# Composition of velocities in relativity

Required math: algebra

Required physics: special relativity

Reference: d’Inverno, Ray, Introducing Einstein’s Relativity (1992), Oxford Uni Press. – Section 2.9 and Problems 2.4, 2.5.

The composition of two velocities in special relativity has a particularly simple derivation using the k-calculus. Suppose we have 3 observers as shown in the diagram:

Observer A is at rest relative to us, while B moves to the right with velocity ${v_{AB}}$ and C also moves to the right with velocity ${v_{AC}}$, with both velocities measured relative to A. Now suppose that A emits two light beams separated by a time interval T. From our k-calculus results, we know that B will receive these beams separated by a time ${k_{AB}T}$. If B then sends these two beams on their way to C, C will receive them at a time interval ${k_{BC}\left(k_{AB}T\right)}$. Thus the overall k-factor from A to C is

$\displaystyle k_{AC}=k_{AB}k_{BC} \ \ \ \ \ (1)$

We already worked out ${k}$ in terms of ${v}$, so we have

 $\displaystyle k_{AC}$ $\displaystyle =$ $\displaystyle \left(\frac{1+v_{AC}}{1-v_{AC}}\right)^{1/2}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{1+v_{AB}}{1-v_{AB}}\right)^{1/2}\left(\frac{1+v_{BC}}{1-v_{BC}}\right)^{1/2} \ \ \ \ \ (3)$

Squaring this equation we get

 $\displaystyle \frac{1+v_{AC}}{1-v_{AC}}$ $\displaystyle =$ $\displaystyle \frac{1+v_{AB}}{1-v_{AB}}\frac{1+v_{BC}}{1-v_{BC}}\ \ \ \ \ (4)$ $\displaystyle v_{AC}\left(1+\frac{1+v_{AB}}{1-v_{AB}}\frac{1+v_{BC}}{1-v_{BC}}\right)$ $\displaystyle =$ $\displaystyle \frac{1+v_{AB}}{1-v_{AB}}\frac{1+v_{BC}}{1-v_{BC}}-1\ \ \ \ \ (5)$ $\displaystyle v_{AC}$ $\displaystyle =$ $\displaystyle \frac{\left(1+v_{AB}\right)\left(1+v_{BC}\right)-\left(1-v_{AB}\right)\left(1-v_{BC}\right)}{\left(1+v_{AB}\right)\left(1+v_{BC}\right)+\left(1-v_{AB}\right)\left(1-v_{BC}\right)}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v_{AB}+v_{BC}}{1+v_{AB}v_{BC}} \ \ \ \ \ (7)$

The composition of two velocities (in the same direction) is therefore less than just the arithmetic sum. In fact, if we start with two velocities, both less than 1 (that is, less than the speed of light), then their sum is also less than 1. We can show this with a little calculus.

Consider the function, defined for ${0 and ${0:

$\displaystyle f(x,y)=\frac{x+y}{1+xy} \ \ \ \ \ (8)$

Taking its two partial derivatives we find

 $\displaystyle \frac{\partial f}{\partial x}$ $\displaystyle =$ $\displaystyle -\frac{y\left(x+y\right)}{\left(1+xy\right)^{2}}+\frac{1}{x+y}\ \ \ \ \ (9)$ $\displaystyle \frac{\partial f}{\partial y}$ $\displaystyle =$ $\displaystyle -\frac{x\left(x+y\right)}{\left(1+xy\right)^{2}}+\frac{1}{x+y} \ \ \ \ \ (10)$

Setting each of these to zero, we get the two conditions

 $\displaystyle y^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (11)$ $\displaystyle x^{2}$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (12)$

Thus there are no maxima, minima, or saddle points anywhere inside the region, and the extreme values of the function must lie on the boundary. The boundaries are

 $\displaystyle f\left(x,1\right)$ $\displaystyle =$ $\displaystyle \frac{1+x}{1+x}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle f(1,y)$ $\displaystyle =$ $\displaystyle \frac{1+y}{1+y}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (16)$ $\displaystyle f(x,0)$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (17)$ $\displaystyle f(0,y)$ $\displaystyle =$ $\displaystyle y \ \ \ \ \ (18)$

Thus the maximum of the function occurs at the point ${\left(x,y\right)=\left(1,1\right)}$ and has the value 1. A plot of the function looks like this:

The nearest corner is the origin, with the point ${\left(1,1\right)}$ lying furthest away.

For velocities ${v\ll1}$, the formula reduces to the Newtonian formula. We can approximate the formula above using a Taylor series:

$\displaystyle \frac{x+y}{1+xy}\simeq\left(x+y\right)(1-xy+\ldots) \ \ \ \ \ (19)$

If we save only up to first-order terms, we get

$\displaystyle \frac{x+y}{1+xy}\simeq x+y \ \ \ \ \ (20)$

or, in terms of velocities

$\displaystyle v_{AC}\simeq v_{AB}+v_{BC} \ \ \ \ \ (21)$

For negative velocities, we can look at the region ${-1 and ${-1. The other two boundaries are

 $\displaystyle f\left(x,-1\right)$ $\displaystyle =$ $\displaystyle \frac{x-1}{1-x}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (23)$ $\displaystyle f\left(-1,y\right)$ $\displaystyle =$ $\displaystyle \frac{y-1}{1-y}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -1 \ \ \ \ \ (25)$

Thus along these boundaries, the extreme value of the function is ${-1}$. The function is actually discontinuous at the two points ${\left(-1,1\right)}$ and ${\left(1,-1\right)}$, where the value tends to ${\pm1}$ depending on how you approach the point. A plot looks like this:

The viewpoint is roughly the same as in the previous plot, with the nearest corner being ${\left(-1,-1\right)}$ and the farthest corner at the top being ${\left(1,1\right)}$.

Another plot rotated about ${90^{\circ}}$ to the left shows the shape a bit more clearly:

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