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Is there something wrong with this reasoning:
First the question:
"A rocket A leave Earth at a speed v at a time where the clocks onboard and those on Earth read 0. After a time T (for the Earth's clock), a second rocket is lauched with a speed u > v. The second rocket catches up with the first one at a time t such that (t - T)u = tv, or equivalently, t = uT/(u-v)."
a) For the clocks on A, when does B leave earth?
MY ANSWER: The time between the events "A leaves" and "B leaves" is proper on earth. It is T on earth; therfor it is given on A by
[tex]t_A_1= T\gamma _{AT} = \frac{T}{\sqrt{1-(v/c)^2}}[/tex]
b) For the clocks on A, when does B catches up with A?
MY ANSWER: By Lorentz,
[tex]t_A_2 = \frac{1}{\sqrt{1-(v/c)^2}}(t-v(vt)/c^2) = t\sqrt{1-(v/c)^2}[/tex]
d) Deduce the speed of B as seen in frame A. Does the result agree with the law of addition of speed?
MY ANSWER: Speed of B = distance traveled in catching up with A / time took to do so
[tex]\Rightarrow v_{B \ in \ A} = \frac{vt_A_2}{t_A_2 - t_A_1}=\frac{vt\sqrt{1-(v/c)^2}}{t\sqrt{1-(v/c)^2} - T/\sqrt{1-(v/c)^2}} = \frac{vt}{t - T/(1-(v/c)^2)}[/tex]
[tex]=\frac{uv}{(u-v)(\frac{u}{u-v}-\frac{1}{1-(v/c)^2})}[/tex]
But according to Mapple (and to me), this is not equal to the predicted
[tex]v_{B \ in \ A} = \frac{u-v}{1-uv/c^2}[/tex]
First the question:
"A rocket A leave Earth at a speed v at a time where the clocks onboard and those on Earth read 0. After a time T (for the Earth's clock), a second rocket is lauched with a speed u > v. The second rocket catches up with the first one at a time t such that (t - T)u = tv, or equivalently, t = uT/(u-v)."
a) For the clocks on A, when does B leave earth?
MY ANSWER: The time between the events "A leaves" and "B leaves" is proper on earth. It is T on earth; therfor it is given on A by
[tex]t_A_1= T\gamma _{AT} = \frac{T}{\sqrt{1-(v/c)^2}}[/tex]
b) For the clocks on A, when does B catches up with A?
MY ANSWER: By Lorentz,
[tex]t_A_2 = \frac{1}{\sqrt{1-(v/c)^2}}(t-v(vt)/c^2) = t\sqrt{1-(v/c)^2}[/tex]
d) Deduce the speed of B as seen in frame A. Does the result agree with the law of addition of speed?
MY ANSWER: Speed of B = distance traveled in catching up with A / time took to do so
[tex]\Rightarrow v_{B \ in \ A} = \frac{vt_A_2}{t_A_2 - t_A_1}=\frac{vt\sqrt{1-(v/c)^2}}{t\sqrt{1-(v/c)^2} - T/\sqrt{1-(v/c)^2}} = \frac{vt}{t - T/(1-(v/c)^2)}[/tex]
[tex]=\frac{uv}{(u-v)(\frac{u}{u-v}-\frac{1}{1-(v/c)^2})}[/tex]
But according to Mapple (and to me), this is not equal to the predicted
[tex]v_{B \ in \ A} = \frac{u-v}{1-uv/c^2}[/tex]
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