In special relativity, to derive the four-velocity from the four-position, one generally needs to do something of this manner:$$\vec V = \frac{d}{dt} \vec X = \left(\frac{d}{d \tau} \vec X \right) \frac{dt}{d\tau}$$Here, $\tau$ is the proper time of the moving object, and $t$ is our measurement of time. Now, in every derivation that I've seen, $\frac{dt}{d\tau} = \gamma = \frac{1}{\sqrt{1-v^2/c^2}}$. However, time dilation takes the factor $\gamma$, so wouldn't we expect $\tau = \gamma t$, which would make $\frac{dt}{d\tau} = \frac{1}{\gamma}$? What am I mixing up?

Question

In special relativity, to derive the four-velocity from the four-position, one generally needs to do something of this manner:$$\vec V = \frac{d}{dt} \vec X = \left(\frac{d}{d \tau} \vec X \right) \frac{dt}{d\tau}$$Here, $\tau$ is the proper time of the moving object, and $t$ is our measurement of time.  Now, in every derivation that I've seen, $\frac{dt}{d\tau} = \gamma = \frac{1}{\sqrt{1-v^2/c^2}}$.  However, time dilation takes the factor $\gamma$, so wouldn't we expect $\tau = \gamma t$, which would make $\frac{dt}{d\tau} = \frac{1}{\gamma}$?  What am I mixing up?

anonymous · Answer

THE DERIVATIVE EQUATION OUTSIDE THE BRACKET. THERE I THINK U HAVE TO WRITE $$d \tau/dt and \not dt/d$$ . but even i m not sure

anonymous · Answer

*$$d \tau/dt$$

anonymous · Answer

not

anonymous · Answer

$$dt/d$$

anonymous · Answer

sorry for posting again and again but thought the eq i posted the first time was not clear

anonymous · Answer

OHH good catch LMAO... since I wrote $\frac{dt}{d\tau}$ instead of $\frac{d\tau}{dt}$, that would explain why I was getting the reciprocal answer.  The correct simplification would be$$\vec V = \frac{d}{dt} \vec X = \left(\frac{d}{d \tau} \vec X \right) \frac{d\tau }{dt} = \boxed{\left(\dfrac{d}{d \tau} \vec X \right)  \gamma}.$$Thanks for catching that silly mistake!

anonymous · Answer

we all do something like that at some point of time glad to help.