On the Definition of an Inertial Frame of Reference

[Eugene Shubert]

$$
\theta^{-1}(t,\,x) \,=\,
\begin{pmatrix} t \,-\, S(x) \\ \\ \\ x \end{pmatrix} \;.
$$​

Composing with the Lorentz transformation

$$
\Lambda(t,\,x) \,=\,
\begin{pmatrix}
\gamma ( t \,-\, vx ) \\ \\ \\ \gamma ( x \,-\, vt )
\end{pmatrix}
$$​

yields

$$
\Lambda \,\circ\, \theta^{-1}(t,\, x) \,=\,
\begin{pmatrix}
\gamma \bigl[ t \,-\, S(x) \,-\, vx \bigr]
\\ \\ \\
\gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr]
\end{pmatrix} \;.
$$​

Finally, $$\theta$$ is given by

$$
\theta^{-1}(t,\,x) \,=\,
\begin{pmatrix} t \,+\, S(x) \\ \\ \\ x \end{pmatrix} \;,
$$​

so the composition $$\theta \,\circ\, \Lambda \,\circ\, \theta^{-1}$$ is given by:

$$
\theta \,\circ\, \Lambda \,\circ\, \theta^{-1} (t,\,x) \,=\,
\begin{pmatrix}
\gamma \bigl[ t \,-\, S(x) \,-\, vx \bigr]
\,+\, S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
\gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr]
\end{pmatrix} \;,
$$​

which is just Eugene's transformation. Surprise again!

Not exactly.

No-one except Eugene thinks his transformation represents anything physical.

Excuse me przyk. While your calculation in post #338 is technically correct, with the exception of at least one typo that I've noticed, you have clearly demonstrated that you don't understand the physical meaning of my generalized Lorentz-equivalent transformation equations.

Equations (54)-(55) connect the event (x,t) in frame $$i$$ having synchronization $$ S_i$$ to the same event recorded as (x',t') in frame $$j$$ having synchronization $$ S_j$$. You assumed that my coordinate-dependent synchronization schemes are identical in all inertial frames of reference. That is a big mistake.

 

[Tach]

Not exactly.




Excuse me przyk. While your calculation in post #338 is technically correct, with the exception of at least one typo that I've noticed, you have clearly demonstrated that you don't understand the physical meaning of my generalized Lorentz-equivalent transformation equations.

LOL. Eugene, your theory is "fucked" , your synchronization function is an absurd function of the prime coordinate $$S_j=S_j(x')$$ resulting into metric coefficients that are both functions of x and x', $$g_{ij}=g_{ij}(x,x')$$. Not only mathematically incorrect but also unphysical.

You assumed that my coordinate-dependent synchronization schemes are identical in all inertial frames of reference. That is a big mistake.

LOL. Your cardhouse is coming down.

 

[Tach]

*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:

$$
\begin{align}
t^{\prime} \,&=\, \gamma \bigl[ t \,-\, vx \,-\, S(x) \bigr]
\,+\, S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
x^{\prime} \,&=\, \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \;.
\end{align}
$$​

You are not paying attention, there are TWO synch functions in Shubert demented theory.

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.

Not my problem that you are a stupid jerk who can't follow my math and insists in misinterpreting the posts. Math speaks a lot clearer, next time follow the math.

 

[przyk]

Equations (54)-(55) connect the event (x,t) in frame $$i$$ having synchronization $$ S_i$$ to the same event recorded as (x',t') in frame $$j$$ having synchronization $$ S_j$$. You assumed that my coordinate-dependent synchronization schemes are identical in all inertial frames of reference. That is a big mistake.

I was wondering what those indices were supposed to represent. But as Alphanumeric explained to Tach several times, no-one is really interested in the details of your transformation. In the most interesting case, you were using the same function S every time, and then $$\theta$$ would have taken the Minkowski metric to a metric of the form I posted earlier, whose components your transformation would have left invariant. If that isn't true, then at best all you have is a special case of the generally covariant formulation of relativity.

