A Linearized metric for GW emitting orbiting bodies

1. Jul 26, 2016

pervect

Staff Emeritus
I'm looking for the linearized metric in the far field for a pair of mutual orbiting bodies that are emitting gravitational waves (GW's). I gather finding this (approximate) metric should be possible using the quadrupole moment of the source.

From Landau & LIfshitz "Classical theory of Fields", pg 377 $110.8. I gather that the metric pertubation $h_{\alpha\beta}$ should be proportional to $$h_{ab} \propto \frac{1}{r} \, \frac{\partial^2}{ \partial t^2 } D_{ab}$$ (If I'm reading R_0 correctly? See more below.) where D_{ab} is the (reduced) quadrupole moment. (I'll note in passing that L&L use a different notation than MTW for the reduced quadrupole moment, and while I'm preferring MTW"s notation, I think I can translate L&L's notation on this point). What I don't think I"m getting is some talk in L&L of "projecting" the quadrupole moment. As a result when I check my attempts at a solution it's not meeting the expected gauge conditions. Is the assumption being made that GW's are emitted radially, and the projection orthogonal to this null outgoing ray? Some additional questions came up from my reading. I gather that the TT gauge condition won't be satisfied by non-GW fields, so I gather those parts of the field are simply not modeled. (for instance, no 1/r term in $g_{00}$) in the solution, so it won't actually be a metric with terms up to order of 1/r in the approach I'm using. Additionally, I was reviewing the PPN formalism as an alternative to calculating the desired metric, and was surprised to see that it didn't seem to have any terms of the appropriate order (I'm expecting 1/r, as I said before) in the metric. So I'm wondering if I"m interpreting this all correctly - the terms of order 1/r due to GW's exist in the far field, but the PPN metric doesn't include them? And likewise, the static term of order 1/r in $g_00$ is not included in the proposed solution? And I'd need to linearize the GW's around a non-flat background metric (rather than the flat metric) to get these terms? While a better understanding of the issues would be good, the main goal is to find the linear-order metric valid in the far field for the purposes of defining the geometry, along with an idea of what approximations were made. 2. Jul 27, 2016 pervect Staff Emeritus The math: Basically I'm looking at exercise 1 in section$110, pg 379 in my edition of Landau & Lifshitz, "The Classical Theory of FIelds". Except that I want the metric pertubations $h_{ik}$ rather than what they want in the exercise.

I won't say what L&L want to determine, as I want to determine something different.

I've used their values for the quadropole moment as is, with their constant factors (different from MTW's) left in. I then simply plug theseesults into 110.8 to get what should be the metric coefficeints.

$$\psi_{ab} = \frac{2k}{c^4 R_0} \, \frac{\partial^2}{\partial t^2} \int \mu x^\alpha x^\beta dV$$

$\psi$ is LL's notation for what's often called $\bar{h}$ in other texts, it's defined as:

$$\psi^k_i = h^k_i - \frac{1}{2} \delta^k_i h$$

and shouldn't be confused with the angle $\Psi$ in the problem statement.

Since In this solution the trace is zero, $h = \Psi$. I'm evaluating the above at retarded time, t-r ( c set to 1).

Using this approach, I get what appears to be a simple looking result:

$$h_{xx} \propto \frac{\cos(2 \omega (t-r))}{r} \quad h_{xy} \propto \frac{\sin(2 \omega (t - r))}{r} \quad h_{yy} \propto -\frac{\cos(2 \omega (t-r))}{r}$$

The proportionality constants are all identical.

The issues is that while $\Box \, \bar{h}_{ik}=0$, I'm getting that ${\partial \bar{h}_{ik}} / {\partial x^k}$ is not zero, i.e. $h_{xx,x} + h_{xy,y}$ is not zero.

Since I'm looking for errors, it's correct to assume that for the spatial terms $h_{ik} = h^i_k = h^{ik}$, correct?

Maybe I'm doing something simple wrong, but I don't see it. I'm not sure if my solution for the metric is wrong, or the manner I'm checking it is wrong, or if I'm missing a term. As far as missing terms go, I think $h_{t*}$ should be zero - but I could be wrong :(.

3. Jul 27, 2016

RockyMarciano

It is the trace reversed perturbation that must meet the gauge condition not the perturbation tensor itself.

4. Jul 27, 2016

pervect

Staff Emeritus
I assume this means that $\bar{h}_{\alpha\beta}$ (what L&L call $\psi^\alpha_\beta$ ) should satisfy the gauge condition (?). But since $h_{yy} = - h_{xx}$, and the other diagonal terms in the proposed solution are zero, the trace of both $\bar{h}$ and h are zero, i.e. $\bar{h}^\alpha_\alpha = h^\alpha_\alpha = 0$, which implies that h = $\bar{h}$.

