# Derive the velocity-additon formula from the Lorentz transformation

In a Euclidian world the sum $s$ of two velocities $v$ and $u$ is so such that $s = v + u$. However, in the world of special relativity that's not the case. Instead, the velcity vector sum $s$ is such that $s = \frac{v + u}{1+vu} \; (c=1)$.

I'm trying to deriving it and down below is my work so far. Any hints that would lead me forwards would be great.

Consider two inertial reference frames $S$ ("the rest frame") and $S'$ ("the moving frame"). In S' we have a 4-velocity vector $\left(c \frac{d {x^0}'}{d \tau}, \frac{d {x^1}'}{d \tau}, \frac{d {x^2}'}{d \tau}, \frac{d {x^3}'}{d \tau}\right)$.

I'm letting ${x^\mu}$ denote $x^\mu$ coordinates in $S$ and ${x^\mu}'$ denote $x^\mu$ coordinates in $S'$.

During $d \tau$, a particle with the velocity $\left(c \frac{d {x^0}'}{d \tau}, \frac{d {x^1}'}{d \tau}, \frac{d {x^2}'}{d \tau}, \frac{d {x^3}'}{d \tau}\right)$ travels from $(0, 0, 0, 0)$ to $\left(c d {x^0}', d {x^1}', d {x^2}', d {x^3}'\right)$.

To simplify the problem we're now going to assume that the relative velocity of $S'$ (the moving frame") is only in the x-direction, and the same is assumed about the 4-velocity vector in $S'$. Thus it simplifies into the event $\left(c d {x^0}', d {x^1}', 0, 0 \right)$.

Now we transform the ${x^0}'$ term into $S$:

$x^0 = \gamma(d {x^0}' - v d {x^1}')$

And the same for the ${x^1}'$ term:

$x^1 = \gamma (d {x^1}' - v d {x^0}')$

So therefore, in $S$, the event is given by $(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$

Now, here's my problem. In order to write an expression for the change in distance over time (that's the velocity) in $S$ coordinates, I'll need to know how to transform $d \tau$ into $S$. I already know ${x^1}$ in $S$ in terms of $S'$ coordinates, but I don't know how ${x^0}$ relates to $d \tau$.

If I knew that, I could simply take $\frac{d {x^1}}{d {x^0}}$ and get an expression for the velocity in $S$.

In other words, I need to know the coordinate of $d \tau$ in $S$.

If I was unclear about something, please don't downvote but leave a comment and ask instead and I'll correct any mistakes or unclear formulations.

• In a Galilean world velocity is additive. – Ryan Unger Aug 19 '15 at 23:38
• There's a really simple way to do this if you think of the Lorentz transformation as a rotation but use hyperbolic trig functions instead of regular $\sin$ and $\cos$. If you do this you find that composing two Lorentz transformations is equivalent to adding the angles associated to each one! – DanielSank Aug 20 '15 at 1:23
• Another way of describing @DanielSank's comment: compose two collinear boosts, by multiplying their matrices. Now find the single velocity parameter that describes the matrix product. You can do this with the matrices in velocity and $\gamma(v)$ form, or you can do it with them as hyperbolic "rotation" matrices with angle (rapidity) $\eta = \mathrm{artanh}(v/c)\Leftrightarrow v = c \tanh\eta$. Now use the hyperbolic angle summation formula $\tanh(\eta_1+\eta_2) = (\tanh\eta_1+\tanh\eta_2)/(1+\tanh\eta_1\,\tanh\eta_2)$. – WetSavannaAnimal Aug 20 '15 at 1:42
• Well asked homework question, BTW. Exactly conforms to the rules. – WetSavannaAnimal Aug 20 '15 at 1:44

Since all the velocities are constant working out $\frac{dx^{1}}{dx^{0}}$ is simple division (no calculus required). Note that the entries in your vector in $S$ are the values you're interested in: $$(dx^0, dx^1, 0, 0)=(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$$.

So we have:

$$\frac{dx^{1}}{dx^{0}} = \frac{d {x^1}' - v d {x^0}'}{d {x^0}' - v d {x^1}'}$$

Dividing the numerator and denominator by $d {x^0}'$ we have: \begin{align} \frac{dx^{1}}{dx^{0}} &= \frac{\frac{d {x^1}'}{d {x^0}'} - v }{1 - v\frac{ d {x^1}'}{d {x^0}'}}\\ &= \frac{u - v }{1 - vu}\\ \end{align}

Where the $-$ signs different to the standard formula is because your two velocities are in the same direction.

You'd pretty much solved the problem yourself, you just hadn't realised it!

