Question on shifting field gradients by an angle

[Goldstone1]

Then let's not so hastily equalize the terms. Let us consider just E=M\phi. If it is Mc^2 then how do you make it relativistic? I saw your last answer and liked it, any others?

 

[Polyrhythmic]

Then let's not so hastily equalize the terms. Let us consider just E=M\phi. If it is Mc^2 then how do you make it relativistic? I saw your last answer and liked it, any others?

No, I think that's all I can contribute so far.

 

[Goldstone1]

No, I think that's all I can contribute so far.

Well thank you. I may have some more questions. These questions as you know are based on that paper I sent here, but was denied. The contributions of the posters will not be disregarded in the paper.

 

[Goldstone1]

Right I do have more questions:

In fact I do have one. Even though you are saying E=M\phi=Mc^2 is technically correct, we are to ignore any contribution to momentum, which seems like a strange and incorrect statement: In our normal notation, it would be:

E=M\Delta\phi(x)

since

\Delta \phi(x) =\phi(x) \rightarrow \phi*(x)

which through substitution, gives a more simplistic form

\Delta E=M\phi(\Lambda^{-1} x)

There is also now the question of what energy has been taken by the shift. If the change in the Hamiltonian of the system was purely gravitational and no other added energies, then the shift mathematically can be given as:

\phi(x) \rightarrow \phi*(x)=\phi(\Lambda^{-1} x)

where the inverse \Lambda^{-1} states that the field has definitely been shifted. If \phi is no longer arbitrary, we can state it is actually the gravitational field, then it's also an energy perturbation of the field as well:

\Delta E_g=M_g\phi(\Lambda^{-1} x)

 

[Polyrhythmic]

Even though you are saying E=M\phi=Mc^2 is technically correct, [...]

I didn't say that this was correct, I provided you with a modified version I could make sense of.

 

[Goldstone1]

You want me to start using the rest symbol M_0