Four-acceleration

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 4; Problems 4.11.

The four-velocity ${\mathbf{u}}$ has the property that, for any object with a non-zero rest mass, ${\mathbf{u}\cdot\mathbf{u}=-1}$ (For a photon, four-velocity cannot be defined.). If we define the four-acceleration as

$\displaystyle \mathbf{a}\equiv\frac{d\mathbf{u}}{d\tau} \ \ \ \ \ (1)$

then if we take the derivative of the square of the four-velocity, we get

 $\displaystyle \frac{d\left(\mathbf{u}\cdot\mathbf{u}\right)}{d\tau}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle 2\mathbf{u}\cdot\frac{d\mathbf{u}}{d\tau}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \mathbf{u}\cdot\mathbf{a}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

Thus for any object with a non-zero rest mass, the four-velocity is orthogonal to the four-acceleration.