# Relativistic kinematics

### Proper velocity and momentum

We have previously seen that the position, velocity and momentum of a particle along world-line in space-time are four-vectors.

If we suppose that the velocity of a particle is changing over time - that is, it can accelerate and decelarate - then the particle's rest frame is not necessarily an inertial reference frame. However, at any given time (either proper time or coordinate time), we can consider a frame that, at that instant, is moving at the same velocity as the particle and has the particle at its origin.

When we talk about the particle frame, we shall really mean this ‘momentarily co-moving reference frame’, and use primed coordinates to refer to it. In this frame, we have $$\begin{matrix}\text{Position } & X^{\mu'} & \rightarrow & (\tau,0,0,0)\\\text{Velocity } & U^{\mu'} = \tfrac{\mathrm{d} X^{\mu'}}{\mathrm{d} \tau} & \rightarrow & (1,0,0,0)\\\text{Momentum } & P^{\mu'} = m U^{\mu'} & \rightarrow & (m,0,0,0)\end{matrix}$$ where

• $\tau = t'$ is the (invariant) proper time associated with the particle frame.
• $m$ is the (invariant) mass of the particle.

We shall refer to the coordinate (unprimed) frame, relative to which the particle moves at Newtonian velocity $\vec{v} = (v_x,v_y,v_z)$, as the stationary frame, and its components are given by $$\begin{matrix}\text{Position } & X^\mu & \rightarrow & (t,x,y,z)\\\text{Velocity } & U^\mu = \tfrac{\mathrm{d} X^\mu}{\mathrm{d} \tau} & \rightarrow & \gamma (1,v_x,v_y,v_z)\\\text{Momentum } & P^\mu = m U^\mu & \rightarrow & m \gamma (1,v_x,v_y,v_z)\end{matrix}$$ where $\gamma = \frac{\mathrm{d} t}{\mathrm{d} \tau} = \frac{1}{\sqrt{1 - v^2}}$ is now related to the instantaneous speed, $v = \sqrt{v_x^2 + v_y^2 + v_z^2}$, of a particle as measured in the stationary frame.

We note that, in either frame the four-velocity is a unit (tangent) vector, since $$U_\mu U^\mu = U_{\mu'} U^{\mu'} = 1$$

### Proper acceleration

We can go further and define the proper acceleration of the particle by $$A^\mu = \frac{\mathrm{d} U^\mu}{\mathrm{d} \tau}$$

We can show that the proper acceleration is orthogonal to the proper velocity. That is, $$U_\mu A^\mu = 0$$

To see why, note that, since $U^\mu$ is a unit vector we must have $$\frac{\mathrm{d}}{\mathrm{d} \tau} \left(U_\mu U^\mu\right) = \frac{\mathrm{d}}{\mathrm{d} \tau}\left(1\right) = 0$$

But, $$\frac{\mathrm{d}}{\mathrm{d} \tau} \left(U_\mu U^\mu\right) = \frac{\mathrm{d} U_\mu}{\mathrm{d} \tau}U^\mu + U_\mu \frac{\mathrm{d} U^\mu}{\mathrm{d} \tau} = A_\mu U^\mu + U_\mu A^\mu = 2 U_\mu A^\mu$$

Thus, as stated, $U_\mu A^\mu = 0$.

In the particle frame, the velocity has components, $U^{\mu'} \rightarrow (1,0,0,0)$. The only way that the acceleration can be orthogonal to the velocity in this frame is for it to have components of the form $$A^{\mu'} \rightarrow (0,A^{1'},A^{2'},A^{3'})$$

In the stationary frame, the time component of the acceleration is $$A^0 = \frac{\mathrm{d} \gamma}{\mathrm{d} \tau} = \frac{\mathrm{d} \gamma}{\mathrm{d} v} \frac{\mathrm{d} v}{\mathrm{d} \tau} = \left(\gamma^3 v\right)\left(\gamma \frac{\mathrm{d} v}{\mathrm{d} t}\right) = \gamma^4 v \frac{\mathrm{d} v}{\mathrm{d} t} = \gamma^4 a v$$ where $a = \frac{\mathrm{d} v}{\mathrm{d} t}$ is the magnitude of the Newtonian acceleration.

The $x$-component in the stationary frame is \begin{align*}A^1 &= \frac{\mathrm{d} U^1}{\mathrm{d} \tau}\\&= \frac{\mathrm{d} \left(\gamma v_x\right)}{\mathrm{d} \tau}\\&= \frac{\mathrm{d} \gamma}{\mathrm{d} \tau}v_x + \gamma \frac{\mathrm{d} v_x}{\mathrm{d} \tau}\\&= \gamma^4 a v v_x + \gamma^2 \frac{\mathrm{d} v_x}{\mathrm{d} t}\\&= \gamma^2 \left(\gamma^2 a v v_x + a_x \right)\end{align*} where $a_x = \frac{\mathrm{d} v_x}{\mathrm{d} t}$ is the $x$-component of the Newtonian acceleration. We can calculate the other stationary components in the same way and summarise them in three-vector terms, $$A^m = \frac{\mathrm{d} \left(\gamma v_m\right)}{\mathrm{d} \tau} = \gamma^2 \left(\gamma^2 a v v_m + a_m \right)$$

Nb. Recall, latin indices imply summation over the spatial components only.

