This is the standard definition.
Not according to Google:
http://www.google.com/search?hl=en&...efine:unit+matrix&aq=f&aqi=&aql=&oq=&gs_rfai=. Which is why I clarified.
It doesn't factorize. Look at its definition, there is no possible factorization. How do you factorize $$S(x,t,S(x,t))$$?.
Where, in Eugene's essay, do you see the expression $$S(x,t,S(x,t))$$? In 1+1 dimensions, $$S \equiv S(x)$$ is just a function of a single parameter.
I have no idea how you got the jacobian to be rank 4 when the Shubert spacetime is 1+1.
I generalised to 1+3 dimensions. If you want to stick to 1+1 dimensions, the Jacobian is even simpler:
$$J_{\theta} \,=\,
\begin{bmatrix}
1 & S^{\prime} \\ \\ \\
0 & 1
\end{bmatrix} \;,
$$
where
S' is the derivative of the function
S. The Jacobian determinant is still 1.
If you wanted the Jacobian in 1+1 dimensions, you'll probably also want the metric after applying $$\theta$$ in 1+1 dimensions. That's also simpler:
$$
\text{d}s^{2} \,=\, -\, \text{d}t^{2}
\,+\, 2 \, S^{\prime} \, \text{d}t \, \text{d}x
\,+\, \bigl( 1 \,-\, (S^{\prime})^{2} \bigr) \, \text{d}x^{2} \;.
$$
(note: I explicitly dropped all the primes, and I'm working in units where
c = 1)
$$S$$ is the synchronization function so I do not understand why you keep insting in sticking $$\partial S$$ into the jacobian
$$\partial_{i} S$$ is shorthand for $$\frac{\partial S}{\partial x^{i}$$.
when the correct elements are obviously $$ \partial x'/ \partial x, \partial x'/\partial t $$ and $$ \partial t'/\partial x, \partial t'/\partial t$$.
That's just the definition of the Jacobian. When you
evaluate those terms, you get derivatives of
S. For
$$
\theta \,:\, (t,\,x) \,\mapsto (t^{\prime},\, x^{\prime}) \,=\, \bigl( t \,+\, S(x), \, x \bigr) \;,
$$
the partial derivatives are
$$
\begin{align}
\frac{\partial t^{\prime}}{\partial t} \,&=\, 1 &
\frac{\partial t^{\prime}}{\partial x} \,&=\, S^{\prime}(x) \\ \\ \\
\frac{\partial x^{\prime}}{\partial t} \,&=\, 0 &
\frac{\partial x^{\prime}}{\partial x} \,&=\, 1 \;.
\end{align}
$$
If you don't know how to construct the jacobian, there is little point in arguing the subject matter.
On the contrary, it looks like I'm a step ahead of you. I did a bit more than just copy and paste the general definition of the Jacobian, you know.
Now I understand why your metric looked so wrong.
Funny, it looks just fine to me. What do you think the metric expression should be after applying $$\theta$$?
Try reading what I wrote. Let's try again:
Flat spacetime -> Riemann Tensor = 0 -> Christoffel symbols=0 -> Covariant derivative = standard derivative.
This last section makes the whole argument moot. If you still don't understand it, we have nothing to talk about.
I can read what you wrote just fine. Now you try reading this:
Things aren't true just because you keep saying they are. You've already been corrected about this several times. prometheus even gave you an explicit counter-example: the Riemann tensor is null in polar coordinates, but, as he showed by direct calculation right in front of you, the Christoffel symbols aren't. That should really settle the issue.