary wrt the vacua that "at rest physics is allowed".

I refer you to those rationalizations of the Twin Paradox (Feynman, Bohm et al) where only in the space ship frame, that actuially accelerated and moved, was the age rate lowered, the earth frame being considered at rest wrt the spwace ship throughout. This is the same as saying only in the accelerated frames do SRT effects occur, which means all the train embankment gedanken must be edited to demand that only the train accelerates, not the embankment, but then there goes the equivalence of intertial frames postulates. The self-slaying cycle of this result is cruel, I know.
Geistkiesel​

 

[Billy T]

...From the frame of reference of someone INSIDE the spacecraft, the frame is inertial. If they were in a windowless room inside the spacecraft, how could they detect their motion?...

No. Only approximately so. One way is to align the axis of a spinning gyroscope with the axis of the space craft - half an orbit later it its axis is transverse with the axis of the space craft.

Another way, harder to do in practice, is to note that a floating ball nearer to Earth than another one is slowing moving towards the front of the space craft as it, in a smaller orbit, takes less time to orbit the Earth.

That is why, when correctly stated, the interior of a space graft is a "micro-gravity" not "zero-gravity" environment. It is not an "inertial frame."

 

[Billy T]

NOTE: The measured speed of light being invariant DOES NOT dictate that a photon's velocity is constant between frames. It in fact can (and most likely) means you are measuring different photons in different frames.

Hence relativity is not indicated from measured invariance of light.

I do not understand what you mean by "DOES NOT dictate that a photon's velocity is constant between frames" Please explain the procedure for measuring "between frames."

The only thing I can guess that you mean is that you think there are grid lines in each frame. say red one in frame 1 and blue ones in frame 2, and you are talking about an experiment that measure the time (and I don't know by whose clocks) required for a photon to go from a red grid line to a blue one. Surely this is not what you are trying to state. Please try to be more clear. Out line the operational procedure by which you measure “between frames.”

Again I am uncertain as to meaning of "indicated" in your last sentence. What I said is that if the speed of light vacuum is same in all frames, then the equations of SR follow as means of transferring observations made in one frame to the other. Perhaps some other set of equations is also consistent with this observation of c being constant that you agree to, but I do not know of any.

Can you show that your non SR theory is consistent with the invariance of the vacuum speed of light? - I strongly doubt you can.

 

[geistkiesel]

by geistkiesel:

"Assume that the two clocks on two inertial frames are zeroed at the same instant as you described. Then have the two ships make slow wide turns until the shps are approaching each other."
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I am a little confused by your example. The clocks were 'zeroed' when they passed. Were they also synchronized to tick at the same rate, and if so, which clock were the synchronized with? The Earth clock? By adjusting the tick rate of each travelling clock, each can be synchronized with the Earth clock, but the they will not be synchronized with each other unless they both have the same relative velocity wrt the Earth clock.
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Only that the two clocks are synchronized wrt a common frame, say the earth. Later we will see that each frame is moving at .2c and .6c in wrt the earth frame and are in a head on collision. Youy are correct, I did confuse the uisue regarding synchrinuization.

by geistkiesel:

"Then have the two ships make slow wide turns until the shps are approaching each other."
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You do realize the ships can't make a turn in an inertial frame, it is then a non-inertial frame. When the ships enter the non-inertial frames, clock synchronization is lost with the Earth clock, and each other.
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If the ships turn symetrically their synchronization will be lost, but can be calculated for analyisis purposes later. This is a minor point as you will see, and a point I no longer feel as important.

by geistkiesel:

"If we assume the two clocks will eject a piece of paper with the current time on each clock at the instant of head on impact of the frames, then A moving at .2c andB moving at .6c, both velocitites wrt their origin on earth,..."
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If one now has a velocity (relative?) of .2c and the other .6c, then the ship with .6c had to accelerate more than the ship with .2c, again changing their synchronization. Why do you believe 'Now it does seem bizaar for anyone to claim that the times printed on both pieces of paper will both show elapsed times less than the other. I understand that this is a statement consistent with SRT,...' is consistent with SRT?
Sorry if you didn't mean for me to answer, geistkiesel.

I say this from the often heard SRT claims that observers in relative inertial frames can concider themsewlves at rest and the other moving , hence the moving frame's clocks should be running slower than the "stationary" frame.

