On the Definition of an Inertial Frame of Reference

[Tach]

Aha - you made an edit!

I should hope so!


The maths is hardly deep!

$$ \eta_{ab} \mapsto g_{ab}, \qquad \partial_a \mapsto \nabla_a $$

So Maxwell's equations can be written

$$ \nabla^a F_{ab} =0 , \qquad \nabla_{[a} F_{bc]} =0 $$

which are clearly invariant under coordinate transformations..

I was afraid that that is what you will write. You realize that "the clearly invariant under coordinate transformations" is true for the Lorentz transforms and general linear transforms but not true for Shubert's cockamamie transforms, right? This is why I asked you to start by calculating the partial derivative wrt t.

Alternatively, use the exterior calculus version of Maxwell's equations and use the fact that pull-backs commute with exterior derivatives

Under what conditions is the above true? Do the pull-backs commute with the exterior derivatives for nonlinear transforms? Are A and $$\Phi$$ four-vectors wrt non-linear transforms? You realize that the tensorial formalism is nothing but a different notation for the classical formalism using partial derivatives, so what is not true for the classical formalism cannot become magically true if one uses four-vectors or tensors.

 

[Guest254]

I was afraid that that is what you will write. You realize that "the clearly invariant under coordinate transformations" is true for the Lorentz transforms and general linear transforms but not true for Shubert's cockamamie transforms, right?

Erm, no. These are tensor equations. They retain the same form under any coordinate transformation. See basic tensor calculus (any notes on GR will contain a thorough introduction).

Under what conditions is the above true? Do the pull-backs commute with the exterior derivatives for nonlinear transforms? Are A and $$\Phi$$ four-vectors wrt non-linear transforms?

Any diffeomorphism $$\phi: M\rightarrow M$$ induces a pull back on the second exterior derivative of the associated cotangent bundle (among other things). Relating to my post, the result is that for $$F \in \Lambda^2 TM$$, we have

$$ \phi^* (\mathrm{d} F) = \mathrm{d} (\phi^* F)$$

But it's probably an idea to look at tensor calculus first.

 

[Guest254]

Another edit - I can't keep up!

You realize that the tensorial formalism is nothing but a different notation for the classical formalism using partial derivatives, so what is not true for the classical formalism cannot become magically true if one uses four-vectors or tensors.

Erm, again, no. In the equations I've written, $$\nabla_a$$ denotes the covariant derivative, which wildly differs from the usual partial derivative unless the Christoffel symbols vanish.

 

[Tach]

Erm, no. These are tensor equations. They retain the same form under any coordinate transformation.

Then you shouldn't have any difficulty in plugging in Shubert's cockamamie transforms and proving that the equations are invariant, right? This is what I've asked you to do.

 

[Tach]

Another edit - I can't keep up!

Erm, again, no. In the equations I've written, $$\nabla_a$$ denotes the covariant derivative, which wildly differs from the usual partial derivative unless the Christoffel symbols vanish.

Feel free to stick with flat spacetime, so partial derivatives are just fine. So, $$ \eta_{ab} $$ and $$ \qquad \partial_a $$ are good enough, no mapping to any covariant derivatives is necessary. Standard SR, please.

 

[Guest254]

Then you shouldn't have any difficulty in plugging in Shubert's cockamamie transforms and proving that the equations are invariant, right? This is what I've asked you to do.

This is a standard exercise (for any coordinate transformation)! This is the whole point of the covariant derivative! The equation transforms like:

$$ \nabla^a F_{ab} \mapsto \frac{ \partial x^c}{\partial \tilde{x}^b} \nabla^a F_{ac} $$

Feel free to stick with flat spacetime, so partial derivatives are just fine. So, $$ \eta_{ab} $$ and $$ \qquad \partial_a $$ are good enough, no mapping to any covariant derivatives is necessary. Standard SR, please.

Look, you clearly don't know what you're talking about. We're doing things on flat space time, in the language of curvilinear coordinates! The whole point of the covariant derivative is that it behaves in a very nice way under general coordinate transformations.

 

[Tach]

This is a standard exercise (for any coordinate transformation)! This is the whole point of the covariant derivative! The equation transforms like:

$$ \nabla^a F_{ab} \mapsto \frac{ \partial x^c}{\partial \tilde{x}^b} \nabla^a F_{ac} $$

That's nonsense , the above is true due to the way $$F_{ab}$$ is constructed from the partial derivatives of the four-vector $$\phi_u=(-A_x,-A_y,-A_z, \phi)$$. The notion of four-vector is defined wrt the Lorentz transforms (any vector that transforms like the prototype (-x,-y,-z,ct) wrt the Lorentz transforms). So, all the properties that you are using for $$F_{ab}$$ are true wrt Lorentz transforms (and trivial linear transforms). You can't use them blindly in the case of non-linear transforms.

