General RelativityPhysics

On Gravitational Waves

In light of the recent experimental results at LIGO, I thought a few words on gravitational radiation are in order.

Gravitational waves are a phenomenon which occurs both in linearised gravity, as well as a solution to the full Einstein equations. They are periodic disturbances in the gravitational field, a time-varying gravitational influence; in the language of General Relativity, they are wave-like perturbations of the geometry of space-time itself, and as such do not require any medium of propagation. For the purpose of this short introduction, we shall limit ourselves to the linearised case; the basic idea here is to describe the metric of space-time as the flat Minkowski metric plus a small perturbation :

(1)   \begin{equation*} \displaystyle{g_{\mu \nu}\equiv \eta _{\mu \nu}+h_{\mu \nu}+O\left ( [h_{\mu \nu}]^2 \right )} \end{equation*}

We can then introduce a rank-2 tensor field ( “gravitational field” ) \displaystyle{\bar{h}_{\mu \nu}}, which is related to the perturbation \displaystyle{h_{\mu \nu}} via

(2)   \begin{equation*} \displaystyle{\bar{h}_{\mu \nu}=h_{\mu \nu}-\frac{1}{2}h\eta _{\mu \nu}} \end{equation*}

There is a freedom here to impose a gauge condition; a suitable choice, such as the Lorentz gauge \displaystyle{\bar{h}{^{\mu \alpha}}_{|\alpha}=0}, reduces the Einstein vacuum equations into the particularly simple form

(3)   \begin{equation*} \displaystyle{\square{\bar{h}_{\mu \nu}}=0} \end{equation*}

The reader will immediately recognise this as a linear, relativistic wave equation; physically, this means that disturbances in the gravitational field propagate as a wave. The simplest solution to the above wave equation is a monochromatic plane wave of the form

(4)   \begin{equation*} \displaystyle{\bar{h}_{\mu \nu}=\Re \left [ A_{\mu \nu} \exp\left ( ik_{\alpha}x^{\alpha} \right ) \right ]} \end{equation*}

In order for this to be a valid solution, two boundary conditions must hold – firstly, the wave vector must satisfy

(5)   \begin{equation*} \displaystyle{k_{\alpha}k^{\alpha}=0} \end{equation*}

which means that it must be a null vector, and hence that the wave must propagate at the speed of light. Secondly, the amplitudes must satisfy

(6)   \begin{equation*} \displaystyle{A_{\mu \alpha}k^{\alpha}=0} \end{equation*}

which physically means that the wave propagates in a direction orthogonal to the disturbance itself. The easiest way to visualise what such a wave might “look like”, is to consider its effect when it passes through a suitable ensemble of test particles. Suppose we have a ring of stationary, non-interacting test masses, oriented such that the plane spanned by the ring is perpendicular to the propagation direction of the gravitational wave; the effect of the passing wave will then be a relative acceleration of these test particles, like so [1] :

The motion here is greatly exaggerated to demonstrate the principle, in reality the acceleration induced will be tiny, even for very strong sources – which is of course why it took so long to directly detect such waves in the laboratory. Once can also plot the above in three dimensions, to show the propagation of the wave itself :

gw-waves-wave

Gravitational waves can be polarised linearly or circularly; in each case, there are two possible polarisation modes. The above two animations depict a linear “plus polarisation”; the second mode, called “cross polarisation”, would be offset by 45 degrees, like so [2] :

 

The effects of a circular polarisation are best shown in three dimensions :

gw-waves-circ

 

The two possible polarisation modes in the circular case are called “left-handed” and “right-handed” – for reasons which should be quite obvious from the above animation.

The source of gravitational radiation is a time-varying quadrupole ( or higher multipole ) moment; this is different from electromagnetic waves, the source of which are dipole moments. Examples of systems with a non-vanishing higher moments would be binary systems, or rapidly rotating bodies which are irregular in shape. In the case of the first LIGO detection, the source was the merger of two black holes with several dozen solar masses each. The signal lasted roughly 0.2 seconds in the detector, and included the initial merger as well as the subsequent ring-down phase. This might be visualised along these lines [3]:

 

We have up to now only considered linear and purely plane waves, for academic purposes; of course, real-world gravitational waves will be neither linear, nor will they ( in general ) be plane. When we go away from the linear approximation and consider the full non-linear Einstein equations, things become very complicated very quickly. For starters, a more general gravitational wave will not be plane, i.e. it will have both a transverse as well as a longitudinal part. Such a disturbance might look like this [4]:

arxiv_org_pdf_1101_2247_pdf

 

Furthermore, due to the non-linearity of the Einstein equations, gravitational waves will self-interact; that means they will interact with themselves, with other waves, and also with background curvature. As a result, you can get effects such as refraction and backscattering, even in regions which are otherwise empty vacuum. This, again, is very unlike the electromagnetic case, which exhibits no such self-interactions ( Maxwell’s equations are linear ). Another subtle and important difference between these two types of radiation is that the energy-momentum “carried” by a gravitational wave is not localisable – it does not make mathematical or physical sense to speak of the “energy in a single wave front” [5]. All one can do is write down the effective energy-momentum averaged over several wave lengths.

