On the Definition of an Inertial Frame of Reference
- Thread starter Eugene Shubert
- Start date
Ah - you keep editing!
You have trouble with the chain rule?
No, but you obviously do. Because you fail to account for the extra terms that show up when you use the Schubert functions. Really, you two make the perfect crackpot pair, help him out to finish his paper, you have all the information that he needs.
Eugene Shubert
Valued Senior Member
I have clearly demonstrated that my nonlinear transformation is invertible, see equation (46), (59) and (60), therefore, if it's differentiable, then the Jacobian exists and is never zero. Please keep in mind however that, in general, my transformation is so general that it's easy to come up with non-differentiable examples.
Yes, I'm sure you are but I mention them because they are examples of non-linear transformations where tensor covariance is demonstrated. I doubt Guest thinks Eugene is onto anything, I certainly don't, he's instead talking about the general concept of coordinate transformations and tensors.
If Eugene's coordinates have non-singular $$J^{a}_{b} = \frac{\partial \tilde{x}^{a}}{\partial x^{b}}$$ then they are 'valid coordinates' and expressing Maxwell's equations in terms of them would still give the form $$\nabla^{a}F_{ab} = 0$$, just F would be different from the F in the Cartesian form.
You said that you think Guest doesn't know what he's talking about. I can assure you that Guest's knowledge in the likes of tensor calculus is formidable, I doubt that many of even the PhDs in physics here would even come close, myself included. I would have thought you might get the hint when he talked about exterior derivatives of things in cotangent bundles.
I wouldn't trust your mathematical abilities as far as I can throw a 2 ton rhino so forgive me if I don't take your word for it and I'm not going to waste my time checking. The point Guest is talking about is entirely independent from the specifics of your nonsense.Eugene said:
That's not the point, anything can be differentiated. What do you get when you do it?
Eugene Shubert
Valued Senior Member
Hopefully Guest, after he's finished playing, will take the time and educate you also.
Nah, he's a crackpot at par with Shubert.
This is why I asked Guest254 to calculate something simple , like the partial derivative wrt to a coordinate of his choice.
That might be but:
-he doesn't understand the underlying physics, this became clear when he maintained that a change of notation changes the equations from being non-covariant to becoming covariant
-he has no proof that the tensor transformation rules apply to implicit functions other than his unsubstantiated claims'
But you get a garbled mess when you try getting the tensor transforms. You claim to be a mathematician (you are definitely not a physicist) but you do not understand the conditions under which the formulas you are citing have been derived. The nonsense that Shubert is doing (S(ax+bt+S(x,t)) is definitely not mainstream , you you seem to believe that it will retain the form of the tensor, while all along refusing to perform any calculations.
I'm afraid I probably think you're a little bit whacko (what I'm having to explain to Tach is largely independent of whatever you've written - I'm talking about any coordinate transformation). On the up side, Tach thinks we have a future together. Call me?
News flash: some coordinate transformations might be messy!
It is not ONLY that they are "messy" , it is that their application does not result into the standard forms of tensor transformation. Alpha says that you are an expert, why are you so reluctant to do a simple calculation to prove it to yourself?
Eugene Shubert
Valued Senior Member
I understand that. I took Ted Frankel's graduate level geometry of physics course when I was a math student at UCSD.
What bit can't you do? You say you're ok with the differentiation, so which bit is too tough?
A coordinate system is a coordinate system is a coordinate system. Any set of diffeomorphisms defines a coordinate change. If you're terribly worried about implicitly defined functions, I'll let you into a secret... the implicit (and inverse) function theorems. Now go yonder and enjoy your new found knowledge!
Good for you.
Good, so you two can write the missing chapter on electrodynamics. While at it, you might collaborate on getting the Doppler effect right and removing the embarrassing chapter on the Twins "paradox".