Why Does Momenergy Have a Magnitude Equal to Mass?

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SUMMARY

The discussion centers on the concept of momenergy, defined as a 4-vector whose magnitude is equal to the mass of a particle. According to Taylor and Wheeler in "Spacetime Physics," the momenergy is calculated using the formula: (mass) × (spacetime displacement) / (proper time for that displacement). The distinction between "proportional to mass" and "equal to mass" is clarified, emphasizing that while the magnitude can vary, it is fundamentally defined to be equal to mass under specific conditions. This unique definition ensures that the four-momentum retains its frame-independent properties.

PREREQUISITES
  • Understanding of 4-vectors in physics
  • Familiarity with spacetime concepts from Einstein's theory of relativity
  • Knowledge of Newtonian momentum
  • Basic grasp of proper time and its significance in relativistic physics
NEXT STEPS
  • Study the derivation of 4-vectors in relativistic physics
  • Explore the implications of spacetime displacement on particle dynamics
  • Investigate the relationship between mass and energy in the context of relativity
  • Review Chapter 7 of "Spacetime Physics" by Taylor and Wheeler for deeper insights
USEFUL FOR

Students and professionals in physics, particularly those focused on relativity, theoretical physicists, and anyone interested in the mathematical foundations of momentum and energy in spacetime.

Ashshahril
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Homework Statement
Why momenergy has magnitude equal to the mass?
Relevant Equations
$$\textbf{(momenergy)} = \text{(mass)} \times \frac{\textbf{(spacetime displacement)}}{\text{(proper time for that displacement)}}$$
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](https://www.eftaylor.com/spacetimephysics/07_chapter7.pdf)

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?

You can find the full contents of the book from [here](https://www.eftaylor.com/spacetimephysics/). Its under creative commons license.
 
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Because this is the relativistic definition of mass - the magnitude of the 4-momentum (which is the usual modern name for what you call momenergy).
 
Ashshahril said:
So, the question: Why it has magnitude equal to the mass?
Taylor and Wheeler note that it makes physical sense for the four-momentum to be proportional to the mass, and they give the simplest definition that satisfies the various conditions they want for the four-momentum. Once you have this definition, then you find that the magnitude of the four-momentum is not only proportional to the mass but is actually equal to the mass. In the reasoning for statement 6, they're saying there is no other way to define the four-momentum in a frame-independent way so it will have these properties. The definition is unique.
 

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