 

[Tach]

I was wondering what those indices were supposed to represent. But as Alphanumeric explained to Tach several times, no-one is really interested in the details of your transformation.

Yet you should, in order to get your math correct. If you did that, you would have recovered the correct answers that I have already posted. As it stands, you just wasted your time.

 

[przyk]

*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:

$$
\begin{align}
t^{\prime} \,&=\, \gamma \bigl[ t \,-\, vx \,-\, S(x) \bigr]
\,+\, S \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \Bigr)
\\ \\ \\
x^{\prime} \,&=\, \gamma \bigl[ x \,-\, vt \,+\, v S(x) \bigr] \;.
\end{align}
$$​

You are not paying attention, there are TWO synch functions in Shubert demented theory.

So? Substitute $$S^{\prime}(y) \,\mapsto\, S^{\prime}_{j}(y)$$ and $$S^{\prime}(x) \,\mapsto\, S^{\prime}_{i}(x)$$ in the derivation I gave you. Same result: $$J_{C_{ij}} \,=\, 1$$.

 

[Tach]

So? Substitute $$S^{\prime}(y) \,\mapsto\, S^{\prime}_{j}(y)$$ and $$S^{\prime}(x) \,\mapsto\, S^{\prime}_{i}(x)$$ in the derivation I gave you. Same result: $$J_{C_{ij}} \,=\, 1$$.

You missed the point that $$g_{ij}=g_{ij}(x,x')$$.
You missed the point that the synch function $$S_j=S_j(x')$$ which is unphysical.

 

[przyk]

Yet you should, in order to get your math correct. If you did that, you would have recovered the correct answers that I have already posted.

What correct answer that you posted? All you had to say about the Jacobian determinant was:

there is no way in hell that the jacobian has a determinant equal to 1

As I've shown, it is 1. Adding indices to S doesn't alter the result. $$J_{C_{ij}} \,=\, 1$$.

 

[przyk]

You missed the point that $$g_{ij}=g_{ij}(x,x')$$.

Where, presumably, x' = x'(x, t). So the metric is still a function of x and t.

You missed the point that the synch function $$S_j=S_j(x')$$ which is unphysical.

Hardly an issue, since I never claimed Eugene's transformation represented anything physical.

 

[Tach]

What correct answer that you posted? All you had to say about the Jacobian determinant was:

Are you daft? The elements of the jacobian are functions of coordinates in both frames of reference : $$g_{ij}=g_{ij}(x,x')$$. Who gives a shit after that?

Hardly an issue, since I never claimed Eugene's transformation represented anything physical.

In other words you agree that the theory is unphysical and that whatever results it produces are irrelevant? Exactly my point , all along.

 

[Eugene Shubert]

I was wondering what those indices were supposed to represent. But as Alphanumeric explained to Tach several times, no-one is really interested in the details of your transformation. In the most interesting case, you were using the same function S every time, and then $$\theta$$ would have taken the Minkowski metric to a metric of the form I posted earlier, whose components your transformation would have left invariant. If that isn't true, then at best all you have is a special case of the generally covariant formulation of relativity.

Actually, my approach to special relativity is more general than the generally covariant formulation of special relativity. My synchronization functions need not be smooth or even continuous. Furthermore, my minimal axiom set permits superluminal coordinate transformations. That's the next big step that I plan to elaborate on in my paper.

 

[Tach]

. My synchronization functions need not be smooth or even continuous.

Don't be an idiot, they need to be differentiable (not that it matters in your case since the transform is unphysical to begin with). How else would you even have a hope of constructing a jacobian.
The problem with your two synch functions is that they are unphysical, as is your whole theory. So, no one gives a shit.

 

[James R]

As I disinterested observer, I'm seeing a lot of chest-puffing here, which is leading to an extended argument over not much.

Transform back into cartesian coordinates and you get null Christoffel symbols. A change of coordinates doesn't change the nature of spacetime. It doesn't make it curved if you go from cartesian to polar.

This is correct, but disingenuous.

