[MikeFontenot (OP)]

_________________________________________________________________

Abstract:

I have previously shown that the exponential version of the gravitational time dilation (GTD) equation (first given by Einstein in 1907) is incorrect, because it is inconsistent with the outcome of the twin paradox. I then gave a corrected version of the GDT equation which IS consistent with the outcome of the twin paradox, and which is also consistent with the co-moving-inertial-frames (CMIF) simultaneity method. In this brief paper, I describe an experimental test of my GTD equation that might be feasible to conduct.

___________________________________________________________________________

Section 1. Einstein’s Exponential GTD Equation

In Einstein’s 1907 paper [ https://einsteinpapers.press.princeton.edu/vol2-trans/319 ], Einstein stated that the GTD equation is

R(g) = exp(g L),

where “g” is the force per unit mass exerted by the uniform gravitational field, and “L” is the constant distance (in the direction of the field) between two stationary clocks. The quantity “R” is the ratio of the tic rates of the two clocks: the clock which is higher in the field (farther from the source of the field) will run “R” times faster than the other clock.

According to the Equivalence Principle, we can then also say that when there are no gravitational fields (i.e., in a Special Relativity scenario), two clocks which are initially unaccelerated, and which are separated by a constant distance “L”, and which are then simultaneously accelerated with an acceleration “A” (in the direction of their separation), then the rate ratio “R” is

R(A) = exp(A L).

The leading clock runs “R” times faster than the trailing clock. Note that for constant “A” and “L”, the rate ratio “R” DOES NOT VARY WITH TIME.

I showed that the above exponential equation is inconsistent with the outcome of the twin paradox. Specifically, if the traveling twin (he) changes his velocity instantaneously at his turnaround, the exponential equation says that the home twin (she) will be INFINITELY old after his turnaround, and when the twins are reunited. That isn’t true: both of their ages are finite at the reunion. Thus the exponential equation for “R” is incorrect. (And the exponential equation is also incorrect for the case where the turnaround is not instantaneous, but is just very quick.)

Section 2. My GTD Equation

The corrected rate ratio equation is R(A) = [ 1 + L A sech_sqrd(theta) ],

where sech_sqrd( ) is the square of sech( ), the hyperbolic secant. (The hyperbolic secant is the reciprocal of the hyperbolic cosine, cosh, which is more likely to be available in tables). Cosh(theta) can also be calculated using the equation

cosh(theta) = { exp^theta + exp^(-theta) } / 2.

The CRC Standard Mathematical Tables book (14th Edition) show plots of the hyperbolic functions on page 527.

The quantity “theta” in the above equation for R(A) is theta(t) = A t,

where “t” specifies how long the constant acceleration “A” has been going on, since it abruptly started from zero acceleration at time zero. So we really should write the “R” equation as

R(A,t) = [ 1 + L A sech_sqrd{ theta(t) } ].

A plot of R(A,t) versus “t” is given in section 4 for the case A = 1 ls/s/s (about 40 g’s) and L = 7.52 ls.

The R(A,t) equation produces quite different qualitative results than those produced by the exponential R(A) equation. The exponential R(A) equation says that if the constant acceleration “A” goes on for an essentially infinite time, the rate ratio NEVER CHANGES ... i.e., the leading clock keeps ticking faster than the trailing clock by the same ratio, forever. And for large “A”, that constant ratio is HUGE! In contrast, the new R(A,t) equation says that, as “t” goes to infinity, R(A,t) approaches 1.0. I.e., the two clocks eventually tic at essentially the same rate. That is quite a qualitative difference, which might be observable experimentally.

Section 3. A Proposed Experimental Test of My GTD Equation

Charged particles can be accelerated to speeds that are a large fraction of the speed of light, by exposing them to very large electric fields. So we can start with a stationary pair of them, separated by the distance “L”, and then switch on a very strong uniform electric field (in the direction of their separation) that will accelerate both of them at the same rate, and maintain their separation at “L”.

But how can we use each particle as a “clock”? We might be able to accomplish that by using UNSTABLE charged particles ... particles that have a known average lifetime (before they decay into uncharged particles that won’t accelerate in the electric field). That way, if the leading particle is ageing faster than the trailing particle, the leading particle will (on average) decay quicker than the trailing particle, which might be observable. That might allow an experimental way to verify or falsify my GDT equation.

[evan_au]

A few practical considerations:
Short-lived radioactive isotopes need to be generated on the spot, at the time they are needed, like the "technetium cow" used to produce imaging radioisotopes.
https://en.wikipedia.org/wiki/Technetium-99m_generator

 The longest linear accelerator in the world is 3km long, which means that the particles will only accelerate for a few microseconds before hitting the end of the line. That means you need a storage ring, rather than a linear accelerator. I am sure that the continual acceleration towards the center of the storage ring will add complexity to the calculations!

