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I'm trying to derive the infinitesimal time dilation relation $dt = \gamma d\tau$, where $\tau$ is the proper time, $t$ the coordinate time, and $\gamma = (1-v(t)^2/c^2)^{-1/2}$ the time dependent Lorentz factor. The derivation is trivial if one starts by considering the invariant interval $ds^2$, but it should be possible to obtain the result considering only Lorentz transformations. So, in my approach I am using two different reference frames $(t,x)$ will denote an intertial laboratory frame while $(t',x')$ will be the set of all inertial frames momentarily coinciding with the observed particle, i.e. the rest frame of the particle. These frames are related by $$t' = \gamma \left(t-\frac{Vx}{c}\right),\quad x' = \gamma \left( x - V t\right),$$ where $V$ is some nonconstant (i.e. time dependent) parameter which is, hopefully, the velocity of the particle in the laboratory frame. Treating $x$, $t$ and $V$ as independent variables (for now) and taking the differential of the above relations, I obtain $$dt' = \gamma \left(dt-\frac{Vdx}{c}\right) - \frac{\gamma^3}{c^2}(x-Vt)dV,$$ and $$dx' = \gamma \left( dx - V dt\right) - \gamma^3 \left(t-\frac{Vx}{c}\right) dV.$$ Imposing either the definition of the rest frame $dx'=0$ or (what should be equivalent) $dx = Vdt$, the only way in which i obtain $dt = \gamma dt'$ is if $dV=0$. So, the derivation breaks badly at some point or I must be wrong in using some of the above equations. Which one is it?

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The Lorentz transformation you used is a Lorentz transformation for a fixed (constant) $V$, the relative velocity between two frames, so by the construction $dV=0$. Frames with non-constant velocities are not "inertial" and they don't belong to special relativity. You may still measure the proper time along an accelerating world line but the coordinate system in which the accelerating observer would always have $x=0$ isn't inertial, so it's not good for a simple formulation of the laws of physics in special relativity. The proper time of an accelerating observer is simply obtained by dividing his world line to infinitesimal pieces and summing (integrating) their proper time. It can't be computed without an integral which is complicated in general. Because you haven't considered any integrals of complicated functions, you haven't been (correctly) calculating the proper time of an accelerating observer, so considering $dV\neq 0$ couldn't have served any useful purpose.

Moreover, the very idea to differentiate the Lorentz transformations is a bit redundant. Without a loss of generality, you could have additively shift the coordinates so that $0=x=t=x'=t'$ around the point you're interested in. When you do so, $dx,dt,dx',dt'$ are really nothing else than $x,t,x',t'$ that are just assumed to be infinitely small. The time dilation is then simply derived by setting e.g. $x=0$ which reduces the first equation to $t'=\gamma t$. It's really the same thing as $dt'=\gamma \cdot dt$ when $t,t'$ are infinitesimal. The $\gamma$ factor is on the proper side of the equation because $x=0$ means that the unprimed coordinates are those for which $x=0$ is the world line of the moving object. That means that $t$ is the time measured in the rest frame of the moving object and indeed, we have $\tau_{\rm proper}=t=t'/\gamma$ where $t'$ is the time measured from some other frame. (Of course, the same calculation may be done passively and/or with the opposite interpretation of primed and unprimed systems – but one isn't allowed to mix these two approaches inconsistently.)

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That's it! Thank you! –  Little Brown One Feb 26 '13 at 15:29

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