# Deriving the Lorentz Velocity Tranformation

1. Oct 4, 2005

### Jibobo

Most textbooks derive this formula by taking the derivative of the Lorentz position transformation. However, I've been presented the problem of deriving it using four velocities and the Lorentz transformation matrix. Unfortunately I've been looking at it for hours and still don't know how to make it come out right.

The exact question:
"In A's frame, B moves to the right with speed v, and C moves to the left with speed u. What is the speed of B, w, with respect to C? In other words, use 4-vectors and the LT to derive the velocity addition formula

w = u + v / (1 + uv/c^2)

More specifically, work only with the four velocities and ask how does v look from C's point of view."

Assigning a velocity vector v = d/dt (ct, x, y, z) = (c, 0, 0, 0) to B in its own frame, I've tried transforming the velocity of B to A's frame, and then transforming that velocity to C's frame, which gives me
w = y_u * y_v * (u + v)
where y_u is 1/(1 - (u/c)^2)^0.5 and y_v is 1/(1 - (v/c)^2)^0.5. This doesn't seem to simplify into the Lorentz velocity transformation.

What's wrong with my logic and what should I do to fix it?

I also considered finding x' and t' and simply dividing x' by t', which does give me the correct answer, but I don't believe that's what they want me to do, since they say specifically to "work only with four velocities."

2. Oct 4, 2005

### robphy

I think you may be missing a certain factor of $$\gamma$$ (y, in your notation).

3. Oct 4, 2005

### Jibobo

Since the y and z components are 0, I'm going to work with truncated 4-vectors.

y_v = 1 / (1 - (v/c)^2)^0.5
y_u = 1 / (1 - (u/c)^2)^0.5
B_v = v/c
B_u = u/c

My work:
From B to A
V_B in B = V_BB = (c, 0)
V_BA = (y_v........B_v*y_v * V_BB = (y_v*c, B_v*y_v*c) = (y_v*c, y_v*v)
............B_v*y_v........y_v)

And then from A to C
V_BC = (something, w)
V_BC = (y_u........B_u*y_u * V_BA = (something, B_u*y_u*y_v*c + y_u*y_v*v)
............B_u*y_u........y_u)
= (something, y_u*y_v*u, y_u*y_v*v)

So I end up with w = y_u*y_v*(u + v)

Last edited: Oct 4, 2005
4. Oct 4, 2005

### robphy

Let me try to TeX this for you:

So, it looks like you essentially transformed a 4-velocity at rest in B to one with relative-velocity v (representing A). Then you transformed this 4-velocity again with relative-velocity u (representing C). You then tried to equate the spatial components.

Admittedly, it "kinda" looks close. But I think there are some inconsistencies in notation and signs.

Allow me to change notation to improve the bookkeeping.

B's 4-velocity in A's reference frame has the form $$\tilde V_{BA}=\gamma_{BA}\left( \begin{array}{c} c \\ v_{BA}\end{array}\right)$$, where $$\gamma_{BA}=(1-(v_{BA}/c)^2)^{-1/2}$$ and $$v_{BA}=v$$.

C's 4-velocity in A's reference frame has the form $$\tilde V_{CA}=\gamma_{CA}\left( \begin{array}{c} c \\ v_{CA}\end{array}\right)$$, where $$\gamma_{CA}=(1-(v_{CA}/c)^2)^{-1/2}$$ and $$v_{CA}=-u$$.

You want $$\tilde V_{BC}=\gamma_{BC}\left( \begin{array}{c} c \\ v_{BC}\end{array}\right)$$ where $$v_{BC}=w$$ is the spatial relative-velocity of B in C's reference frame. [In your last post, you are missing this gamma factor.]

In order to make the calculation a little more transparent, let me introduce the rapidities so that
$$\tilde V_{BA}=\gamma_{BA}\left( \begin{array}{c} c \\ v_{BA}\end{array}\right) =\cosh\theta_{BA}\left( \begin{array}{c} c \\ c\tanh\theta_{BA}\end{array}\right)=c\left( \begin{array}{c} \cosh\theta_{BA} \\ \sinh\theta_{BA}\end{array}\right)$$ where $$\theta_{BA}=\theta_B-\theta_A$$.
(Relative-rapidities are additive... relative-velocities are not.)
For consistency, I probably should have written $$\theta_{BA}=\theta_{BX}-\theta_{AX}$$ where X refers to any inertial reference frame you want to measure this.

Similarly, $$\tilde V_{CA}=c\left( \begin{array}{c} \cosh\theta_{CA} \\ \sinh\theta_{CA}\end{array}\right)$$ and $$\tilde V_{BC}=c\left( \begin{array}{c} \cosh\theta_{BC} \\ \sinh\theta_{BC}\end{array}\right)$$.

So, now: we want to write
\begin{align*} \tilde V_{BC} &= \Lambda_{CA} \tilde V_{BA}\\ c\left( \begin{array}{c} \cosh\theta_{BC}\\ \sinh\theta_{BC} \end{array}\right) &=\left( \begin{array}{cc} \cosh\theta_{CA} & -\sinh\theta_{CA} \\ -\sinh\theta_{CA} & \cosh\theta_{CA} \\ \end{array} \right) c \left( \begin{array}{c} \cosh\theta_{BA} \\ \sinh\theta_{BA} \end{array}\right)\\ \end{align*}

I think this looks notationally consistent. (Someone check!)
I'll let you carry out the matrix multiplications to obtain an expression for $$\sinh\theta_{BC}$$ in terms of $$\theta_{BA}$$ and $$\theta_{CA}$$. Of course, if you divide this expression by $$\cosh\theta_{BC}$$, you get $$\tanh\theta_{BC}$$ on the left hand-side, which is $$\frac{1}{c} v_{BC}$$.

When working with rapidities, one manipulates somewhat familiar trigonometric functions and their identities (and their geometric interpretation) rather than relatively obscure identities involving $$\gamma$$ and $$\beta$$ "factors".

5. Oct 4, 2005

### Jibobo

Thanks a lot for the help. I am actually completely unfamiliar with rapidities and have done very little with hyperbolic trig functions, but I was still able to solve it through some terrible algebra after you told me that I was missing the y_BC term.