Another problem is time dilation (which is what you are trying to measure). Once a particle is traveling at a speed near c, the lifetime of a radioactive particle will increase. However, particles diffuse out of the beam and are lost over time, so the beam gets weaker the longer you run it.

[Bored chemist]

We did this sort of experiment a long time ago.
https://en.wikipedia.org/wiki/Experimental_testing_of_time_dilation#Atmospheric_tests

[Halc]

I have previously shown that the exponential version of the gravitational time dilation (GTD) equation (first given by Einstein in 1907) is incorrect, because it is inconsistent with the outcome of the twin paradox.

I think the rest of the scientific community would have noticed after a century if Einstein was wrong about something as simple as special relativity. The error in your demonstration was pointed out. You continue to assert this despite the errors being identified. This is a good deal of the reason why Kryptid moved this post to the lighter-side of the forum.

Pay attention to the peer review.

In Einstein’s 1907 paper, Einstein stated that the GTD equation is R(g) = exp(g L),

You forgot the τ in the equation, which is important. The equation (in natural units) is something like δ = τ•exp(g L) where δ is the remote duration change, the amount of age change of a hypothetical stationary twin, stationary relative to accelerating reference system ∑.
Einstein did not in any way suggest that equation was relevant to a gravity situation, so calling it a 'GTD' is just plain wrong. The equation is applicable to what is essentially Lass (or radar) Coordinates. What you call the 'GTD' is not applicable to the twins scenario (which for one thing doesn't involve gravity) because the Earth twin is not stationary in ∑, so the equation wouldn't even work in a situation where the turnaround acceleration rate is finite.

According to the Equivalence Principle, we can then also say that when there are no gravitational fields (i.e., in a Special Relativity scenario), two clocks which are initially unaccelerated, and which are separated by a constant distance “L”, and which are then simultaneously accelerated with an acceleration “A” (in the direction of their separation), then the rate ratio “R” is   R(A) = exp(A L).

You omit frame references here, making your statement misleading. Relative to inertial system S (Einstein's designation), two clocks which are separated by a constant distance L, and which are then simultaneously accelerated with a proper acceleration “A” (in the direction of their separation), then the rate ratio “R” relative to S is 1 since they have identical motion in S. Their separation in S remains L after any amount of time. This is effectively Bell's scenario with the string that breaks.
The same is true if both are accelerated at constant coordinate acceleration relative to S.  R remains 1 in S and the clocks remain synced in S. This fact is independent of the acceleration curve so long as the two curves are the same.

On the other hand, relative to the rear of an accelerating rigid object (designated ∑ by Einstein), the separation L in ∑ will remain constant, but the proper acceleration at the rear clock will be greater than at the lead clock. It is in this scenario that the R(A) = exp(A L), but the acceleration is not 'A' at the lead clock, so your specification of both of them accelerating at 'A' seems unclear at best. In ∑, both clocks are stationary so there's no coordinate acceleration at all. Their respective proper accelerations are there, but  they're not identical (*).

Note that for constant “A” and “L”, the rate ratio “R” DOES NOT VARY WITH TIME.

True only for constant proper acceleration. This isn't true for constant coordinate acceleration (relative to S say). You should be clear when screaming assertions like that.

I showed in [in another paper] that the above exponential equation is inconsistent with the outcome of the twin paradox.

Yes. I pointed out the error you made. You ignored that and continue to make this assertion.

Specifically, if the traveling twin (he) changes his velocity instantaneously at his turnaround, the exponential equation says that the home twin (she) will be INFINITELY old after his turnaround.

Again, it says no such thing. As you're written it (specifying R(g) as a rate ratio), it merely says that relative to the twin turning around with infinite acceleration, a distant twin (stationary relative to ∑) ages at an infinite rate, but for zero time. An infinite rate for zero time is undefined, not infinite. That says is that the equation cannot be meaningfully used for a finite impulse with zero duration, and not at all for the twin scenario where the Earth twin doesn't remain a constant distance from the travelling twin.

Section 3. A Proposed Experimental Test of My GTD Equation
...
That way, if the leading particle is ageing faster than the trailing particle, the leading particle will (on average) decay quicker than the trailing particle, which might be observable.

The relative decay rates would be a function of the frame in which that time is measured. Your 'test' description doesn't specify that frame, but in S (the frame of the accelerator), both samples will decay at the same rate regardless of the level of acceleration. If your theory predicts otherwise, you have a problem, but in the absence of a frame reference, your prediction is ambiguous. Either way, relativity predicts what will be measured.



* Note all three spellings of the same homophone in that sentence.

[MikeFontenot (OP)]

In my paper, I said:

[...]
A plot of R(A,t) versus “t” is given in section 4 for the case A = 1 ls/s/s (about 40 g’s) and L = 7.52 ls.
[...]

