High School Theoretical Question On The Twins Paradox and Heart Rate

[Sorcerer]

What do you mean by $\gamma \textbf{u}$?

What I mean is, if you start with, say, a coordinate of some moving object, (cdt, dx), ignoring the y and z directions, of course, and then divide by proper time, written as $\frac{dt}{\gamma\ }$ , you get ($\gamma \textbf{c}$, $\gamma\ u_x$), so that second coordinate, the speed times gamma, is what I mean by $\gamma \textbf{u}$ .

So, if you multiply by invariant mass it becomes γmu, then if you take the time derivative, and then integrate over distance (after doing some manipulations with the differentials and integrating from 0 to u), you end up with a result that is identical to the relativistic kinetic energy equation.

So that's what I meant by all that. I don't know if it's mere coincidence or not, but it seems interesting to me.

 

[PeterDonis]

that second coordinate, the speed times gamma, is what I mean by $\gamma \textbf{u}$.

That's the same thing that @DrGreg means by $\gamma \textbf{v}$.

 

[PeterDonis]

then divide by proper time, written as $\frac{dt}{\gamma\ }$

I think you mean $\frac{dt}{d\tau}$. The 4-velocity vector's components are

$$
U^\mu = \left( \frac{dt}{d\tau}, \frac{d \textbf{x}}{d\tau} \right) = \left( \gamma, \gamma \textbf{v} \right)
$$

(in units where $c = 1$). Multiplying by the invariant mass $m$ then gives the 4-momentum $P^\mu$.

 

[Sorcerer]

I think you mean $\frac{dt}{d\tau}$. The 4-velocity vector's components are

$$
U^\mu = \left( \frac{dt}{d\tau}, \frac{d \textbf{x}}{d\tau} \right) = \left( \gamma, \gamma \textbf{v} \right)
$$

(in units where $c = 1$). Multiplying by the invariant mass $m$ then gives the 4-momentum $P^\mu$.

Well, as I understand it, $\frac{dt}{d\tau}$ = γ, hence proper time must be $\frac{dt}{γ}$ . So dividing dx by that will give $\frac{dx}{\frac{dt}{γ}}$ which is $γ\frac{dx}{dt}$, or $γu$ .But yeah the net result is exactly what you put there. I guess it is indeed a 4-velocity, except I'm ignoring y and z.

So why does that look the same as celerity? Is celerity exactly what the spatial components of 4-velocity are?

 

[PeterDonis]

as I understand it, dtdτdtdτ\frac{dt}{d\tau} = γ, hence proper time must be $\frac{dt}{γ}$ .

If you want to treat differentials that way (which would give many rigorous mathematicians apoplexy, although most physicists wouldn't bat an eye), then yes, you can write $d\tau = dt / \gamma$.

So dividing dx by that will give $\frac{dx}{\frac{dt}{γ}}$ which is $\gamma \frac{dx}{dt}$, or $\gamma u$ .

Yes, which is what @DrGreg was calling $\gamma \textbf{v}$, as I said before (and that's how I wrote it in my post).

So why does that look the same as celerity?

Because, as @DrGreg posted, celerity is equal to $\gamma \textbf{v}$.