Let's see if i can summarize.

The mainstay of SR is that c (the speed of light in vacuum) is the same in all inertial reference frames. This has some interesting implications--basically, nothing else is the same. Distances, time, "simultaneous" differ from one reference frame to another. There are a handful of simple equations that describe how to translate from one reference frame to another; regardless, translations from one reference frame to another are non-intuitive, and are excellent opportunities for introducing mistakes in calculations.

The equations for relativistic doppler shift are different from the equations for classical doppler shift.

SR has no c-regulators; it is in direct opposition to older, "classical" theories that would require a c regulator.

Time dilation is real and has been demonstrated.

It seems that the validity of the relativistic doppler shift formula could be demonstrated; off hand, i don't know of any experiments that have demonstrated this, but i haven't looked for any.

=D> [-X :roll: [-( 8-[ Wow, lots of appropriate emoticons there...

I'm wondering when BA will decide this crap should end. And indeed I would. Wait, did I just answer Sam5's question? ops: One quick extension: in science, certain things are called paradoxes because they appear to be paradoxes, not because they actually are. If a theory actually contained something self-contradictory and unexplained, it would cease to be a theory.

Not too soon, I hope. There is a lengthy reply I wish to write up while off-line tonight. But in case this thread gets locked before then, I will answer Sam5 thusly: There is no paradox, neither in the case where the twins never meet again, nor in the case where they do meet.

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The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

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Which theory is that? Have you provided a reference somewhere before?
Hopefully, at that reference, we'd see some of that proof. Otherwise, the body of evidence seems to favor Einstein's theory.
That's the central idea of relativity.

When Harry Meets Sally--Part One (Vocals)
In this and the following post I'm going to attempt to deal with so-called twin paradox using K-calculus, giving special attention to the role of the two postulates of special relativity, stated earlier by Glom as follows:

1) The laws of physics are the same in all inertial frames.
2) The speed of light is the same as observed in all inertial frames.

The dramatis personae are Harry who stays at home in one inertial frame, and Sally who sallies forth into space at some goodly fraction of the speed of light.

Harry and Sally both have atomic clocks that are made in Lake Wobegon, where all the clocks are of above-average accuracy. During this experiment they will each transmit their times at periodic intervals using an agreed upon wavelength. Furthermore, Sally's spaceship also has a reflector so that Harry will have both Sally's time signals and the echoes of his own transmissions as data.

There being no such thing as clairvoyance or telepathy, the only way that Harry has of reading Sally's clock and vice versa is by the radio signals they transmit of their clock readings. Harry observes that Sally's signals are redshifted to a wavelength K > 1 times the wavelength that they agreed upon. He also notices that the signals arrive, not at the agreed upon interval, but at K times this interval, so that he receives 1/K as many signals as expected. Harry realizes that there is nothing wrong with either Sally's transmitter or her clock, this is just a consequence of redshift due to their relative motion.

Sally likewise sees Harry's transmissions redshifted and receives fewer signals, which she also attributes to redshifting rather than to a fault with Harry's transmitter and clock.

As long as they continue to depart from one another at this fixed velocity, they will continue to observe each other's clocks as running slow when judged by the redshifted signal which is the only information that they have. There is no paradox here. In order to determine which clock is "really" slower we would need to cause both clocks to be transported close enough together for the light-travel time from both clocks to be negligible, thus allowing direct comparison. But this destroys the symmetry of the situation.

Let us now consider the case where Sally goes out some distance at velocity v for a time H/2 measured by Harry's clock and S/2 measured by Sally's clock, then returns at the same velocity so that Harry and Sally meet when Harry's clock reads H and Sally's clock reads S. Can they have the same reading? Read on!

Immediately after Sally turns around, she observes that Harry's signals are now blueshifted to a wavelength K' < 1 times the agreed upon wavelength, so she recieves signals more frequently now. This will this make up for the redshifted signals during the first half of the trip. She sees (S/2)/K signals from Harry during the first half of the trip, and (S/2)/K' signals during the second half for a total of H signals from Harry. In other words:

(1) (S/2)*(1/K + 1/K') = H.

Harry does not see Sally's signals go over to blueshift until some time after H/2. He sees redshifted signals for (1/2+f)*H and blueshifted signals for (1/2-f)*H, where f is greater than or equal to zero and less than or equal to 1/2. Harry accounts for Sally's signals thusly:

(2) H*(1/2+f)/K + H*(1/2-f)/K' = S.

Now for the $64,000 Question: can H = S? We will suppose that it does and get a contradiction.

If H = S then since neither is zero so we can divide both sides of the two equations by one or the other and obtain:

(1') 1/K + 1/K' = 2, and

(2') (1/2)*(1/K + 1/K') + f*(1/K - 1/K') = 1.

Substituting 2 for (1/K + 1/K') in (2') we find:

(3) f*(1/K - 1/K') = 0.