In the above, I wanted to give the reader a feel for the unit of acceleration ls/s/s, since it is not a commonly used unit of acceleration.  I said that 1 ls/s/s is about 40 g's.  But that is VERY wrong!

The more commonly used unit of acceleration in relativistic calculations is ly/y/y.  It just happens that 1 ly/y/y is about 0.97 "g's".

The two units of VELOCITY, ly/y and ls/s, are numerically equal ... one of each of them is just the speed of light: 186,000 miles per second.  But the two units of acceleration, ly/y/y and ls/s/s, are VERY different in magnitude.  1 ls/s/s is greater than 1 ly/y/y by the number of seconds in a year, which, if I've done the arithmetic correctly, is about a factor of 16 MILLION greater.  So 1 ls/s/s is about 16 million "g's", NOT 40 "g's".

Fortunately, that mistake has nothing to do with my results in my paper.  I'm confident that everything I stated in that paper (other than how many "g's" 1 ls/s/s is equivalent to) is correct.

[MikeFontenot (OP)]

the number of seconds in a year, which, if I've done the arithmetic correctly, is about a factor of 16 MILLION greater.

31.5 million actually, or almost exactly πe7 by coincidence.

You're correct, there.  I used 12 hours per day, not 24 hours per day, as I should have.  It was a stupid mistake.  Thanks.

Your equations are not always wrong, but your representation of Einstein's theories are almost always wrong.
[...]

OK, here's a challenge for you:

In his 1907 paper, Einstein said that when two people (each with his own clock) are initially stationary at time zero on the clocks, and not accelerating, and separated by "L" lightseconds (ls), and who then simultaneously start accelerating (according to their accelerometers) at a constant "A" ls/s/s (in the direction of their constant separation), then the leading clock will tic faster than the trailing clock by the factor

  R(A)  =  exp(L A) .

  So here is your challenge:

Do the following sequence of calculations:

In each case, set L = 7.52 ls.  (That just happens to be the separation that I chose in my original paper, so use that value so that we can compare our results.)

In each of the cases, the acceleration "A" and the duration "tau" of the acceleration are chosen such that the product "A tau" is constant for all cases at 1.317 ls/s.  That results in all cases in the velocity of the two clocks, at the end of the acceleration, being 0.866 ls/s.

For the first case, set "tau" equal to 1 second.  Then compute "A" so that

 A tau = 1.317,

i.e., A  =  1.317 / tau  =  1.317 ls/s/s in this first case.

For that first case, compute the reading on the leading clock when the trailing clock reads time "tau".  That reading will be given by

  tau R  =  tau exp(L A).

What do you get for the answer to the first case?

Next, for the second case, set tau = 0.1 second.  (I.e., make tau ten times smaller).  In order to keep the final velocity at 0.866 ls/s, the acceleration "A" must be increased by a factor of 10.  So "A" is now taken to be  A = 13.17 ls/s/s.  What do you get for

    tau R  =  tau exp(L A)

for this second case?

And likewise, do the third case in the analogous way: reduce "tau" by a factor of ten, and increase "A" by a factor of 10.  So for the third case, "tau" = 0.01 second, and "A" = 131.7 ls/s/s.   What do you get for

    tau R  =  tau exp(L A)

for this third case?

Now, look at your results.  Does that sequence of readings on the leading clock appear to be approaching a finite limit as tau goes to zero?  Or does it appear to be diverging to infinity as tau goes to zero? 



 

[MikeFontenot (OP)]

Thank you for responding. It is a bit more clear now as to what you're trying to convey.

In his 1907 paper, Einstein said that when two people (each with his own clock) are initially stationary at time zero on the clocks, and not accelerating, and separated by "L" lightseconds (ls), and who then simultaneously start accelerating (according to their accelerometers) at a constant "A" ls/s/s (in the direction of their constant separation), then the leading clock will tic faster than the trailing clock by the factor   R(A)  =  exp(L A) .

Einstein did not say this. There's several inaccuracies here.
A) You left out specification of coordinate systems / frame-references.
If we use frame S (the inertial frame relative to which the clocks are synced and the clocks are both initially stationary), then the clocks experience identical acceleration relative to S and must remain synced in S by symmetry.

If we use frame ∑, the accelerating reference frame of the rear clock, then L is not constant, the front clock is not stationary in ∑, but it does indeed run slower than the stationary clock at the origin of ∑. Also, there is more than one kind of accelerating coordinate system, and Einstein is not using the one you think he is.

B) Einstein's equation (a related but different one) referenced an entirely different scenario than the Bell's spaceship one which you are describing. This is something I pointed out above, but you're not big on reading comments so you're persisting in this inappropriate scenario.

