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Harold Ellis Ensle: He _cannot_ apply a transformation to himself. (11-Nov-2002) |
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Daryl McCullough news:aqoeu301mem@drn.newsguy.com... > > >> You can't just say "based on the theory of relativity". > >> Tell me which assumption of relativity you used. Tell > >> me how your claim logically follows from that assumption > >> or assumptions. > > > >I am not even sure what more you want. > > All right, let's try again: > Let's look at the example of a twin > (initially 20 years old) who travels out for 5 years > (Earth time) and back 5 years at speed .8c. Let's label > six different events > e1 = the travelling twin blasts off > e2 = the stay-at-home twin's clock shows t = 1.8 years. > e3 = the stay-at-home twin's clock shows t = 5 years. > e4 = the travelling twin turns around > e5 = the stay-at-home twin's clock shows t = 8.2 years. > e6 = the twins reunite > > In the frame of the stay-at-home twin, these events have > the following coordinates: > > e1 = (x=0,t=0) > e2 = (x=0,t=1.8) > e3 = (x=0,t=5) > e4 = (x=4,t=5) > e5 = (x=0,t=8.2) > e6 = (x=0,t=10) > > I claim that in the outgoing reference frame of the > travelling twin, e2 and e4 are simultaneous. My proof > is this: > > 1. Definition: e2 and e4 are simultaneous in the > reference frame of the outgoing twin if the time > coordinate of e2 in that frame is equal to the > time coordinate of e4 in that frame. But how will you determine these coordinates? > 2. By the Lorentz transformations, the time coordinate > in the outgoing twin's frame is given by > > t' = gamma( t - v x/c^2) > > where x and t are the coordinates in the stay-at-home frame. As soon as you use the Lorentz transformations, you have based your result on the view of the stay-at-home. Like I already stated and you snipped. The travelling twin cannot transform his own coordinates as he is at rest in his own frame. > 3. For our example, gamma = 1.66..., v = .8. For > event e2, t = 1.8, x = 0. > > 4. Plugging in these values gives > t' = 1.666 * (1.8 - 0) = 3 years for event e2. > > 5. In our example, gamma = 1.66..., v = .8. For event e4, > t = 5, x = 4. > > 6. Plugging in these values gives > t' = 1.66 * (5 - .8 * 4) > = 3 years for event e4 > > 7. By 4 and 6, the time coordinates of e2 and e4 in the > travelling twin's outward-going frame are both 3 years. > > 8. Therefore, e2 and e4 are simultaneous in the > travelling twin's outward-going frame. > > All right. That's my derivation of the claim that > e2 and e4 are simultaneous in the outward-going > frame of the travelling twin. As soon as you used the Lorentz transformations to represent the travelling twin's view of himself, you made an error. Relative to himself, the travelling twin is at rest (before acceleration), so he _cannot_ apply a transformation to himself. H.Ellis Ensle |
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Index Original post and context: aqoqf3$hs7$1@slb6.atl.mindspring.net |