Transform Coordinates for Torus Metric in Wald

[ergospherical]

I can't figure out how to transform the coordinates to get to the given metric \begin{align*}ds^2 = \cos x (dy^2 - dx^2) + 2\sin x dx dy \end{align*} for a 2-torus. Initially I parameterised it by two angles $\theta$ (around the $z$ axis) and $\phi$ (around the torus axis), to write $\mathbf{r} = ((R + r\cos \phi) \cos \theta, (R + r\cos \phi) \sin \theta, r\sin \phi)$ in cartesians, then
\begin{align*}
d\mathbf{r} &= \begin{pmatrix} -(R + r\cos \phi) \sin \theta d\theta - r\sin \phi \cos \theta d\phi \\
(R + r\cos \phi) \cos \theta d\theta - r\sin \phi \sin \theta d\phi \\
r \cos \phi d\phi
\end{pmatrix} \\ \\
ds^2 &= (R + r\cos \phi)^2 d\theta^2 + r^2 d\phi^2
\end{align*}a little hint would be appreciated?

 

[PAllen]

Just a quick comment you presumably realize - your initial coordinates are orthogonal. The given ones are not, so you presumably want to look at one coordinate ‘line’ spiraling around the torus.

 

[George Jones]

I can't figure out how to transform the coordinates to get to the given metric \begin{align*}ds^2 = \cos x (dy^2 - dx^2) + 2\sin x dx dy \end{align*} for a 2-torus. Initially I parameterised it by two angles $\theta$ (around the $z$ axis) and $\phi$ (around the torus axis), to write $\mathbf{r} = ((R + r\cos \phi) \cos \theta, (R + r\cos \phi) \sin \theta, r\sin \phi)$ in cartesians, then
\begin{align*}
d\mathbf{r} &= \begin{pmatrix} -(R + r\cos \phi) \sin \theta d\theta - r\sin \phi \cos \theta d\phi \\
(R + r\cos \phi) \cos \theta d\theta - r\sin \phi \sin \theta d\phi \\
r \cos \phi d\phi
\end{pmatrix} \\ \\
ds^2 &= (R + r\cos \phi)^2 d\theta^2 + r^2 d\phi^2
\end{align*}a little hint would be appreciated?

Two preliminary comments: 1) The is not a B thread; 2) you have left out important important information. It would have been very useful if you had typed the page #, p. 242. I found the page number by using the Look Inside feature of Amazon.

The question starts with "Let $M$ be the torus $\left(S^1 \times S^1 \right)$ and define the Lorentz metric $g_{ab}$ by (Misner 1963)". Note "Lorentz metric". Later in the question "Show that the closed curves defined by $x = \pi/2$ and $x = 3\pi/2$ are null geodesics". Take the first curve, $x = \pi/2$ is constant and $y$ varies. Then $dx=0$ and $\cos x = 0$. What happens in Wald's metric? What happens in you metric? Your metric is not Lorentz, i.e., it is positive-definite. Also, I think that Wald has chosen $r$ and $R$.

 

[berkeman]

1) The is not a B thread;

Thread prefix changed to "I" for now.

 

[ergospherical]

Yeah, I see the issue with my original approach now. How should I try to approach deriving this metric?

 

[PAllen]

Yeah, I see the issue with my original approach now. How should I try to approach deriving this metric?

I’m not sure what you mean by derive. Before @George Jones posted, I just looked at your metric (the induced metric for the standard smooth embedding of a 2 torus in Euclidean 3 space) and noticed an obvious difference from the given metric. However, in the general case, you can’t derive a metric for a given topology. There are uncountably infinite different geometries, not connected by coordinate transform, that can be imposed on any chosen topology. For example, you can impose a flat Minkowski 2-metric on a 2-torus (with appropriate boundary conditions). Then there could not possibly be a coordinate transform to the given metric, because the geometry is different. So I am not sure there is anything you can do except motivate the given metric by exploration of its properties, showing it is in some way a ‘natural’ geometry for a Minkowskian 2-torus.

