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<x<1} and {0<y<1}:

\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<x<1} and {-1<y<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:

4 thoughts on “Composition of velocities in relativity

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