Read post 319 , I do not have time for your continued idiocies. I should have posted it long ago. Feel free to continue sucking up to the crackpot Shubert.
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On the Definition of an Inertial Frame of Reference
- Thread starter Eugene Shubert
- Start date
Read post 319 , I do not have time for your continued idiocies. I should have posted it long ago. Feel free to continue sucking up to the crackpot Shubert.
Read post 319 , I do not have time for your continued idiocies. I should have posted it long ago. Feel free to continue sucking up to the crackpot Shubert.
Why don't you also suck on this for a while (not that it matters) :
"In this paper we examine the fundamental concept of inertia in the special
theory of relativity, adopting for that purpose the broader scope of the principle
of general covariance. It will be shown that, provided space-time is flat, the
vanishing of the components P/ou(X v) of the Christoffel symbols I represents
a necessary and sufficient condition for the system of coordinates (x v} to
define an inertial frame of reference. Some consequences of this will also be
briefly discussed.
It is today firmly established (even for the special theory) that we must
look upon Einsteinian relativity as the geometry of space-time interpreted
from the standpoint of physics [see, for instance, Synge (1969), p. 34].
Accordingly, we essentially consider special relativity as the theory of fiat space-
time. 2 Indeed, following this approach, one states the geometric characterization
of Minkowski space-time by requiring a vanishing Riemann-Christoffel
curvature tensor everywhere on the relativistic manifold. Physically this means,
of course, that one leaves out genuine gravitational phenomena."
http://www.springerlink.com/content/l81441110u820x31/fulltext.pdf
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Is this you, again, being unable to admit to simple mistakes and trying to bluff a hasty retreat? If I've really made an error, where is it? What do you think the Jacobian of $$\theta$$ is? What do you think the metric after applying $$\theta$$ is? Do you actually have any good answers? Or are you just going to continue hollering insults from the ivory tower you've created for yourself?
You really can't handle when people point out mistakes in your posts, can you?
No, I calculated the jacobian directly from the transforms (54)(55). Had you done the same thing instead of swallowing Shubert's false claims of "decomposition" you would have learned about your error.
You bought Shubert's claim on the decomposition. The claim is FALSE. To prove that his claim is false all you needed to do was to calculate the jacobian from scratch, a very easy thing to do. I have already done it for you.
The calculation of a lazy person that bought the claims of a crook (Shubert) line, hook and sinker about the transform decomposition. This is the price you pay for being lazy and not checking his claims.
Here are the elements of the jacobian calculated correctly:
$$\frac{\partial t'}{\partial t}=\gamma(1-v \frac{\partial S_j}{\partial x'})$$
$$\frac{\partial t'}{\partial x}=\gamma(\frac{\partial S_j}{\partial x'}(1+v \frac{\partial S_i}{\partial x})-\frac{v}{c^2} (1+\frac{c^2}{v}\frac{\partial S_i}{\partial x}))$$
$$\frac{\partial x'}{\partial x}=\gamma(1+v \frac{\partial S_i}{\partial x})$$
$$\frac{\partial x'}{\partial t}=-\gamma v$$
If you took the trouble to examine the completely insane Shubert transforms (eqs. 54-55) you would have noticed that they aren't only nonlinear but that the second clock synchronization function $$S_j$$ depends on x'. How can the clock synchronization function in the unprimed frame depend on the coordinates from the primed frame? Total and utter insanity. This should have clued you in that the transforms are invalid and unphysical.
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Neddy Bate
Valued Senior Member
This thread is a joy to follow. My favorite parts, so far:
Just out of curiosity, what thread are you talking about?
- The many times Tach was shown to be in error
- The many times Tach refused to admit his errors
- The many times Tach told everyone else that they were wrong about his errors
- When Tach got temporarily banned
Tach said:
Just out of curiosity, what thread are you talking about?
Sorry I omitted you in the list of idiots. You have "proved" yourself in another thread. Mea culpa.
The one where you demonstrate your ignorance about the Sagnac effect.
