On the Definition of an Inertial Frame of Reference

I think you might have missed it.

It's been established (see Moller's "Theory of Relativity", page 377) in this thread that:

So since there is a way to choose coordinates which don't account for local curvature and where the frame looks inertial, there is also a choice that does account for curvature in the same frame of reference.

In short, spacetime is curved anywhere mass can accelerate, including "into" itself. Note that you experience non-inertial acceleration on the surface of the earth. You can choose coordinates so acceleration vanishes in a small region, for instance by suspending a material object above the surface--the earth doesn't vanish because you choose coordinates this way.

 

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Just when you think someone must have had enough!

This is perhaps the 10th time I've had to explain this to you. General covariance is the property that physical laws should be representable in a form that is invariant under arbitrary coordinate transformations. Maxwell's equations can be written in the form of a tensor equation, hence obey the principle of general covariance. So for any bonafide coordinate transformation, the form of the equations is retained.

Once again, you demonstrate that you do not understand the principle of general covariance! Just because we're using covariant derivatives etc, does not mean we're doing calculus on curved spaces - we're just using the age old tools from the theory of curvilinear coordinates! We've been through the whole polar coordinate thing with you several times now - it has to land eventually, right?

You can't still be having trouble with the chain rule, surely!

As an interesting aside, I just googled this character "dono" and revealed a rather sad history. It seems this person has been banned from forums (e.g Physics Forums) in the past for talking rubbish and following it up with chest beating. According to the link he's even been banned from Wikipedia! One of more amusing reasons he's banned from posting on Wikipedia because he kept adding his own name to the list of Romanian-Americans!

It seems all this talk of "pretender", "crackpot" and "crank" might be a method of projection.

 

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I can't believe you're in that much denial Tach. You claim Guest and Prom were 'hijacking' the thread, which is clearly nothing more than you attempting to avoid facing up to the fact you have repeatedly been shown to be wrong. How many times do you need your own posts quoted to you? Do you think everyone else just forgets what you've said or that no one can go back a few pages and read it for themselves?

Any reasonable person would realise they've put their foot in their mouth, instead you just keep digging yourself in deeper.

Seriously, you need to go back and work through that GR textbook again because you obviously haven't understood it. But I imagine you won't, if Guest's links are true this isn't the first time you've had your backside handed to you.

 

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And this is the 10th time I explained to you, the Shubert tansforms are not "bona fide". I have shown that to you, yet you still insist in bringing back your rotten red herrings.

In your deluded imagination. It is trivial to prove you a liar, like everything you have posted so far:

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

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


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

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

The jacobian must satisfy the following requirements:

1. Its elements must be well defined, \(g_{22}=g_{22}(x,x')\) isn't by virtue of being dependent on both x and x'
2. \(JJ^T=I\) . This is clearly not happening
3. det(J)=1 . This is happening but it is immaterial given that 1 and 2 are not.

Chew on this.

 

No it's not. You were claiming that the covariant derivative always coincides with the partial derivative in flat space-time. It doesn't. Covariant derivative expressions in polar coordinates will include non-zero Christoffel symbols, even if you're working in flat space. And this isn't some peculiarity of polar coordinates. In general, the Christoffel symbols are non-zero, and the covariant derivative does not coincide with the partial derivative, in coordinate systems in which the metric is inhomogenous. These coordinate systems are precisely those that are related to inertial coordinates by a non-linear transformation. Polar coordinates just happen to be one of the most commonly encountered examples.

How can you say that, when I worked out the metric for you? You were denying this on the basis that Eugene's transformation couldn't be differentiated and his factorisation \(\theta_{j} \,\circ\, \Lambda \,\circ\, \theta^{-1}_{i}\) was incorrect. You've been proven wrong on both counts. In the first case you even disproved yourself by successfully calculating the differential.

 

Wrong, read the post just above yours.

 

But you can express x' in terms of x and t, and vice-versa. So you can write the metric in terms of just one set of coordinates.

This isn't the first time you've had difficulty with variable substitution.

There is no such requirement. Even Lorentz transformations don't satisfy that.

Only if you require that the transformation leaves a certain metric invariant. Eugene's transformation doesn't (unless he always uses the same synchronisation function S), so points 2 and 3 are moot.

 

Wrong. Read the post just below yours.

 

This is the second time you write this idiocy.

What in :

\(t'=\gamma(t-kvt-S_i(x))+S_j(x')\) you don't understand? Is this a valid transform?

Try Moller, page 94.

Of course it does. Wrong again, try Moller, page 95.

 

You're really not up to speed with calculus, are you..

x'=x'(x,t). You can't seriously need that explaining to you? Come on, seriously now!

Erm, obviously we don't need this. We just need the non-vanishing of the determinant of the Jacobian. I hate to highlight your ignorance yet again, but you might like to check the determinant for the map (x,y)->(Rcos t, Rsin t). (these are called polar coordinates by the way).

Erm, obviously we don't need detJ=1. It just needs to be non-vanishing for the inverse function theorem to hold. Do you not remember me giving you this nugget of knowledge - I thought you'd at least google it! Again, you might like to consider the polar coordinates I gave you.

