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.

3 thoughts on “Metric tensor: semi-log coordinates

  1. Pingback: Metric tensor: spherical coordinates | Physics tutorials

  2. Pingback: Covariant derivative in semi-log coordinates | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *