On the Definition of an Inertial Frame of Reference

[Guest254]

If only you would apply the same passion in learning tensor calculus!

[Guest254]

Ah - you keep editing!

Not for IMPLICIT functions of coordinates, you dork. Check your books since you are too lazy to calculate a chain of of derivatives.

You have trouble with the chain rule?

[Tach]

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]

Provided Eugene's coordinates have a nowhere zero Jacobian when expressed in terms of standard Cartesians then they are a valid set of coordinates everywhere.

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.

[Guest254]

Erm, the coordinates are just functions... the derivatives of coordinates are just functions... You don't seriously need to be told how to use the chain rule do you?

[Tach]

Erm, the coordinates are just functions... the derivatives of coordinates are just functions... You don't seriously need to be told how to use the chain rule do you?

have you even looked at his functions, idiot?
did you notice the S(ax+bt+S(S(x,t)) ?

[Guest254]

You can't differentiate that?

[AlphaNumeric]

Agreed, I am very familiar with the transformation between cartesian and polar coordinates.

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 then they are 'valid coordinates' and expressing Maxwell's equations in terms of them would still give the form , 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 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.

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.

[Tach]

You can't differentiate that?

That's not the point, anything can be differentiated. What do you get when you do it?

[Guest254]

Hey, you were worried about derivatives - not me! I know the chain rule.

[Eugene Shubert]

I wouldn't trust your mathematical abilities as far as I can throw a 2 ton rhino so forgive me

Hopefully Guest, after he's finished playing, will take the time and educate you also.

[Tach]

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.

Nah, he's a crackpot at par with Shubert.

If Eugene's coordinates have non-singular then they are 'valid coordinates' and expressing Maxwell's equations in terms of them would still give the form , just F would be different from the F in the Cartesian form.

This is why I asked Guest254 to calculate something simple , like the partial derivative wrt to a coordinate of his choice.

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.

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'

[Tach]

Hey, you were worried about derivatives - not me! I know the chain rule.

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.

[Guest254]

Hopefully Guest, after he's finished playing, will take the time and educate you also.

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?

[Guest254]

But you get a garbled mess when you try getting the tensor transforms. You claim to be a mathematician but you do not understand the conditions under the formulas you are citing have been derived.

News flash: some coordinate transformations might be messy!

[Tach]

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]

(what I'm having to explain to Tach is largely independent of whatever you've written - I'm talking about any coordinate transformation).

I understand that. I took Ted Frankel's graduate level geometry of physics course when I was a math student at UCSD.

[Guest254]

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?

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!

[Guest254]

I understand that. I took Ted Frankel's graduate level geometry of physics course when I was a math student at UCSD.

Good for you.

[Tach]

I understand that. I took Ted Frankel's graduate level geometry of physics course when I was a math student at UCSD.

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".