For reference, see R.C.Tolman , chapter 46 in "Relativity,Thermodynamics and Cosmology" (for example). You can also see the same stuff in Landau and Lifschitz.

Your repeated references the covariant derivatives means absolutely nothing, the formalism expressed in terms of covariant derivatives is nothing but a "repackaging" of the one one already shown above. For your education, see chapter 102 in Tolman. Nowhere in the book there is any such wild claim of invariance wrt non-linear transforms.

Look, you clearly don't know what you're talking about. We're doing things on flat space time, in the language of curvilinear coordinates! The whole point of the covariant derivative is that it behaves in a very nice way under general coordinate transformations.

Prove it, stick in Shubert's transforms , let's see if you get a covariant expression. This is all I've been asking you.

 

[Tach]

To avoid confusion - Maxwell's equation's are not "Eugene Shubert covariant", i.e. the standard Maxwell equations do not take on the same form under his coordinate transformation. Whereas, they are Lorentz covariant, i.e. they do take on the same form under Lorentz transformations.

However, the generally covariant form of Maxwell's equations is obviously invariant under his (and anyone else's) coordinate transformation.

You realize the enormity of your claim? So, a simple change in terms of notation (from classical partial derivatives to tensorial) changes the equations from non-covariant wrt the crackpot Shubert transforms to becoming covariant? No wonder Feynman was so circumspect in his books about the mindless use of the tensor notation. A change in notation cannot change the underlying physics.

 

[Guest254]

That's nonsense

You think tensor calculus is nonsense? Not a big fan of general relativity then?

Your repeated references the covariant derivatives means absolutely nothing, the formalism expressed in terms of covariant derivatives is nothing but a "repackaging" of the one one already shown above. For your education, see chapter 102 in Tolman. Nowhere in the book there is any such wild claim of invariance wrt non-linear transforms.

Right, take a few deep breaths and prepare yourself for some education. "General covariance" means the equations take the same form under changes of coordinates. This is precisely why tensor calculus is so popular - because when you write down an equation in terms of tensors, it transforms to another equation of the same form if you perform a coordinate transformation. For example in

$$ \nabla^a F_{ab} \mapsto \frac{ \partial x^c}{\partial \tilde{x}^b} \nabla^a F_{ac} $$

the F on the right hand side is not the same as that on the left hand side. It is related to the former via

$$ \tilde{F}_{ab} = \frac{\partial x^c}{\partial \tilde{x}^a} \frac{\partial x^d}{\partial \tilde{x}^b} F_{cd} $$

in addition, the Christoffel symbols will have changed in the definitions of the covariant derivatives appearing on the left and right hand sides. However, the form of the equation is the same. This is known as general covariance. Famous example: Einstein's equations. Keeping up?

Prove it, stick in Shubert's transforms , let's see if you get a covariant expression. This is all I've been asking you.

What do you mean "get" a covariant expression? The equations are generally covariant. This follows directly from:

Change in Christoffel symbol under coordinate transformation
Coordinate change of rank 2 covariant tensor

both formulas are given. I would urge you to perform these transformations. I'm certainly not doing them for you, because frankly I think you're a bit of a wolly.

 

[Tach]

You think tensor calculus is nonsense? Not a big fan of general relativity then?

Don't twist my words, I think that what you claim is nonsense.

 

[Guest254]

Don't twist my words, I think that what you claim is nonsense.

You quoted a standard tensor transformation and called it rubbish. I'm not forcing you to make a fool of yourself.

However, read the rest of the post. You might learn something.

 

[Tach]

You quoted a standard tensor transformation and called it rubbish.

No, I quoted your physics claims rubbish. No wonder crackpots like Shubert feel encouraged. Because of idiots like you who manipulate symbols without knowing what they mean.

 

[Tach]

You think tensor calculus is nonsense? Not a big fan of general relativity then?


Right, take a few deep breaths and prepare yourself for some education. "General covariance" means the equations take the same form under changes of coordinates.

This doesn't mean ANY change of coordinates. Just stop and think for a second, you can't turn the Maxwell equations from non-covariant wrt the crackpot Shubert transforms into covariant by a simple change of notation. Think for a second how $$F_{ab}$$ is built out of four vectors. Think about the definition of a four-vector.

Is there such a definition of a four-vector $$\phi_u$$ wrt non-linear transforms, like the crank Shubert transforms? Yes or no?

Can you construct $$F_{ab}$$ in the crackpot Shubert formalism?