[add]It might be clearer to put it this way, using L&L's notation for consistency:

$$\psi_{xx} \propto \frac{\cos(2 \omega (t-r))}{r} \quad \psi_{xy} \propto \frac{\sin(2 \omega (t - r))}{r} \quad \psi_{yy} \propto -\frac{\cos(2 \omega (t-r))}{r}$$

and that $\psi_{xx} = \psi^x_x$ and $\psi_{yy}=\psi^y_y$, so that $\psi^\alpha_\alpha$ = $\psi_{xx} + \psi_{yy} = 0$, therefore $h_{\alpha\beta} = \psi_{\alpha\beta}$

Last edited: Jul 27, 2016
5. Jul 27, 2016

RockyMarciano

But you are mixing two different things here. The first situation is before you fix the transverse-traceless(TT) gauge, and there you have $\bar{h}=-h$, otherwise why bother to use a trace reversed perturbation to be able to implement the Lorenz gauge condition if it is equal to the perturbation tensor?, second step: after this, on top of the Lorenz gauge we fix the TT gauge for the vacuum case modelling of the perturbation as a plane wave solution. This is what LL calls $\psi_{\alpha\beta}$ and in this case indeed there is no trace, the metric perturbation is purely spatial now we are in the vacuum asymptotically flat case, and the former Lorenz gauge makes it also transverse, that's why they call it Transverse-traceless gauge, and of course $\psi_{\alpha\beta}^{TT}=\bar\psi_{\alpha\beta}^{TT}$

6. Jul 27, 2016

RockyMarciano

So I guess you are not calculating right the appliccation of the Lorenz gauge to the TT metric, since it must meet it. Remember that only the cases with components xx and yy on one hand and xy, yx on the other are independent. You seem to be mixing xx and xy above.

7. Jul 28, 2016

pervect

Staff Emeritus
That actually gives me a clue as to the problem. The solution works along the z axis, i.e. if x=y=0, it satisfies the gauge conditions. Apparently that's all it's supposed to do - the original appication was apparently to figure out the radiated power in various directions, and the approach used was to implicitly assume (without making an obvious statement of the assumption) that the solution was transverse. How to modify the solution to do what I want is less clear. As x,y,z are not transverse to r, the solution I'm looking for can't be transverse - at least not in x,y,z coordinates. Switching to spherical coordinates could help - but then the problem is calculating the quadrupole moment, due to the coordinate singularity at r=0.

8. Jul 28, 2016

RockyMarciano

There is no possible way to construct GWs solutions that are not transverse, by construction of the plane-wave solution, this was discussed in other threads along with the harmonic coordinate imposition . You are drawing a distinction between r and z that is not there in any gravitational radiation modelled at arbitrary long distances from the source treatment that I know of.

9. Jul 28, 2016

GeorgeDishman

The plane wave is only an approximation to the real situation. The wave period is half the orbital period of the binary so the z axis of the plane wave approximation differs from the radial direction to the barycentre by an angle of 2λ/r. Of course that too is probably a negligible number locally, comparable to the error in using the plane wave in the first place, but the effect will accumulate round the "equator".

10. Jul 28, 2016

RockyMarciano

Sure, like almost everything in physics, so what? The fact is that it is the only mathematically valid approximation so far. Unless you can come up with a new mathematical approach noone has yet been able to you have to play with the rules that are on the table.
Again, there is simply no mathematical model of GWs producing that sort of accumulation "round the equator". The far source is modelled like a point so there is no equator. Absent the new math that would be required for it you are basically taking "artistic licences" that are fine only if you are aiming towards some nicely illustrated science-fiction work .

11. Jul 29, 2016

GeorgeDishman

My aim has only ever been to provide a graphic illustration of the standard maths, but perhaps I have been naive and my source for that maths is not reliable. My simple source was the two equations given for the Earth-Sun system on Wikipedia.

In that section, the "equator" is described by as θ=π/2 : "For example, if the observer is in the x-y plane then θ=π/2, and cos(θ)=0, so the h× polarization is always zero.".

The original maths (which is beyond my level) is in the next section, Advanced Mathematics, and in particular the solution in spherical coordinates is given in Linear Approximation as

I won't pretend to understand that, but I assume the two simpler equations given earlier for the Earth-Sun system are derived from that, and it is those that I am illustrating. The sources quoted are Kip Thorne [71] and MTW [72].

Last edited: Jul 29, 2016
12. Jul 29, 2016

Staff: Mentor

Physically, this is the plane of the orbit (which is here idealized as a perfectly circular orbit).

