Why Does Momenergy Have a Magnitude Equal to Mass?

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Momenergy is defined as a 4-vector whose magnitude is equal to the mass of a particle, reflecting its spacetime displacement over proper time. This relationship stems from the need for a frame-independent definition of four-momentum, which must satisfy specific physical conditions. The magnitude being equal to mass is a result of using Einstein's concepts of spacetime and proper time instead of classical Newtonian mechanics. Taylor and Wheeler emphasize that this unique definition ensures consistency in relativistic physics. Ultimately, momenergy's magnitude aligns with mass due to its foundational role in describing particle dynamics in a relativistic framework.
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.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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