The book that I'm following, Spacetime Physics by Taylor & Wheeler, says that,

Statement 6: The momenergy 4-vector of the particle is

$$\textbf{(momenergy)} = \text{(mass)} \times \frac{\textbf{(spacetime displacement)}}{\text{(proper time for that displacement)}}$$

Reasoning: There is no other frame-independent way to construct a 4-vector that lies along the worldline and has magnitude equal to the mass.

But where is the derivation?

(The entire contents of this book are now available under a Creative Commons license. Feel free to download, print, share, translate, and adapt, as long as you follow the (minimal) requirements of the license.)

  • $\begingroup$ Please make your question self-contained. $\endgroup$
    – my2cts
    Jul 15, 2021 at 12:53
  • $\begingroup$ Please use mainstream terms or perhaps avoid typos. $\endgroup$
    – my2cts
    Jul 15, 2021 at 12:57
  • $\begingroup$ The book is legit (see eftaylor.com/special.html), and uses precisely that term. I think it is a bit harsh to expect a newbie to jump between books (particularly from such a good one). In any case, I consider the quoted text to be self contained. I just wish I could answer it myself! $\endgroup$
    – m4r35n357
    Jul 15, 2021 at 13:12
  • $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Jul 15, 2021 at 13:24
  • $\begingroup$ Mainstream terminology is energy-momentum 4-vector or 4-momentum for short... But, akin to "space-time", it seems "momentum-energy 4-vector" or "mom-energy 4-vector" is logical, but not mainstream. Given the now more common signature conventions as +--- or -+++ using x0 (over the new less common +++- using x4), "time-space" is more logical compared to "space-time". $\endgroup$
    – robphy
    Jul 15, 2021 at 16:52

1 Answer 1


The assertion sounds a bit weird - by saying that the vector lies along the worldline and has magnitude equal to the mass, you've defined it by giving a direction and magnitude. There is no other way to construct that vector, frame-independent or not, because only one vector fulfills the condition.

Showing that the condition is fulfilled is simple. The momenergy lies along the wordline because as a vector it is proportional to the displacement, which of course is just the direction of the worldline. And you should already have seen that proper time is the "spacetime length" of displacement - so that $\textbf{(spacetime displacement)}/\text{(proper time)}$ is a vector of magnitude one. Multiplying by the mass just gets you a vector of magnitude $m$.

  • 1
    $\begingroup$ (𝐬𝐩𝐚𝐜𝐞𝐭𝐒𝐦𝐞 𝐝𝐒𝐬𝐩π₯𝐚𝐜𝐞𝐦𝐞𝐧𝐭)/(proper time) is a simple version of the "4-velocity", which plays the role of the tangent vector to the worldline (along that worldline's time-axis at that event). [In geometric units, we think of the 4-velocity as a unit-vector. But in other units, we multiply by c.] So, the momenergy 4-vector (commonly known as the 4-momentum) is the [invariant-]mass times the 4-velocity. (By the way, the spatial-component is the relativistic-[3-]momentum and the time-component is the relativistic-energy.) $\endgroup$
    – robphy
    Jul 15, 2021 at 16:39

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.