# Gravitational waves detection

1. Feb 26, 2013

### TrickyDicky

I have a doubt about this, what is actually detected by a GW detector is a specific change in length (δl) in the arms of the interferometer due to the motion of the test masses being affected by the passing wave. But this variation is not of the proper length wich is not uniquely defined in GR but of frame-dependent distance I guess, so it is subject to observer-dependent effects, I don't understand then how would this be considered a physical result of physical radiation.

2. Feb 26, 2013

### haael

Isn't the fact that the background is nearly flat an important assumption?

3. Feb 26, 2013

### Staff: Mentor

No, the variation is of proper length. It's true that there can be multiple geodesics between the same pair of events in curved spacetime; but it is also true that a particular event in spacetime plus a particular tangent vector at that event defines a unique geodesic through the entire curved spacetime. The latter method is how the proper length of the arms of the GW detector is defined: the directions of the two interferometer lasers, firing from the same point (the apex of the L-shaped interferometer), define two spatial directions (and therefore two spacelike tangent vectors) at each firing event, and therefore define two unique spacelike geodesics from that event to the simultaneous events at the mirrors--"simultaneous" in the local inertial frame in which the apex of the interferometer is at rest. Integrating the metric along these geodesics gives proper length.

The mirrors are also at rest in the LIF of the apex, in the "resting" state of the detector, but when a GW passes, the mirrors move in this frame; and the change in apex-mirror distance in the LIF corresponds to a change in proper length of each arm, because within the LIF, curves of constant time correspond to spacelike geodesics as defined above, so length along those curves, which is just spatail distance in the LIF, corresponds to proper length.

4. Feb 27, 2013

### TrickyDicky

Yes, that is the first thing I thought, that the geodesic deviation equation(GDE) described a linearized (LIF), flat background to first order situation since it is explicitly referred to infinitesimally close geodesics, but then this got me even more confused since if we are in a flat background geodesics should remain parallel while in the LIF, and therefore no deviation is possible otherwise we would have a proper acceleration in some of the curves.
And even going to the usual definitions of geodesic deviation like from wikipedia or Carroll's notes my confusion increases.
WP: "if two objects are set in motion along two initially parallel trajectories, the presence of a tidal gravitational force will cause the trajectories to bend towards or away from each other, producing a relative acceleration between the objects. Mathematically, the tidal force in general relativity is described by the Riemann curvature tensor, and the trajectory of an object solely under the influence of gravity is called a geodesic. The geodesic deviation equation relates the Riemann curvature tensor to the relative acceleration of two neighboring geodesics", Carroll's notes: "the geodesic deviation equation expresses something that we might have expected: the relative acceleration between two neighboring geodesics is proportional to the curvature. Physically, of course, the acceleration of neighboring geodesics is interpreted as a manifestation of gravitational tidal forces"
These explanations lead me to think that geodesic deviation is indeed related to curvature of the manifold as expressed by the Riemann tensor, but the Riemann tensor comes from second derivatives of the metric and the Christoffel symbols, that is curvature effects are second order and would be missed by the GDE linearized approach, wouldn't they?
This is really confusing, can someone clarify?

5. Feb 27, 2013

### Staff: Mentor

You're getting confused between the background metric and the full metric including the GW perturbation. You can write a geodesic equation for either one, but they will describe different geodesics. The actual, physical geodesics, the ones we observe, are the ones for the full metric, including the GW perturbation. The geodesics of the flat background by itself are not observable.

This does mean that my use of the term "local inertial frame" is cheating, strictly speaking; the frame in question is more correctly viewed as a perturbed LIF, in which the "time axis" is the geodesic that the apex of the detector follows, and in which perturbations from a completely flat metric are small. I apologize for the confusion.

Geodesic deviation is related to curvature, which is given by second derivatives of the metric; but the term "linearized" as in "linearized GR", the approximation scheme in which GWs are treated, does not refer to the order of derivatives of the metric. It refers to the order to which perturbations about a flat metric are kept. See, for example, here:

http://en.wikipedia.org/wiki/Linearized_gravity

Basically, we assume that perturbations from a flat metric are small enough that we can keep only the terms that are first order (linear) in the perturbations. That makes the equations much simpler. But the perturbations themselves, even to first order, are still proportional to second derivatives of the metric; i.e., we are still working in local coordinates in which all the Christoffel symbols (first derivatives of the metric) are zero. We just don't assume that the second derivatives are all negligible; they have to be small, but they don't have to be negligible.

6. Mar 1, 2013

### TrickyDicky

Thanks for those clear explanations Peter, I was aware of that distinction but I guess only in abstract terms.
This goes probably to the root of my doubts about the whole geodesics and gravitational radiation issue. But linearized gravity is a broad enough subject for another thread.