The one where you carry over your idiocies on the Sagnac effect into your crackpot attempt at clock synchronization in rotating frames.
Neddy Bate
Valued Senior Member
I didn't know you had a list, but I'm honored to be on it. I was asking you what my "idiotic thread on the Sagnac effect" refers to. You mentioned it a couple of times in the thread that got locked, so it must not be that thread. What thread are you referring to?
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Eugene Shubert
Valued Senior Member
You didn't do it correctly.
That's completely wrong. Why do you refuse to learn the chain rule and use it? Also, there is plainly no c in equations (54)-(55).
LOL, you know you have been skewered.
Eugene, you are a pathetic and complete fraud,$$k=1/c^2$$ . how do you feel having your balls in a vise? Why don't you show how you calculate the jacobian for (54)(55)?
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Eugene Shubert
Valued Senior Member
Hmm, you are into S&M, you've had your balls in a vise for quite a while now, you must be enjoying yourself. You are squirming , Eugene, vise too tight around your balls?
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Neddy Bate
Valued Senior Member
That's an old pseudoscience thread created by MacM. I made a few posts in there, that's all. I admit that I didn't understand the Sagnac effect back then. See? That's called admitting one's own errors. You might want to try that some time. People in this thread have asked you to do so.
But you made it sound like I had a whole thread devoted to misconceptions about the Sagnac effect. I was wondering why I didn't remember such a thing. I thought I might have had a case of amnesia, or something.
You admired his "intellectual capabilities"
Based on the idiocies you posted in the latest thread on clock synchronization on a rotating platform you STILL don't understand it NOW.
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Neddy Bate
Valued Senior Member
On the contrary, I was the one who said this:
Neddy Bate said:
and you were the one who said this:
Tach said:
But this thread is not the appropriate place for us to discuss your misconceptions about the Sagnac effect. You might end up getting this thread locked, like you did the other one.
You introduced the subject, if you have nothing to contribute to this thread, go do something useful elsewhere.
Yes, you got your own thread locked due to your own idiocies.
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I've got to say I'm really enjoying this thread. You are so combative and equally wrong it's worse than a train wreck - it's like a train crashing into another train at a level crossing where a lorry full of puppies has just broken down on the tracks, all in glorious slow motion HD.
Also, I've never once talked to Eugene about his ideas and frankly I don't really want to. It's clear from his repeated monty python mantras that he's almost as big an asshole as you are.
Please continue, this is great!
*sigh*. Fine. Step by step. In units where k = 1, Eugene's non-linear transformation, copied almost directly out of page 9 of his essay, is:
$$
\begin{align}
t^{\prime} \,&=\, \gamma \bigl[ t \,-\, vx \,-\, S(x) \bigr]
\,+\, S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
x^{\prime} \,&=\, \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \;.
\end{align}
$$
\begin{align}
t^{\prime} \,&=\, \gamma \bigl[ t \,-\, vx \,-\, S(x) \bigr]
\,+\, S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
x^{\prime} \,&=\, \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \;.
\end{align}
$$
Now, let's calculate the partial derivatives:
$$
\begin{align}
\frac{\partial t^{\prime}}{\partial t} \,&=\,
\gamma \,+\, \frac{\partial}{\partial t}
S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
\,&=\, \gamma \,-\, \gamma v S^{\prime}(y) \;,
\end{align}
$$
where I've set $$y \,\equiv\, \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr]$$. Moving on,\begin{align}
\frac{\partial t^{\prime}}{\partial t} \,&=\,
\gamma \,+\, \frac{\partial}{\partial t}
S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
\,&=\, \gamma \,-\, \gamma v S^{\prime}(y) \;,
\end{align}
$$
$$
\begin{align}
\frac{\partial t^{\prime}}{\partial x} \,&=\, - \, \gamma v \,-\, \gamma S^{\prime}(x)
\,+\, \frac{\partial}{\partial x}
S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
\,&=\, -\, \gamma v \,-\, \gamma S^{\prime} (x)
\,+\, \gamma \bigl( 1 \,+\, v S^{\prime}(x) \bigr) S^{\prime}(y) \;.
