On the Definition of an Inertial Frame of Reference

**Eugene Shubert** · 10-27-10, 07:13 PM

Originally Posted by przyk

The covariant derivative coincides with the partial derivative in inertial coordinate systems.

That, of course, is true for physicists that believe in noncovariant laws of physics and can't generalize the definition of an inertial frame of reference to make the definition generally covariant.

**przyk** · 10-27-10, 07:33 PM

Originally Posted by Tach

Have you read his "masterpiece"?

No. The abstract alone was enough to make me cringe.

**Tach** · 10-27-10, 07:46 PM

Originally Posted by przyk

No. The abstract alone was enough to make me cringe.

Ahh, but this is one of the best crackpot stuff I have ever seen, it is fun to poke holes. Shubert is a case of a very sophisticated crank, complete with paranoia. :-)

**Tach** · 10-27-10, 07:50 PM

Originally Posted by Eugene Shubert

That, of course, is true for physicists that believe in noncovariant laws of physics and can't generalize the definition of an inertial frame of reference to make the definition generally covariant.

Shubert,

Don't flatter yourself, until we explained to u , you didn't even know what covariant means. I am positive that you STILL don't know.
Since you aren't able to remedy the bigger problems with your paper, let's try something really, really simple. Calculate the expression for speed composition in your formalism. You don't have it in your paper but you will need it since it is testable by experiment.

**Guest254** · 10-28-10, 06:04 AM

Why do you insist on digging your hole even deeper!

Originally Posted by Tach

1. We are working in flat spacetime,, with inertial frames ONLY, so the covariant derivatives coincide with the partial ones.

No, the Christoffel symbols vanish if the coordinate system is Cartesian. You've clearly never computed the Christoffel symbols for polar coordinates. Congratulations on once again making a fool of yourself.

Originally Posted by Tach

2. Guest254 agreed with me that the Shubert transforms do not make the classical (differential form) Maxwell laws covariant.

General coordinate transformations will not leave the form of non-tensorial equations unchanged - this should be obvious.

Originally Posted by Tach

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:

My claim is far grander - any coordinate transformation will leave the form of a tensor equation invariant.

Originally Posted by Tach

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

This is cringe-worthy. Polar coordinates a new thing for you then?

Originally Posted by Tach

-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

You're still massively confused as what general covariance means. It is a statement regarding the form of the equations. General covariance means that you can write down the equations in such a way that their form does not change under coordinate transformations. I.e, you can write them down in terms of objects that behave in very special ways under coordinate transformations - these special objects are called tensors.

Originally Posted by Tach

-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.

And to top it off, you admit to not being able to differentiate! But because I'm in a chirpy mood: if f(x,t) = ax + bt + S(cx + dt + S(S(x))), then f_x = a + S'(cx + dt + S(S(x)))(c+S'(S(x))S'(x)). This chain rule thing is magic!

**przyk** · 10-28-10, 06:10 AM

Originally Posted by Eugene Shubert

That, of course, is true for physicists that believe in noncovariant laws of physics and can't generalize the definition of an inertial frame of reference to make the definition generally covariant.

This is silly. In the covariant formalism, inertial coordinate systems are pretty much defined as those in which $g\bigl( \frac{\partial}{\partial x^{\mu}},\, \frac{\partial}{\partial x^{\nu}} \bigr) \,=\, \eta_{\mu\nu}$ . Physicists find it useful to talk about those particular coordinate systems, so they've given them a name. If you want to talk about or use a more general class of coordinates, fine. No-one is stopping you. Just don't call them "inertial" because that particular label has already been claimed for something and has a specific meaning. Unless you're just out to deliberately confuse people or pretend you're doing something more significant than you actually are, call them something else.

**Eugene Shubert** · 10-28-10, 07:55 AM

Originally Posted by przyk

In the covariant formalism, inertial coordinate systems are pretty much defined as those in which $g\bigl( \frac{\partial}{\partial x^{\mu}},\, \frac{\partial}{\partial x^{\nu}} \bigr) \,=\, \eta_{\mu\nu}$ .

I don't believe that that equation is generally covariant. Nevertheless, I will happily defer to the expert on this subject. I only trust mathematicians.

**Tach** · 10-28-10, 10:18 AM

Originally Posted by Guest254

Why do you insist on digging your hole even deeper!