Not quite. "Modelled like a point" here just means that the distance $R$ from the source to the observer is much larger than the radius $r$ of the orbit. It does not mean that the orbit does not exist or does not define a particular plane.

13. Jul 29, 2016

RockyMarciano

Thank you so much for pointing out that an approximation scheme(in this case the plane-wave approx.) does not imply inexistence nor causes stellar orbits and the orbit planes they determine to cease to exist. Many here may have interpreted my remark in that sense so your effort is indeed appreciated. Another possible interpretation, closer to what was intended is that the details about any "accumulation effects of the difference between r and z" in the orbit plane of the binary source orbit are irrelevant in the linearized gravity plane wave approximation used to model detectibilty precisely because in this model there is no effective difference.

14. Jul 29, 2016

RockyMarciano

Your aim is laudable. I simply don't think it is doable in the way you pretend to do it. There is no global picture of source and detection that is compatible with the math.

15. Jul 29, 2016

Staff: Mentor

I'm still not sure what you mean by "the difference between r and z" here. The point of the equations GeorgeDishman gave is that both the polarization and the intensity of GWs that are in principle detectable at a large distance from a binary system depend on the angle $\theta$ between the perpendicular to the system's orbital plane and your line of sight. This dependence on $\theta$ does not go away, no matter how far away you are from the source.

16. Jul 29, 2016

RockyMarciano

I mean this:
And you are saying this angle 2λ/r is the angle $\theta$ between line of sight and the orbital plane perpendicular?

17. Jul 29, 2016

Staff: Mentor

No. I'm not sure what GeorgeDishman is referring to in post #9, I would need more context and some references, but I'm pretty sure it's not the same thing as the angle $\theta$ referred to in post #11, which is the only post I was commenting on. In general, as I already expressed in a previous thread, I think the math required to do properly what GeorgeDishman is trying to do is much more complicated than any equations that have been posted, and in general can only be done numerically.

18. Jul 30, 2016

GeorgeDishman

Most important, my apologies to Pervect as my comment seems to have moved the discussion temporarily away from the original question, hopefully we can return to that once these points are cleared up.

There are two topics getting confused here so first consider just the orbital plane of the binary. A circle round that at large distance from the source is what Pervect described as the "equator". The first image I wanted to do is fairly simple, it shows the strength of the wave effect in the plane and I don't have a problem with this visualisation at all.
This link is originally from a NASA site, now copied into Wikipedia
https://commons.wikimedia.org/wiki/File:Wavy.gif
The vertical displacement illustrating strain is what I would call "artistic licence", I have used colour coding instead:

I don't see why you think that isn't doable or where the error lies in what I've done. The colour coding shows the magnitude of the strain moving out from the centre but also how it creates the illusion that the wave rotates with the binary.
My comment was in reply to #8
Note that the shape of the maximum of the 'ripple' in the NASA illustration is a spiral pattern reminiscent of a spiral galaxy with two arms. This sketch shows just one spiral for simplicity. If we approximate the wave at some point using the plane wave solution, the x-y plane (x shown, red) is a tangent to the spiral so the z direction (red) does not match the r direction of the spherical coordinates (green).

Does that clear up my comment #9?

The second point Peter made is the part of this exercise where I originally had a problem, hopefully now resolved.
The second visualisation I want to do is again to illustrate the strain but over the surface of a sphere and using the same colour coding as in the orbital plane version. The amplitude of the cross polarisation varies with "latitude" which just alters the colours a bit but what I also need to include is the varying direction of the strain as the relative strengths of the two polarisations changes. Some sort of vector field overlay might suffice for that but again that's only work-in-progress, I can experiment with that myself.

P.S. I think I missed a factor of 2π, the angle should be λ/πR, once round the system moves you inwards .

Last edited: Jul 30, 2016
19. Jul 30, 2016

Staff: Mentor

Yes, I see what you're referring to now. However, on its face it seems inconsistent with the equations in the Wikipedia page. The NASA image and your sketch imply that the wave amplitude at a given distance $R$ from the source must depend on $\phi$, the angular coordinate within the equatorial plane; but the Wikipedia equation only shows a dependence of the amplitude at a given $R$ on $\theta$, the angle between the perpendicular to the equatorial plane and the observer's line of sight--i.e., that at a given $R$ (and time $t$) the amplitude at all angles in the equatorial plane would be the same. I suspect that the Wikipedia page and the NASA image are using different, incompatible approximations, but I haven't had time to dig deeper into the underlying math.

20. Jul 30, 2016

GeorgeDishman

The strain at any time depends on $\phi$ as well as the time, but the amplitude of that sinusoidal variation depends only $R$ and $\theta$.
$strain=A(R,\theta)sin(\omega t-2\phi)$
Of course that needs to be expanded for the change of polarisation with $\theta$.