 

[Orodruin]

showing it is in some way a ‘natural’ geometry for a Minkowskian 2-torus.

I am not sure it is very ”natural” in any sense. If I am not fooling myself, it looks like an example of a manifold that is not time-orientable (nor space-orientable).

 

[ergospherical]

Ok - thanks. I was trying to motivate the metric to get a better understanding of what the angular coordinates $x$ and $y$ refer to. We have $g_{yy} = -g_{xx} = \cos{x}$ and $g_{xy} = \sin{x}$. Let me begin by considering the curve $C: y \rightarrow (\frac{\pi}{2}, y)$ parameterised by $y$. Then put $T^a = \frac{\partial x^a}{\partial y} = \delta^a_y$ and write $T_a T^a \big{|}_C = g_{ab} \delta^a_y \delta^b _y \big{|}_C = g_{yy} \big{|}_C = \cos{x} \big{|}_C = 0$, therefore $C$ is null. To check if it is also geodesic,\begin{align*}
T^a \nabla_a T^b = T^a (\partial_a T^b + \Gamma^b_{ac} T^c) = \Gamma^b_{yy} = -\frac{1}{2} g^{bm} \partial_m g_{yy} = \frac{1}{2}g^{bx} \sin{x}
\end{align*}If I'm not mistaken then in fact $|g| = -1$ so $g^{-1} = g$, therefore $T^a \nabla_a T^x \big{|}_C= -\frac{1}{2} \sin{x} \cos{x} \big{|}_C = 0$ whilst $T^a \nabla_a T^y \big{|}_C= \frac{1}{2} \sin^2{x} \big{|}_C = \frac{1}{2} = \frac{1}{2} \delta^y_y$, so the geodesic condition $T^a \nabla_a T^b = \alpha T^b$ does indeed hold with value $\alpha = \frac{1}{2}$. Then I can find an affine parameter $\lambda$ by integrating\begin{align*}
\frac{d\lambda}{dy} = \mathrm{exp} \int \frac{1}{2} dy = e^{\frac{1}{2} y}
\end{align*}which has the 1-parameter family of solutions $\lambda = 2e^{\frac{1}{2} y} + k$. It's then easy to check that $\frac{d\lambda}{dy} = \frac{\lambda}{2} + k'$, and by choosing $k' = 0$ obtain a parameterisation such that $\frac{1}{2} \lambda \frac{dy}{d\lambda} = 1$.

 

[PAllen]

An attempt at an intuitive description of the metric:

Consider an ordinary donut, and call the two angular coordinates "around the donut" and "around the hole", these being orthogonal. Then, for the Wald torus, I would say the x coordinate corresponds roughly to an around the donut angle, and has qualitative features of a closed circle in the x-t plane of Minkowski space (a path of constant y varies periodically between being spacelike, null, and timelike). The y coordinate is then an angular measure "around the hole", but spiralling relative to the constant y curves (this is implied by the non-orgogonality of the metric). Each of these closed spirals (constant x curves) around the the hole has fixed causal classification: closed timelike, closed spacelike, or closed null curve.

This (existence of closed timelike and closed null curves) makes it clear that this torus cannot be embedded in Minkowski 4-space. I thought about the possibility that it could be embedded in the Godel spacetime, but reached no conclusion. At first I thought this was precluded by the fact that constant x closed timelike curves were geodesics, while closed timelike curves in Godel spacetime are not geodesics. However, this, by itself is not definitive - a curve can easily be a geodesic of a submanifold of some manifold, without being a geodesic of the overall manifold. This is true, for example, in the simple case of geodesics of constant cosmological time slices in typical FLRW solutions. The geodesics of the slice are not geodesics of the overall manifold. An even more trivial case is a 2-sphere embedded (standard embedding) in Euclidean 3-space. Its geodesic are obviously not geodesics of the 3-space.