In each of the cases, the acceleration "A" and the duration "tau" of the acceleration

There is no duration of acceleration in Einstein's scenario. It never quits. Just so you know. Einstein describes a particular time interval of that acceleration.
Anyway, duration as measured in which clock? Again, the answer is worldline dependent, but I'm guessing tau on each of the accelerating clocks, not as measured by S where the acceleration wouldn't be constant.

... are chosen such that the product "A tau" is constant for all cases at 1.317 ls/s. That results in all cases in the velocity of the two clocks, at the end of the acceleration, being 0.866 ls/s.

A funny way to say that you want both clocks accelerated to a rapidity of 1.317c, which indeed is a speed relative to S of √3/2 where the Lorentz factor is 2. Nice round number are easy.

For that first case, compute the reading on the leading clock when the trailing clock reads time "tau".

Frame dependent question which has an ambiguous answer. Both simultaneously read 'tau' relative to S.

Judging by your subsequent text, I'm guessing you didn't want it relative to S, but rather relative to the immediate inertial frame S' of the rear clock at time tau. If you want that, you should explicitly say that. Neither of Einstein's equations yield results relative to S', so that guess of mine is also off topic.

That reading will be given by tau R  =  tau exp(L A).

No. That is Einstein's formula for a different situation (and different coordinate system) than the one you are describing, where both clocks maintain a constant proper separation throughout, effectively putting the clocks at opposite ends of a rigid continuously accelerating thing. You're describing Bell's scenario where both clocks undergo identical proper acceleration, and then only temporarily.

What do you get for the answer to the first case?

I admittedly didn't get one since the scenario is complicated, but it is larger than the one you get with your inappropriately applied formula.

Now [after three different insanely high g's], look at your results.  Does that sequence of readings on the leading clock appear to be approaching a finite limit as tau goes to zero?  Or does it appear to be diverging to infinity as tau goes to zero?

OK, this seems to get to the crux of the matter. The two formulas give the same answer for small values of L, but yours is not small, so the equation (30) is inappropriate for the coordinate system Einstein chooses to use. This describes it in more detail:
https://en.wikipedia.org/wiki/Rindler_coordinates#Variants_of_transformation_formulas

Einstein is using his light-signal sync convention, but with an accelerated frame, which yields equation (2i): dt = e^αxdt₀ where α is proper acceleration and dt₀ is your tau. This is Einstein's choice of coordinate system, which is essentially Lass (or radar) coordinates which do indeed tend towards infinity as acceleration increases to really high values.  His equation (30): dt = (1 + αx)dt₀ is seen for the Kottler–Møller coordinates shown in (2c). Einstein is not using Kottler–Møller coordinates, but the two coordinate systems are differ only in second order effects and are similar for reasonable values for α and x, so he chooses to use the simpler formula in his paper.

The third equation (the Rindler one, 2f) is similar, but without the "1 +" part..

So yes, the equation you quote does indeed tend to infinity as acceleration increases, but that's correct for Lass coordinates, and equation (30) works as an approximation of Lass coordinates for unexceptional values such as those one might see in a real situation. Your mistake seems to be an assumption that simultaneity is being determined by frame S' (as described

You have misunderstood almost everything I've said.  Way too many mistakes to respond to separately.

The ONLY reference frame I'm using is the frame in which the two observers (and their clocks) are stationary (during the entire experiment).  I don't care at all what any perpetually-inertial observer anywhere thinks about anything.  And no such inertial observer has specified ANYTHING in the scenario.  The frame I'm using is EXACTLY what Einstein was using.  At the instant (t = 0) when the two observers fire their rockets, they are the same age (say, zero years old), and their clocks are synchronized at zero.  Each of the two observers agree that their separation is constant (at L lightseconds) during the ENTIRE experiment.

All you need to do (for each of the set of experiments) is use the exponential equation to calculate what the leading clock reads at the end of the constant acceleration "A", when the trailing clock reads the selected value of "tau".

[MikeFontenot (OP)]

[...]

I can give you two additional ways to see that the separation "L" between the two accelerating clocks (with the same constant acceleration "A") is constant, according to the two observers co-located with the clocks.

Method 1:  Imagine that the two clocks are connected by a steel beam of length "L".  And suppose that all along that beam are attached rockets and accelerometers that insure that each part of the beam is accelerating with acceleration "A", just like the clocks (with their rockets and accelerometers) are.  So the beam can't expand or contract in length ... it has the length "L" always.

Method 2: Just use the gravitational scenario that is equivalent to the acceleration scenario, as given by the equivalence principle.  In that gravitational scenario, there is a gravitational field that does NOT vary in strength along the direction of the field ... the gravitational field is always and everywhere of strength "g" = "A".  And there are two stationary clocks separated by the distance "L" along the direction of that field.  Since the clocks in that gravitational scenario have a separation "L" that does NOT vary with time, it follows that the clocks in the equivalent accelerational scenario also have a constant separation "L" that does NOT vary with time.

