Anamitra Palit Posted March 6, 2021 Share Posted March 6, 2021 (edited) First we consider the fact that the norm-square of the four acceleration vector is negative or zero. Indeed \[c^2=c^2\left(\frac {dt}{d \tau}\right)^2-\left(\frac {dx}{d \tau}\right)^2-\left(\frac {dy}{d \tau}\right)^2-\left(\frac {dz}{d \tau}\right)^2\] (1) Differentiating both sides with respect to time we obtain \[c^2\frac{d^2 t}{dt^2}\frac{dt}{d\tau}-\frac{d^2 x}{dt^2}\frac{dx}{d\tau}-\frac{d^2 y}{dt^2}\frac{dy}{d\tau}-\frac{d^2 z}{dt^2}\frac{dz}{d\tau}=0\] (2) We transform to an inertial frame where the particle is momentarily at rest. \[c^2\frac{d^2 t}{d\tau^2}\frac{dt}{d \tau}=0\] \[\Rightarrow \frac{d^2t}{d \tau^2}=0\](3) [the above holds since v.v=c^2] Norm square of the acceleration vector in the new frames of reference [that is on transformation ] is negative or zero. Therefore due to the conservation of dot product it is zero or negative in all (inertial) frames of reference: ||a||<=0 [We may prove by alternative techniques that a.a<=0] We now consider an accelerating particle executing a one dimensional motion in the x direction \[v_p=v_x=\frac{dx}{dt}\](4) Proper acceleration component \[a_t=\frac{d^2 t}{d \tau^2}=\frac{d}{d\tau}\left(\frac{dt}{d\tau}\right)=\frac {d \gamma_p}{d \tau}\\=\gamma_p^3 \frac{v_p}{c^2}\frac{d v_p}{d \tau}= \gamma_p^3 \frac{v_p}{c^2}\frac{d }{d \tau}\frac{dx}{dt }\] (5) \[\gamma_p=\frac{1}{\sqrt{1-\frac{v_p^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{v_x^2}{c^2}}}\] Now, \[\frac{d }{d \tau}\frac{dx}{dt}=\frac{d}{d\tau}\left(\frac{dx}{d\tau} \frac{d \tau}{dt}\right)= \frac{d}{d\tau}\left(\frac{dx}{d\tau} \frac{1}{\gamma_p}\right)\\=a_x\frac{1}{\gamma_p}-\frac{1}{\gamma_p^2}\frac{d \gamma_p}{d\tau}\frac{dx}{d\tau}= a_x\frac{1}{\gamma_p}-\frac{1}{\gamma_p^2}\gamma_p^3\frac{v_p^2}{c^2}\frac{d }{d \tau}\frac{dx}{dt}\] (6) From (5) and (6), we obtain, \[\Rightarrow \frac{d }{d \tau}\left(\frac{dx}{dt}\right)\left[1+\gamma_p\frac{v_p^2}{c^2}\right]=a_x\frac{1}{\gamma_p}\] \[\Rightarrow \frac{d }{d \tau}\left(\frac{dx}{dt}\right)=a_x\frac{1}{\gamma_p}\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-1}\] \[a_t=\gamma_p^2 \frac{v_p}{c^2}a_x\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-1}\] (7) For one dimensional motion \[c^2a_t^2-a_x^2 \le 0\](8) \[c^2\left[\gamma_p^2 \frac{v_p}{c^2}a_x\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-1}\right]^2-a_x^2 \le 0\] (9) \[\frac{1}{c^2}\gamma_p^4 v_p^2 a_x^2\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-2}-a_x^2 \le 0\] \[\frac{1}{c^2}\gamma_p^4 v_p^2 \left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-2} \le 1\] \[\frac{1}{c^2}\gamma_p^2 v_p^2 \left[\frac{1}{\gamma_p}+\frac{v_p^2}{c^2}\right]^{-2} \le 1\] (10) With v_p=v_x tending to c the left side of the last inequation given by (10)tends to infinity while the right side stays on unity.There is a breakdown. Edited March 6, 2021 by Anamitra Palit -2 Link to comment Share on other sites More sharing options...
Endy0816 Posted March 6, 2021 Share Posted March 6, 2021 (edited) I think you might have division by zero going on. Edited March 6, 2021 by Endy0816 Link to comment Share on other sites More sharing options...
swansont Posted March 6, 2021 Share Posted March 6, 2021 ! Moderator Note You were asked not to bring this up again. Link to comment Share on other sites More sharing options...
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