Minkowski's equation of motion I'm trying to prove $f^{\mu}U_{\mu}=0$ for four-force $f^{\mu}=c\frac{dP^{\mu}}{ds}$ and four-velocity $U_{\mu}$. I start by using the chain rule, $f^{\mu}=c\frac{dP^{\mu}}{dt}\frac{dt}{ds}=\gamma\frac{dP^{\mu}}{dt}$ since $\frac{dt}{ds}=\frac{\gamma}{c}$. Since four momentum $P^{\mu}=(E/c, \vec{p})$, for energy $E$ and 3-momentum $\vec{F}$. By differetiating with respect to time I find $f^{\mu}=\gamma(0,\vec{F})$ for 3-force $\vec{F}$ in a particular frame. Using the fact that $P^{\mu}=mcU^{\mu}$ I then find $f^{\mu}U_{\mu}=\frac{\gamma}{mc}(0\cdot\frac{E}{c}-\vec{F}\cdot\vec{p})$ which doesnt (necessarily) give zero. Any idea where I've gone wrong here?
 A: 
I'm trying to prove $f^{\mu}U_{\mu}=0$ for four-force $f^{\mu}=c\frac{dP^{\mu}}{ds}$ and four-velocity $U_{\mu}$. I start by using the chain rule, $f^{\mu}=c\frac{dP^{\mu}}{dt}\frac{dt}{ds}=\gamma\frac{dP^{\mu}}{dt}$ since $\frac{dt}{ds}=\frac{\gamma}{c}$. Since four momentum $P^{\mu}=(E/c, \vec{p})$, for energy $E$ and 3-momentum $\vec{F}$. By differetiating with respect to time I find $f^{\mu}=\gamma(0,\vec{F})$ for 3-force $\vec{F}$ in a particular frame. Using the fact that $P^{\mu}=mcU^{\mu}$ I then find $f^{\mu}U_{\mu}=\frac{\gamma}{mc}(0\cdot\frac{E}{c}-\vec{F}\cdot\vec{p})$ which doesnt (necessarily) give zero. Any idea where I've gone wrong here?

With
$$
P^\mu = (E/c, \vec p)\;,
$$
you have to also differentiate the energy E with respect to time. (E depends on p so if p changes E changes). You find:
$$
\frac{dP^\mu}{dt} = (\frac{1}{c}\frac{\partial E}{\partial \vec p}\cdot \dot{\vec p}, \dot{\vec p})\;.
$$
But, by the very definition from Hamiltonian's equations of motion, we also have
$$
\vec v = \frac{\partial E}{\partial \vec p}\;.
$$
I use the definition:
$$
U^\mu = \gamma(c, \vec v)
$$
Thus:
$$
f^\mu U_\mu \propto  (\frac{1}{c}\vec v \cdot \dot{\vec p}, \dot{\vec p}) \cdot {(c, \vec v)}^T
$$
$$
=\vec v \cdot \dot{\vec p} - \dot{\vec p}\cdot \vec v = 0
$$