If we look just at the colliding ships and the current time printed on the papers, both indicating the instantaneous time of arrival, then both numbers cannot be less than the other even with the most complex of analysis.

Whatever is printed on the papers, the time is NOW for both of them, The .6c ship should (or might have) have a number less than the .2c ship, but the .2c ship's number what ever it is or whatever numbers are there, the .2c ships is not indicartive of a lesser time than that printed on the .6c paper.

If both ships started from the same staring point simulataneously and each made the most severe direction changes, velocity changes, whatever and when they finally crashed the numbers on both pieces of paper indicate the same exact time for both space ship frames, whatever the individidual numbers printed on the papers happened to be, and further, the numbers have no direct relationshuip to the passage of time. The numbers, if indeed different, only indicate that various motion changes, accelerations and other force effects regarding the clock instrumentation occurred.

Again, you correctly spoted the gaps in the scenario I was describing, carelessness on my part.. You never need to ask if you can jump into any thread I happen to be engaged in.
Geistkiesel​

 

[Billy T]

...What evidence do you have that the photon you measure in a moving frame is the same photon you measured in the stationary frame?...

None. We agree that one can only observe any particular photon once. I can however measure the speed of light (with different photons) in frame 1 on Mondays, Wenesday, and Fridays and in frame 2 on Thuesday, Thursdays and Saturdays. I am too lazy to work on Sundays

The fact that all such vacuum measurement agree I think we agree on so what is your point?

 

[Billy T]

...In my view photons in one frame do not exist in any other frame....

What nonsense! If I move relative to the camera taking my picture (go into a different frame from the flash bulb attached to the camera) that flash of light will not exist in my frame, so I will not even be a blurry image of me in the photo! according to you.

I am courious: will the photo contain image of the wall behind me as it can not contain my image but something must be there?

ARF ARF ARF arf arf arf...(my dogs are getting tired of laughing at you.)

 

[Billy T]

...The photons in Billy T's embankment frame are not measured by the photo sensors in the train frame....

Of course not. the flash bulb mounted on the side of the train makes zillions of photons, only a few hit any detector.

Again, my scenario does not assume SRT, or even the constancy of light in all frames (which with logic and math is the same thing as assuming SRT equations).

Assume that every frame has a unique speed of light in vacuum if you like. I am not asking you to accept SRT (here). I only ask you to think.

For example, assume that in the ground frame the speed of light is only ten times the speed of sound!

It is still true that on the ground the rear explosion proceeds the one at the front of the train (EVEN MORE SO, AS THE TRAIN WILL MOVE A GREAT DISTANCE WHILE THAT "SLOW LIGHT" IS TRAVELING.)

You are off on another of your "duck and weaves" with this "not the same photons" - that is duck and weave number 15. I wait for duck and weave 16. I am sure you will not fail to create it also.

 

[quadraphonics]

No. Only approximately so. One way is to align the axis of a spinning gyroscope with the axis of the space craft - half an orbit later it its axis is transverse with the axis of the space craft.

Another way, harder to do in practice, is to note that a floating ball nearer to Earth than another one is slowing moving towards the front of the space craft as it, in a smaller orbit, takes less time to orbit the Earth.

That is why, when correctly stated, the interior of a space graft is a "micro-gravity" not "zero-gravity" environment. It is not an "inertial frame."

Right, orbital frames are kind of special, since the force of gravity cancels out the centrifugal force generally felt in rotating frames, so it's a bit more subtle than simply observing a fictitious force like in the merry-go-round example. Nevertheless, it's not an inertial frame.

 

[Billy T]

[Below you say you designed this scnario not using any theory! This seems contradictory here.]

I assume here, Billy T, that the clocks on the embankment frame are synchronized wrt each other. Meaning that instantaneously the embankment clocks read the same time.

Here is some physics that the statement above ignores. When the flash of lights occur the source of the flash has no effect on the speed of the flashes of light. As the frame is moving wrt the embankment, the "vacua", the light will surely reach the on coming light from the rear of the train before the forward moving light will arrive at the forward clock.

One cannot use a relative motion for measuring the absolute motion of light. The light knows not of the moving bombs and clocks, one approaching, the other receding both wrt the motion of the lights.

