On the Definition of an Inertial Frame of Reference

You are aware that Cartesian coordinates are a different coordinate system to polar coordinates right? Come on, this is really basic stuff.

 

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You said that the Riemann tensor = 0 implies the Christoffel symbols = 0. It does not.

 

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Yes, you dork, I am quite aware. Stop hijacking the thread with your irrelevant posts.

 

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I'm a dork again?! That must make you doubly right.

Seriously, this is depressing. How can you not get this?

 

I would have no problem submitting my paper to a reputable journal. But why would any reputable journal want to publish anything meant to be studied in high school?

I did take a minute and did it myself. But it wouldn't mean anything to you as it has already been demonstrated that you don't understand the chain rule. Worse than that, you have revealed that you don't even understand how to evaluate the composition of two functions.

 

How stupid are you? Really? \(R_{iklm}=0\) implies that there is a coordinate system where all the Christoffel symbols are zero->the covariant derivatives are the same as the standard derivatives.

 

You mean that your ego can stand not being published? More like you tried already and the rejections are piling up.

Because it is pure crap.

Let's see your "calculations". Come on, starting from the demented (54)(55) in your "masterpiece".

 

Let me help you out, what you need to do is , starting with your demented equations (54) and (55) to produce the following:

\(\begin{pmatrix}\frac{\partial x'}{ \partial x} & \frac{\partial x'}{ \partial t} \\ \frac{\partial t'} {\partial x} & \frac{\partial t'}{ \partial t} & \end{pmatrix} \)

Once you do that, try calculating its determinant. This is just about at your high school level of math.

 

That's very good. Now use the chain rule on equation (59) to do that easily.

That's even easier. Just multiply the 3 determinants together. The answer is 1 times 1 times 1. If you didn't know, that equals 1.

 

I am not doing your work for you, crackpot. It is for you to work on your crap.

There are no determinants in your demented paper because there are no matrices. Your "factorization" is just bogus.

No, idiot, you need to start from your demented nonlinear expressions. Now. get with the program.

 

In other words, you insist on seeing the computation done without the chain rule. That's understandable. You don't understand the chain rule. And I insist that you learn how to compose nonlinear functions with linear ones. That is precisely what is done in the youtube tutorial.

Let's call it a draw.

 

No, you don't get off the hook so easily, cheap crook. Here are the first computations done for you:

\(\frac{\partial x'}{\partial x}=\gamma(1+v \frac{\partial S_i}{\partial x})\)

\(\frac{\partial x'}{\partial t}=-\gamma v\)

Stop weaseling, you have two more to do. I think that you are starting to realize that you got your balls in a vise.

 

You're back peddling. You were claiming \(R_{abcd} = 0 \Rightarrow \Gamma = 0\) before and now you're changing to "There is some coordinate system where they are zero".

It would seem you realised your mistake and now you've changed what you're claiming. Guest previously quoted you saying "Moller ("Theory of Relativity", page 377) says that you are wrong. If the Riemann tensor is zero, all the Christoffel symbols are also zero. ". No mention of "For at least one set of coordinates", you made a completely general statement. When Promtheus said

you said

You even link to a page on polar coordinates, not mentioning Cartesians at all.

The Riemann curvature tensor has tensorial transform properties. The Christofell symbols are not tensors, they are coordinate choice dependent, as is covered in any introductory course on GR.

At least have the decency to say "Yes, I was wrong about that".

 

This is really sad trout. All you have to do is admit you were wrong and apologize. This isn't physorg and it's not the playground - the ratio of cranks to people with brains is a lot better here and the conversation generally is above the level at which you've pitched it. I'm not claiming to be the worlds expert on GR, but you have made an elementary error and you need to man up to it.

Putting 5 indices on the Riemann tensor is not helping your case either.

 

Its 25% more curved!

 

How many times did Guest tell you that you don't know the chain rule? Didn't you believe Guest when he said that the chain rule is taught in high school? There is no other way to finish this computation. Have you ever thought about taking a refresher course in high school math?

 

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.

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 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)

\(\partial_{i} S\) is shorthand for \(\frac{\partial S}{\partial x^{i}\).

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

\( \begin{align} \frac{\partial t^{\prime}}{\partial t} \,&=\, 1 & \frac{\partial t^{\prime}}{\partial x} \,&=\, S^{\prime}(x) \\ \\ \\ \frac{\partial x^{\prime}}{\partial t} \,&=\, 0 & \frac{\partial x^{\prime}}{\partial x} \,&=\, 1 \;. \end{align} \)​

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.

Funny, it looks just fine to me. What do you think the metric expression should be after applying \(\theta\)?

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.

 

:roflmao:

 

Let's do the last two elements of the jacobian:

\(\frac{\partial t'}{\partial t}=\gamma(1-v \frac{\partial S_j}{\partial x'})\)

\(\frac{\partial t'}{\partial x}=\gamma(\frac{\partial S_j}{\partial x'}(1+v \frac{\partial S_i}{\partial x})-\frac{v}{c^2}(1+\frac{c^2}{v}\frac{\partial S_i}{\partial x}))\)

Combine the above with the other two elements of the jacobian:


\(\frac{\partial x'}{\partial x}=\gamma(1+v \frac{\partial S_i}{\partial x})\)

\(\frac{\partial x'}{\partial t}=-\gamma v\)

and it becomes clear that you have been lying all along, you never calculated the jacobian of your demented transforms. Had you done so, you would have noticed that, contrary to your despicable lying:

-the jacobian is degenerate being dependent on the weird partial derivatives of the two synchronization functions

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

So, your theory is just crap.

 

..and you still have no clue what you are doing.

..but you got the wrong jacobian. You have made this error all along.

Read post 319 , I do not have time for your continued idiocies. I should have posted this long ago. Feel free to continue sucking up to the crackpot Shubert.