Sum of two velocities is smaller than the speed of light

Question

Using the Lorentz transformation from special relativity, we get that the sum of two velocities can be expressed as

$$u=\frac{u'+v}{1+\frac{u'v}{c^2}}.$$

Given that $|u'|,|v| \le c$, I want to prove that $|u| \le c$, ie. that the velocity never exceeds $c$. However I am struggling to produce this bound. I have tried to bound the denominator from above but this produces zero and have tried a case wise approach but this has got me no where either.

Answer 1

With only basic analysis, you can do it by fixing one of the two variables.

You can prove that, for a fixed $u'<c$:

1. The function $f_{u'}(v)=\frac{u'+v}{1+\frac{u'}{c^2}v}$ is a monotonically increasing function.
2. $f_{u'}(c)=c$

From $1$, you can conclude that if $v<c$, $f_{u'}(v)<f_{u'}(c)$,and including $2$, that gives you $f_{u'}(v)<c$.

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This proves that if $u',v$ are both smaller than $c$, that $\frac{u'+v}{1+\frac{u'v}{c^2}}$ will also be smaller than $c$.

Answer 2

Hint: If you're familiar with techniques in optimization, try and choose values of $u'$ and $v$ so as to optimize $u$. If you can prove that the optimal values for making $u$ large yield a value of $u$ at or approaching $c$, then you are done.

Answer 3

Here's an elementary proof which avoids transcendental functions.

Letting $x=u'/c, y = v/c, z = u/c$ the equation for the Einstein sum becomes

$$z = \frac{x+y}{1+x y}\tag{1}$$

and we have to prove that

$$-1\lt z \lt 1\tag{2}$$

for $-1\lt x \lt 1, -1 \lt y \lt 1$.

Now we let

$$x\to \frac{1-r}{1+r},y\to \frac{1-s}{1+s}\tag{3a}$$

or

$$r \to \frac{1-x}{1+x}, s\to \frac{1-y}{1+y}\tag{3b}$$

which transforms $x\in (-1,+1)$ to $r \in (\infty, 0)$ and $y\in (-1,+1)$ to $s \in (\infty, 0)$ .

In other words, we have parametrized $x$ and $y$ with positive parameters $r$ and $s$.

Substituting (3) in (1) gives

$$z = \frac{1-t}{1+t}\tag{4}$$

with $t = r s$. Hence we have $t \in (\infty, 0)$ and from (4) follows (2). QED.

Answer 4

Such relativistic sum of speeds (here denoted as $\oplus$) can be written in terms of a simple pullback: $$ u\oplus v\stackrel{\text{def}}{=}c\cdot\tanh\left(\text{arctanh}\tfrac{u}{c}+\text{arctanh}\tfrac{v}{c}\right)$$ since $\tanh$ is an increasing function with range $(-1,1)$, $\left|u\oplus v\right|< c$ immediately follows.

Answer 5

Let $u'=c-x$ and $v=c-y$ with $x\geq0$ and $y\geq0$. Then we have:

$$ u=\frac{(c-x)+(c-y)}{1+\frac{(c-x)(c-y)}{c^2}}=\frac{2c-x-y}{2-\frac{x}{c}-\frac{y}{c}+\frac{xy}{c^2}}=c\frac{(2-\frac{x}{c}-\frac{y}{c})}{(2-\frac{x}{c}-\frac{y}{c})+\frac{xy}{c^2}}\leq c $$

(since $\frac{xy}{c^2} \geq 0$)

You can similarly set $u'=-c+x$, $v=-c+y$ to show that $u\geq-c$.

Answer 6

Setting $x=u'/c, y = v/c, z = u/c$ as in another answer, i.e. expressing velocities as fractions of the speed of light, you want to show

$$ -1 < \frac{x+y}{1+xy} < 1 \quad\text{for $-1< x,y < 1$}. $$

Note that the denominator is positive, so you want to show

$$ -1-xy < x+y < 1+xy.$$

The first inequality follows from $$ x+y+1+xy = (1+x)(1+y) > 0, $$ the second from $$ 1+xy-(x+y)=(1-x)(1-y)>0. $$