On the Definition of an Inertial Frame of Reference

[Tach]

Keep believing that Sir Knight. You just might prod the pompous to reveal a little backbone.

We all know that your theory is crap. We may have our disagreements as to how we prove that. We all agree that you are a crackpot. There is no disagreement on that.

 

[Eugene Shubert]

We all know that your theory is crap. We may have our disagreements as to how we prove that. We all agree ... There is no disagreement on that.

Thank you Sir Knight for acknowledging that you are united in purpose with my other detractors.

 

[Tach]

Thank you Sir Knight for acknowledging that you are united in purpose with my other detractors.

We aren't detractors, we are debunkers. What you write is bunk.

 

[przyk]

1. I suspect that you started from the Minkowski metric $$ds^2=(cdt)^2-dx^2$$ and you passed it through the Shubert transforms, right?

No, or at least not the full transformation. Eugene claims his transformation factorises as $$C \,=\, \theta \,\circ\, \Lambda \,\circ\, \theta^{-1}$$, where $$\Lambda$$ is a Lorentz transformation, and I took $$\theta$$ to mean

$$
\theta(t,\,x^{i}) \,=\, \bigl( t \,+\, S(x^{i}),\, x^{i} \bigr) \;.
$$​

(the definition of $$\theta$$ as it appears in Eugene's essay has some extra indices, including an index on the t variable. I have no idea what they're supposed to add). I only applied $$\theta$$ to the Minkowski metric. With that in mind, you might want to re-read this part of my earlier post, which should explain why I found this more interesting than applying the entire transformation:

In terms of the decomposition $$\theta \,\circ\, \Lambda \,\circ\, \theta^{-1}$$ of his transformation, $$\theta^{-1}$$ brings you back to the Minkowski metric, the Lorentz transformation $$\Lambda$$ leaves the Minkowski metric invariant, and $$\theta$$ takes you back to the metric above.

so Eugene's transformation $$C$$ leaves invariant the metric expression I posted earlier. Obviously if I'd just applied $$C$$ directly on the Minkowski metric it wouldn't leave its components invariant. It would take the Minkowski metric to some other metric.

A quick examination shows that you missed the fact that $$S_i=S_i(x,t,S_j(x,t)$$.

As Guest pointed out earlier, if you think there's a problem with "differentiating S with respect to itself" - which I take to mean you think differentiating Eugene's transformation produces terms like $$\frac{\partial S}{\partial S}$$ - then you've got a misconception about the chain rule. Specifically, it sounds like you're not clear on the distinction between differentiating a function with respect to its parameters, and evaluating the function or differential at a point.

Differentiating something like $$S \bigl( S (x) \bigr)$$ produces a term like

$$\frac{\partial S}{\partial x} \bigl( S(x) \bigr) \;.$$​

In words: the partial derivative of S with respect to its parameter (here called x, not to be confused with the spatial coordinate also called x) and evaluated at S(x).

Firstly, you cannot prove that the jacobian is non-degenerate.

The Jacobian matrix of $$\theta$$ is just

$$
J_{\theta} \,=\,
\begin{bmatrix}
1 & \partial_{1} S & \partial_{2} S & \partial_{3} S \\
0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1
\end{bmatrix} \;.
$$​

The determinant of this matrix is just 1. The Jacobian determinant of the full non-linear transformation is

$$\det \bigl[ J_{C} \bigr]
\,=\, \det \bigl[ J_{\theta} \,\circ\, \Lambda \,\circ\, J_{\theta^{-1}} \bigr]
\,=\, \det \bigl[ J_{\theta} \bigr] \, \det \bigl[ \Lambda \bigr] \,
\det \bigl[ J_{\theta^{-1}} \bigr] \,=\, 1 \;,$$​

so clearly the Jacobian matrix of the transformation is non-degenerate. This shouldn't be surprising, since non-degeneracy of the Jacobian matrix everywhere is equivalent to invertibility of the transformation, and Eugene explicitly gave the inverse of $$\theta$$ in his essay.

Secondly, in order for the covariance of the metric to be preserved under transformation, the jacobian needs to be a unit matrix[1] , a condition that is clearly not satisfied.

