neilparker62
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Does this solution mean anything ?
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It means something, but it has at least two issues:Does this solution mean anything ?
You do realize that these are the same thing, right? (I assume that by ##j## you mean the imaginary unit.) ##\sin (\omega t) = ( e^{i \omega t}  e^{ i \omega t} ) / 2 i##. Sines and cosines are just different ways of writing exponentials with imaginary arguments. If you treat the full velocity vector instead of just a single component, in polar coordinates it is indeed of the exact form ##e^{i \omega t}##.I thought the solution might be of the form c * e^(jwt) rather than sin(wt).
The equation you wrote down is not valid in the presence of gravity. Your equation is only valid in special relativity; gravity requires general relativity.But the frequency looks interesting  very plainly it is ridiculously low for (say) an object in a gravitational field.
Reference, please?Einstein always emphasized that the notation [itex]μ=\frac{m}{\sqrt{1\frac{v^{2}}{c^{2}}}}[/itex] is physically meaningless.
First of all, this would not be a different "reference frame"; it would be a different convention for what the symbol ##m## means.it would be more natural to describe a "m" using different reference frame parametrized by [itex]γ=\frac{1}{\sqrt{1\frac{v^{2}}{c^{2}}}}[/itex] to conserve physical laws.(with m not sticked with γ )
We don't do this to deal with acceleration; we do it to deal with gravitymore precisely, with *tidal* gravity.Moreover, in relativity, to deal with acceleration, we need to use curved spacetime structure to replace the "force" concept .
Oh, thanks. Do you mean different reference "frames"?First of all, this would not be a different "reference frame"; it would be a different convention for what the symbol ##m## means.
Second, you can't "conserve physical laws" by adopting this definition for ##m##; some of the laws still have to change form from their Newtonian versions (I assume what you mean by "conserve physical laws" is "all the laws look exactly the same as their Newtonian versions").
We don't do this to deal with acceleration; we do it to deal with gravitymore precisely, with *tidal* gravity.