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, 20214 yr by Anamitra Palit

I think you might have division by zero going on.

Edited March 6, 20214 yr by Endy0816

!

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You were asked not to bring this up again.