No, the Christoffel symbols vanish if the coordinate system is Cartesian. You've clearly never computed the Christoffel symbols for polar coordinates. Congratulations on once again making a fool of yourself.

General coordinate transformations will not leave the form of non-tensorial equations unchanged - this should be obvious.

This is cringe-worthy. Polar coordinates a new thing for you then?

For education, polar coordinates TRANSFORMATION is a SPACE transformation.
The pertinent transforms are spaceTIME transforms, like the Lorentz (retain covariance) vs. Galilei/Crackpot_Shubert (do NOT retain covariance).Why do you insist of making such an ass of yourself every opportunity you have?

My claim is far grander - any coordinate transformation will leave the form of a tensor equation invariant.

I am not a psychiatrist, I can't cure your delusions of grandeur but why don't you try to prove your grand claims by applying them to the Shubert transforms (more about this challenge to your mathematical prowess in the last two paragraphs below).

You're still massively confused as what general covariance means. It is a statement regarding the form of the equations. General covariance means that you can write down the equations in such a way that their form does not change under coordinate transformations. I.e, you can write them down in terms of objects that behave in very special ways under coordinate transformations - these special objects are called tensors.

Did I tell you anything different in this thread? Or you can only listen to yourself?

And to top it off, you admit to not being able to differentiate!

No, idiot, this is NOT what I said. What I said is that the form of the Shubert crackpot transform renders it useless for retaining the covariance of the classical Maxwell equations under his transformations. As an example, I challenged you to calculate the operator $\delta/\delta t$

But because I'm in a chirpy mood: if f(x,t) = ax + bt + S(cx + dt + S(S(x))), then f_x = a + S'(cx + dt + S(S(x)))(c+S'(S(x))S'(x)). This chain rule thing is magic!

Congratulations! After much proding you did it. Now try using it in showing that the wave equation is covariant.

**Guest254** · 10-28-10, 10:35 AM

Originally Posted by Tach

For education, polar coordinates TRANSFORMATION is a SPACE transformation.
The pertinent transforms are spaceTIME transforms, like the Lorentz (retain covariance) or Galilei/Crackpot_Shubert (do NOT retain covariance).Why do you insist of making such an ass of yourself every opportunity you have?

Oh come on - at least try to defend your "Christoffel symbols vanish in flat space" comments! This is no fun otherwise.

Originally Posted by Tach

No, idiot, this is NOT what I said. What I said is that the form of the Shubert crackpot transform renders it useless for retaining the covariance of the classical Maxwell equations under his transformations. As an example, I challenged you to calculate the operator $\delta/\delta t$

Really? I thought you were having trouble because, and I quote, "you get derivatives of S with respect to itself". Or have you overcome the hurdle that is the chain-rule now?

Originally Posted by Tach

Congratulations! After much proding you did it. Now try using it in showing that the wave equation is covariant.

Don't congratulate me - congratulate the high school education system! But you'd like a generally covariant form of the wave equation - here we go:

$\nabla^a \nabla_a \phi =0$

You're really having trouble with this notion of general covariance, aren't you?

**Tach** · 10-28-10, 10:38 AM

Originally Posted by Eugene Shubert

I don't believe that that equation is generally covariant. Nevertheless, I will happily defer to the expert on this subject. I only trust mathematicians.

Come on , Shubert, do the simple exercise, find out the formula for speed composition. Are you keeping silent because you realized you can't do it? Any paper attempting to dethrone master Einstein has to have such a formula. Yours doesn't, did you realize you can't do even such a simple thing in your crank formalism?

**Tach** · 10-28-10, 10:40 AM

Originally Posted by Guest254

Oh come on - at least try to defend your "Christoffel symbols vanish in flat space" comments! This is no fun otherwise.

I still see that you are still hung up on your brilliant discovery that Christoffel symbols are nonzero for the case of polar transforms.
I see that you still can't tell spacetime transforms from space transforms.

**Eugene Shubert** · 10-28-10, 11:00 AM

Originally Posted by Guest254

You're really having trouble with this notion of general covariance, aren't you?

What did you expect? Tach is very proud of announcing triumphantly that he doesn't discern the velocity addition formula for generalized Lorentz-equivalent transformation equations in my paper. Obviously it is equation (64).