[Halc]

You have misunderstood almost everything I've said.  Way too many mistakes to respond to separately.

Translation, you can't refute anything I've said, mostly because your replies  show that apparently you've not even read most of it.
I've responded point by point to your comments. I even invited you to point out explicitly how any given comment of mine is wrong, and apparently you can't.

The ONLY reference frame I'm using is the frame in which the two observers (and their clocks) are stationary (during the entire experiment).

Well if you actually read my comments, you'll see that there is no such frame, except S, and then only at t=0. In any other frame after t=0, the two observers are increasing their proper distance, hence the string breaking in Bell's spaceship scenario, which, again if you had actually read my comments, is what you're describing.

I don't care at all what any perpetually-inertial observer anywhere thinks about anything.

That's good because neither did I. OK, I mentioned and dismissed the inertial frame S since Einstein does, but none of the measurements are taken from it. You seem to be interested in an accelerating frame of the rear observer, but a different accelerating coordinate system than the one Einstein is specifying.

The frame I'm using is EXACTLY what Einstein was using.

Einstein is using the Lass frame of the rear observer. Are you? Einstein has his observers moving very differently from what you've specified.

At the instant (t = 0) when the two observers fire their rockets, they are the same age (say, zero years old), and their clocks are synchronized at zero.

Synced to zero in the S frame, and you said you didn't care about any inertial frame. Now you contradict that.
Nevertheless, this is close enough to what Einstein says, yes.

Each of the two observers agree that their separation is constant (at L lightseconds) during the ENTIRE experiment.

Nonsense. Read up on Bell's spaceship 'paradox'. You are exactly describing that and now what Einstein describes. The separation grows and the string (which represents constant proper length) breaks.
Pretend, just for a couple minutes, that I actually have a point, and that maybe, just possibly, you've made a mistake here. The ships cannot maintain constant proper separation if the do what you describe. That knowledge has been around for over 70 years, and actually well before Bell published his thing explicitly pointing this out.

All you need to do (for each of the set of experiments) is use the exponential equation to calculate what the leading clock reads at the end of the constant acceleration "A"

I did that, if you actually read my reply.

Why are you resisting this? You're painting yourself as a troll, when I know you're better than the typical nonsense slinger we get here. So why the refusal of the constructive criticism?
Why did you post you stuff in a forum if you don't want discussion about it? We have rules about just evangelizing your pet blog or whatever, and this seems to be your only motive. Actually writing something without errors is not a goal for you.
Admittedly, there's good money to be had in writing fallacious stuff that shows Einstein to be wrong. There are crank sites that gobble up such articles. Maybe conspiracyoflight would be a platform more to your liking.

Imagine that the two clocks are connected by a steel beam of length "L".  And suppose that all along that beam are attached rockets and accelerometers that insure that each part of the beam is accelerating with acceleration "A", just like the clocks (with their rockets and accelerometers) are.  So the beam can't expand or contract in length ... it has the length "L" always.

This is Newtonian thinking. A rigid accelerating object must accelerate more at the rear than at the front, due to relativity of simultaneity. I was hoping you knew this.
Think about it. After that one second at the rear, several seconds have passed up front, and if it had been accelerating at the same rate as the rear, it would be moving at many times the speed (celerity) of the front, which would increase the length between them.

So given rigid motion and your 3rd example where tau at the rear is 0.01 second (acceleration of over 3 billion g), the acceleration at the front would be about 420k g, or about 1/7500th as much. In that manner, both ends of the rigid object end up moving at the same rate (stopped) in the accelerating frame of the object.

Method 2: Just use the gravitational scenario that is equivalent to the acceleration scenario, as given by the equivalence principle.

Same thing. Gravity would be 3 billion g at the rear and 0.42 million g at the front. Only in Newtonian mechanics is it possible for a gravitational field to be the same strength (acceleration) everywhere. Under relativity, a uniform field is defined as one without tidal forces, and that has the same gravity dropoff over distance as the rigid acceleration case, else 'the string/beam would break', due to tidal forces.

In that gravitational scenario, there is a gravitational field that does NOT vary in strength along the direction of the field ... the gravitational field is always and everywhere of strength "g" = "A"

This is impossible under relativity. Under Newtonian physics, it can be approached with a sufficiently large plane of material (a sheet of aluminum foil say). Under relativity, spacetime would be curved, making the strength (coordinate acceleration) of the field 1/r, far different than the flat strength of Newtonian gravity.

[MikeFontenot (OP)]

MikeFontenot previously said:
Imagine that the two clocks are connected by a steel beam of length "L".  And suppose that all along that beam are attached rockets and accelerometers that insure that each part of the beam is accelerating with acceleration "A", just like the clocks (with their rockets and accelerometers) are.  So the beam can't expand or contract in length ... it has the length "L" always.

