# Thread: On the Definition of an Inertial Frame of Reference

1. Originally Posted by Tach
Are you daft?
No. Are you?

It clearly talks about inertial frames.
I'll take that as a "no" then. As I said, didn't think so.

Meaning that it is perfectly applicable to this thread.
No, that's just you retroactively changing the subject. Everyone who's posted in this thread already knows that the Christoffel symbols vanish in inertial coordinate systems. If you'd recognised that your article was specifically about inertial frames then you wouldn't have brought it up in the first place, since it's redundant and only says what people in this thread have already told you.

He clearly constructs inertial frames, just with a weird clock synchronization scheme.
No, he claims his frames are inertial. This is something he's wrong about. Not that this diversion helps you.

No, you haven't. You just said
That would have been true if $S_j=S_j(x,t)$ not for $S_j=S_j(x-kvt-S_i(x))$.Since this is not the case, you need to pay attention before you start gloating.
which doesn't help you. $S_j(x-kvt-S_i(x))$ is an expression that defines a function of x and t.

2. Originally Posted by AlphaNumeric
Did you even read the post? I wasn't talking about the Christoffel symbols, I was talking about Jacobians, about how they don't have to satisfy $JJ^{\top}=I$. I gave an explicit counter example, as has Guest.
Sure they do, I even gave you the line and paragraph in the Moller book.
I have also pointed out to both you and przyk that, contrary to your claims, the Lorentz transforms satisfy that condition and both of you came back with some very lame answers.

At least read my posts rather than simply skipping them and making up what you think they say.
I read your posts, I am tired on your harping on the same nonsense so I gave you a little challenge that should have taken you 5 minutes to complete. You have been posting for 1.5 hours but nothing on the challenge.

Rather than trying to change the subject why don't you try actually discussing things?
I am not changing the subject, I am just tired about your regurgitating the same stuff, so I decided to post a challenge. We can work in parallel, I will answer to your tripe while you are still struggling with answering my challenge.

I couldn't care less about Eugene's stuff,

3. Originally Posted by przyk

No, he claims his frames are inertial. This is something he's wrong about.
Prove it.

4. Originally Posted by Tach
I have already. It took me less than 5 minutes.
Then what the hell do you need us to do it for?

You mean, you can't. If you could, you would have done it.
Nope. "The result does not interest me" is a perfectly valid reason for me not to do an exercise.

5. Originally Posted by przyk
Then what the hell do you need us to do it for?
To see if you can do it. Al three of you are adept at masturbating the jacobian of the cartesian->polar coordinates transform but when faced with a little problem you all resort to diversions.

Nope. "The result does not interest me" is a perfectly valid reason for me not to do an exercise.
So, you can't. Figures.

6. Originally Posted by AlphaNumeric
I don't think Eugene is onto anything, no one here is defending him.

I have been attacked and I am still being attacked for incredibly stupid reasons. Obviously, I have rattled a lot of cages. And please note this: Not one mathematician has attempted to argue that my concepts or math are incorrect.

7. Originally Posted by Tach
Prove it.
Er, by definition, an inertial coordinate system is one in which the space-time metric is
$\text{d}s^{2} \,=\, -\, \text{d}t^{2} \,+\, \text{d}x^{2} \,+\, \text{d}y^{2} \,+\, \text{d}z^{2} \.$
The coordinate transformations which leave this metric invariant are the ones we call Lorentz transformations. Eugene's transformation is not a Lorentz transformation and it does not leave the metric above invariant. Therefore no more than one of the coordinate systems he's using can satisfy the definition of "inertial frame".

8. Originally Posted by przyk
Er, by definition, an inertial coordinate system is one in which the space-time metric is
$\text{d}s^{2} \,=\, -\, \text{d}t^{2} \,+\, \text{d}x^{2} \,+\, \text{d}y^{2} \,+\, \text{d}z^{2} \.$
The coordinate transformations which leave this metric invariant are the ones we call Lorentz transformations. Eugene's transformation is not a Lorentz transformation and it does not leave the metric above invariant.
Good. So the transformations do not leave the Maxwell equations invariant either. See my post 46. Are we done with this stuff?