**Tach** · 10-28-10, 11:07 AM

Originally Posted by Eugene Shubert

What did you expect? Tach is very proud of announcing triumphantly that he doesn't discern the velocity addition formula for generalized Lorentz-equivalent transformation equations in my paper. Obviously it is equation (64).

Shubert,

Eq(64) is the SR velocity addition , derived from the Lorentz transforms.
I am asking you for your theory formula, derived from the Shubert crackpot transforms. Get with the program. Try again, no cheating this time.

**Eugene Shubert** · 10-28-10, 11:19 AM

In other words, you are unable to discern that equation (62) is my generalized Lorentz-equivalent transformation equation from frame $i$ with synchronization $S_i$ to frame $j$ with synchronization $S_j$ ?

**Tach** · 10-28-10, 11:47 AM

Originally Posted by Eugene Shubert

In other words, you are unable to discern that equation (62) is my generalized Lorentz-equivalent transformation equation from frame $i$ with synchronization $S_i$ to frame $j$ with synchronization $S_j$ ?

Eugene,

Stop being pathetically dishonest, here is your homework: use your crackpot transform (62) in order to derive your crackpot speed composition formula. Stop the weaseling and get with the program.

**Eugene Shubert** · 10-28-10, 11:55 AM

$\mbox{If }\begin{pmatrix} x \\ t \end{pmatrix}' = {\Theta_j}^{-1} L(\nu_1) \Theta_i \begin{pmatrix} x \\ t \end{pmatrix} \mbox{ and } \begin{pmatrix} x \\ t \end{pmatrix}'' = {\Theta_k}^{-1} L(\nu_2) \Theta_j \begin{pmatrix} x \\ t \end{pmatrix}' \hspace{.62 in}\mbox{ then } \begin{pmatrix} x \\ t \end{pmatrix}'' = {\Theta_k}^{-1} L\left(\frac{\nu_1 + \nu_2}{1 + k \nu_1 \nu_2}\right) \Theta_i \begin{pmatrix} x \\ t \end{pmatrix}$

**Tach** · 10-28-10, 12:01 PM

Originally Posted by Eugene Shubert

$\mbox{If }\begin{pmatrix} x \\ t \end{pmatrix}' = {\Theta_j}^{-1} L(\nu_1) \Theta_i \begin{pmatrix} x \\ t \end{pmatrix} \mbox{ and } \begin{pmatrix} x \\ t \end{pmatrix}'' = {\Theta_k}^{-1} L(\nu_2) \Theta_j \begin{pmatrix} x \\ t \end{pmatrix}' \hspace{.62 in}\mbox{ then } \begin{pmatrix} x \\ t \end{pmatrix}'' = {\Theta_k}^{-1} L\left(\frac{\nu_1 + \nu_2}{1 + k \nu_1 \nu_2}\right) \Theta_i \begin{pmatrix} x \\ t \end{pmatrix}$

I didn't ask you about the Einstein speed composition rule.
So, you don't know how to derive if from your crackpot transforms. Another strike against your crackpot theory.

**Eugene Shubert** · 10-28-10, 12:09 PM

$\mbox{If }\begin{pmatrix} x \\ t \end{pmatrix}' = {\Theta_j}^{-1} L(\nu_1) \Theta_i \begin{pmatrix} x \\ t \end{pmatrix} \mbox{ and } \begin{pmatrix} x \\ t \end{pmatrix}'' = {\Theta_k}^{-1} L(\nu_2) \Theta_j \begin{pmatrix} x \\ t \end{pmatrix}' \hspace{.62 in}\mbox{ then } \begin{pmatrix} x \\ t \end{pmatrix}'' = {\Theta_k}^{-1} L\left(\frac{\nu_1 + \nu_2}{1 + k \nu_1 \nu_2}\right) \Theta_i \begin{pmatrix} x \\ t \end{pmatrix}$

The trick is to factor the generalized Lorentz transformation equations (44)-(45) into 3 function compositions. That's high school math.

**przyk** · 10-28-10, 12:25 PM

Originally Posted by Eugene Shubert

I don't believe that that equation is generally covariant. Nevertheless, I will happily defer to the expert on this subject. I only trust mathematicians.

Do you even know what the equation means?