TL;DR

In the theory of exact plane gravitational waves, we can write the Rosen metric as ##\mathrm{d}s^2 = 2\,\mathrm{d}U\,\mathrm{d}V + e^{2\beta} \left[ e^{2\gamma} (\cos\alpha\,\mathrm{d}x + \sin\alpha\,\mathrm{d}y)^2 + e^{-2\gamma} (-\sin\alpha\,\mathrm{d}x + \cos\alpha\,\mathrm{d}y)^2 \right]##. But how to understand the transformation with the angle of rotation ##\alpha##?

The metric can be written as ##
\mathrm{d}s^2 = 2\,\mathrm{d}U\,\mathrm{d}V + e^{2\beta}
\begin{pmatrix} \mathrm{d}x & \mathrm{d}y \end{pmatrix}
R^T(\alpha)
\begin{pmatrix} e^{2\gamma} & 0 \\ 0 & e^{-2\gamma} \end{pmatrix}
R(\alpha)
\begin{pmatrix} \mathrm{d}x \\ \mathrm{d}y \end{pmatrix}=
\mathrm{d}s^2 = 2\,\mathrm{d}U\,\mathrm{d}V + e^{2\beta}
\begin{pmatrix} \mathrm{d}x & \mathrm{d}y \end{pmatrix}
\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}
\begin{pmatrix} e^{2\gamma} & 0 \\ 0 & e^{-2\gamma} \end{pmatrix}
\begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix}
\begin{pmatrix} \mathrm{d}x \\ \mathrm{d}y \end{pmatrix}##

The matrix term is clearly a rotational transformation, but I have problems to understand how the transformation looks like and what vectors we transform.

If we have a sandwich plane gravitational wave on the flat background, then in front of the wave, the space-time is flat, so the metric is Minkowskian and the basis vectors ##e_x## and ##e_y## are ortonormal to each other in the flat metric. Then the wave comes and let's say that it is X-polarised. Since the Rosen metric is comoving with rest particles, the basis vectors ##e_x## and ##e_x## will be stretched with the space-time.

The metric in which the basis vectors point in the direction of the principle axes of the ellipse (shown in the picture below) is ##\mathrm{d}s^2 = 2\,\mathrm{d}U\,\mathrm{d}V + e^{2\beta}
\begin{pmatrix} \mathrm{d}x' & \mathrm{d}y' \end{pmatrix}
\begin{pmatrix} e^{2\gamma} & 0 \\ 0 & e^{-2\gamma} \end{pmatrix}
\begin{pmatrix} \mathrm{d}x' \\ \mathrm{d}y' \end{pmatrix}##. And here comes my confusion - the transformation matrices ##R## and ##R^T## rotates the basis, in which the metrix is diagonal, back to the basis, in which the metric ##g_{ij}##,where ##i,j=1,2##, can have non-diagonal components. But the "diagonal basis" must have perpendicular vectors so that these vectors point in the same diretion as the principal axes of the ellipse. But the rotation must preserve its perpendicularity. But the "non-diagonal basis" is oblique as shown in the picture, so there is a contradiction.

The general metric can have non-diagonal components, that is the metric with respect to the "oblique" basis, the stretched basis. But its rotation doesn't ensure that its two vectors point in the same direction as the principle axes. So, what basis do we rotace exactly? Because this looks like we rotate the "perpendicular flat basis" by ##\alpha##. But that doesn't make sense because then we can't get the oblique basis.

I tried to visualise my thoughts in this picture. I hope it clarifies what I mean.

Lotto said:

The matrix term is clearly a rotational transformation

Really? It looks like more than a rotation to me.

Dale said:

Really? It looks like more than a rotation to me.

You mean it is also a stretching by ##e^{2\beta}##? But ##e^{2\beta}## could be given inside the matrix and then it is only a rotation. Right?

Ibix said:

I suspect Dale is referring to the ##\mathbf{R}^T\mathbf{MR}## structure, which looks like a rotated "stretch and squish" to me

Oh, you mean ##e^{2\gamma}## and ##e^{-2\gamma}##? Ok, that characterisises the anisotropy of the space-time. So ok, the basis of the flat space-time is stretched and squinshed and then rotated. But that means that the basis vectors should be perpendicular to each other, because rotation doesn't change angles between vectors. But our basis is oblique...

Lotto said:

Do you mean this metric?

If not, can you give a reference?

Last edited: Aug 8, 2026

PeterDonis said:

Do you mean this metric?

If not, can you give a reference?

No, that is not my metric. My metric describes plane gravitational waves and can be written in many forms. The only source for the form I am using is from this work https://arxiv.org/pdf/2606.25792v1 in the equation (35).

Ibix said:

No, ##\mathbf{R}^T\mathbf{MR}## is a rotation, a stretch/squish, and an inverse rotation. That doesn't preserve perpendicularity, as you can easily see by multiplying out the matrix - it has off-diagonal elements that will give you a ##dx\, dy## term.

