# Trying to understand what a gravitational wave is

It seems to me that there is a great deal of interest in the possibility of gravitational waves. Several gravitational-wave detectors have been built, and there is even a branch of science with that specific goal in mind, which is gravitational-wave astronomy.

What I don’t understand is the difference between gravitational wave as produced by, say an exploding supernova, versus the change in gravitation at a fixed point in space by some other effect such as a planet (or some other dense body) passing by. Wouldn’t a passing dense body produce a single wave pulse of stronger, and then weaker gravitation, which would travel in a wave to infinity?

For example isn't the effect of Jupiter moving along its orbit producing more or less a gravitational wave which could be detected by sufficiently sensitive instruments millions of miles away?

Accelerating masses generate gravitational waves, much like accelerating charges generate electromagnetic waves.

Masses in orbit are continually accelerating, so you're right, Jupiter will generate gravitational waves by orbiting around the Jupiter-Sun centre of mass. In this case there's also gravitational radiation from the sun orbiting around the centre of mass of the Jupiter-Sun system (which is at about the radius of the sun). However the power of the wave emitted is tiny.

Wikipedia gives the following formula for the gravitational radiation of two masses in orbit around each other:

$P = -\dfrac{32}{5} \dfrac{G^4}{c^5}\dfrac{(m_1m_2)^2(m_1+m_2)}{r^5}$

For the Jupiter-Sun system this gives about 200 Watts - less power than my fridge uses!

For two stars orbiting very closely this power can be much higher - if the two objects are both neutron stars of one solar mass and are separated by 1.9$\times$10$^8$ m, then the power radiated is around 1$\times$10$^{28}$ W, which is quite a bit higher.

Using the inverse square law, the two gravitational wave sources will have the same intensity if their distances are in the ratio 1:10$^{14}$. Since we're about 1.5$\times$10$^{11}$ metres away from the sun, a binary neutron pair's gravitational waves will overwhelm that of the Sun if they're closer than around 10$^{25}$ m away, or about 1 billion light years!

• What does $C$ represent above? The speed of light? I'm used to seeing it lower-case if that's what it is. Commented Jul 20, 2015 at 14:13
• The speed of light, that's right. I've edited it to c. Commented Jul 20, 2015 at 17:51
• That equation speaks volumes. G is very small in the SI system (but not in natural units, of course), while c is large and r will usually be large as well... and all three of them are raised to large powers. One should expect gravity waves to be very weak indeed. Commented Jul 20, 2015 at 19:16
• Must be a really old fridge, or enormous. Current models use well under 100W average, even for double door auto-defrost with icemaker. Order of magnitude is right and very interesting, thanks. Commented Jul 21, 2015 at 3:04
• Thanks to everyone who contributed for excellent answers. There was more than one, so I choose this one to close out the question because the main thing I was missing was that a mass needs to accelerate in order to create gravitational waves. Commented Jul 21, 2015 at 4:23

By a wave we normally mean a plane wave.

Possibly this is simpler to understand if you consider electromagnetic waves, as we're all familiar with them. If you're near some arrangement of charges there will be an electric field in your vicinity, and if the charges are moving this field will be time dependent. But we wouldn't normally describe the changing electric field as a wave because it's localised to the charges i.e. if we go far enough away we stop detecting the field. By contrast a plane wave, e.g. a radio wave or a light wave, will in principle travel indefinitely in a vaccum and can be detected at any distance.

The same argument applies to gravitational waves. Near any system of masses there will be a gravitational field that usually changes with time. However this gravitational field isn't propagating i.e. if we go far enough away from the masses we can no longer detect it. However a plane gravitational wave propagates without needing any nearby masses, and will propagate through a vacuum indefinitely.

• I thought that a stationary or slowly changing static electric field is thought to consist of “virtual” photons, while the radiating wave source (i.e. due to accelerated electrons) consist of “real” photons. Would it be correct to think of static gravitation versus gravitational waves in a similar manner? Commented Jul 20, 2015 at 12:10
• @pigdog627: You could probably get away with thinking about it this way for sufficiently weak gravitational fields (i.e., the Newtonian-gravity limit). The main problem is that we don't have a good way to describe the way gravitons behave beyond "tree level" (i.e., more complicated Feynman diagrams), and it's not even 100% clear that the graviton picture is the correct way of thinking about gravity once you go beyond this level. Commented Jul 20, 2015 at 18:06

In principle, your understanding seems correct: it is simply a question of the relative magnitudes of all the waves in question. Bodies in mutual orbit do emit gravitational waves and indeed this phenomenon gives us our main indirect evidence for gravitational waves to date: the observed rate of spindown of the Hulse-Taylor binary star system since 1974 is precisely as calculated from general relativity.

When we get to bodies in the Solar System, the emitted gravitational wave powers are truly miniscule. For the Earth-Sun mutual orbit, the emitted power is about 200 watts, in a quadripole radiation pattern. This power level means waves that would be well below levels that can currently be detected. See the Wikipedia page for Gravitational Wave; this page has most of the information you need.

• I see. So a gravitational quadrupole (or higher) is required to radiate gravitational energy. The case that I was thinking about is like a dense planet sized object moving freely in space, and not in orbit with other bodies. A small test mass would feel the gravitational tug as the dense bodies passes by, and work would be done (the momentum of the small test mass would change) but now I think that means that a quadrupole moment exists as the two masses gravitationaly interact. Is that correct? Commented Jul 20, 2015 at 21:33

In most situation, we can make a distinction between a gravitational wave and the quasi-static gravitational field by noting that a gravitational wave propagates in space on it's own. Once two black holes collapse, the gravitational wave emerging from the event can travel for years and be completely independent of what is currently happening now at the location of the black hole merger. In a certain sense, it is carrying the picture of the merger at the very moment it left but otherwise is a completely independent field excitation.

This is very much similar to say radio (electromagnetic) waves: Somewhere in space there is a radio wave carrying a speech by John F. Kennedy even though John F. Kennedy is long gone. But the case of an electrostatic field is quite different: there can be no electrostatic field if the charge is long gone from sight. The quasi-static field can be understood to be "dragged along" the sources whereas a gravitational wave is "running wild".

If Jupiter passes around and we feel it's quasi-static field, the effects of the field are not fundamentally different from the effect of the gravitational wave apart from the fact that we can argue from observation that the force is clearly dragged along with Jupiter. But in practice there is a very clear distinction because the time-scales of vibrations from gravitational waves are expected to be much shorter. The variation of gravity could be years at best for Jupiter because it takes $\approx 12$ years to go around the Sun. On the other hand, the detectable gravitational waves are expected to have periods of at most tens of seconds, but more like tenths or hundredths of seconds.

• there can be no electrostatic field if the charge is long gone from sight.  But suppose an electron is annihilated by a positron (let's say, moving near the speed of light from the opposite direction, so we can ignore it's own electric field). The electrostatic field still exists outside the light cone of the annihilation event, right? Commented Jul 21, 2015 at 1:24