- #31
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Here is my try. There are two ways. One is to use coordinate time and the other is to work with the manifestly covariant equations of motion. For the latter way we use the normalized four-velocity and assume that the constant acceleration is in ##x^1##-direction. So we can work in (1+1)-dimensional Minkowski space.
The normalized four-velocity then reads
$$(u^{\mu})=\begin{pmatrix} \cosh y \\ \sinh y \end{pmatrix},$$
where ##y## is the (momentum-space) rapidity, which parametrization works in the constraint ##u_{\mu} u^{\mu}=1##.
The equations of motion are
$$\mathrm{d}_{\tau} u^0=\alpha/c u^1, \quad \mathrm{d}_{\tau} u^1=c \alpha/c u^0.$$
The two equations are not independent because of the constraint. So we use the 2nd equation with the parametrization using the rapidity,
$$\cosh y \mathrm{d}_{\tau} y =\alpha/c \cosh y \; \Rightarrow \; y=\frac{\alpha \tau}{c}+y_0.$$
A direct check shows that indeed also the first eom is fulfilled so we have
$$(u^{\mu})=\begin{pmatrix} \cosh(\alpha \tau/c+y_0) \\ \sinh(\alpha \tau/c+y_0) \end{pmatrix}.$$
The spacetime vector is then defined by
$$\mathrm{d}_{\tau} x^{\mu}=c u^{\mu}$$
from which
$$x^{\mu}=\begin{pmatrix} \frac{c^2}{\alpha} \sinh(\alpha \tau/c+y_0) \\ x_0^{1} + \frac{c^2}{\alpha} [\cosh(\alpha \tau/c + y_0)-1]. \end{pmatrix}$$
I've chosen the initial time such that ##t_0=t(0)=0##.
In terms of coordinate time we have
$$x^1(t)=x_0^1 + \frac{c^2}{\alpha} \left [\sqrt{\alpha^2 t^2/c^2+1}-1 \right]$$
and for the velocity
$$v(t)=\mathrm{d}_t x^1=\frac{\alpha t}{\sqrt{\alpha^2 t^2/c^2+1}}.$$
The normalized four-velocity then reads
$$(u^{\mu})=\begin{pmatrix} \cosh y \\ \sinh y \end{pmatrix},$$
where ##y## is the (momentum-space) rapidity, which parametrization works in the constraint ##u_{\mu} u^{\mu}=1##.
The equations of motion are
$$\mathrm{d}_{\tau} u^0=\alpha/c u^1, \quad \mathrm{d}_{\tau} u^1=c \alpha/c u^0.$$
The two equations are not independent because of the constraint. So we use the 2nd equation with the parametrization using the rapidity,
$$\cosh y \mathrm{d}_{\tau} y =\alpha/c \cosh y \; \Rightarrow \; y=\frac{\alpha \tau}{c}+y_0.$$
A direct check shows that indeed also the first eom is fulfilled so we have
$$(u^{\mu})=\begin{pmatrix} \cosh(\alpha \tau/c+y_0) \\ \sinh(\alpha \tau/c+y_0) \end{pmatrix}.$$
The spacetime vector is then defined by
$$\mathrm{d}_{\tau} x^{\mu}=c u^{\mu}$$
from which
$$x^{\mu}=\begin{pmatrix} \frac{c^2}{\alpha} \sinh(\alpha \tau/c+y_0) \\ x_0^{1} + \frac{c^2}{\alpha} [\cosh(\alpha \tau/c + y_0)-1]. \end{pmatrix}$$
I've chosen the initial time such that ##t_0=t(0)=0##.
In terms of coordinate time we have
$$x^1(t)=x_0^1 + \frac{c^2}{\alpha} \left [\sqrt{\alpha^2 t^2/c^2+1}-1 \right]$$
and for the velocity
$$v(t)=\mathrm{d}_t x^1=\frac{\alpha t}{\sqrt{\alpha^2 t^2/c^2+1}}.$$
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