On the Definition of an Inertial Frame of Reference

Idiot, you get derivatives of S wrt itself. Did you even read Shuber's folly?
If you spent the time doing the calculations rather than arguing in ignorance, you would have found out that you get a nonlinear expression in terms that have NOTHING in common with any coordinates.

S is not a coordinate, S represents the clock synchronization FUNCTION. Why don't you read the lousy paper instead of blowing air out of your behind.

 

Log in or Sign up to hide all adverts.

But you're having too much fun playing with Tach. Enjoy.

 

Log in or Sign up to hide all adverts.

This thread appears to be suffering from a serious bout of "plonkerism".

 

Log in or Sign up to hide all adverts.

Why don't you post your attempt and we can help?

So? If it's used in the definition of a bona fide coordinate transformation, then it doesn't matter one bit.

 

Help with what , idiot ? You can't calculate the form for yourself and see its defects?
Try it for yourself:

\(t'=at+bx+S(a_1t+b_1x+S(t,x,S(x,t)))\)

Shubert claims that it is invertible as well, maybe you can work with him to verify that claim as well.

 

Well, the reason I asked is because you said you ended up "differentiating S with respect to itself", which indicates you're not fully at grips with the chain rule.

Well it's fairly obvious that we have to assume S is such that the thing is invertible and sufficiently smooth! They are defined as coordinates, after all!

 

You're new here so perhaps you don't know but he's a postdoc in mathematics somewhere, clearly those people whose job is it to know good research think he's capable of it.

Where did he say that? He talked about changing from an equation which wasn't covariant to one which is. When he said \(\eta \to g\) he didn't mean change notation he meant when you construct, from the ground up, a formalism where you don't assume the metric is the Minkowski one then you obtain a result which is of the same form as the flat space version but instead of partial derivatives you have covariant ones, which include connection terms. \(\partial \to \nabla\) isn't a change of notation, its an entire change of formalism to say "I no longer ignore connections", which is what you do if you only use partial derivatives. Yes, when you then say "Having formalised this for the general case I now consider the special case of the Minkowski metric" then \(\nabla = \partial\) but this is because \(\nabla = \partial + \Gamma\) with \(\Gamma = 0\) if \(g = \eta\). He clearly wasn't talking about a change of notation but a change of formalism.

\(\partial^{a}F_{ab}\) is only covariant if you view \(\partial\) as \(\nabla\) with the connection manifestly set to zero by the metric choice. Saying \(\partial \to \nabla\) is not a change of notation from that respect, few people competent at relativity would view it in such a way.

The tensor structure invariance under general coordinate transformations do not require you to have an explicit analytic expression for the transformation, an implicit one still defines some transformation and provided you can use the implicit demonstration to demonstrate a non-singular Jacobian then the coordinate transformation is valid. J is defined by some \(\tilde{x}^{\mu} = \tilde{x}^{\mu}(x^{\nu})\) and how that dependency is expressed is irrelevant provided the definition of \(\tilde{x}^{\mu}\) allows you to demonstrate a valid Jacobian.

I don't think I used the word 'expert', I said he's likely better than many, if not all, of the PhDs here in certain things and while that's no small thing that doesn't automatically equate to 'expert'. Being a postgrad or postdoc doesn't make one a genius, as plenty of said postgrads and postdocs will readily admit.

And Guest's abilities can be discerned without having to see him do simple algebra, he's able to hold his own in discussions, as are any of the other regular posters here who say they have formal education beyond degree level.

 

It doesn't seem to me that Guest is claiming it is a valid coordinate definition but rather that it could be, being implicitly defined doesn't present an automatically problem. Also, there's nothing wrong with that form in principle, its possible to take partial derivatives wrt t and x of t'.

Whether the S Eugene specifically uses leads to valid changes of coordinates it another matter and one which seems beside Guest's point.

Or am I misunderstanding you Guest?

 

Precisely. My statement was (and still is!) that the equations \(\nabla^a F_{ab}=0\) etc. obey the principle of general covariance, i.e. it takes on the same form in any coordinate system. Whether the coordinates are defined implicitly or not - it doesn't matter.

Tach's main confusion seems to be (as stated here) that he doesn't realise tensor transformation laws hold for arbitrary changes of coordinates. Well, that and he doesn't seem to be familiar with the basics of tensor calculus. Our discussion is entirely independent of the document in the opening post, since my points hold for any coordinate transformation.

 

Is it now time to see if Guest has the courage to correct misguided physicists?

 

Regardless of what Tach thinks of Guest I think its safe to say all of us think you are misguided in your claims.

You claim you've got 'flawless' work and yet you missed something as basic as the errors I pointed out, the kind of thing children studying algebra can spot and yet you didn't.

 

I would be pleased if Guest confirmed your claim of basic errors in algebra.

 

Why? Do you need help to demonstrate that if \(g(\mu_{ij}) = \epsilon \pm \frac{\sqrt{1+k\mu_{ij}^{2}}}{\mu_{ij}}\) then \(g(\mu_{ij})^{2} \neq \frac{1}{\mu_{ij}^{2}}+k\) for general \(\epsilon\)?

Guest's argument with Tach has nothing to do with the validity (or not) of your claims, it was a more general point about coordinate transformations and tensors. Guest's point is valid regardless of whether your claims are valid. And no one other than you thinks that they are valid.

 

You obviously did not do the COMPLETE chain of derivatives. I thought you were a mathematician, you are just a lame pretender.

 

But this is the whole point of the argument, the lazy bum did not even read the paper, the Shubert crackpot transforms don't lend themselves to a standard transformation of tensors at the change of coordinates. They are nothing like the change from cartesian to polar coordinates. The recursive definition prevents them from working, try calculating the partial derivative wrt t or x, whichever you prefer, and you'll see for yourself.

 

I don't think you have been following the argument.Here is a recap.

1. We are working in flat spacetime,, with inertial frames ONLY, so the covariant derivatives coincide with the partial ones.
2. Guest254 agreed with me that the Shubert transforms do not make the classical (differential form) Maxwell laws covariant.
3. The disagreement starts over the general covariant expression (tensor notation). Guest says that the change of variable proposed by Shubert is legit, I am saying that it isn't. For two reasons:

-in flat spacetime the difference between 2 and 3 is indeed just a change in notation (connection coefficients are zero)

-it is not possible for a change in notation to change the underlying physics, i.e. both 2 and 3 are not covariant, you can't have 2 non-covariant and 3 covariant

-the transformations rules for tensors don't work with the Shubert crackpot coordinate transforms (this is why 2 fails the covariance) but the lazy bum (Guest) refuses to even try to calculate the simplest expression in order to convince himself.

Well, on this one he flunked.

 

Why do I need help with claims that I have never asserted?

 

I understand what Guest said. I forgot the definition of a tangent bundle, pull-backs and exterior derivatives but what does that have to do with misguided physicists being confused by freshman level physics?

 

Er, no, that isn't right. The covariant derivative coincides with the partial derivative in inertial coordinate systems. It doesn't, generally, in non-inertial coordinate systems. Flat space-time just means that the curvature tensor is zero everywhere.

In light of the above, you might want to reconsider that one.

 

The frames in Shubert's crank theory are inertial, no accelerated frames. Have you read his "masterpiece"?