Either f or (1/K-1/K') (or both) must equal zero. But K is >= 1, K' <= 1, so the only way that the parenthesized expression can be zero is if K = K' = 1, that is, no redshift or blueshift, which is only true if Sally is not moving with respect to Harry, in contradiction to our assumptions. We must then have f=0. But if f=0, then Harry starts getting blueshifted signals right at the moment Sally turns around, in other words, the speed of light must be infinite, thus we have a contradiction and therefore H<>S via reductio ad absurdum.

Let's take a moment to look at what we've done: we have destroyed any notion that there can be such a thing as "universal" or "absolute" time that can be read off a clock. Furthermore, we have only used the first postulate of special relativity to do this. We have not said anything about the values of K and K' apart from the fact that they represent a redshift and a blueshift respectively. To calculate these values and answer the question of how much H differs from S, we must use the second postulate, which we shall do in the second post.

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Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

I now officially condemn CM's skits as smartaleck, ignorant, sophomoric, inflammatory and and a poor reflection on the level of discussion in BAUT. -- Bob Angstrom

When Harry Meets Sally--Part Two (Instrumental)
In this and the previous post I attempt to deal with so-called twin paradox using K-calculus, giving special attention to the role of the two postulates of special relativity, stated earlier by Glom as follows:

1) The laws of physics are the same in all inertial frames.
2) The speed of light is the same as observed in all inertial frames.

In order to calculate the various red- and blue-shifts referred to in the previous post we will have to use the second postulate of special relativity and Harry's reflected signal. (There was a reason for that reflector on Sally's spaceship!)

For Harry's signals that are received during the first half of Sally's trip, Sally (and her reflector) sees them as redshifted, so if the signals came at hourly intervals, Sally and her reflector would receive them every K hours. In turn, considered as a periodic signal, the reflected signal is received by Harry every K*K hours. If Harry emits a signal at a time T and it is reflected when Sally is outbound, he will receive the echo at K*K*T, and the elapsed time between emission and reception is (K*K-1)*T.

Let t be time taken for Harry's signal to reach Sally's reflector. Now we invoke the second postulate. The light signal travels a distance of c*t from Harry to Sally, and the same distance in the same time back again, so that (K*K-1)*T = 2*t. As Harry sees it, Sally has traveled for a total time of T+t to reach the point where the signal reaches her, for a distance of v*(T+t). In other words:

(1) c*t = v*(T+t).

We may solve for t in terms of T:

(1') (c-v)*t = v*T, define beta = v/c, and divide both sides by c:

(1'') t = (beta/(1-beta))*T.

Combining with (K*K-1)*T = 2*t, we have:

K*K-1 = 2*beta/(1-beta),

K*K = (1+beta)/(1-beta),

Therefore, K = sqrt( (1+beta)/(1-beta) ).

Notice that despite the appearance of c-v in equation (1'), nothing is actually moving at that velocity. This is just a formal manipulation in the solution for t in terms of T.

Another thing to note is that for small beta (velocities we're likely to encounter everyday) that K = 1 + beta + O(beta^2). Therefore it is approximately equal to the classical Doppler shift value for small values of beta.

Now, what is the value of S (Sally's time measured by her clock) in terms of H (Harry's time measured by his clock)?

From the previous post we have S*(1/K+1/K') = 2*H. K' = 1/K = sqrt( (1-beta)/(1+beta) ). You may calculate this yourself as an exercise, or you can be lazy and get it from the formula for K by substituting -beta for beta. Then

S * ( sqrt(1-beta)/sqrt(1+beta) + sqrt(1+beta)/sqrt(1-beta) ) = 2*H,

S * ( (1-beta)/sqrt(1-beta*beta) + (1+beta)/sqrt(1-beta*beta) ) = 2*H,

2*S/sqrt(1-beta*beta) = 2*H,

therefore, S = H * sqrt(1-beta*beta).

One last comment: in Euclidean geometry, one never speaks of a "triangle paradox". Each side of a triangle is less than (or equal to in a degenerate case) the sum of the other two sides. No one tries to consider two sides to be one straight line. In special relativity, we have a similar situation in the experiment considered above, except that here the "non-straight" trajectory has less proper time rather than more distance.

I think I'll have what Sally is having!

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

I now officially condemn CM's skits as smartaleck, ignorant, sophomoric, inflammatory and and a poor reflection on the level of discussion in BAUT. -- Bob Angstrom

Exactly what I was referring to Sean. Thanks for the clarification!

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. . . My moustache is touching my brain!!!!

good work fellas

I am not here to tell you what it is! I only want to tell you that you MUST FIGURE OUT ONLY WHAT YOU NEED AND WANT TO UNDERSTAND.

1. Einstien was not reading the mind of god when he created this theory.

2. It was just his ABSTRACTION of the real world.

3. Its good abstraction as long as it works ..otherwise we figure out others.

4. Don't make him god PLEASEEE..

enjoy!

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i hate god