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.

2 thoughts on “Metric tensor: sinusoidal coordinates

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