The clocks on the moving frame are also synchronized within the moving frame. There is no rational manipulation of the independence postulate that will allow the lights to arrive at the two clocks simultaneously in the moving frame.​

Up to the last sentence here you are correct. I don't understand why you made that sentence. The flash bulb is equally distant from bomb triggers at front of train and at rear of train. If in the train frame it does not travel these equal distances in equal time, please tell which takes longer and why.

Whatever is the perception of an observer at the midpoint of the two clocks, the light is moving at a speed c wrt to the vacua, and seen from the embankment, Vce - Vte = Vlt, which say the speed of light wrt to the embankment, minus the speed of the train, also wrt the embankment, provides the realtive velocity of frame aand photon, as seen from the embankment.

You are assuming much more than me. All I want is that in the train frame the speed of light headed to the front is the same as the speed of light headed to towards the rear. I am NOT calculating the speed of light in the ground frame as you are (erroneously) and it can be only 10 times the speed of sound for all I care, but again I want the "down the track" speed to be the same as the "up the track" speed.

That is, I do not calculate any speeds or assume any particular relationship between ground speed of light and train speed of light. I do not assume that there is an observer at the center of the train watching the explosions or that any observer any where is watching in any frame. Weeks later the times recorded will be read, and the report written. All I assume is that speed of light is not a function of which way it is going and that wrt to the station, the train is moving. (I do not even tell which is moving wrt your "Absolute Reference Frame", ARF, as that does not exist, but since last time we discussed, you did believe in ARF, and were offering several "proofs" of it, assume half of the motion wrt ARF is train's and half is the station's or any other ratio you like. )

I do not, as you put it later, "impose the independence postulate." The rest of your post is filled with other false understanding of what I am requiring, so can not respond, other than to ask you to read more carefully, instead of assuming I am (here) defending SRT etc.​

 

[2inquisitive]

No. Only approximately so. One way is to align the axis of a spinning gyroscope with the axis of the space craft - half an orbit later it its axis is transverse with the axis of the space craft.

Another way, harder to do in practice, is to note that a floating ball nearer to Earth than another one is slowing moving towards the front of the space craft as it, in a smaller orbit, takes less time to orbit the Earth.

That is why, when correctly stated, the interior of a space graft is a "micro-gravity" not "zero-gravity" environment. It is not an "inertial frame."

I agree with everything you have said, Billy T. I have argued for a long time that GPS frames are not inertial, therefore Special Theory does not apply. No frame on the face of the Earth is truely inertial, not at particle accelerator labs or anywhere else. No frame in the solar system is inertial, just varying degrees of non-inertial. There is no straight-line motion in the solar system, only elipses (orbits). The only way to have an inertial frame is to IGNORE gravitational effects. The speed of light does not have to be invariant to every observer in accelerated frames (non-inertial). Thanks for agreeing with me!

 

[Billy T]

Right, orbital frames are kind of special, since the force of gravity cancels out the centrifugal force generally felt in rotating frames, so it's a bit more subtle than simply observing a fictitious force like in the merry-go-round example. Nevertheless, it's not an inertial frame.

Glad you agree.

However, I must note that "centrifugal force" does not exist. If it "cancelled out gravity," then with zero net force acting on you, your trajectory is a straight line, not an orbit.

I have beaten that non existent "frame effect" to death in another thrread, so will not repeat it all here, except to note that if it did exist, it would violate Newton's third law as there is not "equal and opposite" force on the agent applying the centrifugal "force."

Only your unaccustomed feeling of the force required to make you not travel in a straight line (in accord with Netwon's first law) is acting. - I.e. gravity is action on you and the Earth is feeling the 3d law "equal and opposite" force as you act on it and both co orbit about your common center of mass.

 

[MacM]

Mac, there is no region in the observable universe that is free of microgravity. Gravity can only 'balance' in all directions at some specific interstellar location, but it is still there and, technically, there are still 'tidal forces' due to all distant gravitating bodies.
What does this have to do with the LISA mission or the Sagnac effect?

Edit: And before you bring it up, yes, I know gravity 'balances' at some of the LaGrange points within the solar system. But it has nothing to do with the LISA mission.

I was not addressing LISA, Sagnac or LaGrange Points. Only noting for others the technical fact that there is no such thing as "Zero" gravity. It is and must always be "Micro-"gravity and that even free fall is not technically inertial due to tidal forces inherent in the non-parallel pull of generally spherical massive bodies.

