Coordinate transformation for a comoving frame for abitrary trajectory

Let's suppose there is a particle moving through space time. It moves along the x-direction with a time dependent velocity. Seen from an inertial frame it has the velocity $$v(t)$$ where $$t$$ is the coordinate time of the inertial frame. Let $$t \in [0,T]$$.

I want to get the coordinate transformation for the comoving frame of the particle. By that I mean a coordinate system in which the particle is at rest for all times. I tried to get this transformation by assuming the particle moves for short time periods $$\Delta t$$ with constant velocity. For constant velocities we can use a normal Lorentz transformation as a coordinate transformation. If I now split the time interval $$[0,T]$$ into $$n$$ intervals of with length $$\Delta t = T/N$$ it is possible to find a LT for each time interval of length $$\Delta t$$.

For $$t \in [n \Delta t,(n+1) \Delta t]$$ we get

$$t' = \gamma_n (t- v_n x)+ a_n$$

The $$a_n$$ is unknown but it needs to be connected to the previous time interval $$[n \Delta t,(n+1) \Delta t]$$ by

$$a_n = a_{n-1} + n \Delta t(\gamma_{n-1}-\gamma_{n-2})+ \left(\sum_{m=0}^{n}v_m \Delta t\right) (\gamma_{n} v_n - \gamma_{n-1}v_{n-1})$$

Now by iteration and converting the sums into integrals I get

$$a_n =: a(t)= \int_0^t \tilde{t} \frac{\mathrm d\gamma}{\mathrm d\tilde{t}} +r(\tilde{t})\frac{\mathrm d(\gamma v)}{\mathrm d\tilde{t}} \mathrm d\tilde{t}$$

where $$r(t)$$ is the position of the particle seen from the inertial frame with $$r(0)=0$$.

So in total we get for the transformed time:

$$t' = \gamma(t) (t- v(t) x)+ \int_0^t \tilde{t} \frac{\mathrm d\gamma}{\mathrm d\tilde{t}} +r(\tilde{t})\frac{\mathrm d(\gamma v)}{\mathrm d\tilde{t}} \mathrm d\tilde{t}$$

Something similar can be done for the space coordinate.

Now my question is if this is the right approach? Or did I make a mistake somewhere? In my new coordinate system the particle should be at rest for all $$t'$$. Are there other ways to find the comoving coordinate system ? And is there a formula without the integral?

• This is not an inertial frame, does something not look right to you? Are you getting a counter intuitive result? – ggcg Dec 12 '18 at 17:54
• @ggcg For the twin paradox I get the right answer. But in that case you just have two velocities and you don't need the iteration and integration. But I was just wondering if the result above can be found in the literature or if there is something similar to be found. – yasalami Dec 12 '18 at 20:00
• Most texts on GR will do the uniform accelerating frame. I think you can find it in Misner Thorne and Wheeler (MTW), Gravitation. – ggcg Dec 12 '18 at 21:06
• You can at least check that case against your result. – ggcg Dec 12 '18 at 21:07
• @yasalami, I see you're right about the $t'$. I made a mistake with the measure of integration when I switched to $\tau$. Your $t'$ is indeed equal to $\tau$ along the trajectory of the particle. I deleted my answer. – octonion Dec 12 '18 at 23:28

Seen from an inertial frame it has the velocity v(t)

$$\def\D#1#2{{d#1 \over d#2}}$$ Define $$x(t)$$ such that $$dx/dt=v$$, then $$h(\tau)$$ so that \eqalign{ \D x\tau &= \sinh h(\tau) \cr \D t\tau &= \cosh h(\tau).\cr} Note that $$\tanh h = v$$, $$\cosh h = \gamma$$, $$h$$ is rapidity ($$\tau$$ is proper time.)

The transformation you're looking for is \eqalign{ x &= \int_0^\xi \sinh h(\xi')\,d\xi' + \eta\,\cosh h(\xi) \cr t &= \int_0^\xi \cosh h(\xi')\,d\xi' + \eta\,\sinh h(\xi).\cr} You may see that for $$\eta=0$$ \eqalign{ dx = \sinh h(\xi)\,d\xi \cr dt = \cosh h(\xi)\,d\xi \cr} so that motion of a point fixed at $$\eta=0$$ is just the one assigned for $$x(t)$$ with $$\xi=\tau$$.

In coordinates $$(\xi,\eta)$$ the metric is $$[1 + \eta\,h'(\xi)]^2 d\xi^2 - d\eta^2$$ showing that $$(\xi,\eta)$$ are generalized Rindler coordinates.

Other properties easily follow, especially for motion of points with $$\eta=\mathrm{const.}\ne0$$.

• Not sure, but in the first two equations, the argument of the hyperbolic functions shouldn't be $h(\tau)$? – Marco81 Dec 13 '18 at 21:47
• @Marco81 You're absolutely right. Corrected. Thank you. – Elio Fabri Dec 14 '18 at 8:59

I agree with the suggestion to look into the MTW. I really like the section on uniformly accelerating objects.

I don't think you need the complicated step that involves the Lorentz transforms, and taking the continuum limit.

In any inertial frame, a world-line of a moving object can be expressed as $$x^\mu=\left(ct,\, \mathbf{r}(t)\right)^\mu$$, so $$dx^\mu=(c, \mathbf{v}(t))^\mu dt$$, and $$dx^2=(c^2-v^2)dt^2$$, but also, by definition, $$dx^2=c^2 d\tau^2$$ ($$\tau$$=proper time), so $$dt/d\tau=\gamma=\left(1-v^2/c^2\right)^{-1/2}$$, even if velocity is time-dependent.

If you know the expression for the time-dependent velocity $$v=v(t)$$, simply integrate $$d\tau=dt\sqrt{1-v(t)^2/c^2}$$ to get the proper time ($$\tau$$) in terms of lab-time time ($$t$$).

Once you know how proper time relates to lab-time i.e. $$\tau=\tau\left(t\right)$$ you can write down the four-velocity $$u^{\mu}(\tau)=\frac{dx^{\mu}}{d\tau}=\frac{d\tau}{dt}\cdot\frac{dx^{\mu}}{dt}=\frac{d\tau}{dt}\cdot\left(c, \mathbf{v}(t(\tau))\right)^{\mu}$$

Four-velocity will give you the direction of the temporal axis for the inertial frame in which your object is at rest at that specific time. You then have an arbitrary choice on how to span the 3 spatial dimensions that are perpendicular to four-velocity.