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> I just wrote: > > "I have already demonstrated that the relation > a' (or b, as you like) = a/(g(1-vV/c^2)) is false." > > I think it's worth repeating the demonstration, because > it is not refutable: > > The "time" LT is t' = g(t-xv/c^2). > When x = a+Vt, this LT becomes, according to SR (not to me), > t' = g(t-(a+Vt)v/c^2) = gt(1-Vv/c^2) - gav/c^2. > > The "position" LT is x' = g(x-vt). > By replacing x by a+Vt, it becomes, also according to SR, > x' = g(a+Vt-vt) = ga + gt(V-v). > > But x' can also be expressed by the relation > x' = a' (or b, if you prefer) + V't', > where V' = (V-v)/(1-Vv/c^2). > > Hence, > x' = a/(g(1-vV/c^2)) + [(V-v)/(1-Vv/c^2)] * g(t-(a+Vt)v/c^2). > > This relation is valid for any value of V and t. > Thus, it is valid when V = 0, t = 0. > Then, x' = a/g + gav^2/c^2, a correct solution according to SR. > > Or, x' = ga + gt(V-v) gives another solution, i.e. x' = ga. > > Only a false theory gives two different solutions to the same problem. > This amply demonstrates that SR is wrong. > > As I told you many times, the correct value of t' is obtained by > replacing x by x-a = Vt in the Lorentz transformation > t' = g(t-xv/c^2). Then, t' = gt(1-Vv/c^2), and the term -gav/c^2, > on which the so-called relativity of simultaneity is based, > simply disappears. > Otoh, x' = a' + V't' becomes x' = ga + gt(V-v), which reduces, > as shown above, to x' = ga when t = 0. No contradiction anymore ! > > Marcel Luttgens Sorry, x' = ga + gt(V-v) is of course equivalent to x' = a/(g(1-vV/c^2)) + [(V-v)/(1-Vv/c^2)] * g(t-(a+Vt)v/c^2) Forget my so-called proof! Marcel Luttgens |
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Index Original post and context: b45b8808.0206081308.19f93c6a@posting.google.com |