I'm trying to analyse non-inertial motion in special relativity. First I'll start off with inertial motion. In my reference frame, my particle has coordinates
$x^\mu = (t, x)$
And in the particle's reference frame, it has coordinates
$x'^\mu = (\tau, 0)$
My understanding is that since we are in flat (affine) space, the particle's position transforms as a vector, and I can boost it by its velocity $v$ to give
$\begin{pmatrix}\gamma & v \gamma\\ v \gamma & \gamma\end{pmatrix} \begin{pmatrix} \tau \\ 0 \end{pmatrix} = \begin{pmatrix} t \\ x \end{pmatrix}$
Which tells me that $x = v\gamma\tau$ and $t = \gamma\tau$, as usual. It's clear that in both frames we have $\dot{x}^\mu\dot{x}_\mu = -1$, from explicit calculation and from the fact that $\dot{x}^\mu\dot{x}_\mu$ is a Lorentz scalar, and our two frames are related by Lorentz boost.

I start to have problems when I move to non-intertial motion. If I take my reference frame to be the same, and the particle's reference frame to now be its instantaneous reference frame, from what I can see nothing should change, apart from the particle's velocity is now some (as yet unspecified) function of time, and the Lorentz boost becomes time dependent. In my new non-intertial scenario my particle still has $x = v\gamma\tau$ and $t = \gamma\tau$ from the same analysis above, and it's clear that it still has $\dot{x}^\mu\dot{x}_\mu = -1$ in its own instantaneous reference frame. By the fact that this is a Lorentz scalar, I expect this quantity to be the same in my frame.

By explicit calculation in my reference frame, I find that
$\dot{x}^\mu = (\gamma^3 v \dot{v}\tau + \gamma, \gamma^3 v^2 \dot{v} \tau + \gamma \dot{v}\tau + \gamma v) $
Which gives
$\dot{x}^\mu\dot{x}_\mu = -1 + \gamma^4\dot{v}^2\tau^2$
Clearly this not acceptable.

My general relativistic intuition tells me that my problem lies in the fact that I've treated the manifold as a vector space, and that I should only use the Lorentz boost on elements of the tangent space. So maybe I should start with the particle's velocity 2-vector instead of its position. But I can't see why this approach should work for inertial motion and not for non-intertial motion, and my understanding is that because Minkowski spacetime is flat, it should be OK to treat elements of the manifold as vectors.

Any ideas?


2 Answers 2


Short answer: the Lorentz transformation does not give the correct transformation from an inertial frame to an accelerating frame, even if you let the velocity vary with time as you suggest.

To convince you that this is so, recall that

  1. An inertial observer see light travel along straight lines.
  2. A uniform gravitational field bends light.
  3. A uniform gravitational field is equivalent to a uniformly accelerating observer.

Put these together to conclude that a uniformly accelerating observer will see light bend. Since the Lorentz transformation maps straight lines to straight lines, it will not transform a light ray to the appropriate bent curve as necessary.

Technical Note: even a time-dependent Lorentz transformation maps straight lines to straight lines: \begin{align} \begin{pmatrix} \gamma(t) & -\gamma(t)\beta(t) \\ -\gamma(t)\beta(t) & \gamma(t) \end{pmatrix} \begin{pmatrix} t \\ t \end{pmatrix} = \begin{pmatrix} \gamma(t)(1-\beta(t))t \\ \gamma(t)(1-\beta(t))t \end{pmatrix} \end{align}

  • $\begingroup$ A truly physical approach. +1. See my comment to Ben Crowell's answer for another argument. $\endgroup$ Nov 12, 2014 at 7:00
  • $\begingroup$ I am convinced. So the correct transformation between the two reference frames would be the transformation to Rindler coordinates? $\endgroup$
    – Jojo
    Nov 12, 2014 at 16:28
  • $\begingroup$ @Joe For an observer undergoing constant proper acceleration, yes. $\endgroup$ Nov 12, 2014 at 17:00

In the inertial frame K momentarily comoving with the particle, it is not true that its own coordinates are $(\tau,0)$. Here $\tau$ would be the accumulated time on a clock that has been moving along with the particle since some earlier time. This clock reading depends on the previous history of the particle, and is not equal to the time coordinate in Minkowski coordinates in the inertial frame K.

  • $\begingroup$ "Momentarily comoving" concept requires stopping time. This means there is no movement, and not "comovement". Therefore there is nothing to compare, nothing to apply transforms to. No go. $\endgroup$ Nov 12, 2014 at 7:02
  • $\begingroup$ Ok great, thanks. So the Lorentz transformation is only valid for transforming elements of the tangent space (eg. the velocity of the particle), and it's just a coincidence in the zero-acceleration case that we can use the Lorentz transform on the particle's position, because in that case integrating and differentiating with respect to $\tau$ commute with the (constant) Lorentz transformation? When I work with $\dot{x}^\mu = (1, 0)$ in the particle's frame and transform this, I get a much better result $\endgroup$
    – Jojo
    Nov 12, 2014 at 16:26

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