The components of acceleration are somewhat more involved than velocity or momentum. However, in the limit where $v$ is very small, we would have $\gamma \approx 1, v v_x \approx 0, v v_y \approx 0, v v_z \approx 0$ and so \begin{align*}A^0 &\rightarrow 0\\A^m &\rightarrow (1^2)\left(1^2\cdot a \cdot 0 + a_m\right) = a_m\end{align*}

Thus, for small velocities, the spatial components of the proper acceleration, as measured in the stationary frame, are just the three components of Newtonian acceleration.

If we consider the simplest case, which is motion restricted to the $x$-axis, then $v_x = v, a_x = a$ and $v_y = v_z = a_y = a_z = 0$, and so the $x$-component becomes \begin{align*}A^x &= \gamma^2 \left(\gamma^2 v^2 + 1\right) a\\&= \gamma^2 \left(\frac{v^2}{1 - v^2} + 1\right) a\\&= \gamma^2 \left(\frac{v^2 + 1 - v^2}{1 - v^2}\right) a\\&= \gamma^2 \left(\frac{1}{1 - v^2}\right) a\\&= \gamma^2 \left(\gamma^2\right) a\\&= \gamma^4 a\end{align*}

The stationary frame components of the of the four proper vectors we have so far considered are then $$\begin{matrix}\text{Position } & X^\mu & \rightarrow & (t,x,0,0)\\\text{Velocity } & U^\mu = \tfrac{\mathrm{d} X^\mu}{\mathrm{d} \tau} & \rightarrow & \gamma (1,v,0,0)\\\text{Momentum } & P^\mu = m U^\mu & \rightarrow & m \gamma (1,v,0,0)\\\text{Acceleration } & A^\mu = \tfrac{\mathrm{d} U^\mu}{\mathrm{d} \tau} & \rightarrow & \gamma^4 a (v,1,0,0)\end{matrix}$$ and the particle frame components are $$\begin{matrix}\text{Position } & X^{\mu'} & \rightarrow & (\tau,0,0,0)\\\text{Velocity } & U^{\mu'} & \rightarrow & (1,0,0,0)\\\text{Momentum } & P^{\mu'} & \rightarrow & (m,0,0,0)\\\text{Acceleration } & A^{\mu'} & \rightarrow & (0,\alpha,0,0)\end{matrix}$$ where we have set $A^{1'} = \alpha$ (not necessarily constant).

### Constant proper acceleration

Suppose that a particle (or rocket) can maintain a constant acceleration locally, where $\alpha$ is constant.

Since the proper acceleration is a four-vector, then we can use the corresponding Lorentz transformation to calculate its $x$-component in the unprimed frame, $$A^1 = \gamma \left(v A^{0'} + A^{1'}\right) = \gamma \alpha$$

Thus, using the result from the previous section , we have $$\gamma \alpha = A^1 = \gamma^4 a$$ or $$a = \gamma^{-3} \alpha \lt \alpha$$ so the Newtonian acceleration decreases as the speed increases, even though the local acceleration is constant.

If we write $a = \frac{\mathrm{d} v}{\mathrm{d} t}$ so that $$\alpha = \gamma^3 \frac{\mathrm{d} v}{\mathrm{d} t}$$

Multiply both sides by $\mathrm{d} t$ to get $$\alpha \mathrm{d} t = \gamma^3 \mathrm{d} v = \frac{\mathrm{d} v}{\left(1 - v^2\right)^{3/2}}$$ and then integrate, $$\alpha \left(t - t_0\right) = \int_{t_0}^{t} \alpha \mathrm{d} t = \int_{v_0}^{v} \frac{\mathrm{d} v}{\left(1 - v^2\right)^{3/2}} = \frac{v}{\sqrt{1 - v^2}} - \frac{v_0}{\sqrt{1 - v_0^2}}$$ where $v_0 = v(t_0)$

If our rocket starts out at rest (that is, $v_0 = 0$ at $t_0 = 0$) then $$\alpha t = \frac{v}{\sqrt{1 - v^2}}$$

So, under constant proper acceleration, $\alpha$, the time $t$ it takes to reach the speed $v$ is $$t(v) = \frac{1}{\alpha}\frac{v}{\sqrt{1 - v^2}}$$

We can solve this for $v$ to get the speed of the particle $v$ at time $t$, $$v(t) = \frac{\alpha t}{\sqrt{1 + \alpha^2 t^2}}$$