So why is all this so important ? Well, for one thing, being able to directly examine gravitational waves will give us a new tool with which we can observe the universe around us, and which provide us with new insights not obtainable from the EM spectrum. More importantly though, the wave that was detected had precisely the form and characteristics we expected it to have – this is a strong indicator that our model of classical gravity ( General Relativity ) is indeed a good and valid description within its domain of applicability. The gravitational wave manifested precisely as predicted, and the dynamics of the source ( binary black holes ) likewise is precisely as predicted, meaning it matched exactly the template obtained from numerical calculations of such scenarios, based on General Relativity.

References

[1] https://upload.wikimedia.org/wikipedia/commons/b/b8/GravitationalWave_PlusPolarization.gif

[2] https://upload.wikimedia.org/wikipedia/commons/b/b8/GravitationalWave_CrossPolarization.gif

[3] https://cbsdetroit.files.wordpress.com/2016/02/einstein-gravitational-wave.gif

[4] http://arxiv.org/pdf/1101.2247.pdf

[5] Misner/Thorne/Wheeler, Gravitation, §35.7, page 955 onwards

0 votes

10 thoughts on “On Gravitational Waves

  1. Yes, your reasoning makes sense, but take a rare cold gas where collisions are rare. If after a short wave packet passage the atoms do not return to the rest, they already took some energy from the wave packet, even before collisions (before dissipating this energy into heat). I am not sure they return to the rest.

    1. I would consider a cold gas such as this a weakly interacting system – as such, there would be a transfer of energy from the wave to the gas, and the atoms would gain an average kinetic energy. I would expect this transfer to be of a very, very small magnitude, though ( don’t have any figures at hand, sorry ).

  2. No. Let us admit that there is a reference frame from which we observe a probe particle and a narrow wave packet directed to this particle. After influencing the particle position, the wave packet goes away, but the particle does not return to the rest – it acquires some energy-momentum and it moves after interaction. In other words, I mean the final kinetic energy of the probe particle.

    1. Actually, the particle would indeed come to rest. What the gravitational wave does is change the distance between any set of two particles, in very specific ways; however, in the rest frame of each particle, there is no proper acceleration measured at all when the wave passes, so no force acts on them.

      My guess is that in order to extract energy-momentum from the passing wave fronts, you will need an ensemble of many particles which are mutually bound – a solid body, in other words. The passing wave will deform such bodies, leading to stresses and internal heating. I don’t think a ring of non-interacting test particles will do the trick, since each particle will continue to trace out a geodesic at all times, even when under the influence of the wave front.

        1. Yes, you are correct, inertial motion in flat space-time by definition traces out a geodesic.

          Basically, what I was trying to explain is this – if you have a ring of non-interacting (!) test particles as in the animation on my post, then each of these particles will be in a state of free fall, i.e. it will trace out a geodesic. To see why, imagine you attach an accelerometer to each particle – you will find that all the accelerometers will continue to read exactly zero, at all times, throughout the experiment. And that includes those periods when the wave front passes, and the ring gets distorted. Now, if there is no proper acceleration, there cannot be a force, and hence no work is being performed anywhere by the wave.

          That is the situation for free, non-interacting particle ensembles. But now imagine what happens when we connect the particles with bits of wire, so that they are no longer in free fall – the distortion still happens, but different parts of the circle get distorted in different ways at different times, so the wire between the particles will counteract the distortion, and hence we now measure proper acceleration on some of the accelerometers. The individual test particles no longer trace out geodesics, and work is performed on other overall system – that work goes into mechanical deformation of the wires, as well as into heat loss. The wave is now transferring energy-momentum into our system.

          Does this make sense ? Basically, all I am trying to point out is that in order to extract energy-momentum from the wave, you need a system of bound mass points, like a continuum body – you cannot extract energy from the wave by using an ensemble of non-interacting masses in free fall, since no work is performed on such masses. A passing gravitational wave only effects the separation between them.

  3. Hi Markus,

    Do you know any calculation result of affecting a still point mass with the “wave” that has been recently registered? It seems to me the still probe particle must start moving after the “wave” passage; thus, eating out some energy from the wave. How much?

    1. I am not quite sure I understand your question, I’m afraid; the passage of the wave induces a relative acceleration between neighbouring test particles, but within their individual rest frames, no proper acceleration is measured. On the other hand, within a continuum, the individual particles are not free, but interact with one another – under these circumstances, the wave will induce stresses and strains, and will generate a small amount of heat. Is this transferred energy what you are asking about ?

      1. If one integrates the force over the time, one obtains the final velocity => energy and momentum. It may be finite and determine the energy loss experienced with the wave packet after crossing the space filled with otherwise non interacting (neutral) particles (bodies).

Comments are closed.