# Gravitational Waves and Entropy

1. Mar 14, 2012

### pawprint

I'm learning about gravitational waves and associated matters (detectors etc.). I don't do much maths, but physics is in my bones. Totally at home with classical physics and SR; as for GR- well, I'm O.K. as long as you stay on the 'friendly' side of event horizons...

I'm comfortable with the Wikipedia article 'Gravitational waves' until halfway down the page (just below this image...)

(Wikipedia commons)

...where it states without qualification that "gravitational waves can pass through any intervening matter without being scattered significantly." I seek clarification of the term 'scattered significantly'.

The above image probably reflects a neutron star binary system with an orbital period of about 4 seconds, resulting in a wavelength around 2 light seconds and frequency of 0.5 Hertz.

Let me place the Earth in that image near the plane of greatest gravitational wave distortion. Also allow that the binary system is 40 times more rapid than shown above: It's now inspiralling significantly and has a wavelength close to one Earth diameter and a frequency around 20 Hertz.

The Earth itself is being distorted by the wave. It is immersed within and conforming to 4 dimensional spacetime, as it must. In our (The Earth's) frame of reference our clocks and rulers are being distorted but so are we as observers; we see nothing of this directly.

If the radiated spacetime wave is of sufficient magnitude an observer whose frame of reference is distant from the orbital plane of the radiating system should be able to measure these distortions. The observer might, for example, observe Doppler shifts in Earthly EM transmissions. Our Sun would show slightly broadened emission and absorbtion lines.

Since I'm stepping into a new arena I have probably not expressed myself well. I welcome suggestions and corrections.

My questions-

1) Does the above require clarification? Is anything about it inherently impossible or wrong?
2) If not does the Earth's interaction with the spacetime wave 'significantly scatter' the wave?
3) The base question- Does the gravitational interaction increase entropy?

Last edited by a moderator: Apr 19, 2017
2. Mar 15, 2012

### edwinhubble

Hi pawprint,

I think I understand your questions, and I thought I could probably chime in on a couple points.

I work in the GW (Gravitational Wave) community (not on any of the theory stuff, I'm more of a materials science person), but it seems as a matter of course, we're all educated on what's going on with GWs.

I'm going to be a little hand-wavy because I don't want to take the time to look it up, but maybe somebody else can back me up here. The reason that GWs aren't scattered or absorbed 'significantly' is sort of for two reasons:

First, GW interactions with matter aren't very big. If you've read the wikipedia page, you know that GWs are quadupolar, which means that the interactions they have with matter will also be quadupolar, which is much more difficult to do than most other forces' dipole interactions. So the energy imparted by a GW isn't very big.

Second, GWs have huge amounts of energy, and I mean huge. As the wikipedia page gives an example, the inspiral of two neutron stars radiates 1.38×1028 watts, or 100 times the output of the sun.

Combining these two means that any energy that the Earth absorbs from the GWs would be tiny in comparison. If we consider scattering as gravitational waves that are absorbed and then re-emitted, that will be tinier still (by a large amount). So, GWs are not significantly scattered by matter.

Finally, to answer your third question, yes, GWs do contribute to entropy--everything does. I mean, the conversion of orbital energy into GWs in the case orbiting neutron stars is an example of entropy, by creating heat in the stars, and also by increasing the stochastic background in the gravitational field. If you want a more 'classical' way that GWs contribute to entropy, think of the energy absorbed by the Earth in your example. It might not be much, but it causes oscillating distortions in the body of the Earth. When the GW passes, the Earth won't keep ringing forever, the energy will be converted to heat through mechanical loss--more entropy.

3. Mar 15, 2012

### pawprint

Thank you, edwinhubble. Your answer is at least as clear as my question. Furthermore I'd just about given up hope on ANY answers. The forum areas I have been inhabiting can get half a dozen replies per hour.

I think I can safely infer that a gravitational wave passing through ultradense matter would be more 'scattered' than one passing through the Earth.

Last edited: Mar 15, 2012
4. Mar 20, 2012

### colin456

This is an interesting discussion and I have nothing to add but a question. So what you are saying edwinhubble is that gravitational waves carry a lot of energy, but do not dissipate that energy very well? Does this result come directly from theory (i.e. GR) or is it set by the inability of current experiments to detect gravitational waves (such that current experiments can set an upper limit to how efficiently gravitational waves dissipate energy)?

5. Mar 20, 2012

### pawprint

I posed the question because I am one who believes that the sublime silence of LIGO is itself a scientific datum. I am working on an entirely different approach to the detection of gravitational waves; one which has less reliance on entropic interactions. Some information is on my website, accessible through my username link.

6. Mar 31, 2012

### edwinhubble

It definitely comes from theory. Even before LIGO was at design sensitivity, it was predicted that while detecting GWs with initial LIGO was possible, there would probably be only one signal loud enough every 100 years, so it wasn't likely. Nobody who knows what they're talking about is surprised that LIGO hasn't detected anything. It's the next generation of detectors, like Advanced LIGO, which comes online sometime in 2014-2016, which should make regular detections. If Advanced LIGO is silent, then we need to start re-thinking the theory.

7. Apr 1, 2012

### pawprint

Welcome to PF edwinhubble. Your first two posts have brought smiles to my face:{)

There appear to be differences of opinion within LIGO as to how it works. Some say the result will be detected by the movement of the 'test masses' to which the end mirrors are attached. Others expect the data to show up as lengthening or shortening of the arms themselves. This seems a fundamental dichotomy.

In the first case a powerful GW would be required to move a 50+ kg test mass while passing at the speed of light. This detection method is reminiscent of a Weber bar in that it requires a resonant response from the test mass. Any induced real motion in the test mass must result in an increase in local entropy.

In the second case movements of the test mass are noise (e.g. Luna), not signal. The shortening of the arm is not 'mechanical'; rather it results from a spacetime distortion travelling at c along the arm. It is not at all obvious to me that this interaction alters net entropy.

Enlightenment would be welcome.

Last edited: Apr 1, 2012
8. Apr 1, 2012

### zhermes

That's not entirely accurate. The current rate-estimates are entirely consistent with initial LIGO/Virgo having not detected anything, and still a high probability that advanced will---but when they first became operational, that was not the case. Initial LIGO/Virgo probably wouldn't have been built had people known there was such a low probability of detection...

These are actually the same effects. The 'movement' of the test masses is the measurable effect of the detector arms changing length. There are no sensors to detect the position of every element of the detector arms (and it would be impossible to make those sensors accurate enough).

To linear order the mass of the test-mass doesn't matter, having a larger mass makes it easier to damp out noise. Detection requires no resonance with the mass, unlike a weber-bar. The resonant interaction between a GW and a mass is still uncertain, but most general relativists aren't hopeful that its a plausible means of detection... none-the-less, there is a currently active network of Weber-Bar detectors.