• Thank you! Let me just ask you two questions: why don't I need to know d𝛕 in terms of S coordinates? Why are the signs negative instead of positive? – Madde Anerson Aug 20 '15 at 10:49
• @MaddeAnerson For the minus signs its probably easier to think about the Galilean version. If I'm going at $u\text{ms}^{\text{-1}}$ in the $x$ direction relative to you and you see something going at $v\text{ms}^{\text{-1}}$ in the same direction as me then I see it as going at $(v-u)\text{ms}^{\text{-1}}$. If on the other hand you see the other object going at $v\text{ms}^{\text{-1}}$ in the $-x$ direction then I'll see it as going at $(u+v)\text{ms}^{\text{-1}}$ in that direction. The $-$ signs just relate to the directions of the velocities. – or1426 Aug 20 '15 at 10:57
• @MaddeAnerson: What we're actually trying to do here is take our knowledge of how $S^\prime$ is moving relative to $S$ and how the object in question is moving relative to $S^\prime$ and find an expression for how the object is moving relative to $S$ in terms of that information. To repeat myself we have to write things in terms of the primed coordinates (and $v$) because those are the bits of information we start with. You can derive the formula in a similar way if you start by knowing two (4)velocities in frame $S$ and try to work out how one of them appears from the frame of the other. – or1426 Aug 20 '15 at 11:01

If you have two vectors you can project one onto the other. It you orthogonally project a spacetime vector onto your unit tangent then the length of that projection is how much time you observe separating the two events. That might be enough to answer your question right there.

So to get an average velocity between two events first compute the vector A between the two events. Then project that A onto your unit tangent W to get the projection P. Then the vector A equals P+R with the rejection R orthogonal to the projection P. We get R from R=A-P. And if we divide R by the length of P we get the velocity of A relative to W. All I did was write everything in terms of the geometry so that the basis didn't matter. Let's do it in a nice basis so we see what is going on.

Let W=(1,0,0,0) that's the basis that makes things look nice, the comoving frame of W. So if the difference between the two events is A=(a,b,c,d) then the projection is (a,0,0,0) so subtracting it gives (0,b,c,d) and since the time between them is $a=\sqrt{(a,0,0,0)\cdot(a,0,0,0)}$ we get the velocity is (0,b,c,d)/a. which has a speed equal to its length $\sqrt{(b^2+c^2+d^2)/a^2}.$

So if the vector is (a,b,c,d) and your unit tangent is (1,0,0,0) then the projection is (a,0,0,0) so subtracting it gives (0,b,c,d) and so the velocity you observe is (0,b,c,d)/a. This might seem painfully easy. But the same applies to someone moving like (1,-v,0,0) they observe you moving at speed $v$ and you observe someone moving like (1,u,0,0) as moving at speed u. So how does the one of the left see the one on the right? Project, subtract and divide (in geometric algebra this is basically one step).

First step: project the right onto the left. (1,u,0,0)$\cdot$(1,-v,0,0) (1,-v,0,0)/$(1-v^2)$ (that's $\vec a\cdot \vec b \vec b / (\vec b\cdot \vec b)$ to project $\vec a$ onto $\vec b$). So we get $\frac{1+uv}{1-v^2}$(1,-v,0,0).

Next step: subtract that projection. (1,u,0,0)-$\frac{1+uv}{1-v^2}$(1,-v,0,0) =

($\frac{1-v^2-1-uv}{1-v^2}$,$\frac{u-v^2u+v+v^2u}{1-v^2}$,0,0)

= ($\frac{-v^2-uv}{1-v^2}$,$\frac{u+v}{1-v^2}$,0,0).

This is a vector orthogonal to (1,-v,0,0), just like (0,b,c,d) was orthogonal to (1,0,0,0).

Let's check(1,-v,0,0)$\cdot$($\frac{-v^2-uv}{1-v^2}$,$\frac{u+v}{1-v^2}$,0,0)=$\frac{-v^2-uv}{1-v^2}$-(-v)$\frac{u+v}{1-v^2}$=0. Check.

The person on the left thinks the person on the right traveled P=$\frac{1+uv}{1-v^2}$(1,-v,0,0) in time (which to the person on the left is a pure time direction) and traveled a spatial displacement ($\frac{-v^2-uv}{1-v^2}$,$\frac{u+v}{1-v^2}$,0,0) (which to the person on the left is a pure spatial direction.) We could check that they add up to the full spacetime displacement, but we did get one from subtracting the other. And we know the final answer so we don't need check our work all the time.

So let's find the length of that time displacement (this is finding a). We want the length of P=$\frac{1+uv}{1-v^2}$(1,-v,0,0) which is $\frac{1+uv}{\sqrt{1-v^2}}.$ So we divide by that to get the relative velocity vector.

Final step: divide. The relative velocity is $\frac{\sqrt{1-v^2}}{1+uv}(\frac{-v^2-uv}{1-v^2}$,$\frac{u+v}{1-v^2}$,0,0).

Simplify: $\frac{\sqrt{1-v^2}}{1+uv}(\frac{-v^2-uv}{1-v^2}$,$\frac{u+v}{1-v^2}$,0,0) =

$\frac{u+v}{(1+uv)\sqrt{1-v^2}}$(-v,1,0,0) which has length (speed) (u+v)/(1+uv).

• Some attention to math formatting would make this easier to read. Note that using double dollar signs centers the math and makes it bigger. I also find inline fractions with / much easier to read than the way it's done here. – DanielSank Aug 20 '15 at 4:27