9. Originally Posted by Tach
To see if you can do it.
What difference does it make whether I can do it?

but when faced with a little problem you all resort to diversions.
Because the problem itself is a diversion. It is irrelevant to anything I, Guest, or Alphanumeric have claimed in this thread. We'd be jumping through a hoop you set for us for the sole purpose of proving we could jump through that particular hoop you set for us.

So, you can't. Figures.
If you want to believe that, go ahead.

10. Originally Posted by Tach
Sure they do, I even gave you the line and paragraph in the Moller book.
I have also pointed out to both you and przyk that, contrary to your claims, the Lorentz transforms satisfy that condition and both of you came back with some very lame answers.
Then it appears you are not only apathetic for not reading what I said but you're also unable to do basic calculus. In a specific representation the Lorentz transforms can satisfy $\Lambda \cdot \Lambda^{\top} = I$ but not in all represnetations. You made a general statement, which is false. The standard representation of Lorentz transformations don't satisfy them.

This seems to be your standard mistake, you make broad statements which are infact only true for particular cases. You did it for Christoffel symbols too.

Then there's Guest's example, $(x,y) = (R\cos t,R\sin t)$. If $JJ^{\top}=I$ then det(J) = $\pm 1$. The determinant of the Jacobian of that example has determinant R (or 1/R, depending which way you're going). No representation is going to satisfy $JJ^{\top}=I$ for general R.

Its trivial to construct such counter examples. For instance, $(x')^{a} = \lambda x^{a}$ in $\mathbb{R}^{N}$ will have a Jacobian determinant of $\lambda^{N}$. If $\lambda \neq 1$ (or zero, to be valid) then its a counterexample to your claim.

You clearly fail to realise the book you're reading is talking about specific cases, not all cases.

Originally Posted by Tach
I read your posts, I am tired on your harping on the same nonsense so I gave you a little challenge that should have taken you 5 minutes to complete. You have been posting for 1.5 hours but nothing on the challenge.
Yes, you're tired of me harping on about your mistakes so you're trying to make up an excuse to ignore me. Can I be bothered to do your little challenge? No. Does that mean your mistakes are not mistakes? No.

Originally Posted by Tach
I am not changing the subject, I am just tired about your regurgitating the same stuff, so I decided to post a challenge. We can work in parallel, I will answer to your tripe while you are still struggling with answering my challenge.
I don't need to prove myself by jumping through your hoops. Do you think I'm unfamiliar with coordinate transforms, Jacobians, covariance etc? If you're Trout from PhysOrg then you know I'm more than capable when it comes to those things, so your "Why don't you do my challenge!" is just an excuse to avoid facing up to your mistakes.

Given your mistakes when it comes to covariance, Jacobians and coordinate transforms I seriously wonder if you can even do such things. If you're the person banned from editing Wikipedia Guest linked to then clearly I'm not the first person to have misgivings about your abilities.

Originally Posted by Tach
So no one should ever point out a mistake in someone's post if that mistake isn't directly to do with the original post? Oh please.

The reason this thread is so off topic is because you couldn't just say "Fair point, I was mistaken about that". Instead it takes 4 or 5 pages of several people explaining your mistake half a dozen different ways before you realise you're wrong and then you try to change the subject. If you'd just faced up to your mistake there'd have been no need to reiterate it again and again.

11. Originally Posted by Tach
Good. So the transformations do not leave the Maxwell equations invariant either. See my post 46. Are we done with this stuff?
They still leave the generally covariant expression of Maxwell's equations invariant in form. That's not very interesting, because any arbitrary transformation will leave the generally covariant formulation of Maxwell's equations intact. Hence why the nicest thing anyone has had to say about Eugene's transformation is "trivial".

12. Originally Posted by Eugene Shubert
Obviously, I have rattled a lot of cages.
That's the self deluding spirit, if someone says "That's crap" and takes no notice its not because its crap but because you've 'rattled their cage'.

Originally Posted by Eugene Shubert
Not one mathematician has attempted to argue that my concepts or math are incorrect.
Well according to Tach we're all useless. Besides, you've been peddling this stuff for years and you've not taken any notice of anything anyone says. It'd be wasted effort clearly.