\end{align}
$$
The partial derivatives of $$x^{\prime}$$ are easier to evaluate:\begin{align}
\frac{\partial t^{\prime}}{\partial x} \,&=\, - \, \gamma v \,-\, \gamma S^{\prime}(x)
\,+\, \frac{\partial}{\partial x}
S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
\,&=\, -\, \gamma v \,-\, \gamma S^{\prime} (x)
\,+\, \gamma \bigl( 1 \,+\, v S^{\prime}(x) \bigr) S^{\prime}(y) \;.
\end{align}
$$
$$
\begin{align}
\frac{\partial x^{\prime}}{\partial t} \,&=\, -\, \gamma v
\\ \\ \\
\frac{\partial x^{\prime}}{\partial x} \,&=\, \gamma
\,+\, \gamma v S^{\prime}(x) \;.
\end{align}
$$
Easy, right? There should be no arguments here, since apart from setting c = 1 these turn out to be the same partial derivatives you found in post #319. Congratulations, it turns out you can apply the chain rule correctly after all! You had us all worried there, with your talk about differentiating S with respect to itself and all that.\begin{align}
\frac{\partial x^{\prime}}{\partial t} \,&=\, -\, \gamma v
\\ \\ \\
\frac{\partial x^{\prime}}{\partial x} \,&=\, \gamma
\,+\, \gamma v S^{\prime}(x) \;.
\end{align}
$$
Now onto the part you didn't bother to calculate: the Jacobian determinant. It's given by
$$
J_{C} \,=\, \frac{\partial t^{\prime}}{\partial t} \,
\frac{\partial x^{\prime}}{\partial x} \,-\,
\frac{\partial t^{\prime}}{\partial x} \,
\frac{\partial x^{\prime}}{\partial t} \;.
$$
The first part is:J_{C} \,=\, \frac{\partial t^{\prime}}{\partial t} \,
\frac{\partial x^{\prime}}{\partial x} \,-\,
\frac{\partial t^{\prime}}{\partial x} \,
\frac{\partial x^{\prime}}{\partial t} \;.
$$
$$
\begin{align}
\frac{\partial t^{\prime}}{\partial t} \,
\frac{\partial x^{\prime}}{\partial x}
\,&=\, \gamma^{2} \bigl( 1 \,-\, v S^{\prime}(y) \bigr)
\bigl( 1 \,+\, v S^{\prime}(x) \bigr)
\\ \\ \\
\,&=\, \gamma^{2} \Bigl[ 1 \,+\, v S^{\prime}(x) \,-\, v S^{\prime}(y)
\,-\, v^{2} S^{\prime}(x) S^{\prime}(y) \Bigr] \;.
\end{align}
$$
The second part is:\begin{align}
\frac{\partial t^{\prime}}{\partial t} \,
\frac{\partial x^{\prime}}{\partial x}
\,&=\, \gamma^{2} \bigl( 1 \,-\, v S^{\prime}(y) \bigr)
\bigl( 1 \,+\, v S^{\prime}(x) \bigr)
\\ \\ \\
\,&=\, \gamma^{2} \Bigl[ 1 \,+\, v S^{\prime}(x) \,-\, v S^{\prime}(y)
\,-\, v^{2} S^{\prime}(x) S^{\prime}(y) \Bigr] \;.