 

[Guest254]

No, I quoted your physics claims rubbish. No wonder crackpots like Shubert feel encouraged. Because of idiots like you who manipulate symbols without knowing what they mean.

Well that's not particularly nice. I'm assuming you've now read my post and are feeling a little sheepish.

 

[Tach]

Well that's not particularly nice. I'm assuming you've now read my post and are feeling a little sheepish.

No, I think that you don't understand Maxwell equations and they you are writing nonsense.

 

[Guest254]

This doesn't mean ANY change of coordinates. Just stop and think for a second, you can't turn the Maxwell equations from non-covariant wrt the crackpot Shubert transforms into covariant by a simple change of notation. Think for a second how $$F_{ab}$$ is built out of four vectors. Think about the definition of a four-vector.

I can't believe I'm having to send you to wikipedia again, but here goes...

General Covariance is the invariance of physical laws under arbitrary differentiable coordinate transformations...

Had enough?

 

[Tach]

I can't believe I'm having to send you to wikipedia again, but here goes...

General Covariance is the invariance of physical laws under arbitrary differentiable coordinate transformations...

Had enough?

You can't turn the Maxwell equations from non-covariant wrt the crackpot Shubert transforms into covariant by a simple change of notation. Think for a second how $$F_{ab}$$ is built out of four vectors. Think about the definition of a four-vector.

Is there such a definition of a four-vector $$\phi_u$$ wrt non-linear transforms, like the crank Shubert transforms? Yes or no?

Can you construct $$F_{ab}$$ in the crackpot Shubert formalism? No?

Can you construct any invariant in the Shubert crackpot formalism? No?

To help you out read here the EXACT conditions under which general covariance can be applied: http://en.wikipedia.org/wiki/Local_Lorentz_covariance. Doesn't mean a free for all, all the constructs are STILL based on LORENTZ invariance, not the Shubert cockamamie invariance (which doesn't even exist).

Doesn't mean ANY transforms. Does not include non-linear transforms. Does not include the crackpot Shubert transforms. No wonder that crackpots like Shubert are encouraged.

 

[Guest254]

You can't turn the Maxwell equations from non-covariant wrt the crackpot Shubert transforms into covariant by a simple change of notation. Think for a second how $$F_{ab}$$ is built out of four vectors. Think about the definition of a four-vector.

Is there such a definition of a four-vector $$\phi_u$$ wrt non-linear transforms, like the crank Shubert transforms? Yes or no?

Can you construct $$F_{ab}$$ in the crackpot Shubert formalism?

This is getting worse and worse. It seems I gave you too much credit - I thought this bit was obvious:

Maxwell's equations in standard $$ (t,x,y,z) $$ coordaintes are $$\partial^a F_{ab} =0$$ etc. With respect to these coordinates and $$g_{ab} = \eta_{ab}$$, all the Christoffel symbols vanish, so this equation is the same as $$ \nabla^a F_{ab}=0$$ etc. However, this equation, because it is a tensor equation, retains it's form under arbitrary, smooth coordinate changes. This is why it is a "generally covariant" form of Maxwell's equations. Use whatever coordinate transformation you like, and you will end up with something else of the form

$$ \frac{ \partial x^c}{\partial \tilde{x}^b} \tilde{\nabla}^a \tilde{F}_{ac} $$

I've even put lots of tildes in to make things super-duper clear for you.

To help you out read here the EXACT conditions under which general covariance can be applied: http://en.wikipedia.org/wiki/Local_Lorentz_covariance

Erm, you've linked to a page on Lorentz invariance. Is all this becoming a little too much?

 

[Tach]

This is getting worse and worse. It seems I gave you too much credit - I thought this bit was obvious:

Maxwell's equations in standard $$ (t,x,y,z) $$ coordaintes are $$\partial^a F_{ab} =0$$ etc.

...provided you understand how $$F_{ab}$$ is constructed. Something that you obviously don't.

With respect to these coordinates and $$g_{ab} = \eta_{ab}$$, all the Christoffel symbols vanish, so this equation is the same as $$ \nabla^a F_{ab}=0$$ etc. However, this equation, because it is a tensor equation, retains it's form under arbitrary, smooth coordinate changes.

Prove that this is the case for the Shubert transforms. I'll give you a hint, start with the partial derivative wrt t. Put up or shut up.


T

 

[Guest254]

Prove that this is the case for the Shubert transforms. I'll give you a hint, start with the partial derivative wrt t. Put up or shut up.

Now, now. I've explained how to do this, giving you links to wikipedia pages (although, for the long term, you should get a book). I've also explained to you that I shan't be doing it for you, because you've proven yourself to be a bit of a wolly.