Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 ${U'}^i/{U'}^t$.

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

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.

share|improve this answer
Thank you for a really good anwser. –  71GA Dec 10 '12 at 21:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.