You're on a roll now - any more clangers you'd like to drop?

 

It is not my fault that you are a lazy bum who refuses to read Shubert's paper and talks through his ass.


\(t'=\gamma(t-kvt-S_i(x))+S_j(x')\)

so, t'=t(x,t, AND x') !

False, read Moller page 94.

False again. Read Moller page 95 and maybe you'll learn something.It is a good book, you should invest in buying it. Consult it before you make a fool of yourself.

 

That you need the same point explained to you multiple times reflects poorly on you, not me.

Yes. It's perfectly valid, because \(x^{\prime} \,=\, \gamma \bigl[ x \,-\, vt \,+\, v S_{i}(x) \bigr]\), so

\( t^{\prime} \,=\, \gamma \bigl[ t \,-\, kvx \,-\, S_{i}(x) \bigr] \,+\, S_{j} \Bigl( \gamma \bigl[ x \,-\, vt \,+\, v S_{i}(x) \bigr] \Bigr) \;. \)​

Do you really need variable substitution explained to you?

I don't own that reference. What does Moller say on page 94?

What are you talking about? Lorentz transformations don't satisfy \(\Lambda^{T} \Lambda \,=\, I\). They're not rotation matrices. The defining property of Lorentz transformations is \(\Lambda^{T} \eta \Lambda \,=\, \eta\) where \(\eta\) is the Minkowski metric. Surely you know this, right? And even that isn't a requirement in order for a transformation to be "valid". It's just a requirement if you want to leave the Minkowski metric invariant, which is not required in the generally covariant formulation.

 

I will not deal with this idiocy again.

It says exactly what I told you, that you must have \(\Lambda^{T} \Lambda \,=\, I\)

It is very easy to show that you are wrong. Let's start with something simple, in only two dimensions:


\(\begin{pmatrix}\gamma & -i \gamma \beta \\ i \gamma \beta & \gamma & \end{pmatrix} \)

 

Please explain, precisely, what is so idiotic about variable substitution - a technique normally learned early on in highschool level math. Or is this you bluffing a disingenuous and hasty retreat again?

That's not the only valid way of representing a Lorentz transformation. You really are narrow minded if you think the particular way you learned to do things is the only valid way. Representing time as an imaginary variable is a trick sometimes used in relativity, but it doesn't generalise well to the generally covariant case. Personally I don't see it used much.

And as I told you, general transformations are not required to leave a metric invariant.

 

You are sinking lower and lower. Quit while you are still ahead.

This was just a trivial example to show that you are wrong. Feel free to try it out with the full 4x4 matrix. If you don't get the same result it means that you botched your calculations. Ad-hominens will not help your cause. Buy the book, you might learn something.

The condition doesn't only leave the metric invariant, it is instrumental in the invariance of Maxwell's equations.

 

Only by your extremely warped standards.

Nothing to say about your earlier condemnation of variable substitution, by the way?

I'm not disputing that you can represent Lorentz transformations as complex rotation matrices if you represent time as an imaginary variable (ie. you perform a Wick rotation). But that's not the only or even the most common way of representing Lorentz transformations. See the definition here for example, which presents the much more familiar

\( \Lambda^{\alpha}_{\gamma} \, \Lambda^{\beta}_{\delta} \, \eta_{\alpha\beta} \,= \, \eta_{\gamma\delta} \;. \)​

Then it's a good thing I don't need them. I've already explained your errors.

On a related note, you've just explained to yourself why dismissing other people's patient explanations as "idiocy" does not count as a valid rebuttal. Can we hope you will adhere to your own advice from now on?

Read the introductory part of any text on general relativity, which discusses the generally covariant formulation as a prerequisite to doing relativity in curved space. You might learn something.

But not their general covariance. I can't believe this still hasn't sunk in with you.

 

Sour grapes for being proven wrong, eh?

Let me remind you that Shubert's theory is flat spacetime. There is no "relativity in curved spacetime" in his paper. Try to stay on topic.

 

Evidently, you do need it explaining to you! Wow. Read the following slowly if necessary: if x'=x'(x,t), then S(x') = S(x'(x,t))... which is a function of... drum roll... (x,t)! Hurray!

Is it really a good idea to once again quote results from a book you've already shown you don't understand? Or have we already forgotten the "Christoffel symbols vanish" fiasco! But I'm happy to help out again: please quote, verbatim, the result in this book to which you are referring.

 

No. You're picking an infrequently used convention and passing it off as the definitive definition of the Lorentz transformation. You are proving yourself dogmatic, not right.

I'm still waiting for you to explain what you've got against variable substitution, by the way.

For heaven's sake, how many times do you have to be told that the generally covariant formulation is just as valid in flat space-time as in curved space-time? It is necessary for working in curved space-time. That is why you can safely expect to find it covered in any text on general relativity. But it is still perfectly applicable in the special case where the curvature tensor is zero. You've already had this explained to you multiple times by several people here. You really have no excuse.

 

While the above is true, you can continue masturbating the null Christoffel symbols. I posted a paper for you about inertial frames in flat spacetime. You obviously did not study it. Adios.