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Mar 2, 2017 at 21:23 history edited Tjow CC BY-SA 3.0
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Mar 2, 2017 at 20:19 comment added Tjow But I think this can be overcame by the fact that these components of Tau(x',y,z,t) are still operating under that same function, they may only be treated as separate functions within Tau(x',y,z,t) so long as they contain variables in the input (e.g. Tau[x',0,0,t+(x'/c-v)]). So when the three components contain Tau(0,0,0,0) in the approximations, I treat each Tau(0,0,0,0) as equal to the other two Tau(0,0,0,0) in the other approximations since the variables are removed and so the value is no longer a composite and may go back to being considered a value of Tau(x',y,z,t)
Mar 2, 2017 at 20:04 comment added Tjow So I took the Taylor approximation of each component at the origin (0,0,0,0), which I assumed Einstein would've chosen the origin, to me it just makes sense to choose the origin. And instead of using Tau(x',y,z,t) for the approximations I treated each Tau within equation (1) as separate functions rather than different values of the same function (that function again being Tau(x',y,z,t)). This could be problematic when simplifying being that this would mean Tau(o,o,o,o) in each approximation has distinct values and therefore cannot be combined when you plug them back into equation (1)
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Mar 2, 2017 at 18:56 history answered Tjow CC BY-SA 3.0