Derivation of Rindler coordinates for general motion

Question

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1. What the problem is asking

We are asked to derive the line element

[ \boxed{ds^{2}= \Bigl(1+\frac{a_{x’}x’}{c^{2}}+\frac{a_{y’}y’}{c^{2}}+\frac{a_{z’}z’}{c^{2}}\Bigr)^{2}c^{2}\,dt’^{2} -(dx’^{2}+dy’^{2}+dz’^{2})} \tag{1} ]

that is claimed to describe the space‑time as seen by an observer whose proper acceleration (\mathbf a(t’)=(a_{x’}(t’),a_{y’}(t’),a_{z’}(t’))) is an arbitrary (i.e. not necessarily constant) function of the observer’s proper time (t’). In other words, we must show how the flat Minkowski metric looks in the non‑inertial coordinates ((t’,x’,y’,z’)) that are attached to a non‑rotating, arbitrarily accelerated reference frame.

The derivation uses the Fermi‑normal (or Fermi‑Weyl) coordinates built around the world‑line of the observer.
Below every step is written out explicitly.

2. Step‑by‑step derivation

2.1 Set up the inertial background

Start with ordinary inertial Cartesian coordinates ((t, x^{i})\;(i=1,2,3)) in flat space‑time.
The metric is the Minkowski one (signature (+—))

[ ds^{2}=c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2} \equiv \eta_{\mu\nu}\,dX^{\mu}dX^{\nu}, \qquad \eta_{\mu\nu}= \mathrm{diag}(c^{2},-1,-1,-1). ]

Greek indices (\mu,\nu) run over (0,1,2,3) with (X^{0}=ct).

2.2 The observer’s world‑line and its tetrad

Let the observer move along a timelike world‑line

[ X^{\mu}=x^{\mu}(\tau),\qquad \tau\equiv t’ ]

where (\tau) is the proper time measured by the observer.

Define the observer’s four‑velocity and four‑acceleration

[ u^{\mu}\equiv \frac{dx^{\mu}}{d\tau},\qquad a^{\mu}\equiv \frac{du^{\mu}}{d\tau}. ]

Because (\tau) is proper time, [ u^{\mu}u_{\mu}=c^{2},\qquad u^{\mu}a_{\mu}=0 . ]

Introduce an orthonormal tetrad ({e_{(0)}^{\mu},e_{(i)}^{\mu}}) that is momentarily comoving with the observer:

(e_{(0)}^{\mu}=u^{\mu}/c) (unit time‑like vector);
(e_{(i)}^{\mu}\;(i=1,2,3)) are three space‑like unit vectors orthogonal to (u^{\mu}): [ e_{(i)}^{\mu}e_{(j)\,\mu}=-\delta_{ij},\qquad e_{(i)}^{\mu}u_{\mu}=0 . ]

We demand that the spatial basis does not rotate with respect to the local inertial frames; this is achieved by Fermi–Walker transport:

[ \frac{De_{(i)}^{\mu}}{d\tau}\equiv \frac{de_{(i)}^{\mu}}{d\tau}

\Gamma^{\mu}{\;\alpha\beta}u^{\alpha}e{(i)}^{\beta} = (a_{\nu}e_{(i)}^{\nu})\frac{u^{\mu}}{c^{2}} . \tag{2} ]

In flat space the connection (\Gamma^{\mu}_{\;\alpha\beta}=0), so (2) reduces to

[ \boxed{\displaystyle \frac{de_{(i)}^{\mu}}{d\tau} =\frac{a_{\nu}e_{(i)}^{\nu}}{c^{2}}\,u^{\mu}} . \tag{3} ]

Define the proper‑acceleration components measured in the comoving frame

[ a_{i}(\tau) \equiv -\,a_{\mu}e_{(i)}^{\mu}\quad\Longrightarrow\quad a_{\mu} = -a_{i}e_{(i)\,\mu}. ]

(The minus sign comes from the space‑like character of (e_{(i)}).)
With this definition (3) becomes simply

[ \frac{de_{(i)}^{\mu}}{d\tau}= \frac{a_{i}}{c^{2}}\,u^{\mu}. \tag{4} ]

2.3 Defining the accelerated coordinates

Take a point (P) that is near the observer. Its coordinates in the inertial frame are written as a Taylor expansion about the observer’s world‑line:

[ X^{\mu}=x^{\mu}(\tau)+e_{(i)}^{\mu}(\tau)\,\xi^{i}, \tag{5} ]

where (\xi^{i}) are the spatial coordinates measured in the instantaneously comoving inertial frame.
We shall identify

[ t’ \equiv \tau ,\qquad x’ \equiv \xi^{1},\qquad y’ \equiv \xi^{2},\qquad z’ \equiv \xi^{3}. ]

Equation (5) is the definition of Fermi normal coordinates (sometimes called “Rindler‑like coordinates for arbitrary acceleration”).

2.4 Compute the differential (dX^{\mu})

Differentiate (5) while remembering that the basis vectors depend on (\tau):

[ \begin{aligned} dX^{\mu} &= \frac{dx^{\mu}}{d\tau}\,d\tau

\frac{de_{(i)}^{\mu}}{d\tau}\,\xi^{i}\,d\tau
e_{(i)}^{\mu}\,d\xi^{i} \[2mm] &= u^{\mu}d\tau +\Bigl(\frac{a_{i}}{c^{2}}u^{\mu}\Bigr)\xi^{i}d\tau
e_{(i)}^{\mu}d\xi^{i} \qquad\text{[using (4)]}\[2mm] &= u^{\mu}\Bigl(1+\frac{a_{i}\xi^{i}}{c^{2}}\Bigr)d\tau
e_{(i)}^{\mu}d\xi^{i}. \end{aligned} \tag{6} ]

2.5 Insert into the Minkowski line element

Now evaluate

[ ds^{2}= \eta_{\mu\nu}\,dX^{\mu}dX^{\nu}. ]

Using the orthonormality of the tetrad,

[ \eta_{\mu\nu}u^{\mu}u^{\nu}=c^{2},\qquad \eta_{\mu\nu}u^{\mu}e_{(i)}^{\nu}=0,\qquad \eta_{\mu\nu}e_{(i)}^{\mu}e_{(j)}^{\nu}=-\delta_{ij}, ]

the cross‑terms in (6) vanish and we obtain

[ \begin{aligned} ds^{2} &= c^{2}\Bigl(1+\frac{a_{i}\xi^{i}}{c^{2}}\Bigr)^{2} d\tau^{2} -\delta_{ij}\,d\xi^{i}d\xi^{j}\[2mm] &= \Bigl(1+\frac{a_{x’}x’}{c^{2}}+\frac{a_{y’}y’}{c^{2}} +\frac{a_{z’}z’}{c^{2}}\Bigr)^{2} c^{2} dt’^{2} -(dx’^{2}+dy’^{2}+dz’^{2}). \end{aligned} \tag{7} ]

Equation (7) is exactly the metric quoted in the problem statement.

2.6 Remarks on the

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