- #1

Wox

- 70

- 0

[tex]

U=\gamma(c,\bar{u})

[/tex]

but I get

[tex]

U=\gamma(1,\frac{\bar{u}}{c})

[/tex]

Consider the timelike curve [itex]\bar{w}(t)=(ct,\bar{x}(t))[/itex] with velocity [itex]\bar{v}(t)=(c,\bar{x}'(t))\equiv (c,\bar{u}(t))[/itex] and the arc-length (proper time)

[tex]

\tau\colon I\subset \mathbb{R}\to \mathbb{R}\colon t\mapsto \int_{t_{0}}^{t}\left\| \bar{v}(k)\right\|dk

[/tex]

for which (by First Fundamental Theorem of Calculus (1), the Minkowskian inner product (2) and the definition of the Lorentz factor (3) )

[tex]

\Leftrightarrow \frac{d\tau}{dt}=\left\| \bar{v}(t)\right\|=\sqrt{c^{2}-\bar{u}^{2}(t)}\equiv \frac{c}{\gamma}

[/tex]

then the velocity of the curve after arc-length (proper time) parameterization, is given by

[tex]

\bar{v}(\tau)=\frac{d\bar{w}}{d\tau}=\frac{d\bar{w}}{dt}\frac{dt}{d\tau}=\frac{\bar{v}(t)}{\left\| \bar{v}(t)\right\|}=\frac{(c,\bar{u}(t))}{\frac{c}{\gamma}}=\gamma(1,\frac{\bar{u}(t)}{c})

[/tex]

I would think that my [itex]\bar{v}(\tau)[/itex] is the four-velocity but in fact [itex]\bar{v}(\tau)=\frac{U}{c}[/itex] where U the four-velocity as defined in textbooks. What am I missing?