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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.

user258521
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    Suggestion: Exclude the case $u'v=-c^2$ where the denominator is zero. E.g. by assuming that $|u'|,|v| < c$. – Qmechanic Oct 24 '18 at 17:03

6 Answers6

19

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 $f_{u'}(v)<c$.

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 $c$.

5xum
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  • This is very nice ! – user258521 Oct 24 '18 at 12:40
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    @user258521 Thanks. It's a little strange in the sense that you "break the symmetry" of the original expression - you view one variable as a variable, and the other as a parameter, and then solve the problem for each value of the parameter. Some call it ugly, some like it, it depends. But so long as it works :) – 5xum Oct 24 '18 at 12:42
  • I think it is perfectly fine, nothing wrong with this. Also in 1. it should have $c^2$ on the denominator, a small edit. – user258521 Oct 24 '18 at 12:47
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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.

Jack D'Aurizio
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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.

Dr. Wolfgang Hintze
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    The interval notation $(a,b)={x\mid a<x<b}$ is supposed to have $a<b$; otherwise $(a,b)={}$ is empty. You should have $(0,\infty)$ instead of $(\infty,0)$. – mr_e_man Oct 24 '18 at 16:47
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    @mr_e_man : For real variables, this is the usual convention. HOWEVER -- be aware that it is not uncommon to think of $(a,b)$ as the directed interval starting at $a$ and ending at $b$, i.e., with the parameterization ${ta+(1-t)b\mid 0<t<1}$. This is more often seen in the complex setting, so that this is a directed line segment in the plane (endpoints can be included/excluded by using $\leq$ and square bracket(s) as needed). For example, $[1-i,1+i]$. In the end, I agree with you, +1. – MPW Oct 24 '18 at 18:43
  • @MPW -- I think that should be ${(1-t)a+tb}$, if we want $a$ when $t=0$. – mr_e_man Oct 24 '18 at 18:48
  • @mr_e_man : Yep. I always write that backwards for some reason. – MPW Oct 24 '18 at 19:09
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    @mr_e_man Thank you for your remark. You might have noticed that my notation not just described an interval but a mapping of intervals, and I wanted to keep the order. – Dr. Wolfgang Hintze Oct 29 '18 at 08:06
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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 $$ 1+xy-(x+y)=(1-x)(1-y)>0. $$

Carsten S
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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 $u\geq-c$.

Joe K
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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.

Cade Reinberger
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  • The only optimisation techniques I can think of are derivatives so not too sure if this will be useful? Relating to your answer, $u'=v=c$ is a possible value but not sure how the argument follows? – user258521 Oct 24 '18 at 12:33
  • Well, you're optimizing over a square, which importantly is compact. Explicitly, first prove the problem on the boundary (that is, when at least one of $u'$ or $v$ is $c$). Then you know if there is any value above c, it must be on the interior, and thus must be a simultaneous zero of both partial derivatives. – Cade Reinberger Oct 24 '18 at 12:38