# Metric tensor: semi-log coordinates

Required math: algebra

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 5; Problem 5.4.

As an example of a different coordinate system, we can define the semi-log system by introducing coordinates ${p}$ and ${q}$ defined as:

 $\displaystyle p$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (1)$ $\displaystyle q$ $\displaystyle =$ $\displaystyle e^{by} \ \ \ \ \ (2)$

where ${b}$ is a constant. If ${x}$ and ${y}$ have units of length, then ${b}$ must have units of ${\mbox{length}^{-1}}$. Note that this means that ${p}$ and ${q}$ have different units, with ${p}$ having the units of length and ${q}$ being dimensionless.

The curves of constant ${p}$ are just the same as the curves of constant ${x}$, that is, vertical lines. The curves of constant ${q}$ are defined by ${e^{by}=k}$ or ${y=\frac{\ln k}{b}}$. These are horizontal lines, although the spacing for equal steps in ${k}$ will translate into the variable spacing seen on semi-log plots.

If an object has an acceleration ${\mathbf{a}=a^{i}}$ in the rectangular system, then we can find its acceleration in the semi-log system in the usual way.

 $\displaystyle a^{p}$ $\displaystyle =$ $\displaystyle \frac{\partial p}{\partial x}a^{x}+\frac{\partial p}{\partial y}a^{y}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{x}\ \ \ \ \ (4)$ $\displaystyle a^{q}$ $\displaystyle =$ $\displaystyle \frac{\partial q}{\partial x}a^{x}+\frac{\partial q}{\partial y}a^{y}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle be^{by}a^{y} \ \ \ \ \ (6)$

The units of ${a^{p}}$ are still those of acceleration, but the units of ${a^{q}}$ are ${\mbox{length}^{-1}\cdot\mbox{acceleration}}$.

We can work out the metric of the semi-log system from the rectangular metric by direct calculation:

$\displaystyle g_{ij}^{\prime}=g_{kl}\frac{\partial x^{k}}{\partial x^{\prime i}}\frac{\partial x^{l}}{\partial x^{\prime j}}=\left[\begin{array}{cc} 1 & 0\\ 0 & \frac{1}{\left(bq\right)^{2}} \end{array}\right] \ \ \ \ \ (7)$

The length squared of ${\mathbf{a}}$ is an invariant, since

 $\displaystyle a^{2}$ $\displaystyle =$ $\displaystyle g_{ij}^{\prime}a^{\prime i}a^{\prime j}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(a^{x}\right)^{2}+\left(\frac{1}{bq}\right)^{2}\left(be^{by}a^{y}\right)^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(a^{x}\right)^{2}+\left(a^{y}\right)^{2} \ \ \ \ \ (10)$

The basis vectors in the semi-log system have lengths obtainable from the the metric:

 $\displaystyle \mathbf{e}_{p}\cdot\mathbf{e}_{p}$ $\displaystyle =g_{pp}=$ $\displaystyle 1\ \ \ \ \ (11)$ $\displaystyle \left|\mathbf{e}_{p}\right|$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (12)$ $\displaystyle \mathbf{e}_{q}\cdot\mathbf{e}_{q}$ $\displaystyle =g_{qq}=$ $\displaystyle \frac{1}{\left(bq\right)^{2}}\ \ \ \ \ (13)$ $\displaystyle \left|\mathbf{e}_{q}\right|$ $\displaystyle =$ $\displaystyle \frac{1}{bq} \ \ \ \ \ (14)$

Incidentally, the question part (e) as written in Moore’s book doesn’t make sense; he asks for the length of the basis vector ${\partial x}$. If we want to use partial derivatives as basis vectors, we need to define a curve along which to take the derivative. Since Moore doesn’t even mention the use of partial derivatives as a basis in the text, I can only assume this is a typo.