# Why momenergy has magnitude equal to the mass?

The mom-energy of a particle is a 4-vector: Its magnitude is proportional to its mass, it points in the direction of the particle's spacetime displacement, and it is reckoned using the proper time for that displacement. How are these properties combined to form momenergy? Simple! Use the recipe for Newtonian momentum: mass times displacement divided by time lapse for that displacement. Instead of Newtonian displacement in space, use Einstein's displacement in spacetime; instead of Newton's "universal time," use Einstein's proper time."

Statement from: Spacetime Physics; Wheeler & Taylor; Chapter 7

From the same chapter:

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.

In the first statement, it says, "magnitude is proportional to its mass" and in the second one, it says, "magnitude equal to the mass". Why two different statements?

Proportional to the mass means,  sometimes, it can be equal to the mass. But in all cases? Can it be true? You may say, the author said,

$$\textbf{(momenergy)} = \text{(mass)} \times \frac{\textbf{(spacetime displacement)}}{\text{(proper time for that displacement)}}$$ As the value of spacetime displacement and proper time for that displacement are the same, so, "magnitude equal to the mass".

But remember, he said "magnitude equal to the mass" to establish this formula.

So, the question: Why it has magnitude equal to the mass?

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• By magnitude, do you mean dimension? If so, it's because we nondimensionalize speed viz. $c=1$.
– J.G.
Jul 18, 2021 at 14:12