Have you submitted your work to a reputable journal, to people whose business it is to evaluate the merit of your work? If not, why not? If so, what did they say? Since you're here peddling your work I'd guess you haven't got your work published. Let me guess, you 'rattled their cage' and that's why you were rejected?

13. Originally Posted by AlphaNumeric
Have you submitted your work to a reputable journal, to people whose business it is to evaluate the merit of your work?

According to popular mythology, ghosts have a hard time realizing that they are ghosts and they are stuck in a loop, continually rethinking and recycling their strongest fixations. Please understand that you're dead from Sir Knight's sword through your head and that you have already asked me these questions and that I have already answered.

14. Originally Posted by przyk
Hence why the nicest thing anyone has had to say about Eugene's transformation is "trivial".

Let's understand this fact. That's an endorsement of my position. I agree that my nonlinear transformation equations are trivial.

15. [QUOTE=arfa brane;2643135]I think you might have missed it.

I'm not sure I missed it?

JamesR said the following in post #353

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

That's why I responded. Do you think the Christoffels are non-zero just because you choose to describe an inertial frame of reference with polar coordinates? I would suspect you don't agree with that but I'll ask incase I'm wrong.

I have no disagreement with anything else you said concerning choice of coordinate systems.

16. Originally Posted by brucep
Do you think the Christoffels are non-zero just because you choose to describe an inertial frame of reference with polar coordinates?
There are physical systems and frames of reference in them. The two don't necessarily commute (see for instance, Eugene's "paper"), which is why choosing the "right" coordinate systems is important.

The only truly inertial physical system is one with no mass in it.
Christoffel symbols are just a mathematical way to describe local curvature, which is known as gravity in Newtonian frames of reference, which frames are non-relativistic. This doesn't mean you can't use Christoffel symbols in a coordinate sytem which is non-relativistic.

But don't quote me on any of that, will you?

17. Originally Posted by przyk
They still leave the generally covariant expression of Maxwell's equations invariant in form.
We've been over this already. This is not the subject being discussed in post 46.

That's not very interesting, because any arbitrary transformation will leave the generally covariant formulation of Maxwell's equations intact. Hence why the nicest thing anyone has had to say about Eugene's transformation is "trivial".
This is why the general covariance of Maxwell's laws is not a test of the validity of Shubert transforms. On the other hand, the test on the speed composition IS. Five hours after I posted the challenge and none of you three managed to solve it. Telling....

18. Originally Posted by AlphaNumeric

Then there's Guest's example, $(x,y) = (R\cos t,R\sin t)$.
Would you give it a rest? The cartesian->polar coordinate transforms do not maintain the invariance of the Minkowski metric. Therefore it is not expected to satisfy the condition $JJ^T=I$. It isn't even a spacetime transform, it transforms only spatial coordinates. Why do you keep bringing up this IRRELEVANT counterexample? On the other hand , the Lorentz transform maintains the metric invariance, and, it satisfies the condition $\Lambda \Lambda ^T=I$ (see the exercise I gave you and przyk).
Now, how about you worked on my challenge a little? Five hours of struggling with it is a little excessive, don't you think?

19. Originally Posted by Tach
This is why the general covariance of Maxwell's laws is not a test of the validity of Shubert transforms.

It's nice to see that you are going to stop beating that dead horse. Not me, the nonsense about the alleged crime of expressing Maxwell's equations in terms of arbitrary coordinates.

Originally Posted by Tach
On the other hand, the test on the speed composition IS.

It's obvious Sir Knight that my nonlinear transformations are the product of 3 function compositions and that equation (64) is physically correct.

20. Originally Posted by Eugene Shubert
It's nice to see that you are going to stop beating that dead horse. Not me, the nonsense about the alleged crime of expressing Maxwell's equations in terms of arbitrary coordinates.
Post 46 says that you are full of shit. You've always been.

It's obvious Sir Knight that my nonlinear transformations
...are invalid

are the product of 3 function compositions and that equation (64) is physically correct.
It is also obvious that your crackpot theory fails any velocity addition test. Being a self-proclaimed "mathematician" I do not expect you to know which test falsifies your theory, this question is left for physicists. The challenge is for the three musketeers to figure out, they are still struggling, five hours after the question was put to them. Makes you wonder....are they physicists as they claim or just pretenders like you.