 

[Billy T]

...The speed of light does not have to be invariant to every observer in accelerated frames (non-inertial). Thanks for agreeing with me!

General Relativity is too hard for me, so I make no strong claims about what can or does happen in accelerating frames, but I do believe that the speed of light does not change if observed from or measured in an accelerated frame.

I believe this because any given accelerating frame is approximately the same as the inertial frame that for that instant has the same velocity. The only difference I see is that the accelerating frame has some false gravity, but I do not think gravity affects the speed of light, only its color. If you measure very quickly, it would seem to me that the speed of light in two frames, one accelerating and one inertial, should measure to be the same, if both are essentailly briefly moving at essentailly the same speed.

 

[quadraphonics]

Glad you agree.

However, I must note that "centrifugal force" does not exist. If it "cancelled out gravity," then with zero net force acting on you, your trajectory is a straight line, not an orbit.

Right, centrifugal force is a "fictitious" force, which arises in non-inertial frames (rotating ones in this case). The presense of such frame forces is the usual indicator that a frame is non-inertial. 2inquisitive seemed to be thinking that its absense here implied that the frame was inertial, but it does not. Orbital motion is somewhat more subtle, because the speed of orbiting objects (and thus their total momentum) is constant.

 

[MacM]

However, I must note that "centrifugal force" does not exist. If it "cancelled out gravity," then with zero net force acting on you, your trajectory is a straight line, not an orbit.

And if it were ficticious and did not exist to balance the pull of gravity then the centriputal force of gravity would have you for breakfast (long before lunch).

 

[Billy T]

...Orbital motion is somewhat more subtle, because the speed of orbiting objects {IN CIRCULAR MOTION} (and thus their total momentum {MAGNITUDE}) is constant.

NOTE MY INSERTS IN CAPS. Now it is true statement, at least from the POV of an Earth orbit center.

Many people, myself included until a few months ago when janus58 set me straight, think the the moon orbits the Earth, but this is only an Earthling's view of it.

The sun's gravity on the moon is much stronger than the Earth's and the moon, like the Earth, is in nearly circular orbit about the sun. It distance from the sun varries between very slightly more and very slightly less than that of the earth a little more than 12 cycles each Earth year. It's trajectory, like that of the Earth is ALWAYS concave towards the sun.

 

[DaleSpam]

If we look just at the colliding ships and the current time printed on the papers, both indicating the instantaneous time of arrival, then both numbers cannot be less than the other even with the most complex of analysis.

Whatever is printed on the papers, the time is NOW for both of them, The .6c ship should (or might have) have a number less than the .2c ship, but the .2c ship's number what ever it is or whatever numbers are there, the .2c ships is not indicartive of a lesser time than that printed on the .6c paper.

The elapsed time (the numbers printed on the papers) have little to do with the relative or apparent clock rate (the number of papers printed per second). One is the value of a function and the other is its derivative. Both observers will agree whose number is smaller, but both will claim that the other one is printing them out at a slower rate immediately before the collision (according to relativity).

-Dale

 

[DaleSpam]

General Relativity is too hard for me, so I make no strong claims about what can or does happen in accelerating frames, but I do believe that the speed of light does not change if observed from or measured in an accelerated frame.

I believe this because any given accelerating frame is approximately the same as the inertial frame that for that instant has the same velocity. The only difference I see is that the accelerating frame has some false gravity, but I do not think gravity affects the speed of light, only its color. If you measure very quickly, it would seem to me that the speed of light in two frames, one accelerating and one inertial, should measure to be the same, if both are essentailly briefly moving at essentailly the same speed.

You may be correct for linearly accelerating frames, but rotating frames definitely violate the constancy of the speed of light. That may be related to the fact that a rotating frame has both a gravity-like centrifugal frame force and a weird Coriolis frame force. As I mentioned earlier, if you rotate at 1 cycle/s then the moon is traveling faster than c in your frame.

-Dale

 

[MacM]

but rotating frames definitely violate the constancy of the speed of light. -Dale

I don't believe this statement is true. If you are think Sagnac affect it is accounted for in the frames of referance and does not represent an invariance of light itself.

If not what are you referring to?