(Halc responded):
This is Newtonian thinking. A rigid accelerating object must accelerate more at the rear than at the front, due to relativity of simultaneity. I was hoping you knew this.

(MikeFontenot responds):
No.  All of the rockets are instructed to maintain an acceleration of exactly "A" ls/s/s at their location on the beam, as shown on their accelerometers.  So the acceleration IS exactly "A" ls/s/s everywhere along the beam, and also at both clocks (which also have their own accelerometers, and their own rockets which precisely maintain a constant "A" ls/s/s).

MikeFontenot previously said:
Method 2: Just use the gravitational scenario that is equivalent to the acceleration scenario, as given by the equivalence principle.   In that gravitational scenario, there is a gravitational field that does NOT vary in strength along the direction of the field ... the gravitational field is always and everywhere of strength "g" = "A".  And there are two stationary clocks separated by the distance "L" along the direction of that field.  Since the clocks in that gravitational scenario have a separation "L" that does NOT vary with time, it follows that the clocks in the equivalent accelerational scenario also have a constant separation "L" that does NOT vary with time.
[/quote]

(Halc responds):
Same thing. Gravity would be 3 billion g at the rear and 0.42 million g at the front.

(MikeFontenot responds):
No.  By the equivalence principle, the gravitational scenario that is equivalent to the accelerational scenario which has the same constant acceleration "A" at both separated clocks (with a a constant separation), MUST have the same gravitational field at each (fixed) location of the two clocks.

(Addendum by MikeFontenot]):
Most of your (Halc's) statements, about the accelerations and the distances that differ from my statements, differ because you are using the conclusions about the scenario ACCORDING TO SOME PERPETUALLY-INERTIAL OBSERVER.  I'm NOT interested AT ALL in the conclusions of any perpetually-inertial observer in my acceleration scenario.  And there are no perpetually-inertial observers who are exerting any CONTROL over the rockets ... the rockets are controlled by the accelerometers to produce an acceleration of the accelerating observers of "A" ls/s/s.  Perpetually-inertial observers are irrelevant to my arguments.

[Bored chemist]

No.  All of the rockets are instructed to maintain an acceleration of exactly "A"

From whose point of view?

[MikeFontenot (OP)]

No.  All of the rockets are instructed to maintain an acceleration of exactly "A" ls/s/s.

From whose point of view?

From the point of view of the accelerometer stationed with the rocket.

[MikeFontenot (OP)]

The acceleration varies along the length of a rigid beam.

No it doesn't.  The "Method 1" example I gave you in an earlier post clearly produces constant acceleration, and a constant separation "L" of the accelerating observers (and their clocks).  I'll repeat it here:

Method 1:  Imagine that the two clocks are connected by a steel beam of length "L".  And suppose that all along that beam are attached rockets and accelerometers that insure that each part of the beam is accelerating with acceleration "A", just like the clocks (with their rockets and accelerometers) are.  So the beam can't expand or contract in length ... it has the length "L" always.

[Halc]

From whose point of view?

Mike has specified proper acceleration, which unlike coordinate acceleration, is frame independent. Proper acceleration is what the accelerating thing measures.
One can maintain constant proper acceleration indefinitely (by any clock), but coordinate acceleration only for finite time. For instance, 1g of constant coordinate acceleration cannot be maintained (from a stop) for an entire year.

No.  All of the rockets are instructed to maintain an acceleration of exactly "A" ls/s/s at their location on the beam, as shown on their accelerometers.

Well you can't have your cake and eat it too. They're connected by a beam, so either the rockets are strong enough to break the beam (Bell's scenario), or the front one can't pull away from the rear one (which is stationary in the rear ship frame) due to the tension in the beam.

So the acceleration IS exactly "A" ls/s/s everywhere along the beam, and also at both clocks (which also have their own accelerometers, and their own rockets which precisely maintain a constant "A" ls/s/s).

That's quite impossible given the premises of relativity. It's as I said, having your cake and eating it too. The acceleration varies along the length of a rigid beam. Only one ship is needed if you have the beam. The front one can say pull the entire object given a pre-stressed cable/beam. The rear will measure more acceleration than would the ship at the front pulling it. This is spelled out in countless websites, notably the on on Rindler coordinates which I linked.

No.  By the equivalence principle, the gravitational scenario that is equivalent to the accelerational scenario

Not true. The EP is only a local principle, and this is not a local scenario. The acceleration case is flat Minkowskian spacetime, and the frame of any point along the beam comprises a Rindler frame which is just different coordinates assigned to spacetime which is still flat. The gravity case is quite different. Gravity under relativity is spacetime curvature, and thus by definition not Minkowskian. It is locally flat, sure, but there cannot be relativistic gravity without curvature.

Most of your (Halc's) statements, about the accelerations and the distances that differ from my statements, differ because you are using the conclusions about the scenario ACCORDING TO SOME PERPETUALLY-INERTIAL OBSERVER.

