Four force giving me headaches

[GarageDweller]

I attempted to derive a fully expanded expression for the force four vector

The four velocity being --> (γ,γv) (Only considering one spatial component here)
The four acceleration is thus, (I've used κ for proper time as I cannot find the tau symbol)

(∂γ/∂κ,∂γv/∂κ) --> γ(∂γ/∂t,∂γv/∂t)

Now from here, it turns into a miasma of terms, since both v and γ and functions of the coordinate time (generally), we would have spatial velocity and acceleration mixed together and making it virtually impractical in calculations.

 

[Bill_K]

Heck, it's not all that bad. As you said, the 4-velocity is v = (γc, γv).

The 4-acceleration is a = dv/dτ = (cγ dγ/dt, γ²a + γ dγ/dt v)
Use dγ/dt = γ³ (v·a)/c², so you get a = (γ⁴(v·a)/c, γ²a + γ⁴(v·a)/c² v)

The 4-force, of course, is m times the 4-acceleration.

 

[DrGreg]

GarageDweller, if you are familiar with hyperbolic functions, how to differentiate them, and the hyperbolic versions of the trig identities, an easy approach (for the one-spatial-dimension case), is to substitute$$\begin{align}<br /> v &= c \tanh \phi \\<br /> \gamma &= \cosh \phi \\<br /> \gamma v &= c \sinh \phi<br /> \end{align}$$