Is there a way to connect 4-velocity to equations for adding speeds? I know 4-velocity $U^\mu$ is derived like this:

\begin{equation} \begin{split} P^\mu &= m U^\mu \Longrightarrow U^\mu = P^\mu \frac{1}{m} = \begin{bmatrix} p_x\\ p_y\\ p_z\\ \frac{W}{c} \end{bmatrix} \frac{1}{m} = \begin{bmatrix} \frac{mv_x \gamma(v)}{m}\\ \frac{mv_y \gamma(v)}{m}\\ \frac{mv_z \gamma(v)}{m}\\ \frac{mc^2 \gamma(v)}{c m}\\ \end{bmatrix} = \begin{bmatrix} v_x \gamma(v)\\ v_y \gamma(v)\\ v_z \gamma(v)\\ c \gamma(v)\\ \end{bmatrix} \end{split} \end{equation}

Or like this:

\begin{equation} \begin{split} U^\mu = \frac{d X^\mu}{d \tau} = \frac{d}{dt}X^\mu \gamma(v)= \frac{d}{dt} \begin{bmatrix} d x\\ d y\\ d z \\ c d t \end{bmatrix} \gamma(v) = \begin{bmatrix} v_x\gamma(v)\\ v_y\gamma(v)\\ v_z\gamma(v)\\ c \gamma(v) \end{bmatrix} \end{split} \end{equation}

And i know how to derive Lorentz transformations for velocities $\perp$ to relative velocity $u$ (which is in direction of $x$, $x'$ axis) and $\parallel$ to $u$. It goes like this:

a.) $\parallel$ to $u$: \begin{equation} \begin{split} v_y' &= \frac{dy'}{dt'}=\frac{d y}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \\ &= \frac{\frac{dy}{dt}}{\gamma \left(\frac{dt}{dt} - \frac{dx}{dt} \frac{u}{c^2} \right)}\\ &\boxed{v_y' = \frac{v_y}{\gamma \left(1 - v_x \frac{u}{c^2} \right)}} \end{split} \end{equation}

b.) $\perp$ to $u$: \begin{equation} \begin{split} v_x' &= \frac{dx'}{dt'}=\frac{\gamma (d x - u d t)}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \\ &= \frac{d x - u d t}{d t - d x \frac{u}{c^2}} = \frac{\frac{d x}{d t} - u \frac{d t}{d t}}{\frac{d t}{d t} - \frac{d x}{d t} \frac{u}{c^2}}\\ &\boxed{v_x' = \frac{v_x - u}{1- v_x \frac{u}{c^2}}} \end{split} \end{equation}


Where is the connection among 4-velocity and equations derived under a.) and b.)? How can i show the connection?


1 Answer 1


Yes, from the transformation law for the four-velocity, we can explicitly derive the transformations of three-velocities parallel to and perpendicular to a given boost.

First, we need to talk about what the transformation law is for the four-velocity with respect to a boost. Let the four-velocity be $U = (U^t, U^x, U^y, U^z)$. Let's boost this along the x-direction by a speed $\mu c$. Like any four-vector, the four-velocity transforms under Lorentz transformations like so:

$$\begin{align*} {U'}^t &= W(U^t - \mu U^x) \\ {U'}^x &= W(U^x - \mu U^t) \\ {U'}^y &= U^y \\ {U'}^z &= U^z\end{align*}$$

where $W = 1/\sqrt{1-\mu^2}$ is the Lorentz factor of the boost.

Now, break down the original four-vector $U$ as $U= \gamma c(1, \beta^x, \beta^y, \beta^z)$. We can find the components of $U'$ as

$$\begin{align*} {U'}^t &= W\gamma c(1 - \mu \beta^x) \\ {U'}^x &= W \gamma c (\beta^x - \mu) \\ {U'}^y &= \gamma c \beta^y \\ {U'}^x &= \gamma c \beta^z\end{align*}$$

You can then find the components of three-velocity in the primed frame by taking ${U'}^i/{U'}^t$.

$$\begin{align*} \frac{{v'}^x}{c} &= \frac{{U'}^x}{{U'}^t} = \frac{\beta^x - \mu}{1 - \mu \beta^x} \\ \frac{{v'}^y}{c} &= \frac{{U'}^y}{{U'}^t} = \frac{\beta^y}{W(1-\mu \beta^x)} \\ \frac{{v'}^z}{c} &= \frac{{U'}^z}{{U'}^t} = \frac{\beta^z}{W(1-\mu \beta^x)} \end{align*}$$

These are algebraically the same as the formulas you posted in (a) and (b). To me, this is much simpler than churning through velocity-addition. The transformation laws of four-vectors are simple to learn, and the amount of manipulation needed to find the new three-velocity is minimal.

*Note: your terminology on parallel vs. perpendicular to the boost seems confused. Nevertheless, I believe my results here capture what you intended. The boost velocity is in the direction of the x-axis.

  • $\begingroup$ Why are the 3 velocities unitless? $\endgroup$
    – Señor O
    Commented Sep 11, 2017 at 14:58
  • $\begingroup$ @SeñorO $\beta$ and $\mu$ are dimensionless, and I choose to express the equations in terms of these dimensionless quantities for convenience. You'll notice factors of $c$ running around that wouldn't be there if you wrote the equations in terms of dimensioned quantities. $\endgroup$
    – Muphrid
    Commented Sep 15, 2017 at 19:01
  • $\begingroup$ thank you. It appears I can recover velocity in $\frac{distance}{time}$ if I multiply $v^{'i}$ by $c$? $\endgroup$
    – Señor O
    Commented Sep 15, 2017 at 21:07
  • $\begingroup$ @SeñorO Yes, you're correct, and I've corrected the answer to reflect that. $\endgroup$
    – Muphrid
    Commented Sep 16, 2017 at 3:51

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