I'm not familiar with the term "unit matrix". If you mean a matrix whose determinant is of absolute value 1, then this is true and Eugene's transformation satisfies this requirement, as shown above.

4. Not if $$\theta^{-1}$$ does not exist since the jacobian cannot be proven non-degenerate.

As stated above, Eugene explicitly gave $$\theta^{-1}$$ in his essay. It's easily seen to be

$$
\theta^{-1}(t,\, x^{i}) \,=\, \bigl( t \,-\, S(x^{i}),\, x^{i} \bigr) \;.
$$​

6. Here you made another error. The spacetime being flat, the Riemann tensor is null so the Christoffel symbols are all null. [2]. The covariant derivatives reduce to standard derivatives.

I've already told you this isn't correct. The Christoffel symbols are not a measure of curvature. Intuitively, they measure the pseudo-forces present in non-inertial coordinate systems, eg. centrifugal force or the gravitational "force" present in accelerating frames. It's easy to prove that the Christoffel symbols are non-zero in any coordinate system in which the metric components are inhomogenous. This is the case for the metric I gave in my earlier post, unless the function S, and therefore Eugene's entire transformation, happens to be linear or affine.

 

[Tach]

I'm not familiar with the term "unit matrix". If you mean a matrix whose determinant is of absolute value 1,

This is the standard definition.

No, or at least not the full transformation. Eugene claims his transformation factorises as $$C \,=\, \theta \,\circ\, \Lambda \,\circ\, \theta^{-1}$$, where $$\Lambda$$ is a Lorentz transformation, and I took $$\theta$$ to mean

$$
\theta(t,\,x^{i}) \,=\, \bigl( t \,+\, S(x^{i}),\, x^{i} \bigr) \;.
$$​

It doesn't factorize. Look at its definition, there is no possible factorization. How do you factorize $$S(x,t,S(x,t))$$?. You need to check Shubert's claims.

The Jacobian matrix of $$\theta$$ is just

$$
J_{\theta} \,=\,
\begin{bmatrix}
1 & \partial_{1} S & \partial_{2} S & \partial_{3} S \\
0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1
\end{bmatrix} \;.
$$​

I have no idea how you got the jacobian to be rank 4 when the Shubert spacetime is 1+1. I don't think I will waste my time on this obvious error.
$$S$$ is the synchronization function so I do not understand why you keep insting in sticking $$\partial S$$ into the jacobian when the correct elements are obviously $$ \partial x'/ \partial x, \partial x'/\partial t $$ and $$ \partial t'/\partial x, \partial t'/\partial t$$. I won't waste my time debugging this error either. If you don't know how to construct the jacobian, there is little point in arguing the subject matter. Now I understand why your metric looked so wrong.

As Guest pointed out earlier, if you think there's a problem with "differentiating S with respect to itself" - which I take to mean you think differentiating Eugene's transformation produces terms like $$\frac{\partial S}{\partial S}$$ - then you've got a misconception about the chain rule. Specifically, it sounds like you're not clear on the distinction between differentiating a function with respect to its parameters, and evaluating the function or differential at a point.

S is a direct function of x,t and it is also an iimplicit function of (x,t) through S(x,t).

then this is true and Eugene's transformation satisfies this requirement, as shown above.

No, it isn't since you haven't calculated the jacobian correctly. I will not waste more of my time with you on this subject.

I've already told you this isn't correct. The Christoffel symbols are not a measure of curvature.

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.

 

[prometheus]

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.

What are the Christoffel symbols for the following metric?

$$ds^2 = -dt^2 + dr^2 + r^2 \left(d\theta^2 + \sin^2 \theta d\phi^2 \right) $$

 

[Eugene Shubert]

It doesn't factorize. Look at its definition, there is no possible factorization.

The factorization of my nonlinear transformation equations (54)-(55) is clearly evident in 3 easy steps. You, Sir Knight, simply don't understand how to compose functions. Consider a few elementary examples of function composition and then try to follow the composition in equation number (51), (52) and (53).