As has been clearly pointed out to you, Tach, the Christoffel symbols are non-zero in polar coordinates, even in flat spacetime.

You clearly made a general statement that flat spacetime implies that Christoffel symbols are zero in all coordinate systems.

Now, do you want to keep thumping your chest, or admit that you made a mistake? An inability to admit when you're wrong only has the effect of lowering your academic standing in most people's eyes, you know.

So: apology to prometheus and others who corrected you? Or most egotistical posturing? What's it to be?

 

[James R]

LOL. Eugene, your theory is "fucked"...

LOL. Your cardhouse is coming down.

Oh, and another word of advice, Tach.

Ridicule is not a strong form of argument. It just makes you look small and young. Try dealing with the substance. Attack the argument, not the man. You'll look a lot more mature and gain more respect that way.

 

[brucep]

As I disinterested observer, I'm seeing a lot of chest-puffing here, which is leading to an extended argument over not much.



This is correct, but disingenuous.

As has been clearly pointed out to you, Tach, the Christoffel symbols are non-zero in polar coordinates, even in flat spacetime.

You clearly made a general statement that flat spacetime implies that Christoffel symbols are zero in all coordinate systems.

Now, do you want to keep thumping your chest, or admit that you made a mistake? An inability to admit when you're wrong only has the effect of lowering your academic standing in most people's eyes, you know.

So: apology to prometheus and others who corrected you? Or most egotistical posturing? What's it to be?

Read the paper linked in post #322 and explain how real gravitational effects are introduced into an inertial frame of reference as an artifact of choice of coordinate system?

 

[arfa brane]

brucep: explain how a material particle can experience a force acting on it in an inertial frame of reference.

I think Tach is up the creek again, with the Christoffel symbols all being zero thing. In fact, most of the components of the Christoffel symbols can be zero, but the symbols themselves don't "vanish", that doesn't make much sense if you think about it. It would mean there is a flat spacetime where no acceleration is "allowed" (??)

 

[Tach]

As has been clearly pointed out to you, Tach, the Christoffel symbols are non-zero in polar coordinates, even in flat spacetime.

Let me make it clear to you, the fact that the Christoffel symbols are non-zero in polar coordinates has absolutely nothing to do with the discussion in this thread, it is just a red herring introduced by prometheus. I called him on it, yet he persisted banging on the pot. On the other hand, the paper cited earlier that shows that zero Christoffel symbols is a necessary and sufficient condition for describing inertial frames in flat spacetime has everything to do with the discussion since it shows that Guest254 hijacked the thread with yet another red herring that I've exposed as such. may I suggest that you read the whole thread before passing judgement? The whole diversion into the general covariance is the ultimate example of thread hijack.Eventually the thread settled to my original claim, that Shubert's theory does not have a valid metric. Heck, the theory can't even calculate a velocity composition formula. These are valid points, the stuff that prometheus and Guest254 isn't.

 

[brucep]

brucep: explain how a material particle can experience a force acting on it in an inertial frame of reference.

Uh maybe something bumps it and then the frame is accelerated?

 

[arfa brane]

The only spacetime where acceleration doesn't occur is Minkowski with lightlike geodesics. That is, one where the only particles have constant velocity and zero mass.

The real universe contains material particles with mass, however. Therefore the Christoffel symbols can't disappear if acceleration is possible--note the italics--for such particles; Tach is up shit creek and I think he's lost the paddle.

 

[brucep]

The only spacetime where acceleration doesn't occur is Minkowski with lightlike geodesics. That is, one where the only particles have constant velocity and zero mass.

The discussion isn't whether flat spacetime actually exists in this universe. It's perfectly ok to approximate the area tangent to the gravitational field as flat. It's called Special relativity. The special case for a more general theory of gravity where the effect of gravity can be ignored. Read the paper linked in post #322 and explain how real gravitational effects are introduced into an inertial frame of reference as an artifact of choice of coordinate system? Please use the actual definition of the inertial reference frame not some hybred, so called, real frame where accelerations must occur at some time.