Talk-Polywell.org

i've already posted the math to prove what i'm saying. the case was closed a long time ago - decades ago, in fact.

it's a gradual phenomena, not discontinuous. the interference pattern always occurs, just at varying phases, amplitudes, scales, etc, as a function of the slit separation among other things. it is always interference.

Happyjack. I'll happily read your posted math but I don't remember anything relevant on this thread.

When you say the case was closed, are you claiming that my statements are contrary to what most physicists and experimentalists believe, or that you believe mainstream thought then and since has ignored obvious arguments and/or data?

I suspect you are misunderstanding what I've said. But the conclusion, that the speed-changing twin comes back slightly younger than his sibling, is both accepted theoretically, and proved experimentally, many times. I've posted detailed explanations (the most detailed from Tom Van Flandern). No-one has challenged this.

I'll do the easier one!

A is stationary twin, B is twin who travels as speed v (relative to A) to alpha centauri and then at speed v back. Suppose B clock shows elapsed time T during journey from earth to alpha centauri.

LT transforms between A and B clocks.

Since A frame stays constant (fixed newtonian frame) we can use A's transformation of B clock consistenctly through journey to compare times.

Note we can't do the reverse so simply since B's frame changes, which means a correction must be made at the frame change to B's transformation of A's clock.

we know that A clock changes by T/gamma during outward journey as B clock changes by T (the subjective journey time).

gamma = 1/(sqrt(1-v^2/c^2).

Similarly on the return journey. Although the speed and therefore LT is different, the coefficient between A and B times is the same , and there is no change in the B clock time as transformed into A frame during the turnaround.

Therefore total elapsed time on clocks when they meet is:
B: 2T
A: 2T/gamma > 2T

Note I'm assuming here that v is high enough for this effect to be much larger than any GR effects due to gravity well (which would make A clock read slightly slower than B clock).

This proves my statement about proper time of "bent" path being less than proper time of stright path between same two points with timelike separation. (the points here are the B leaving A and B returning to A events).

Best wishes, Tom

ok, now do it from B's reference frame.

tom, here is the math i posted.

Sure. But which reference frame? LT applies only to reference frames which do not accelerate (inertial frames), so there are two distinct frames for B, B1 and B2. (B1 outgoing, B2 incoming).

In B1 A's clock is slowed down by gamma (symmetrical and opposite from the previous argument. Similarly in B2.

The problem is that B1 and B2 are different, and therefore the transformed times of A's clock in the two frames are very different. Call this time shift of the A clock between B1 and B2 Ts.

In B's instantaneous frames for an elapsed time of 2T on B's clock we therefore have an elapsed time of 2Tgamma+Ts on A's clock.

We need to calculate Ts. It will be precisely enough to make this calculation consistent with the previous one (obviously). So we should have:

Ts = 2T(gamma - 1/gamma) = [2T*gamma][1-(1/gamma)^2] =

2T*gamma*v^2/c^2

But before we do that, I'd better deal with any objection you have to this framework, and the fact that the LT time of A's clock in B1 and B2 is different at the point of the turnaround, when B is furthest from A. Note that any such Ts will be proportional to the distance between A and B and therefore will be zero when A and B are in the same place, as at the start and end of the twin paradox journey.

EDIT - corrected eqns above.

This applies directly for what A thinks is the time on B's clock (the LT transformed time of B's clock).

It does not apply to what B thinks is the time on A A's clock. The LT transformed time (really just a calculated time) of A's clock in the two B frames is different, so at the point where B canges frame this results in a correction.


It should come as no surprise that A clock time depends on which reference frame B is in, since in relativistic world changing relative velocity changes the time coordinate of distant events (you get this from the LT).

utterly pointless.

Possibly it is. But your criticism of my (I hope clear) explanation does not get us very far, because it does not exist. Your maths posted above assumes that B stays in a single inertial frame, and therefore does not correspond to the problem.

I think your misconception is the idea that the LT of A's clock in B's frame is the same for B outgoing or ingoing. That of course would make the situation symmetrical.

It is not, because of the time shift Ts.

from LT:
t' is time in B1 or B2 frame when B reverses
t1 is time of A clock for B1 time = t'
t2 is time of A clock for B2 time = t'

t' = gamma (t1 + xv/c^2)
t' = gamma(t2 - xv/c^2)

Ts = t2 - t1 = 2*v*x/[c^2*gamma]

But x = vT =>

Ts = 2*v^2T/(c^2*gamma)

Well, you can see it is of the right form, but I've been careless somewhere with gamma, 1/gamma. I'll return to it when happyjack has admitted that Ts is non-zero (at the moment I think he does not understand why it should exist).

Just to knock this on the head. SR says that the laws of physics should be the same in any inertial frame. In this problem A's trajectory (note compliant use of jargon) is inertial. B's trajectory is not inertial.

That is not a relative statement, it is an absolute one. Neither SR nor GR says that a non-inertial trajectory is equivalent to an inertial one.

GR says that constantly accelerating trajectories are equivalent to inertial ones in a gravity field. In this case because the acceleration is at only one point in time, you would need a gravity field that switched on and off in a most unusual way, and the equivalence would be between the inertial trajectory with the switchable gravity field, and the non-inertial trajectory. Hence A and B remain non-equivalent.

Best wishes, Tom

Tom, you base your reasoning in that for twins to reunite one (or both) has to abandon his initial inertial FOR, and if only one of them does so that asymmetry justifies the age difference when they meet.

But what if we make a different experiment (without twins), one in which no one abandon their initial inertial FORs at any moment?

An observer OB stays in its own inertial FOR all the time. A ship S1 pass it at high relative speed, at the moment of maximum approximation S1 synchronizes an on board clock with another in OB.

S1 never changes speed. Some time later it crosses with another ship S2, inbound toward OB. At the moment of maximum approximation S2 synchronizes its clock with the one in S1.

When S2 gets near OB their respective clocks are compared (without decelerating).

What's the result? Do both clocks, the one that remained in OB and the one that came in S2, give the same time?

_________________
"The problem is not what we don't know, but what we do know isn't so" (Mark Twain)

Good question. That is the one that shows it can't be acceleration that is significant here.

S1 and S2 have different frames. therefore they will calculate the time of a distant clock (that of OB) differently. The time shift correction must be made when changing from S1 to S2.

There is no problem for the clocks on S1 and S2, they can be synchronised at the moment S1 & S2 pass with no ambiguity, since the time shift correction only applies to distant clocks. Also there is no problem synchronising with OB when it passes.

Thus the S1/S2 clock time is identical to the previous B time, and shorter than the OB time. However you cannot calculate the elapsed time of OB clock from S1/S2 without a time shift correction, just the ame as in the B case.

This makes it clear that it is the change of frame that requires the correction, not the physical acceleration. And also that it is only calculating the time of a distant object that changes when there is chnage of frame.

Best wishes, Tom

Sorry Tom, I don't follow you there.

Do you mean that when they compare OB's and S2's clocks they say the same time, or that S2's clock is behind?

_________________
"The problem is not what we don't know, but what we do know isn't so" (Mark Twain)

S2 is behind.

Look at it from OB's frame, the S1/S2 situation is the same as the B situation in my A/B example, and the S1/S2 clocks are behind.

Look at it from S1/S2 frames (which is why you introduced S1/S2). The two journeys are as before, with OB clock slightly behind S1 and also S2. However in changing from S1 frame to S2 frame the same time shift happens as before with B, which makes S1/S2 time slightly behind OB time.

This frame change is only an issue when calculating OB time that corresponds to S1 or S2 time using LT. It is what turns a symmetrical calculation (the two journeys) into an asymmetrical one.