We can integrate the velocity over time $$\int_{0}^{t} v(t) \mathrm{d} t = \int_{0}^{t} \frac{\alpha t}{\sqrt{1 + \alpha^2 t^2}} \mathrm{d} t = \left[\frac{1}{\alpha}\sqrt{1 + \alpha^2 t^2}\right]_{0}^{t}$$ to get the distance, $x$ travelled by the particle at time $t$ is $$x(t) = \frac{1}{\alpha}\left(\sqrt{1 + \alpha^2 t^2} - 1\right)$$

Finally, recall that $$\frac{\partial t}{\partial \tau} = \gamma = \frac{1}{\sqrt{1 - v^2}}$$ into which we can substitute the expression for $v$ as a function of coordinate time $t$, to get $$\frac{\partial t}{\partial \tau} = \sqrt{1 + \alpha^2 t^2}$$

By re-arranging and integrating, we get $$\tau(t) = \int_0^t \frac{\mathrm{d} t}{\sqrt{1 + \alpha^2 t^2}} = \frac{1}{\alpha} \sinh^{-1}(\alpha t)$$ which we can solve for the coordinate time, $t$, as a function of the proper time, $\tau$, $$t(\tau) = \frac{1}{\alpha} \sinh(\alpha \tau)$$

### Example

Suppose a rocket left earth and was able to maintain a constant local acceleration equivalent to the gravitational acceleration at the Earth's surface, $$g = 9.8 \frac{\text{metres}}{\text{second}^2}$$

In order to use the equations we've derived above, we need to convert $g$ into the correct units, by equating $$1 \text{ metre} = \frac{1}{3 \times 10^8} \text{ seconds}$$ so that now the units of both time and space are seconds although we refer to the spatial units as light-seconds.

Thus, the acceleration in these units is $$g = \frac{9.8}{3 \times 10^8} \frac{\text{light-second}}{\text{second}^2} = \frac{9.8}{3 \times 10^8} \frac{1}{\text{second}}$$

For our example, years and light-years are more appropriate, hence we want to substitute years for seconds using the conversion $$1 \text{ year} = 3.15 \times 10^7 \text{ seconds}$$

Thus, finally $$g = \frac{9.8 \times 3.15 \times 10^7 }{3 \times 10^8} \frac{\text{light-year}}{\text{year}^2} = 1.03 \frac{1}{\text{year}}$$

#### How long would it take for the rocket to reach a speed of 99.9% of the speed of light?

We take $v = 0.999$ and $\alpha = g = 1.03 \text{ years}^{-1}$. Then, \begin{align*}\text{Coordinate time } & t(0.999) = \frac{1}{1.03}\frac{0.999}{\sqrt{1 - 0.999^2}} \text{ years } = 21.7 \text{ years}\\\text{Proper time } &= \tau(21.7) = \frac{1}{1.03} \sinh^{-1}(1.03 \times 21.7) \text{ years } = 3.69 \text{ years}\end{align*}

#### How long would it take for the rocket to reach Alpha Centauri (with speed zero) and then come back?

Alpha Centauri is approximately $4.36 \text{ light-years}$ from Earth.

To calculate the coordinate time as function of distance we need to solve $x(t) = \frac{1}{\alpha}\left(\sqrt{1 + \alpha^2 t^2} - 1\right)$ for $t$, which gives $$t(x) = \frac{\sqrt{(1 + \alpha x)^2 - 1}}{\alpha}$$

For a comfortable journey, we could accelerate until we reach half the distance, then decelerate (at $g = -1.03$) until we reach Alpha Centauri at speed $v = 0$. We could carry on decelerating back to Earth until we reached halfway, where we would start accelerating again in order to get back to Earth at zero speed.

It's reasonable to assume, by symmetry, that each of the four portions of the journey ($x = 2.18 \text{ light-years}$) would take an equal length of time \begin{align*}\text{Coordinate time } & t(2.18) = \frac{\sqrt{(1 + 1.03 \times 2.18)^2 - 1}}{1.03} \text{ years } = 3.00 \text{ years}\\\text{Proper time } &= \tau(3.00) = \frac{1}{1.03} \sinh^{-1}(1.03 \times 3.00) \text{ years } = 1.79 \text{ years}\end{align*}

So, in total, the proper time experienced by the rocket crew would be $4 \times 1.79 = 7.17\text{ years}$, compared to the $4 \times 3.00 = 12.0\text{ years }$ that passed at home.

#### How long would it take for the rocket to reach the centre of the galaxy?

The centre of the galaxy is approximately $27,000 \text{ light-years}$ from Earth.

The difference between the two times is now huge, \begin{align*}\text{Coordinate time } & t(27,000) = \frac{\sqrt{(1 + 1.03 \times 27,000)^2 - 1}}{1.03} \text{ years } = 27,000 \text{ years}\\\text{Proper time } &= \tau(27,000) = \frac{1}{1.03} \sinh^{-1}(1.03 \times 27,000) \text{ years } = 10.6 \text{ years}\end{align*}

Nb. If $\alpha x \gg 1$, then $t(x) \approx x$