Deriving the Lorentz Velocity Tranformation

**Jibobo** · Oct4-05, 03:21 AM

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."

**robphy** · Oct4-05, 06:58 AM

Can you show your work for your expression for w?
I think you may be missing a certain factor of $LaTeX Code: \\gamma$ (y, in your notation).

**Jibobo** · Oct4-05, 08:22 AM

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)

**robphy** · Oct4-05, 03:23 PM

Let me try to TeX this for you:

Originally Posted by Jibobo

"From B to A"
$LaTeX Code: \\tilde V_{BB}=\\left( \\begin{array}{c} c\\\\ 0 \\end{array} \\right)$

$LaTeX Code: \\tilde V_{BA}= \\left( \\begin{array}{cc} \\gamma_v & \\gamma_v\\beta_v \\\\ \\gamma_v\\beta_v & \\gamma_v \\\\ \\end{array} \\right)\\tilde V_{BB} =\\left( \\begin{array}{cc} \\gamma_v & \\gamma_v\\beta_v \\\\ \\gamma_v\\beta_v & \\gamma_v \\\\ \\end{array} \\right) \\left( \\begin{array}{c} c\\\\ 0 \\end{array} \\right)= \\left( \\begin{array}{c} \\gamma_v c\\\\ \\gamma_v\\beta_v c \\end{array} \\right)= \\left( \\begin{array}{c} \\gamma_v c\\\\ \\gamma_v v \\end{array} \\right)$

"And then from A to C"
$LaTeX Code: \\tilde V_{BC}=\\left( \\begin{array}{c} ? \\\\ w \\end{array} \\right)$

$LaTeX Code: \\tilde V_{BC}= \\left( \\begin{array}{cc} \\gamma_u & \\gamma_u\\beta_u \\\\ \\gamma_u\\beta_u & \\gamma_u \\\\ \\end{array} \\right)\\tilde V_{BA} =\\left( \\begin{array}{cc} \\gamma_u & \\gamma_u\\beta_u \\\\ \\gamma_u\\beta_u & \\gamma_u \\\\ \\end{array} \\right) \\left( \\begin{array}{c} \\gamma_v c \\\\ \\gamma_v v \\end{array} \\right)= \\left( \\begin{array}{c} ?single-quote & \\gamma_u\\beta_u \\gamma_v c + \\gamma_u\\gamma_v v \\end{array} \\right) = \\left( \\begin{array}{c} ?single-quote & \\gamma_u \\gamma_v u + \\gamma_u\\gamma_v v \\end{array} \\right) $

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

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 $LaTeX Code: \\tilde V_{BA}=\\gamma_{BA}\\left( \\begin{array}{c} c \\\\ v_{BA}\\end{array}\\right)$ , where $LaTeX Code: \\gamma_{BA}=(1-(v_{BA}/c)^2)^{-1/2}$ and $LaTeX Code: v_{BA}=v$ .

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

You want $LaTeX Code: \\tilde V_{BC}=\\gamma_{BC}\\left( \\begin{array}{c} c \\\\ v_{BC}\\end{array}\\right)$ where $LaTeX Code: v_{BC}=w$ 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
$LaTeX Code: \\tilde V_{BA}=\\gamma_{BA}\\left( \\begin{array}{c} c \\\\ v_{BA}\\end{array}\\right) =\\cosh\\theta_{BA}\\left( \\begin{array}{c} c \\\\ c\\tanh\\theta_{BA}\\end{array}\\right)=c\\left( \\begin{array}{c} \\cosh\\theta_{BA} \\\\ \\sinh\\theta_{BA}\\end{array}\\right)$ where $LaTeX Code: \\theta_{BA}=\\theta_B-\\theta_A$ .
(Relative-rapidities are additive... relative-velocities are not.)
For consistency, I probably should have written $LaTeX Code: \\theta_{BA}=\\theta_{BX}-\\theta_{AX}$ where X refers to any inertial reference frame you want to measure this.

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

So, now: we want to write
$LaTeX Code: \\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 $LaTeX Code: \\sinh\\theta_{BC}$ in terms of $LaTeX Code: \\theta_{BA}$ and $LaTeX Code: \\theta_{CA}$ . Of course, if you divide this expression by $LaTeX Code: \\cosh\\theta_{BC}$ , you get $LaTeX Code: \\tanh\\theta_{BC}$ on the left hand-side, which is $LaTeX Code: \\frac{1}{c} v_{BC}$ .

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

**Jibobo** · Oct4-05, 08:38 PM

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.