Accelerating observer

We will examine an observer accelerating at a constant rate with respect to some stationary observer. We can still describe this case using special relativity, but some things are more complicated.

We start by defining 4-acceleration as the proper time derivative of 4-velocity. This is the same formulation we used when considering the electric and magnetic fields.

\begin{align*} \fv a = \frac{\d \fv u}{\d\tau}. \end{align*}

We know that $\fv u \cdot \fv u = -c^2$ , which is a constant, so

\begin{align*} \frac{\d}{\d\tau} (\fv u \cdot \fv u) = \fv a \cdot \fv u + \fv u \cdot \fv a = 2\fv a \cdot \fv u = 0. \end{align*}

Consider the accelerating observer $\O'$ . At some moment in time, there exists a reference frame that is stationary with respect to $\O'$ . This is the momentarily comoving reference frame, let’s call it $\O_\mcrf$ . In the MCRF, the motion of the accelerating observer looks particularly simple.

\begin{align*} u_\mcrf^t &= c, \quad\quad u_\mcrf^{x,y,z} = 0, \quad\quad \d\tau = \dt_\mcrf \\ a_\mcrf^\mu &\compeq \mat{0 \\ \d u_\mcrf^x / \dt \\ \d u_\mcrf^y / \dt \\ \d u_\mcrf^z / \dt} \end{align*}

The acceleration measured in the MCRF is the same as the acceleration measured in $\O'$ . This is what we mean when we say an observer is accelerating at a certain rate.

From the invariants we know about $\fv u$ and $\fv a$ we can relate the quantities measured in the MCRF. For this derivation we assume the observer has acceleration $a$ in the $x$ direction.

\begin{align*} \fv u \cdot \fv u &= -c^2 &&\implies& -(u^t)^2 + (u^x)^2 &= -c^2 \\ \fv u \cdot \fv a &= 0 &&\implies& -u^t \frac{\d u^t}{\d\tau} + u^x \frac{\d u^x}{\d\tau} &= 0 \\ \fv a \cdot \fv a &= a^2 &&\implies& -\left(\frac{\d u^t}{\d\tau}\right)^2 + \left(\frac{\d u^x}{\d\tau}\right)^2 &= a^2 \end{align*}

These equations can be solved directly, but instead let’s take a different approach. We will consider the motion of the accelerating frame in terms of its rapidity $\tanh \phi = v$ .

\begin{align*} a &= \frac{\d v}{\d\tau} = \frac{\d v}{\d \phi}\frac{\d \phi}{\d\tau} = \frac{1}{\cosh^2\phi} \frac{\d\phi}{\d\tau}. \end{align*}

In the MCRF at the instant of comotion ( a = frac{dphi}{dtau}. )

And because we have assumed $a$ is a constant, we can integrate this equation to find $\phi = a \tau$ at this instant. This makes sense based on what we know about rapidity. Rapidities add normally, and so it makes sense for the total rapidity after some proper time $\tau$ of proper acceleration $a$ to be $a\tau$ .

This allows us to find the position and time of the accelerating frame in the stationary frame.

\begin{align*} \frac{\d t}{\d\tau} &= \gamma = \frac{1}{\sqrt{1-\tanh^2(a\tau)}} = \cosh a\tau \\ t(\tau) &= \frac1a \sinh (a\tau). \end{align*}

We can similarly find the position

\begin{align*} \frac{\d x}{\d\tau} \frac{\d\tau}{\d t} &= v = \tanh (a\tau) \\ \frac{\d x}{\d\tau} \frac{1}{\cosh a\tau} &= \frac{\sinh a\tau}{\cosh a\tau} \\ x(\tau) &= \frac1a \cosh a\tau. \end{align*}

Here we have set the integration constants to zero. These represent our starting coordinates.

We can now compare these results to the constraints found above. By differentiating $x(\tau)$ and $t(\tau)$ with respect to proper time, we find

\begin{align*} u^x(\tau) &= \sinh a\tau \\ u^t(\tau) &= \cosh a\tau. \end{align*}

These values satisfy the constraints found above.