C5 - more 4-vectors

velocity

define the 4-velocity:

u^{α} = \frac{d}{d τ} x^{α}

$x^{α}$ is a known valid 4-vector, $d / d τ$ is an invariant scalar, hence $u^{α}$ is a calid 4-vector

u^{α} = γ \frac{d}{d t} x^{α} = γ (c, \vec{v})

where, $\vec{v}$ is the Newtonian velocity

this can again form an invariant scalar:

u^{α} u_{α} = γ (c, \vec{v}) \cdot γ (c, \vec{v}) = γ^{2} (c^{2} - v^{2}) = c^{2}

momentum

define the 4-momentum:

p^{α} = m u^{α} = m γ (c, \vec{v})

where, $m$ is the rest mass, an invariant scalar

p^{α} = (γ m c, γ m \vec{v}) = (\frac{γ m c^{2}}{c}, \vec{p})

this gives the notion of a relative mass $(γ m)$ , which is typically not adjusted for the frame as $m^{'}$ , but rather left as rest mass and include $γ$

force

define the 4-force:

F^{α} = \frac{d}{d τ} (p^{0}, \vec{p})

the spatial part:

F^{1, 2, 3} = \frac{d}{d τ} \vec{p} = γ \frac{d \vec{p}}{d t} = γ \vec{F}

the time-like part:

F^{0} = γ \frac{d p^{0}}{d t} = γ \frac{d}{d t} (γ m c)

if accelerating, $γ (v)$ changes
consider $u_{α} f^{α}$ , which is a scalar invariant:

u_{α} F^{α} = u_{α} \frac{d}{d τ} (m u^{α}) = \frac{m}{2} \frac{d}{d τ} (u_{α} u^{α})

$u_{α} u^{α} = c^{2}$ and $\frac{d}{d τ} (c^{2)} = 0$ , therefore:

u_{α} F^{α} = 0

u_{0} F^{0} + u_{1} F^{1} + u_{2} F^{2} + u_{3} F^{3} = 0 ‘

using the Minkowski metric:

u^{0} F^{0} - u^{1} F^{1} - u^{2} F^{2} - u^{3} F^{3} = 0

γ c F^{0} = γ v_{x} γ F_{x} + γ v_{y} γ F_{y} + γ v_{z} γ F_{z} = γ^{2} \vec{v} \cdot \vec{F}

this is the definition of power, $P :$

\vec{v} \cdot \vec{F} = P = \frac{d U}{d t}

f^{α} = γ (\frac{1}{c} \frac{d U}{d t}, \vec{F})

back to 4-momentum:

F^{0} = \frac{d}{d τ} p^{0} \frac{γ}{c} \frac{d U}{d t} = γ \frac{d}{d t} p^{0} ∴ p^{0} = \frac{U}{c}

p^{α} = (\frac{U}{c}, \vec{p})

$U$ is the relativistic energy
this give ss new invariant:

p_{α} p^{α} = \frac{U^{2}}{c^{2}} - p^{2} = constant

in the rest frame, $\vec{p} = 0$ , and $U = m c^{2} :$

U^{2} = m^{2} c^{4} + p^{2} c^{2}

and 4-acceleration:

a^{α} = \frac{d}{d τ} u^{α} = \frac{d}{d τ} (γ_{c}, γ_{\vec{v}}) = γ \frac{d}{d t} [γ (c, \vec{v})]

if $γ$ is cosntant, $a^{α} = γ^{2} (0, \vec{a})$
this applies to circular motion, where $| v |$ and $v^{2}$ are constant, thus $γ$ is constant
if $γ$ is not constant:

\begin{aligned} a^{0} & = γ \frac{d}{d t} (γ c) = \frac{c}{2} \frac{d}{d t} (γ^{2}) \\ = \frac{c}{2} \frac{d}{d t} {(1 - \frac{v^{2}}{c^{2}})}^{- 1} \\ = \frac{c}{2} (- 1) (1 - \frac{v^{2}}{c^{2}}) (- \frac{2 v}{c^{2}}) \frac{d v}{d t} \\ = γ^{4} \frac{\vec{v} \cdot \vec{a}}{c} \end{aligned}

a^{1} = γ^{2} \frac{d v_{x}}{d t} + \frac{v_{x}}{2} \frac{d}{d t} (γ^{2}) = γ^{4} a_{x} + γ^{4} \frac{\vec{v} \cdot \vec{a}}{c^{2}} v_{x}

a^{α} = (γ^{4} \frac{\vec{v} \cdot \vec{a}}{c}, γ^{2} \vec{a} + γ^{4} \frac{\vec{v}}{c^{2}} (\vec{v} \cdot \vec{a}))

consider:

u_{α} a^{α} = u_{α} \frac{d}{d τ} (u^{α}) = \frac{1}{2} \frac{d}{d τ} (u_{α} u^{α}) = \frac{d}{d τ} (c^{2}) = 0

so, the 4-acceleration is always orthogonal to the 4-velocity