Twins Paradox Paper

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21 years 1 month ago #6956 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by kc3mx</i>
<br />Is Einstein right or wrong? I am not sure what the paper says on this point. It seems to say that LR is more practical than Einsteins theory. But that is not a reason to change textbooks.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

The paper was not about the validity of SR, but rather its physical interpretation. SR has now been falsified by a showing that the speed of gravity strongly exceeds the speed of light in forward time, as shown in other papers on that subject. This paper is concerned with showing that SR is not an internally contradictory theory, but does do violence to our intuitions of what space and time are all about.

The first thing that must go is our intuition about a "universal instant of <i>now</i>". Then one must accept "time slippage" as a real, physical effect. I do not support either concept because I conclude in MM that a "universal instant of now" does exist and that time slippage does not. But I conclude here that SR can only be falsified by experiment and not by claimed contradictions. The latter are claimed by people who, IMO, have not delved beneath the surface and truly understood SR because they are unable or unwilling to suspend their disbelief about its physical implications.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">There seems to be a mistake. The two solutions for time given in the paper are inconsistent. They are not inversely related although they appear to be so on the surface. So I think this needs to be clarified since the notation was unclear to me.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

You might wish to explain what you mean. When I substitute one set of equations into the other, I get an identity. Did you fail to recognize gamma^2 when you did this? -|Tom|-

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21 years 1 month ago #7122 by kc3mx
Replied by kc3mx on topic Reply from Harry Ricker
This last comment is what I was thinking. But I am not sure I understand the notation. I use the traditional unprimed and primed notation for S and S'. Then using g for the gamma factor, I interpret the equations as t'=gt and t'=t/g. These do not produce an idenity.To get an idenity you need to write t'=gt and t=t'/g. These equations are inversely related and give the idenity upon substitution.I interpret the first two as the equations used in the paper not the last two. That is why I thought that there was an error. Perhaps I misunderstood the notation.

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21 years 1 month ago #7013 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by kc3mx</i>
<br />That is why I thought that there was an error. Perhaps I misunderstood the notation.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

You didn't misunderstand the notation. You skipped the "time slippage" term, which was one of the chief points of the article, and is absolutely essential to any hope of understanding SR. (Again, I don't agree with SR, but I do now understand it.)

Ask yourself why g, which is always greater than or equal to unity, is in the numerator of your first equation [t'=gt] despite the fact that t' clocks must be ticking <i>slower</i> that t clocks as viewed from frame S.

Switching to your notation, the first Lorentz equation is not just t' = gt, as you had it. It is t' = g(t - vx). (Note: To keep this simple for ascii purposes, I'll drop c^2 wherever it appears. It just sets the units. So g = 1/sqrt(1-v^2). You can put the c^2 back if you wish. Nothing in what follows will be affected.)

Next, use the first "length contraction" equation: x' = g(x-vt).

Solve this for x: x = x'/g + vt.

Substitute for x in the first time equation: t' = g(t-vx'/g - v^2 t).

Simplify this to: t' = gt - vx' - v^2 gt = gt (1-v^2) - vx'.

Use the definition of g in the form: 1/g^2 = (1-v^2).

Substitute this for (1-v^2) in the time equation: t' = t/g - vx'.

Rearrange this to be an equation for t in terms of t': t = g(t'+vx').

Finally, substitute -v for v because, from frame S', the velocity is in the opposite direction: t = g(t'-vx').

This is the second time equation just as I listed it. SR requires that it look just like the first time equation; i.e., the time transformation law must look exactly the same no matter what frame it is used in. My article discusses the physical implications of the vx (or vx') "time slippage" term, and why g switches from denominator to numerator during this process. After you have mastered the above derivation, perhaps the article will start to make more sense to you. So might some of Einstein's early writing. -|Tom|-

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21 years 1 month ago #7021 by kc3mx
Replied by kc3mx on topic Reply from Harry Ricker
I am afraid I didn't miss it. There is no such thing as time slippage. But I intend to keep an open mind concerning this point. What you call time slippage is the traditional method of obtaining a contracted time or distance interval using the Lorentz transformation. Einstein introduced this in his 1905 paper to prove that moving clocks run slow. But he really showed that the clock runs fast. Since the time slippage as you call it, makes the clock run fast, then this compensates for the clocks in orbit running slow. But this is really based on changing the time scale of the GPS satellite to compensate the effect of the orbital motion on clock rate.

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21 years 1 month ago #7028 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by kc3mx</i>
<br />There is no such thing as time slippage.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

If you have understood anything I said, how can you say that? By using the time transformation equation, I showed the amazing but indisputable fact of SR that clocks in any inertial frame with a relative motion will run slower than one's own clocks, yet frame time in that same inertial frame with a relative motion goes by faster than our own frame time. ["Frame time" means time always measured at the nearest point in any passing frame.]

For example, if v = 0.99c, then gamma = 7. As we look into that fast-moving frame, each and every clock within in runs 7 times slower than our clocks after allowing for light-time propagation. And it is a given that all clocks in that frame were synchronized within the frame. But as seen from our frame, time shown on a succession of passing clocks in the fast-moving frame will elapse 7 times faster than time in our own frame. And if we accept SR premises, there is nothing contradictory about these statements! I chose to call this phenomenon "time slippage". You may not care for my term, but I don't see how you can deny the phenomenon without denying SR.

Did you follow the derivation I gave? (If the math is beyond you, don't be embarrassed to say so. Math is not eveyone's bag.) If you did follow, then you surely see the essential role of the vx term in showing why gamma is sometimes in the numerator and sometimes in the denominator -- and correspondingly, why frame time differs from time as shown on any single clock within a frame having relative motion.

Is your goal to understand SR the way others do, or to justify your own critique of Einstein's papers? I'm afraid the latter is not going to survive this kind of scrutiny. But if you've showed your idea elsewhere, you already know that. -|Tom|-

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21 years 1 month ago #7439 by Jim
Replied by Jim on topic Reply from
Here we have a situation where one side is talking about a model and the other is talking about the real stuff. Both arguments are right and that fact seems to elude both sides.

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