 

[Tach]

The factorization of my nonlinear transformation equations (54)-(55) is clearly evident in 3 easy steps. You, Sir Knight, simply don't understand how to compose functions. Consider a few elementary examples of function composition and then try to follow the composition in equation number (51), (52) and (53).

Look at the insane equations (54)(55). Hard as you try, non-linear function compositions do not result into matrix multiplication, Shubert.

 

[Tach]

What are the Christoffel symbols for the following metric?

$$ds^2 = -dt^2 + dr^2 + r^2 \left(d\theta^2 + \sin^2 \theta d\phi^2 \right) $$

Zero. See Moller (previous citation). This is nothing but the Minkowski metric expressed in polar coordinates. The spacetime is flat, so the Riemann tensor is zero, therefore the Christoffel symbols are all null. I think Guest254 tried the same silly game before. You can always check it out here. If you think that you got anything else, you have done your calculations incorrectly.

 

[Eugene Shubert]

Look at the insane equations (54)(55). Hard as you try, non-linear function compositions do not result into matrix multiplication, Shubert.

As capable as you are Sir Knight in creating your own reality, it is easy to compose nonlinear functions with linear ones. That is precisely what is done in the youtube tutorial. Furthermore, the current edition of my quintessence paper does not contain the word matrix. And you will not find the word matrix in any paper containing my title.

 

[Tach]

As capable as you are Sir Knight in creating your own reality, it is easy to compose nonlinear functions with linear ones.

This is not the point, crackpot. The point is, when you do that, you do not get the associated matrices to multiply each other. You may have fooled przyk who took your "math" for granted, you did not fool me.

 

[Eugene Shubert]

you have done your calculations incorrectly.

Even though you don't know how to compute Christoffel symbols in polar coordinates, you can at least google the problem and see that the entire mathematical world disagrees with you. Here's one explicit entry, page 85 of Pavel Grinfeld's book Tensor Calculus.

 

[Eugene Shubert]

The point is, when you do that, you do not get the associated matrices to multiply each other.

Matrices represent linear functions. What is the definition of the associated matrix of a nonlinear function? I am certainly not using whatever mathematical technique that you imagine. How do you compute the associated matrix of the nonlinear function in the youtube tutorial?

 

[Tach]

Matrices represent linear functions. What is the definition of the associated matrix of a nonlinear function?

That was the whole point, crackpot, you managed to fool przyk into thinking that there are.

 

[Eugene Shubert]

That was the whole point

Exactly Sir Knight. Being deluded as you are, you have no idea what you're talking about.

 

[Tach]

Exactly Sir Knight. Being deluded as you are, you have no idea what you're talking about.

Have you started submitting your "masterpiece" to journals? Have the rejections started piling up? How do you cope with the fact that your "work" will get ignored forever and that the only place where you get any attention is the crackpot forums? Will you still be bitching about the "conspiracy of Einstein worshipers" to your grave? Say "hi" to your "nemesis"

 

[Green Destiny]

Oh tach, a new bait with the usual suspects.

This place is monotonous for the same old, seeing who can piss the furthest. I don't know about anyone else, but I find it, tiring.

 

[prometheus]

Zero. See Moller (previous citation).

I'm going to say this once, and there is no argument about it - you are wrong and the Christoffel symbols for the metric I wrote above are not all zero

the definition of the Christoffel symbols:

$$\Gamma^\rho_{\mu \nu} = \frac{g^{\rho \lambda}}{2} \left(\partial_\nu g_{\lambda \mu} +\partial_\mu g_{\lambda \nu}- \partial_\lambda g_{\mu \nu} \right)$$

take for example $$\Gamma^r_{\theta \theta}$$:

$$\Gamma^r_{\theta \theta} = \frac{g^{r \lambda}}{2} \left(\partial_\theta g_{\lambda \theta} +\partial_\theta g_{\lambda \theta}- \partial_\lambda g_{\theta \theta} \right)\\
= \frac{g^{r r}}{2} \left(\partial_\theta g_{r \theta} +\partial_\theta g_{r \theta}- \partial_r g_{\theta \theta} \right)\\
= -\frac{g^{r r}}{2} \partial_r g_{\theta\theta} \\
= - \frac{1}{2 r^2} \partial_r r^2\\
= -\frac{1}{r} $$

This is nothing but the Minkowski metric expressed in polar coordinates. The spacetime is flat, so the Riemann tensor is zero

Indeed. I agree with this - if the Riemann tensor is zero then the spacetime is flat. If you do the calculation properly and compute all of the components of the Riemann tensor using the non zero christoffel symbols you will find exactly that.

therefore the Christoffel symbols are all null.

