Astronomy Answers
Relativistic Travel

\(\def\|{&}\DeclareMathOperator{\D}{\bigtriangleup\!} \DeclareMathOperator{\d}{\text{d}\!}\)

\( \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \def\d{\,\text{d}} \)

According to the Special Theory of Relativity, measures of space and time depend on the observer. Suppose an observer S stays home in an inertial system (i.e., without accelerating or decelerating or changing direction) and an observer R gets in a rocket and travels through space. How do both observers see the time and distance covered by R?

We assume that the traveller stays far away from stars and planets in empty space, so that effects of gravity can be ignored and the speed of light is equal to that in vacuum.

We use the following symbols:

At a certain moment \( τ \), R has velocity \( \vec{v}(τ) \) compared to S. An infinitesimal interval \( \d τ \) later, the velocity of R has increased (according to R) by \( \vec{α}\d τ \). According to S (and the Special Theory of Relativity) the increase in the velocity of R is only \( 1 - \frac{v^2}{c^2} \) times as great, so \( \vec{v} \) increases with

\begin{equation} \d \vec{v} = \left( 1 - \frac{v^2}{c^2} \right) \vec{α}\d τ ⇔ \d \vec{β} = \frac{\vec{α}}{γ^2c} \d τ \end{equation}

so

\begin{equation} \dfrac{\d \vec{v}}{1 - \dfrac{v^2}{c^2}} = \vec{α} \d τ ⇔ γ^2 \d \vec{β} = \frac{\vec{α}}{c} \d τ \end{equation}

In some cases a direct formula for \( \vec{v}(τ) \) can be derived from this.

The time \( t \) that S measures is tied to the time \( τ \) that R measures, according to

\begin{equation} \d t = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} \d τ ⇔ \d t = γ \d τ \label{eq:dt} \end{equation}

The position of R according to S is determined by

\begin{equation} \d \vec{r} = \vec{v} \d t = γ \vec{v} \d τ \label{eq:dr} \end{equation}

According to R it is

\begin{equation} \d \vec{ρ} = \vec{v} \d τ \label{eq:drho} \end{equation}

In some cases the above formulas can be integrated.

1. Acceleration Along a Straight Line

If the acceleration is along a straight line, then we find

\begin{equation} ∫ \dfrac{1}{1 - \dfrac{v^2}{c^2}} \d v = ∫ α \d τ ⇔ ∫ γ^2 \d β = ∫ \frac{α}{c}x \d τ \label{eq:dvrecht} \end{equation}

We define

\begin{equation} θ(τ) ≡ ∫^τ_{τ_0} \frac{α(τ′)}{c} \d τ′ \label{eq:ups} \end{equation}

where the integration is from a time \( τ_0 \) at which the speed of R relative to S was equal to zero. \( θ \) is the total speed difference that R has experienced from a Galilean perspective, measured in units of the speed of light. Integrate formula \eqref{eq:dvrecht} between \( τ_0 \) and \( τ \) to find

\begin{align} \| \artanh(β) = θ \\ \| ⇒ β = \tanh(θ) \label{eq:beta} \\ \| ⇒ γ = \cosh(θ) \label{eq:gamma} \end{align}

Equations \eqref{eq:dt} and \eqref{eq:gamma} yield

\begin{equation} \d t = \cosh(θ) \d τ \end{equation}

Equations \eqref{eq:dr} and \eqref{eq:gamma} yield

\begin{equation} \d r = γ c \tanh(θ) \d τ = c \sinh(θ) \d τ \end{equation}

Equations \eqref{eq:drho} and \eqref{eq:beta} yield

\begin{equation} \d ρ = c \tanh(θ) \d τ \end{equation}

2. Constant Acceleration Along a Straight Line

For constant acceleration along a straight line, equation \eqref{eq:ups} yields

\begin{equation} θ = θ_0 + \frac{ατ}{c} \end{equation}

We define

\begin{equation} u ≡ θ(t) = u_0 + \frac{αt}{c} \end{equation}

and find

\begin{align} β \| = \tanh(θ) = \frac{u}{\sqrt{1 + u^2}} \\ γ \| = \cosh(θ) = \sqrt{1 + u^2} \\ \d t \| = \cosh(θ) \d τ ⇒ t = \frac{c}{α} \sinh(θ) ⇔ τ = \frac{c}{α} \arsinh(u) \\ \d r \| = c \sinh(θ) \d τ ⇒ r = \frac{c^2}{α} (\cosh(θ) - 1) = \frac{c^2}{α} \left( \sqrt{1 + u^2} - 1 \right) \\ \d ρ \| = c \tanh(θ) \d τ ⇒ ρ = \frac{c^2}{α} \ln(\cosh(θ)) \end{align}

3. Trip

R travels from S to some far-away location that is in rest compared to S. R first accelerates during a period \( τ_0 \) from \( τ = t = 0 \) at a constant rate \( α \) and then decelerates at a constant rate \( -α \) during a period of the same length.

