# Identify the 4-force

## Homework Statement

Use the 4-dimensional version of Newton's 2nd law to identify the 4-force as $$f^a=\gamma(\mathbf{u}\cdot\mathbf{F},\mathbf{F})$$, where F is the force acting on the particle.

## Homework Equations

$$f^a=\frac{dp^a}{d\tau}$$$$p^a=m_ou^a$$$$u^a=\gamma(1,\mathbf{u})$$

## The Attempt at a Solution

I define the acceleration vector as $$\mathbf{a}=(a_1,a_2,a_3)=(\frac{du_1}{d\tau},\frac{du_2}{d\tau},\frac{du_3}{d\tau})$$
and the force vector as $\mathbf{F}=(F_1,F_2,F_3)=(m_oa_1,m_oa_2,m_oa_3)$.
$$f^a=m_o(\frac{d\gamma}{d\tau},\frac{d({\gamma}u_1)}{d\tau},\frac{d({\gamma}u_2)}{d\tau},\frac{d({\gamma}u_3)}{d\tau})$$$$\frac{d\gamma}{d\tau}=-\frac{1}{2}(1-u^2)^{-\frac{3}{2}}\frac{d(1-u^2)}{d\tau}$$I substitute in $u_1^2+u_2^2+u_3^2$ for $u^2$ and $\gamma^3$ for $(1-u^2)^{-\frac{3}{2}}$.$$\frac{1}{2}\gamma^3(\frac{du_1^2}{d\tau}+\frac{du_2^2}{d\tau}+\frac{du_2^2}{d\tau})=\gamma^3(u_1a_1+u_2a_2+u_3a_3)$$
Here, I use the product rule since both $\gamma$ and $u_n$ are functions of $\tau$.$$f^a=m_o(\frac{d\gamma}{d\tau},u_1\frac{d\gamma}{d{\tau}}+\gamma\frac{du_1}{d{\tau}},u_2\frac{d\gamma}{d{\tau}}+\gamma\frac{du_2}{d\tau},u_3\frac{d{\gamma}}{d{\tau}}+\gamma\frac{du_3}{d\tau})$$$$f^a=m_o(\gamma^3\mathbf{u}\cdot\mathbf{a},\gamma^3u_1\mathbf{u}\cdot\mathbf{a}+{\gamma}a_1,\gamma^3u_2\mathbf{u}\cdot\mathbf{a}+{\gamma}a_2,\gamma^3u_3\mathbf{u}\cdot\mathbf{a}+{\gamma}a_3)$$$$f^a=\gamma(\gamma^2\mathbf{u}\cdot\mathbf{F},{\gamma}^2u_1\mathbf{u}\cdot\mathbf{F}+F_1,\gamma^2u_2\mathbf{u}\cdot\mathbf{F}+F_2,\gamma^2u_3\mathbf{u}\cdot\mathbf{F}+F_3)$$So I've got 2 things that make my solution different from $f^a=\gamma(\mathbf{u}\cdot\mathbf{F},\mathbf{F})$.
One, I've got this $\gamma^3$ term that is not present in the book's solution.
Two, I've got the $\gamma^2u_n\mathbf{u}\cdot\mathbf{F}$ term in each of the space coordinates that the book's solution does not have.
Does anyone see where I'm messing up?