- #456
Dale
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I don't know what you are talking about here. There are no first principles involved, and I have no idea what you mean by "equivalence" or "relevant basis" in this context. You don't derive the transform, you simply define the transform. There are no "correct" or "incorrect" transforms as long as they are smooth and 1-to-1.Austin0 said:Don't you agree that to assert an equivalence between coordinates or values between two frames in relative motion you need to transform the values from one frame to the other.
If in fact you do not already have the correct transform functions, the T,X,T' and X' in your generalization you cannot simply assume the equivalence between some values in both frames and derive a valid transform from that . There has to be some relevant basis for the equivalence from first principles to justify such an identity and substitution.
Wouldn't you agree??
If you are worried about the 100's then you don't have to simplify, it won't change anything except for making the expressions messier. The limits will still be the same. I leave that as an exercise for you.Austin0 said:Is the 100 in the A frame equivalent to the 100 in the Z frame?
No, it isn't axiomatic. It is derived for a specific class of transforms known as the Lorentz transforms.Austin0 said:isn't it axiomatic that if these events are simultaneous in the A frame that they cannot be simultaneous in the Z frame?? It follows that the distances , the spatial intervals in the two frames cannot be congruent also Yes??
I cannot believe that you actually wrote this. This is by far the silliest objection that you have come up with yet, and it really makes you look extremely unreasonable.Austin0 said:how do you justify the equivalence [itex]100-vt=100/2^n[/itex] over time when the systems are not only in relative motion but one of them is non-linear??
The transitivity of equality is one of the most basic elements of math and you should have learned it in grade school. Since [itex]d=100-vt[/itex] and [itex]d=100/2^n[/itex] then by transitivity of equality [itex]100-vt=100/2^n[/itex]. You cannot get any stronger justification than that.
Again, I don't know what you mean by "equivalence" in this context. Any function can be used as a valid transformation provided that it is smooth and one-to-one. The transform presented meets both of those criteria, so it is a valid transform. See p. 37 here for details: http://arxiv.org/abs/gr-qc/9712019Austin0 said:It appears to me that to make this assumption of equivalence is unfounded and circular. I.e.,,to determine if these are equivalent requires a valid transformation so to use them to derive a transformation then makes them equivalent circularly.