The Riemann tensor depends on the Christoffels, not the other way around. The fact that you use the word "null" here is not a good sign either, because in relativity null means a vector that is lightlike, ie $$v.v=g_{\mu \nu}v^\mu v^\nu=0$$

I think Guest254 tried the same silly game before.

I don't have guest's patience, or the love of watching the train wreck unfold.

You can always check it out here.

You do realise that your link gives the correct non zero Christoffel symbols don't you? I presume not, otherwise you wouldn't have stupidly linked to it, or we simply wouldn't be having this dumb conversation because you'd know that the Christoffels aren't zero.

Look at numbers (43) through (48) and feel deeply embarrassed.

If you think that you got anything else, you have done your calculations incorrectly.

Irony. Gotta love it.

 

[Tach]

I'm going to say this once, and there is no argument about it - you are wrong and the Christoffel symbols for the metric I wrote above are not all zero

the definition of the Christoffel symbols:

$$\Gamma^\rho_{\mu \nu} = \frac{g^{\rho \lambda}}{2} \left(\partial_\nu g_{\lambda \mu} +\partial_\mu g_{\lambda \nu}- \partial_\lambda g_{\mu \nu} \right)$$

take for example $$\Gamma^r_{\theta \theta}$$:

$$\Gamma^r_{\theta \theta} = \frac{g^{r \lambda}}{2} \left(\partial_\theta g_{\lambda \theta} +\partial_\theta g_{\lambda \theta}- \partial_\lambda g_{\theta \theta} \right)\\
= \frac{g^{r r}}{2} \left(\partial_\theta g_{r \theta} +\partial_\theta g_{r \theta}- \partial_r g_{\theta \theta} \right)\\
= -\frac{g^{r r}}{2} \partial_r g_{\theta\theta} \\
= - \frac{1}{2 r^2} \partial_r r^2\\
= -\frac{1}{r} $$

Well, you goofed. Both Moller and Wolfram says you are wrong. See here.

Indeed. I agree with this - if the Riemann tensor is zero then the spacetime is flat. If you do the calculation properly and compute all of the components of the Riemann tensor using the non zero christoffel symbols you will find exactly that.

Moller ("Theory of Relativity", page 377) says that you are wrong. If the Riemann tensor is zero, all the Christoffel symbols are also zero.

The Riemann tensor depends on the Christoffels, not the other way around. The fact that you use the word "null" here is not a good sign either, because in relativity null means a vector that is lightlike, ie $$v.v=g_{\mu \nu}v^\mu v^\nu=0$$

Let's try again: if $$R_{iklm}=0$$ then all the Christoffel symbols are ZERO.

You do realise that your link gives the correct non zero Christoffel symbols don't you? I presume not, otherwise you wouldn't have stupidly linked to it, or we simply wouldn't be having this dumb conversation because you'd know that the Christoffels aren't zero.

It is not my fault that you can't read. BTW, why are you going on this wild goose chase? What does your nonsense have to do with the analysis of Shubert's errors?

 

[Guest254]

This has become difficult to watch! But it's a beautiful day and I'm feeling charitable...

Moller ("Theory of Relativity", page 377) says that you are wrong. If the Riemann tensor is zero, all the Christoffel symbols are also zero.

Let's try again: if $$R_{iklm}=0$$ then all the Christoffel symbols are ZERO.

Now I don't have a copy of this particular book, so perhaps it would be an idea for you to quote, verbatim, the result you are referring to and I will explain it to you. I suspect the author proves a very standard result: if the Riemann tensor vanishes, then there exists a set of coordinates for which the Christoffel symbols vanish.