I'm not. I'm doing almost everything in the frame of one ship or the other. I specify my frames when it matters. 'Identical proper separation of ships' means the separation is constant in the frame of either ship.

The acceleration varies along the length of a rigid beam.

From https://en.wikipedia.org/wiki/Born_rigidity#Stresses_and_Born_rigidity

If the endpoints of a body are accelerated with constant proper accelerations in rectilinear direction, then the leading endpoint must have a lower proper acceleration in order to leave the proper length constant so that Born rigidity is satisfied.
...
However, if a different acceleration profile is chosen by which the endpoints of the body are simultaneously accelerated with same proper acceleration as seen in the external inertial frame, its Born rigidity will be broken.

The above quote (my bold) agrees with all my posts: Either the beam holds the observers at constant separation with differing accelerations (the 1st case), or the proper accelerations are the same and the beam/cable breaks (2nd case).

I invite you to find something supporting your position, preferably better than say quora, in which one can find support for pretty much any position no matter how absurd.

No it doesn't.  The "Method 1" example I gave you in an earlier post clearly produces constant acceleration, and a constant separation "L" of the accelerating observers (and their clocks).  I'll repeat it here

Actually, you've repeated this assertion many times now, all without backing it up. I don't need to have you do it again. Your assertions lead to direct contradictions except in Newtonian physics, which is apparently where your intuitions lie if it 'clearly' works.

I am using mostly inertial frames here since you're observers are all inertial all the time except for one second each.

So let's examine your specific complicated scenario. Two ships (Fred in front, Ralph in rear) stationary in S, which simultaneously accelerate at 1.317 ls/s (about 40 million g) for one second each as measured by their respective clocks. They go inertial after that, each with a speed of 0.866c (γ = 2) relative to S.
I'm going to start with both ships at time -1.315 seconds in S, at which point they sync their clocks to -1 seconds. Ralph is a x=-.76 and Fred at x=+6.76
The reason for these numbers is to put Ralph right at x=0, t=0 in all frames at the end of his acceleration, making the frame rotations easier.

So both ships take off at that moment and cease acceleration at t=0 on their local clocks, which is also t=0 in S, but now they're both moving at 0.866c in S, with Ralph at x=0 and Fred at x=7.52.
Let's do a Lorentz transform to S' which is the inertial frame a v=.866c relative to S with the origin at Ralph at 0,0. The event where Fred shuts off his engine at

t' = γ (t - vx) = 2 (0 -.866•7.52) = -13.025 seconds

which means that Fred will experience 14.025 seconds to Ralph's 1 second, a different value than given by any equation I've seen from you, which answers your 'challenge' question posed in post 6.

x' = γ (x - vt) = 2 (7.52 - 0) = 15.04 ls

The separation has doubled. Both ships are stationary in S' after the acceleration of each ceases, so this value doesn't subsequently change in S over time. 'Clearly' the mathematics shows your intuitions to be incorrect.
Moreover, in the accelerating frame ∑ of the beam attached to the front of Ralph's ship, the beam has been accelerating the entire time that Ralph does. Fred on the other hand does all his acceleration in the 1st 14th of the duration in that frame and coasts the other 13/14th of the time. Of course he's going to outrun the front of the beam.

Please let me know which of the calculations above are wrong and why. Please do the calculation correctly, since you know deep down that you are right and the Einstein, Bell, and all those other physicists don't know what they're talking about. But if you're just going to repeat the same assertions, I will assume you're just a mathematics denier.

[MikeFontenot (OP)]

[...]
I am using mostly inertial frames here since your observers are all inertial all the time except for one second each.
[...]

(The red highlighting above was added by me [Mike Fontenot]).

Your statement (in red) is incorrect.  NONE of my observers are inertial, except at the instant before they turn on their rockets. They are accelerating with a constant acceleration, as indicated by accelerometers attached to them.

[pzkpfw]

It's a bit like your experiment needs you to move faster than light. You are relying on something that can't be done.

Halc has shown your multiple accelerometers and rockets won't work the way you expect.

[MikeFontenot (OP)]

It's a bit like your experiment needs you to move faster than light. You are relying on something that can't be done.

There isn't any faster than light motion in my scenario, or in my proposed experimental test.

 

[pzkpfw]

It's a bit like your experiment needs you to move faster than light. You are relying on something that can't be done.

There isn't any faster than light motion in my scenario, or in my proposed experimental test.

 

I didn't say there was. Note the use of the word "like" in my post. I was trying to get you to see you are relying on something impossible, to show something else.

Before you can test your "Gravitational Time Dilation Equation", you need to take a step back and show your method would work.

[Halc]

your observers are all inertial all the time except for one second each.

Your statement (in red) is incorrect.  NONE of my observers are inertial, except at the instant before they turn on their rockets. They are accelerating with a constant acceleration, as indicated by accelerometers attached to them.

