A particle is moving in a system of reference $S$. In its proper system of reference, say $S'$, the particle is still and it is described by the event $(c\tau,0,0,0)$. In the inertial frame $S$, the same event is described by $(ct,x,y,z)$.

First attempt

Using the metric $\mathrm ds^2 = -c^2\mathrm dt^2 + \mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2$ it is easy to work out the derivative $\mathrm dt/\mathrm d\tau$ which is $$\frac{\mathrm dt}{\mathrm d\tau} = \gamma(u)$$ where $u$ is the velocity of the particle in $S$.

Second attempt

The Lorentz transformations from $S'$ to $S$ are $$ t = \gamma(u)(t' + ux'/c^2), \qquad x = \gamma(u)(x' + ut') $$ Applying them to the events considered, we get $$ t = \gamma(u)\tau, \qquad x = \gamma(u)u\tau $$ and working out the derivative of the first with respect to $\tau$ one gets $$ \frac{\mathrm dt}{\mathrm d\tau} = \gamma(u)$$


I feel that the second argument is wrong but I cannot understand why. Can someone explain to me the fallacy? Of course, if the argument is right, can someone explain to me why it is right?


Watching the second equation of the LT ($x=\gamma(u)u\tau$) this formula suggests that, according to frame $S$, the particle is moving with uniform motion, but this is not the case. I do not understand, however, why.


1 Answer 1


The only thing that’s somewhat “wrong” about the latter is that it chooses a specific axis for the $x$-axis which is different from the possibly arbitrary choice of $x$ in the problem description, and likewise it assumes a common spacetime origin between $S$ and $S'$ whereas we are not given that in the problem statement. Other than those minor quirks, the second is a perfectly good derivation.

Put in a somewhat more general light, it is correct because all of its essential points are correct; doing a Poincaré transform of the point $[c\tau, 0, 0, 0]$ the path is indeed written in $S'$ as the moving 4-position $$\begin{align} c~t(\tau) &= c~t_0 + \gamma~c~\tau\\ x(\tau) &= x_0 + \gamma~v_x~\tau = x_0 + v_x~(t - t_0)\\ y(\tau) &= y_0 + \gamma~v_y~\tau = y_0 + v_y~(t - t_0)\\ z(\tau) &= z_0 + \gamma~v_z~\tau = z_0 + v_z~(t - t_0) \end{align}$$and indeed the derivative of $t(\tau)$ with respect to its argument is $\gamma.$

In other circumstances where one starts with some path $[w(u), x(u), y(u), z(u)]$ this derivative is not sufficient because $u$ might not be the proper time but might be some general path-progress coordinate; you would then indeed need to calculate $$\left[c~\frac{d\tau}{du}\right]^2 = [w'(u)]^2 - [x'(u)]^2 - [y'(u)]^2 - [z'(u)]^2$$ and only after that could you write $$\frac{dt}{d\tau} = w'(u)~\left[c~\frac{d\tau}{du}\right]^{-1}.$$But in this case we have guaranteed by construction that $d\tau/du = 1,$ which is why we used the symbol $\tau$ for $u$ in the first place. And we know that this is true by construction because we know that the defining property of Lorentz transforms like we are using, is that they preserve this 4-norm.

  • $\begingroup$ Thank you so much for your answer: I really forgot to think that, since the primed reference is the proper one, the axes are in general moving and cannot be considered parallel to the non-primed system. However, I have still not understood why the derivative of t with respect to tau is gamma. Isn’t gamma a function of t (and therefore of tau)? Why is that derivative so easy? $\endgroup$
    – Logos
    Aug 14, 2019 at 23:23
  • $\begingroup$ @Logos Is $\gamma$ a function of $t$ or $\tau$? What is that function? $\endgroup$
    – CR Drost
    Aug 15, 2019 at 4:38
  • $\begingroup$ $\gamma$ is a function of the velocity $u$, which is the derivative of $\sqrt{x^2 + y^2 + z^2}$ which are functions of $t$. However, this means also that they are functions of $\tau$, for $\tau$ amd $t$ are functionally linked. $\endgroup$
    – Logos
    Aug 15, 2019 at 9:28
  • $\begingroup$ Isn't $u$ the velocity? $\endgroup$
    – PM 2Ring
    Aug 15, 2019 at 12:52
  • $\begingroup$ @Logos I mean that is not quite true ($u= dr/dt=d/dt\sqrt{x^2+y^2+z^2}$ is a statement that motion is purely radial, and hence that $x_0=y_0=z_0=0$) but I encourage you to calculate out what the actual velocity is for a concrete problem, and see what function of time it is. These things are better learned by example. $\endgroup$
    – CR Drost
    Aug 15, 2019 at 13:17

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