\end{align}
$$
$$
\begin{align}
\frac{\partial x^{\prime}}{\partial t} \,
\frac{\partial t^{\prime}}{\partial x}
\,&=\, \gamma^{2} v \Bigl[ v \,+\, S^{\prime}(x) \,-\, S^{\prime}(y)
\,-\, v S^{\prime}(x) S^{\prime}(y) \Bigr]
\\ \\ \\
\,&=\, \gamma^{2} \Bigl[ v^{2} \,+\, v S^{\prime}(x) \,-\, v S^{\prime}(y)
\,-\, v^{2} S^{\prime}(x) S^{\prime}(y) \Bigr]
\end{align}
$$
The Jacobian determinant -- **drumroll** -- is just the difference of these two terms:\begin{align}
\frac{\partial x^{\prime}}{\partial t} \,
\frac{\partial t^{\prime}}{\partial x}
\,&=\, \gamma^{2} v \Bigl[ v \,+\, S^{\prime}(x) \,-\, S^{\prime}(y)
\,-\, v S^{\prime}(x) S^{\prime}(y) \Bigr]
\\ \\ \\
\,&=\, \gamma^{2} \Bigl[ v^{2} \,+\, v S^{\prime}(x) \,-\, v S^{\prime}(y)
\,-\, v^{2} S^{\prime}(x) S^{\prime}(y) \Bigr]
\end{align}
$$
$$
\begin{align}
J_{C} \,&=\, \gamma^{2} ( 1 \,-\, v^{2} )
\\ \\ \\
\,&=\, 1 \;.
\end{align}
$$
Surprise!\begin{align}
J_{C} \,&=\, \gamma^{2} ( 1 \,-\, v^{2} )
\\ \\ \\
\,&=\, 1 \;.
\end{align}
$$
Er, wrong again:
$$
\theta^{-1}(t,\,x) \,=\,
\begin{pmatrix} t \,-\, S(x) \\ \\ \\ x \end{pmatrix} \;.
$$
Composing with the Lorentz transformation\theta^{-1}(t,\,x) \,=\,
\begin{pmatrix} t \,-\, S(x) \\ \\ \\ x \end{pmatrix} \;.
$$
$$
\Lambda(t,\,x) \,=\,
\begin{pmatrix}
\gamma ( t \,-\, vx ) \\ \\ \\ \gamma ( x \,-\, vt )
\end{pmatrix}
$$
yields\Lambda(t,\,x) \,=\,
\begin{pmatrix}
\gamma ( t \,-\, vx ) \\ \\ \\ \gamma ( x \,-\, vt )
\end{pmatrix}
$$
$$
\Lambda \,\circ\, \theta^{-1}(t,\, x) \,=\,
\begin{pmatrix}
\gamma \bigl[ t \,-\, S(x) \,-\, vx \bigr]
\\ \\ \\
\gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr]
\end{pmatrix} \;.
$$
Finally, $$\theta$$ is given by\Lambda \,\circ\, \theta^{-1}(t,\, x) \,=\,
\begin{pmatrix}
\gamma \bigl[ t \,-\, S(x) \,-\, vx \bigr]
\\ \\ \\
\gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr]
\end{pmatrix} \;.
$$
$$
\theta^{-1}(t,\,x) \,=\,
\begin{pmatrix} t \,+\, S(x) \\ \\ \\ x \end{pmatrix} \;,
$$
so the composition $$\theta \,\circ\, \Lambda \,\circ\, \theta^{-1}$$ is given by:\theta^{-1}(t,\,x) \,=\,
\begin{pmatrix} t \,+\, S(x) \\ \\ \\ x \end{pmatrix} \;,
$$
$$
\theta \,\circ\, \Lambda \,\circ\, \theta^{-1} (t,\,x) \,=\,
\begin{pmatrix}
\gamma \bigl[ t \,-\, S(x) \,-\, vx \bigr]
\,+\, S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
\gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr]
\end{pmatrix} \;,
$$
which is just Eugene's transformation. Surprise again!\theta \,\circ\, \Lambda \,\circ\, \theta^{-1} (t,\,x) \,=\,
\begin{pmatrix}
\gamma \bigl[ t \,-\, S(x) \,-\, vx \bigr]
\,+\, S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
\gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr]
\end{pmatrix} \;,
$$
Now, you were calling someone lazy...?
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Er, in case you didn't notice, the moderator who locked that thread is the same moderator who banned you a few days ago.
I missed this edit:
No-one except Eugene thinks his transformation represents anything physical.