On the Definition of an Inertial Frame of Reference

[Tach]

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.

Yes, this is precisely what he's showing. So, can we return now to Shubert's theory and your misconceptions about his errors? His theory is all about flat spacetime so there is no point in your introduction of the red herring (covariant derivatives and general covariance). You and your sidekicks wasted a lot of time on this wild goose chase that has nothing to do with the OP. Shubert's theory, though expressed in 1+1 FLAT spacetime DOESN'T even have a metric.Try to stay on topic.

 

[Guest254]

Yes, this is precisely what he's showing.

So do you understand how stupid you looked when you kept insisting that you knew what you were talking about and at the same time insisting that the Christoffel symbols will always vanish on flat space?

You might like to apologise to prometheus and przyk.

 

[Tach]

So do you understand how stupid you looked when you kept insisting that you knew what you were talking about and at the same time insisting that the Christoffel symbols will always vanish on flat space?

Where did you see always you dishonest idiot?
Not as stupid as you are by insisting on introducing general covariance into Shubert's theory. Now , that you dragged everybody on a wild goose chase in order to show your abilities in tensor calculus, how about we returned to the analysis of Shubert's theory?

 

[Guest254]

Where did you see always you dishonest idiot?
Not as stupid as you are by insisting on introducing general covariance into Shubert's theory.

Is it really a good idea to start calling people stupid, when you've just made such an enormous fool of yourself? Come on, think about it.

 

[Tach]

Is it really a good idea to start calling people stupid, when you've just made such an enormous fool of yourself? Come on, think about it.

Not my fault that you can't read. Not my fault that you are dishonest <shrug>

 

[Tach]

To avoid confusion - Maxwell's equation's are not "Eugene Shubert covariant", i.e. the standard Maxwell equations do not take on the same form under his coordinate transformation.

This is the post where you introduced your red herring. We wasted a lot of time on your nonsense. You hijacked the thread at post 46, time to bring the thread back. I propose that we return to the post prior to this and we agree that Maxwell's equations are not covariant wrt the Shubert transforms. End.

 

[Guest254]

Come on man, have a little dignity.

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.

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.

Have you no shame?!

 

[Tach]

Come on man, have a little dignity.

You wasted 200+ posts on your nonsense. Enough.

 

[Guest254]

Come on now, remember what I said? Have a little dignity, suck up the embarrassment, gracefully apologise to those who spent considerable time explaining basic tensor calculus to you and move on.

This is not the time for you to continue to beat your chest.

 

[Tach]

Come on now, remember what I said? .

..that you are intelectually dishonest and that you hijacked the thread at post 46?

To avoid confusion - Maxwell's equation's are not "Eugene Shubert covariant", i.e. the standard Maxwell equations do not take on the same form under his coordinate transformation.

This is what my point was all about, this is what the thread was all about prior your hijack, not about your tensor calculus masturbations. Time to bring the thread back. Enough.

 

[Guest254]

STFU

More LOUD NOISES! Will you ever learn?

EDIT: Ah - you edited again! Perhaps you are learning...

 

[prometheus]

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

At this point you say "Proof: ..."

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

That website says the Christoffel symbols are not zero! Proof:

 

[Eugene Shubert]

Shubert's theory, though expressed in 1+1 FLAT spacetime DOESN'T even have a metric. Try to stay on topic.

Why Sir Knight are you condemning me for what I haven't yet written? Shouldn't you try to cope with what is already there in my 13 pages?

 

[Tach]

Why Sir Knight are you condemning me for what I haven't yet written? Shouldn't you try to cope with what is already there in my 13 pages?

Because you wrote enough idiocies for a lifetime in your 13 pages.

 

[Eugene Shubert]

Because you wrote enough idiocies for a lifetime in your 13 pages.

But because you are unable to resolve your misunderstanding of what is written there, you have to imagine what isn't written and then try to refute that by imagining that you are proficient in higher math?

 

[prometheus]

Gone very quiet on the whole "Christoffel symbols are zero," thing, eh Tach?

 

[Tach]

Let's cut the nonsense from post 46:

To avoid confusion - Maxwell's equation's are not "Eugene Shubert covariant", i.e. the standard Maxwell equations do not take on the same form under his coordinate transformation. Whereas, they are Lorentz covariant, i.e. they do take on the same form under Lorentz transformations.

Good, you agree with me.

However, the generally covariant form of Maxwell's equations is obviously invariant under his (and anyone else's) coordinate transformation.

Nonsense, for (at least) two reasons:

1. The spacetime for Schubert's crackpot theory is flat, not curved, so there is a system of coordinates in which the covariant derivatives reduce to standard derivatives. This means, according to the point above, that Maxwel's equations aren't covariant anymore.

2. The notion of general covariance makes no sense in flat spacetime, so, after 200+ posts, your red herring is starting to rot.

3. If you tried to calculate the jacobian for Schubert's crackpot transforms

x'=x'(x,t)
t'=t'(x,t)

you would have found out that you cannot prove that it is non-degenerate.

 

[Green Destiny]

It's like watching punch and judy at a sunday stroll along the promenade.

 

[Tach]

Gone very quiet on the whole "Christoffel symbols are zero," thing, eh Tach?

And your stinking red herring is relevant to the discussion (absence of the covariance of Maxwell's equations in Shubert's crackpot formalism) , how?

 

[Green Destiny]

Have a little dignity, suck up the embarrassment.

Not sticking up for him for his posts but...

How disengenuous considering you were the one setting him up for all this. Hence why someone said here you like to see the train wreck.