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.

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