Perhaps I read your scenario wrong. I was under the impression the acceleration ceased for both ships after one second, based on the following statement, my bold showing where I got that impression:

In each of the cases, the acceleration "A" and the duration "tau" of the acceleration are chosen such that the product "A tau" is constant for all cases at 1.317 ls/s.  That results in all cases in the velocity of the two clocks, at the end of the acceleration, being 0.866 ls/s.

You said the duration of the acceleration was one second, which gave me the impression that the acceleration ceased after that. The second part I bolded talks about the end of the acceleration, also suggesting that the acceleration ends.

You posted no correction, so I can only assume that perhaps you meant the acceleration to last indefinitely, but you want to talk about the specific moment after Ralph has accelerated for that first second.  I can do number for that case as well. At least Fred will be accelerating the whole time, solving one of my complaints.
I will also introduce a 3rd observer Bob the beamer who is at the front of the 7.52 ls beam that is always everywhere stationary in Ralph's accelerating reference frame. Bob is initially in the same location as Fred when both start their acceleration. According to you, Bob and Fred are always together, of course that cannot be.


Same initial conditions as before, so Ralph is at t=t'=0 ,x=x'=0 after his first second of acceleration.  I will consider the same inertial frame S' in which Ralph is momentarily stationary at that event.  Where are Bob and Fred at t'=0?
Bob is easy. At t'=0 he's at x=7.52 and is stationary by definition relative to the beam.

Fred, at the event where his own clock reads zero is stationary in S' at:

t' = γ (t - vx) = 2 (0 -.866•7.52) = -13.025 seconds
x' = γ (x - vt) = 2 (7.52 - 0) = 15.04 ls

We computed that at the prior post. This is the only moment he is stationary in S' since we're going to have Fred continue his acceleration forever this time. We want to compute his location at S' time t' = 0, and we also want to know what his clock says at that event.

I plugged in 13.025 more seconds (as measured in S') into my calculator and got  x=12.29 additional ls of distance on top of the 15.04 totalling 27.33 ls, almost 4x further away from Ralph than the end of the beam where Bob is.  Fred's clock reads +2.685 seconds and he's moving at 0.998c relative to Ralph and Bob and the beam, hardly stationary as you assert.

If I once again got the scenario wrong, then tell me clearly what the scenario is instead of just saying I didn't guess correctly.
No matter the scenario, it is clear (despite your intuitions) that Fred isn't going to stay put at the end of the beam with Bob. Bob is accelerating at around 4.7 million g in order to stay with the end of the beam, where Fred and Ralph are accelerating at over 40 million g.

I have no numbers from you. Just assertions. You're supposed to show the correct mathematics because obviously mine is all wrong.

[MikeFontenot (OP)]

Perhaps I read your scenario wrong. I was under the impression the acceleration ceased for both ships after one second,[...]

It actually doesn't matter whether the acceleration ends at "tau" or later on ... what's important is what we calculate to be the state at the given time "t" = "tau" that we have selected for the end of a given experiment (with the acceleration starting at "t" = zero).  Because of the principle of causality, what happens later has no effect on the state at "tau".  I think I DID word it with the acceleration ending after "tau", but it doesn't really matter if it goes on longer or not ... it's the state at "tau" that we want to calculate.  And then, we want to do a sequence of experiments, always with the product ("A" "tau") = 1.317 ls/s, but with each iteration of the experiment reducing "tau" by a factor of ten, and increasing "A" by a factor of ten.  That gives us a sequence which makes it clear what happens to (tau R) as "tau" goes to zero ... i.e., what happens to the reading on the leading clock when there is an instantaneous change of velocity from zero to 0.866 ls/s.

____________________________________________

Previously, I've given two arguments for the conclusion that the separation of the two clocks (with their attending observers) is constant (my "Method 1" and "Method 2").  There is also a third argument that I will give now:

Method 3:

Einstein gave us the exponential equation

  R(L, A) = exp(L A)

that tells us how much faster the leading clock is ticking, compared to the trailing clock's tic rate, GIVEN THAT the distance between the two clocks is "L" and the acceleration of both clocks is "A".  So "L" and "A" are what we call "the independent variables", and "R" is "the dependent variable", of the equation.  If you specify the values of "L" and "A", we can calculate "R".

But if "L", as you claim, VARIES during the acceleration, it can't be one of the independent variables, which are constant during the experiment.  If Einstein had intended that "L" vary during the acceleration, he would have needed to SAY how "L" varies during the acceleration ...  there is no way to carry out the computation of the exponential equation without knowing what "L" and "A" are equal to. Einstein gave no such information about "L" varying, nor about HOW it varies at a function of "t" during the acceleration.  Clearly, Einstein regarded the separation "L" of the clocks to be constant during the entire acceleration.