# Metric tensor: sinusoidal coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 5; Problem 5.5.

As another example of a different coordinate system, we can define a sinusoidal system by introducing coordinates ${u}$ and ${w}$ defined as:

 $\displaystyle u$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (1)$ $\displaystyle w$ $\displaystyle =$ $\displaystyle y-A\sin\left(bx\right) \ \ \ \ \ (2)$

where ${A}$ and ${b}$ are constants. If ${x}$ and ${y}$ have units of length, then ${b}$ must have units of ${\mbox{length}^{-1}}$. If ${A}$ has units of length, then ${u}$ and ${w}$ both have units of length.

The curves of constant ${u}$ are just the same as the curves of constant ${x}$, that is, vertical lines. The curves of constant ${w}$ are defined by ${y=k+A\sin\left(bx\right)}$. These are parallel horizontal sine curves displaced vertically by the constant ${k}$.

The inverted relations are

 $\displaystyle x$ $\displaystyle =$ $\displaystyle u\ \ \ \ \ (3)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle w+A\sin\left(bu\right) \ \ \ \ \ (4)$

We can work out the metric of the sinusoidal system from the rectangular metric by direct calculation

$\displaystyle g_{ij}^{\prime}=g_{kl}\frac{\partial x^{k}}{\partial x^{\prime i}}\frac{\partial x^{l}}{\partial x^{\prime j}}=\left[\begin{array}{cc} 1+\left[Ab\cos\left(bu\right)\right]^{2} & Ab\cos\left(bu\right)\\ Ab\cos\left(bu\right) & 1 \end{array}\right] \ \ \ \ \ (5)$

Incidentally, we can check this by calculating the reverse transformation metric:

$\displaystyle g_{ij}=g_{kl}\frac{\partial x^{\prime k}}{\partial x^{i}}\frac{\partial x^{\prime l}}{\partial x^{j}} \ \ \ \ \ (6)$

For example (using ${x=u}$):

 $\displaystyle g_{xx}$ $\displaystyle =$ $\displaystyle 1+\left[Ab\cos\left(bx\right)\right]^{2}-\left[Ab\cos\left(bx\right)\right]^{2}-\left[Ab\cos\left(bx\right)\right]^{2}+\left[Ab\cos\left(bx\right)\right]^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (8)$

Similar calculations give the other components of the rectangular metric.

Now suppose we have an object moving with velocity ${\mathbf{v}}$ such that ${v^{x}=v}$ and ${v^{y}=0}$. In the sinusoidal system, we get

 $\displaystyle v^{u}$ $\displaystyle =$ $\displaystyle \frac{\partial u}{\partial x^{i}}v^{i}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle v\ \ \ \ \ (10)$ $\displaystyle v^{w}$ $\displaystyle =$ $\displaystyle \frac{\partial w}{\partial x^{i}}v^{i}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -Ab\cos\left(bu\right)v\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -Ab\cos\left(bvt\right)v \ \ \ \ \ (13)$

where the last line arises because ${u=x=vt}$, where ${t}$ is the time.

The square is

 $\displaystyle v^{2}$ $\displaystyle =$ $\displaystyle g_{ij}^{\prime}v^{\prime i}v^{\prime j}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+\left[Ab\cos\left(bu\right)\right]^{2}\right)v^{2}-\left[Ab\cos\left(bu\right)\right]^{2}v^{2}-\left[Ab\cos\left(bu\right)\right]^{2}v^{2}+\left[Ab\cos\left(bu\right)\right]^{2}v^{2}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle v^{2} \ \ \ \ \ (16)$

Thus the square of the velocity is invariant.

The unit vector ${\mathbf{e}_{u}}$ is tangent to the constant-${w}$ curves and points in the direction of increasing ${u}$. Since the constant-${w}$ curves are sine curves, this unit vector starts off horizontal (when ${bu=\pi/2}$), then slopes downward as ${bu}$ heads towards ${3\pi/2}$ at which point it is horizontal again. Then it slopes upwards as ${bu}$ heads towards ${5\pi/2}$ and so on. Its magnitude is given by ${\left|\mathbf{e}_{u}\right|=\sqrt{g_{uu}}=\sqrt{1+\left[Ab\cos\left(bu\right)\right]^{2}}}$.

The other unit vector ${\mathbf{e}_{w}}$ is tangent to the constant-${u}$ curves, so it always points up, and always has a length of 1. Note that the off-diagonal elements of ${g_{ij}^{\prime}}$ are zero when ${bu}$ is an odd multiple of ${\pi/2}$, which is where ${\mathbf{e}_{u}}$ is horizontal, and thus perpendicular to ${\mathbf{e}_{w}}$, so ${g_{uw}^{\prime}=g_{wu}^{\prime}=\mathbf{e}_{u}\cdot\mathbf{e}_{w}=0}$.

The reason that ${v^{w}}$ is not constant, even though the velocity itself is constant is because the unit vector ${\mathbf{e}_{u}}$ oscillates in both magnitude and direction. Since ${\mathbf{e}_{w}}$ is a constant, the component multiplying this unit vector (that is, ${v^{w}}$) must also oscillate to compensate for the non-constant ${u}$ component.

In the rectangular system, the acceleration is zero (since ${v}$ is constant). In the sinusoidal system, ${dv^{u}/dt=0}$, but ${dv^{w}/dt\ne0}$. Thus the magnitude of ${a^{2}}$ would not be invariant under the transformation of coordinates, so this cannot be the correct way of calculating derivatives in a general coordinate system.