In the meantime a couple of further questions:

I think here is one point that might need clarification, I know that with both an initial event and an initial velocity we define a unique geodesic unlike the case when we don't define an initial velocity vector (there is a formulation for this, it is called the generalized Jacobi equation or nonlinear Jacobi equation) but then again velocities in relativity are coordinate dependent and therefore to some extent the choice of this vector is arbitrary, no? This was part of the issue in my OP.

This might be an arbitrary assumption, due to the initial velocity frame dependence I mentioned above. But I realize that it is an automatic assumption in the context of the"linearized gravity" formulation.

7. Mar 1, 2013

### Staff: Mentor

We're talking about the velocity 4-vector here, not a 3-vector. It's not coordinate-dependent.

8. Mar 1, 2013

### TrickyDicky

Yes, it's the 4-velocity, but I'm not sure what you mean, in what sense is it not frame dependent?
Edit:
Ok, I see what you mean, the magnitude is constant

Last edited: Mar 1, 2013
9. Mar 1, 2013

### TrickyDicky

We need to differentiate between 4-velocities and 4-accelerations in SR versus GR.
Otherwise note that this leads to a problem here, if you consider that the 4-velocity is not coordinate-dependent (and I agree in SR it isn't), then you cannot call the 4-acceleration that is computed in the geodesic deviation equation a coordinate acceleration.

10. Mar 1, 2013

### Staff: Mentor

No, that's not what I mean (although the magnitude is indeed constant). The direction of the four-velocity is also frame-independent; it's tangent to the worldline, and this remains true even though the numerical values of the components may be expressed differently in different coordinate systems.

11. Mar 1, 2013

### TrickyDicky

Ok, but then why the four-acceleration that is the derivative wrt proper time of the 4-velocity is considered coordinate-dependent?

12. Mar 1, 2013

### Staff: Mentor

"Initial velocity" in the rule you state should more precisely be called "initial 4-velocity". You have an event and a tangent vector at that event. The tangent vector is not coordinate-dependent; its *spatial component* may be (that is the ordinary "velocity" or 3-velocity), but the tangent vector itself is not.

13. Mar 1, 2013

### TrickyDicky

Ok, I was identifying the relative acceleration vector of the geodesic deviation equation with a 4-acceleration, but apparently it isn't, right?

14. Mar 1, 2013

### Staff: Mentor

Four-accelerations aren't coordinate-dependent either. Just like four-velocity and four-momentum, the numerical values of the individual coordinates may vary with the choice of coordinate systems, but the four-vector itself is coordinate independent.

15. Mar 1, 2013

### TrickyDicky

Got it now, thanks, I was mixing two different acceleration vectors.

16. Mar 1, 2013

### TrickyDicky

Ok, so I'd like to post a couple of things I draw as conclusions to wrap up this thread:
First, this thread was started by me as a spin-off of another thread among various similar ones where the concept of geodesic motion in GR was debated in relation with gravitational radiation. Specifically the different points of view wrt whether extended objects in GR can be considered to follow geodesics like test particles do, and if this was compatible with radiating GWs.
I'm afraid that centering on the detection of GWs is of no use to decide this because even though the emmision of GWs from binary BHs or neutron stars binary systems is a strong-field problem that requires numerical relativity, at the detection end the problem is a weak-field one and it can be dealt with using linearized gravity perturbative methods. So it doesn't solve the possible geodesic motion nature of the source.
So I guess there is no final answer, we have experts like Sam Gralla and SAs like PAllen with one view while other SAs like PeterDonis and Bill_K have a different one, I myself am turning my opinion to neutral, after having been in both extremes thru the last two years :tongue2:

Notwithstanding this, for me the thread was not a waste of time, since I have disentangled a couple of misunderstandings of my own in the meantime.

17. Mar 1, 2013

### Staff: Mentor

It's not the 4-acceleration of either worldline individually, if that's what you mean. But it is a coordinate-independent 4-vector if defined properly.

18. Mar 1, 2013

### Staff: Mentor

I agree with this; the detection problem is a much easier one than the source problem.

Just to be clear, Sam Gralla's opinion is much better informed than mine, since he is much more familiar with the details of how the source models are constructed.

19. Mar 1, 2013

### TrickyDicky

Hmm, both Carroll's notes and wikipedia define it as the "relative acceleration vector" $A^μ$. How is a relative acceleration coordinate-independent?

20. Mar 1, 2013

### Staff: Mentor

In this context, the relative acceleration is a change in the separation vector between between two worldlines, so is defined in a coordinate-independent way.