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Old 07-03-2009   #42 (permalink)
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AnssiH
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Re: An “analytical-metaphysical” take on Special Relativity!

Just a quick reply...
Oh btw, I started to comment on Dennet and Consciousness Explained, but it went totally off-topic so quickly that I decided to not post it here. Anyhow, have not read the book, but been meaning to many times.

Quote:
Originally Posted by Doctordick View Post
You were dividing through (through the whole equation) by the coefficient of \delta : i.e., the factor \sqrt{1-\left(\frac{v}{v_?}\right)^2}v_?.

You just did the square root first instead of after. Perhaps I confused you by my comment, "divide through by the coefficient of \delta...", as I should have said "divide through by the coefficient of \delta^2...".
Hehe, actually it was just another silly ambiguity that got me there. \delta was also a coefficient by itself so I interpreted "divide through by the coefficient of \delta" as "divide through by the coefficient \delta", i.e. I attempted to divide through with \delta itself. That didn't get me anywhere!

Quote:
You just proved to yourself (by that common denominator move) that

\delta= \left(\frac{v}{v_?}\right)\frac{1}{v_?\sqrt{1-\left(\frac{v}{v_?}\right)^2}}

which implies (multiplying both sides by v_?^2 that

v_?^2\delta= \frac{v}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}

but we have also just shown that

\alpha=\frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}

Those two equations taken together imply that v_?^2\delta=\alpha v.
Ah, I see it now.

Quote:
If you look back at the OP I think you will find that [\gamma=\alpha \frac{\beta}{v_?^2\delta}] is one of the original four equations we started with.
Ah, spotted it now

Back to OP:

Quote:
Originally Posted by Doctordick View Post
Finally, since \beta=\alpha v and v_?^2\delta=\alpha v, it is quite obvious that \gamma=\alpha \frac{\beta}{v_?^2\delta} clearly implies \gamma=\alpha.
So now I understand that bit.

Quote:
At this point, we have solved the problem; from the above it is quite clear the only possible relationship which can exist between moving coordinate system (moving at constant velocity v) is given by;

x'=\frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}[x-vt]\quad \quad t'= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}\left[t-\left(\frac{v}{v_?}\right)\left(\frac{x}{v_?}\right)\right]
Right, so transformation for x' is:
x'=\alpha x -\beta t

And
\alpha= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}
\beta=\alpha v

So clearly
x'=\frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}[x-vt]

I was struggling with t', and after looking at it for a while, it seems like it was due to an error earlier in the OP;

Quote:
...it is fairly easy to show that the transformation from one coordinate system to the other can be no more complex than x'=\alpha x -\beta t and t'=\gamma x -\delta t
I think that should be t'=\gamma t -\delta x in there. That's how it's used in the subsequent steps in the OP, including the final result (me thinks).

So, assuming I'm right:
t'=\gamma t -\delta x

And
\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}}
\delta = \left(\frac{v}{v_?}\right)\frac{1}{v_?\sqrt{1-\left(\frac{v}{v_?}\right)^2}}

Then
t'= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}} t - \left(\frac{v}{v_?}\right)\frac{1}{v_?\sqrt{1-\left(\frac{v}{v_?}\right)^2}} x

And to tidy it up:
t'= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}} t - \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}} \left(\frac{v}{v_?}\right) \frac{x}{v_?}

t'= \frac{1}{\sqrt{1-\left(\frac{v}{v_?}\right)^2}} \left [t - \left(\frac{v}{v_?}\right) \left (\frac{x}{v_?}  \right ) \right ]

Which is your result. Yup, looks valid.

Quote:
Just for convenience, one can define sine\theta = v/v_? as this makes the square root in the above equations equal to cos(\theta) yielding a simpler representation. If that constant velocity v? were to be c, those would become exactly the standard relativistic transformations.
Yup.

Quote:
I did this derivation in detail for one very simple reason: most publications merely publish the results and imply that their truth is support for Einstein's theory of special relativity. I prefer to view it as nothing more than the result of requiring a very specific symmetry: namely that some specific velocity must be the same in any inertial coordinate system. These relations are exactly the standard Lorentz transformations Einstein's theory of special relativity was concocted to explain.
Yup, so we've drawn out - with explicit logical steps - that the transformation required to maintain a specific constant velocity between moving coordinate systems, must (unsurprisingly) affect the way that all the data is laid down in a given coordinate system. And that the transformation is, unsurprisingly, Lorentz transformation.

And from that I also know that if the data is plotted in a coordinate system where "t" is also expressed as an axis of its own, the transformation can be intuitively understood as a straightforward scale procedure.

And in this context the transformation is simply a requirement of the explanation to remain self-coherent when it's referring to the same data expressed, only expressed in different (moving) coordinate systems.

Quote:
The fact that my model requires them for internal consistency implies that my model actually requires any conceivable universe to satisfy the relations associated with special relativity.
Yup, certainly looks that way.

Quote:
Unless you have more questions, I think we are done.
I think I'll still walk through the bits about definition of simultaneity carefully, albeit the result there seems to me to be a very obvious consequence of everything we've talked thus far. But at least I might spot some more typos

Also I'll do those clarifying animations for all the lurkers, and have a stab at explaining what all this means as it appears there is quite a bit of confusion there still, and I think I know where that confusion is (people try to view this result as if it says something about the ontological reality of nature, as otherwise they don't understand how could any result be meaningful)

Quote:
I appreciate the time you have put into this very much. I really didn't realize how bad my presentation was until you took the trouble to try and understand it. I hope I can do a better job of presenting the GR deductions. The math there is nowhere near as complex as Einstein's Rieman geometry but neither is it "easy". The calculus required includes use of the Euler–Lagrange equation which is rather advanced mathematics.
Okay, as it is with SR, I'm not really familiar with the math of GR. I just look at it in a kind of "intuitive" way, imagining a spacetime transformation in my mind. But I'm sure if I'm careful enough and ask enough questions, I can walk through it as well.

Quote:
I hope I can drag you through it.
Certainly. A new thread?

-Anssi
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