For the acceleration phase \( 0 ≤ τ ≤ τ_0 \) and

\begin{align} θ \| = \frac{ατ}{c} \\ v \| = \tanh(θ) \\ t \| = \frac{c}{α} \sinh(θ) ⇒ t_0 = \frac{c}{α} \sinh\left( \frac{ατ_0}{c} \right) \\ r \| = \frac{c^2}{α} (\cosh(θ) - 1) ⇒ r_0 = \frac{c^2}{α} \left( \cosh\left( \frac{ατ_0}{c} \right) - 1 \right) \\ ρ \| = \frac{c^2}{α} \ln(\cosh(θ)) ⇒ ρ_0 = \frac{c^2}{α} \ln\left( \cosh\left( \frac{ατ_0}{c} \right) \right) \end{align}

For the deceleration phase \( τ_0 ≤ τ ≤ 2τ_0 \) and

\begin{align} θ \| = \frac{α}{c} (2τ_0 - τ) \\ v \| = c \tanh(θ) \\ t \| = 2t_0 - \frac{c}{α} \sinh(θ) \\ r \| = 2r_0 - \frac{c^2}{α} (\cosh(θ) - 1) \\ ρ \| = 2ρ_0 - \frac{c^2}{α} \ln(\cosh(θ)) \end{align}

The whole trip has lasted \( 2τ_0 \) according to R and \( \frac{2c}{α} \sinh\left( \frac{ατ_0}{c} \right) \) according to S, and R has then traversed distance \( \frac{2c^2}{α} \left( \cosh\left( \frac{ατ_0}{c} \right) - 1 \right) \) according to S, but \( \frac{2c^2}{α} \ln\left( \cosh\left( \frac{ατ_0}{c} \right) \right) \) according to R.

4. A Journey to a Place at a Known Distance

What are the characteristics of such a journey to a place at distance \( D \)? Then \( D = 2 r_0 \) so

\begin{align} q \| = \frac{D α}{2 c^2} + 1 \\ θ_\text{max} \| = \arcosh(q) \\ v_\text{max} \| = c \tanh(θ_\text{max}) = c \sqrt{1 - \frac{1}{q^2}} \\ τ \| = \frac{2c}{α} θ_\text{max} \\ t \| = \frac{2c}{α} \sqrt{q^2 - 1} \\ ρ \| = \frac{2c^2}{α} \ln(q) \end{align}

5. Small Speeds

If the maximum speed remains much less than the speed of light (so \( Dα ≪ c^2 \)) then we can approximate:

\begin{align} v_\max \| = \sqrt{D α} \\ θ_\max \| = \sqrt{\frac{D α}{c^2}} = \frac{v_\max}{c} \\ τ \| = 2\sqrt{\frac{D}{α}} - \frac{\sqrt{α} D^{3/2}}{12 c^2} = \frac{2 v_\max}{α} - \frac{v_\max D}{12 c^2} \\ t \| = 2\sqrt{\frac{D}{α}} + \frac{\sqrt{α} D^{3/2}}{4 c^2} = \frac{2 v_\max}{α} + \frac{v_\max D}{4 c^2} \\ ρ \| = D - \frac{α D^2}{4c^2} = D - \frac{D v^2_\max}{4c^2} \\ t - τ \| = \frac{\sqrt{α} D^{3/2}}{3 c^2} = \frac{v_\max D}{3 c^2} ≈ \frac{t v^2_\max}{3c^2} \\ D - ρ \| = \frac{α D^2}{4c^2} = \frac{D v^2_\max}{4c^2} ≈ \frac{t v^3_\max}{4c^2} \end{align}

These relativistic effects are negligible in our daily life. If someone makes a similar journey in a car, with acceleration and deceleration during 10 seconds each, with a maximum speed of 100 km/h, then that person wins \( t - τ \) = 3 × 10⁻¹⁴ seconds compared to someone who remained at rest.

6. Table

\[ \frac{ατ}{c} = 1.033 \frac{α}{g} \frac{τ}{\text{year}} \]

and

\[ \frac{D α}{c^2} = 1.033 \frac{D}{\text{ly}} \frac{α}{g} \]

where \( g \) is the acceleration of gravity at the equator of the Earth (9.81 m/s²), "year" a Julian year of 365.25 days, and "ly" a lightyear based on a Julian year.

Below, I provide some values for a constant acceleration equal to 0.968 times the gravity of the Earth so that \( ατ/c = 1 \) when \( τ \) is equal to a Julian year.

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Last updated: 2021-07-19