So we can essentially say that the diagonal matrix corresponds to the metric with basis vectors ##e_x## and ##e_y## pointing in the principle axes of the ellipse. And by transforming the metric with ##R## and ##R^T## we rotate these axes by ##\alpha##, which with the final metric creates the oblique basis?

I tried to visualise my thoughts:

Lotto said:

But that means that the basis vectors should be perpendicular to each other, because rotation doesn't change angles between vectors.

Rotation, stretching, and rotation is not equivalent to just rotation. Angles do change.

Lotto said:

My metric describes plane gravitational waves

Yes, I see that from the paper, thanks for the reference.

Key things to keep in mind: the coordinates ##U## and ##V## are null. The waves propagate along null lines of constant ##V##; ##U## is the null coordinate along the wave vector, i.e., the null wave vector is collinear with ##\partial_U##. So, as the paper notes, the parameters ##\beta##, ##\gamma##, and ##\alpha## are all functions of ##U##. The ##x##, ##y## plane is the plane transverse to the waves, and the part of the metric involving ##dx## and ##dy## describes a combination of shear (##\gamma##) and rotation (##\alpha##) in that plane.

I think that my mistake was that I tried to give the angle ##\alpha## a physical meaning. The angles between the oblique and the transformed basis don't have to be ##\alpha##. This angle is just a way to parametrize the metric. For a X-polarised wave the angle would be ##45^\circ## and the picture would be just a rotated ellipse in the Euclidean plane. We can visualise the angle in the Euclidean plane as an angle between a rotated and non-rotated orthogonal Euclidean base. Is this idea correct?

Lotto said:

I think that my mistake was that I tried to give the angle ##\alpha## a physical meaning.

I don't think so. I think your mistake was trying to give ##\alpha## the wrong physical meaning.

As I noted in post #11, ##\alpha## is a function of ##U##. That means it changes along the wave. It is not a constant. Your visualizations appear to me to assume that it is.

The physical meaning of ##\alpha## is that it describes the phase, changing along the wave, of the rotation in the ##x## - ##y## plane. The paper you referenced calls this the "polarization angle".

PeterDonis said:

I don't think so. I think your mistake was trying to give ##\alpha## the wrong physical meaning.

As I noted in post #11, ##\alpha## is a function of ##U##. That means it changes along the wave. It is not a constant. Your visualizations appear to me to assume that it is.

The physical meaning of ##\alpha## is that it describes the phase, changing along the wave, of the rotation in the ##x## - ##y## plane. The paper you referenced calls this the "polarization angle".

I know that it can change with ##U##, but my initial idea was that it is the angle between the oblique basis and the "principal axis basis". But I suppose that is not true in general. It is just an angle between Euclidean basis vectors.

Lotto said:

my initial idea was that it is the angle between the oblique basis and the "principal axis basis".

It's the angle, which changes along the wave, between the ##x## - ##y## basis and the principal axis of polarization of the wave.

Lotto said:

It is just an angle between Euclidean basis vectors.

I'm not sure what you mean by this. Unless ##\gamma = 0## (no shear), the geometry of the ##x## - ##y## plane is not Euclidean.

PeterDonis said:

Unless ##\gamma = 0## (no shear), the geometry of the ##x## - ##y## plane is not Euclidean.

I am pretty sure that the ##x’,y’## sub manifold has a Riemann curvature tensor all 0.

Dale said:

I am pretty sure that the ##x’,y’## sub manifold has a Riemann curvature tensor all 0.

Hm. When I plug the metric into Maxima, I do get a Riemann tensor with no components purely in the ##x## - ##y## submanifold. However, I also get a nonzero Einstein tensor, which doesn't make sense if this is supposed to be a gravitational plane wave; such a spacetime should be vacuum, with only Weyl curvature.

I am not familiar with the metric. But I guess with proper choice of the null coordinates it may be possible to make the spatial submanifolds flat.

Dale said:

I am not familiar with the metric.

It's equation (35) in the paper that was referenced in post #8 in the thread.

Dale said:

I guess with proper choice of the null coordinates it may be possible to make the spatial submanifolds flat.

The null coordinates used in the paper seem to me to be the obvious (and commonly used) ones for a plane wave: ##U## is the affine parameter along each wave propagation worldline, and ##V## labels which worldline it is.

If my Maxima result is correct, then in these coordinates the purely spacelike ##x## - ##y## submanifolds are indeed flat, and my statement at the end of post #15 was incorrect. But I need to double check because I was expecting that metric to have zero Einstein tensor and Maxima says it doesn't.

PeterDonis said:

It's equation (35) in the paper that was referenced in post #8 in the thread.

Yes, I just mean I have not worked with this metric before this thread. So I don’t have any of the sort of understanding that one gets through experience