# Four force giving me headaches

1. Jul 20, 2012

### 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.

2. Jul 20, 2012

### 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.

Last edited: Jul 20, 2012
3. Jul 20, 2012

### 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} v &= c \tanh \phi \\ \gamma &= \cosh \phi \\ \gamma v &= c \sinh \phi \end{align}